Suppose the same curve is described by \(X(\theta), Y(\theta)\) for \(\theta \in [c, d]\), with \(x(t) = X(\theta(t))\) and \(y(t) = Y(\theta(t))\), where \(\theta : [a,b] \to [c,d]\) is differentiable and increasing with \(\theta(a) = c\) and \(\theta(b) = d\). By the chain rule, \(x'(t) = X'(\theta)\,\theta'(t)\) and \(y'(t) = Y'(\theta)\,\theta'(t)\). Since \(\theta'(t) \ge 0\),

by the substitution rule. (A decreasing \(\theta\) traverses the curve backwards; the same computation, with \(|\theta'|\), gives the same length.)

\(\blacksquare\)

Let \(\zeta\) be the translation and \(\gamma(t) = (x(t), y(t))\). Then \(\zeta(\gamma)\) is \((x(t) + h, y(t) + k)\). Since \(h\) and \(k\) are constants, \(\big(x(t) + h\big)' = x'(t)\) and \(\big(y(t) + k\big)' = y'(t)\), so

\(\blacksquare\)

Write a vector as \(\begin{pmatrix} r\cos\alpha \\ r\sin\alpha \end{pmatrix}\). Then

This is the vector of the same length at angle \(\alpha + \theta\), so \(\rho_\theta\) is the anticlockwise rotation by \(\theta\).

\(\blacksquare\)

The rotation by \(\theta\) sends \(\gamma(t) = (x(t), y(t))\) to \(\big(x\cos\theta - y\sin\theta,\ x\sin\theta + y\cos\theta\big)\). Since \(\theta\) is constant, the derivatives are \(x'\cos\theta - y'\sin\theta\) and \(x'\sin\theta + y'\cos\theta\). Squaring and adding, the cross terms cancel:

\(\blacksquare\)

Given \(X = (x_0, y_0)\) and \(Y = (x_1, y_1)\), the translation by \(\begin{pmatrix} x_1 - x_0 \\ y_1 - y_0 \end{pmatrix}\) sends \(X\) to \(Y\), and it is an isometry by Statement 2.2.1.

\(\blacksquare\)

Write \(X - A = (X-A)\cdot u \, u + n\), where \(n\) is perpendicular to \(u\). The reflection keeps the component along \(l\) and reverses the perpendicular one, so the image of \(X\) is \(A + (X-A)\cdot u\, u - n\). Since \(n = (X-A) - \big((X-A)\cdot u\big)u\), this is \(A + 2\big((X-A)\cdot u\big)u - (X-A)\).

\(\blacksquare\)

Reflection in the \(x\)-axis sends \((x(t), y(t))\) to \((x(t), -y(t))\), which clearly preserves \(\sqrt{x'^2 + y'^2}\). A reflection in a line through the origin at angle \(\beta\) is \(\rho_\beta \rho_r \rho_{-\beta}\): rotate the line onto the \(x\)-axis, reflect, and rotate back. This is a composite of isometries (Statement 2.2.3), hence an isometry.

If \(l'\) does not pass through the origin, let \(l\) be the parallel line through the origin and \(h\) the distance between them. Reflection in \(l'\) is reflection in \(l\) followed by a translation by \(2h\) perpendicular to \(l\), again a composite of isometries. Each step uses that a composite of isometries is an isometry: if \(f\) and \(g\) preserve every length, so does \(g \circ f\).

\(\blacksquare\)

Reflection in the line at angle \(\beta\) is \(R_\beta = \rho_\beta\,\rho_r\,\rho_{-\beta}\), and multiplying out gives

Let \(A = 2\beta\) and \(B = 2\alpha\). The composite \(R_\beta R_\alpha\) has entries \(\cos A\cos B + \sin A\sin B = \cos(A-B)\), \(-\cos A\sin B + \sin A\cos B = \sin(A-B)\), and so on. The matrix is \(\rho_{A-B} = \rho_{2(\beta-\alpha)}\).

\(\blacksquare\)

Let \(n\) be a unit normal to the lines. Reflection in \(l_1\) sends \(X\) to \(X - 2\big((X - a)\cdot n\big)n\) for a point \(a\) on \(l_1\). Reflection in \(l_2\), a line at distance \(h\) further along \(n\), has the same form with \(a\) replaced by a point \(a + hn\). Composing, the normal component of every point changes by exactly \(2h\) and the parallel component by nothing. That is a translation by \(2hn\) (up to sign, the direction from \(l_1\) to \(l_2\) doubled).

\(\blacksquare\)

For any \(z\), \(|f(z)| = |z|\) and \(|f(z) - 1| = |z - 1|\), so \(f(z)\) lies on the circle about \(0\) through \(z\) and on the circle about \(1\) through \(z\). These two circles meet only at \(z\) and \(\bar z\), so \(f(z) \in \{z, \bar z\}\). In particular \(f\) fixes every real number.

Suppose \(f(z_0) = z_0\) for some non-real \(z_0\), and let \(w\) be any point. If \(f(w) = \bar w \ne w\), then \(|\bar w - z_0| = |f(w) - f(z_0)| = |w - z_0|\), so \(z_0\) is equidistant from \(w\) and \(\bar w\), which forces \(z_0\) to be real, a contradiction. So \(f(w) = w\) for every \(w\). Otherwise \(f(z) = \bar z\) for every non-real \(z\), and so for every \(z\).

\(\blacksquare\)

Let \(g(z) = e^{i\theta} z\), a rotation, so \(g\) and \(g^{-1}\) are isometries fixing \(0\). Then \(g^{-1} \circ f\) is an isometry fixing \(0\) and \(1\), since \(g^{-1}(e^{i\theta}) = 1\). By Statement 2.5.1, either \(g^{-1} f(z) = z\), giving \(f(z) = e^{i\theta} z\), or \(g^{-1} f(z) = \bar z\), giving \(f(z) = e^{i\theta} \bar z\).

\(\blacksquare\)

Let \(f(0) = c\) and \(T(z) = z + c\). Then \(f_1 = T^{-1} \circ f\) is an isometry with \(f_1(0) = 0\), so \(|f_1(1)| = 1\) and \(f_1(1) = e^{i\theta}\) for some \(\theta\). By Statement 2.5.2, \(f(z) = e^{i\theta} z + c\) or \(f(z) = e^{i\theta} \bar z + c\).

Case 1: \(f(z) = e^{i\theta} z + c\). If \(e^{i\theta} = 1\), \(f\) is a translation. Otherwise \(f\) has the fixed point \(a = \dfrac{c}{1 - e^{i\theta}}\), and \(f(z) - a = e^{i\theta}(z - a)\). So \(f\) is the rotation by \(\theta\) about \(a\).

Case 2: \(f(z) = e^{i\theta} \bar z + c\). Let \(u = e^{i\theta/2}\), and split \(c = c_\parallel + c_\perp\) with \(c_\parallel\) along \(u\) and \(c_\perp\) perpendicular to it. Since \(e^{i\theta}\overline{c_\perp} = -c_\perp\), the map \(z \mapsto e^{i\theta}\bar z + c_\perp\) fixes every point of the line \(\tfrac{c_\perp}{2} + \mathbb{R}u\), so it is the reflection in that line. Then \(f\) is that reflection followed by the translation by \(c_\parallel\) along the line. If \(c_\parallel = 0\), \(f\) is a reflection; otherwise it is a glide reflection.

\(\blacksquare\)

A translation by \(v\) is the composite of reflections in two lines perpendicular to \(v\), at distance \(\tfrac{|v|}{2}\) apart (Statement 2.4.2). A rotation about \(a\) by \(\theta\) is the composite of reflections in two lines through \(a\) at angle \(\tfrac{\theta}{2}\) (Statement 2.4.1). A glide reflection is a reflection followed by a translation along its mirror, which is the reflection composed with two more reflections. By Statement 2.5.3, every isometry is one of these.

\(\blacksquare\)

Since \(T\) is an isometry, \(|u - v|^2 = |T(u) - T(v)|^2\). Putting \(v = 0\) gives \(|u|^2 = |T(u)|^2\), and similarly for \(v\). Expanding the first identity and cancelling gives \(u\cdot v = T(u)\cdot T(v)\).

\(\blacksquare\)

Case 1: \(P = (x_0, y_1)\) and \(Q = (x_0, y_2)\) lie on a vertical line, with \(y_1 \le y_2\). For any curve from \(P\) to \(Q\), \(|y'(t)| \le \sqrt{x'(t)^2 + y'(t)^2}\), so

The vertical segment has length exactly \(y_2 - y_1\), so it is shortest.

Case 2: otherwise, translate \(P\) to the origin, rotate so that \(Q\) lies on the vertical axis, and translate back. This composite is an isometry (Statements 2.2.1 and 2.2.3), so it does not change any lengths. After it, \(P\) and \(Q\) are on a vertical line and Case 1 applies; the shortest path is the segment \(PQ\).

\(\blacksquare\)

Let \(g\) be a shortest curve from \(P\) to \(Q\) and \(\phi\) an isometry, so \(\ell(\phi(g)) = \ell(g) = d(P,Q) = d(\phi(P), \phi(Q))\). If \(\phi(g)\) were not shortest, there would be a curve \(\gamma\) from \(\phi(P)\) to \(\phi(Q)\) with \(\ell(\gamma) \lt \ell(\phi(g))\). Then \(\phi^{-1}(\gamma)\) joins \(P\) to \(Q\) and has length \(\ell(\gamma) \lt d(P,Q)\), which is impossible. Applying this to every short piece of a geodesic proves the statement.

\(\blacksquare\)

Let \(ABC\) be the triangle, and let \(\ell\) be the line through \(A\) parallel to \(BC\). It exists and is unique by the parallel postulate. The angle between \(\ell\) and \(AB\) on one side of \(A\) is equal to \(\beta\), since they are alternate angles across the transversal \(AB\). Likewise the angle between \(\ell\) and \(AC\) on the other side is equal to \(\gamma\), by alternate angles across \(AC\).

The three angles at \(A\), namely the two just found and \(\alpha\), together fill the straight angle along \(\ell\). So \(\alpha + \beta + \gamma = \pi\).

Note. The step that uses alternate angles is where the parallel postulate enters. Without it, the angle sum is not fixed, and in Part 3.8 we will see that it is less than \(\pi\) in the hyperbolic plane.

\(\blacksquare\)

Clearly \(d \ge 0\), and \(d = 0\) if and only if \(x_2 - x_1 = 0\) and \(y_2 - y_1 = 0\), that is, if and only if the points are equal. Symmetry holds because \(|x_2 - x_1| = |x_1 - x_2|\) and \(|y_2 - y_1| = |y_1 - y_2|\). For the triangle inequality, take a third point \((x_3, y_3)\). By the triangle inequality for real numbers,

\(\blacksquare\)

The first two axioms are immediate: a sum of squares is zero only when both squares are zero, and \((x_2 - x_1)^2 = (x_1 - x_2)^2\). For the triangle inequality, let \(a = x_2 - x_1\), \(b = x_3 - x_2\), \(c = y_2 - y_1\) and \(d = y_3 - y_2\). We need

Both sides are non-negative, so we may square. After cancelling \(a^2 + b^2 + c^2 + d^2\) from both sides, this becomes

which is the Cauchy–Schwarz inequality.

\(\blacksquare\)

The maximum of two non-negative numbers is zero exactly when both are zero, so \(d = 0\) if and only if \(x_1 = x_2\) and \(y_1 = y_2\). Symmetry is clear. For the triangle inequality,

and in the same way \(d(P_1,P_2) + d(P_2,P_3) \ge |y_1 - y_3|\). So the left side is at least the larger of the two, which is \(d(P_1, P_3)\).

\(\blacksquare\)

The first two axioms are as before: \(d_p(x,y) = 0\) if and only if every \(|x_i - y_i| = 0\), and \(|x_i - y_i| = |y_i - x_i|\). The triangle inequality is Minkowski's inequality, \(\left(\sum |a_i + b_i|^p\right)^{1/p} \le \left(\sum |a_i|^p\right)^{1/p} + \left(\sum |b_i|^p\right)^{1/p}\), applied with \(a_i = x_i - y_i\) and \(b_i = y_i - z_i\). When \(p = 2\) this is the Euclidean metric on \(\mathbb{R}^n\), and Minkowski's inequality reduces to Cauchy–Schwarz as in Statement 1.3.

\(\blacksquare\)

Write \(\varphi(t) = \dfrac{t}{1+t} = 1 - \dfrac{1}{1+t}\) for \(t \ge 0\), so \(d' = \varphi \circ d\). Then \(\varphi(t) = 0 \iff t = 0\), so \(d'(x,y) = 0 \iff d(x,y) = 0 \iff x = y\), and symmetry is inherited from \(d\).

For the triangle inequality we use two facts about \(\varphi\). It is increasing, because \(\dfrac{1}{1+t}\) decreases. And it is subadditive: for \(a, b \ge 0\),

Since \(d(x,z) \le d(x,y) + d(y,z)\), we get \(d'(x,z) = \varphi(d(x,z)) \le \varphi\big(d(x,y) + d(y,z)\big) \le d'(x,y) + d'(y,z)\).

Finally, \(d'(x,y) = 1 - \dfrac{1}{1 + d(x,y)} \lt 1\), so \(0 \le d' \lt 1\) and \(d'\) is bounded.

\(\blacksquare\)

The first two axioms are inherited from \(d\), since \(\sqrt{t} = 0\) if and only if \(t = 0\). For the triangle inequality, \(\sqrt{a+b} \le \sqrt a + \sqrt b\) for \(a, b \ge 0\), because squaring both sides gives \(a + b \le a + b + 2\sqrt{ab}\). Since \(\sqrt{\ }\) is increasing,

More generally, \(\varphi \circ d\) is a metric whenever \(\varphi\) is increasing, subadditive and satisfies \(\varphi(t) = 0 \iff t = 0\), as in Statement 1.6.

\(\blacksquare\)

For non-negative numbers, \(\max(a,b) \le a + b\). So \(d(x,z) \le \max\big(d(x,y), d(y,z)\big) \le d(x,y) + d(y,z)\).

\(\blacksquare\)

Let \(d(x,y) = \alpha\), \(d(y,z) = \beta\) and \(d(x,z) = \gamma\), and suppose for contradiction that the largest is strictly bigger than the other two. Without loss of generality \(\alpha \gt \beta\) and \(\alpha \gt \gamma\). But the strong triangle inequality gives \(\alpha = d(x,y) \le \max\big(d(x,z), d(z,y)\big) = \max(\gamma, \beta) \lt \alpha\), a contradiction.

\(\blacksquare\)

By definition \(|x - y|_p \ge 0\), with equality exactly when \(x - y = 0\). Since \(\nu_p(-w) = \nu_p(w)\), we have \(|x - y|_p = |y - x|_p\).

The key step is \(|w + z|_p \le \max\big(|w|_p, |z|_p\big)\) for all \(w, z \in \mathbb{Q}\). If \(w\), \(z\) or \(w + z\) is zero this is immediate, so assume none is. Write \(w = p^{a} m\) and \(z = p^{b} n\), where \(m\) and \(n\) are rationals whose numerators and denominators are coprime to \(p\), and without loss of generality \(a \le b\). Then

The bracket is a rational whose denominator is coprime to \(p\), so its valuation is at least \(0\). Hence \(\nu_p(w+z) \ge a\), which means \(|w+z|_p \le p^{-a} = |w|_p = \max\big(|w|_p, |z|_p\big)\). When \(a \lt b\) the bracket's numerator is also coprime to \(p\), so in fact \(|w + z|_p = |w|_p\).

Now apply this with \(w = x - z\) and \(z' = z - y\): \(|x - y|_p = |(x - z) + (z - y)|_p \le \max\big(|x - z|_p, |z - y|_p\big)\), that is, \(d(x,y) \le \max\big(d(x,z), d(z,y)\big)\).

\(\blacksquare\)

An ultrametric satisfies \(d(x,x) = 0\), symmetry and the strong triangle inequality, so it is an ultrapseudometric. Conversely, an ultrapseudometric is an ultrametric exactly when it satisfies the first axiom of a metric, that is, when \(d(x,y) = 0\) forces \(x = y\), since symmetry is already part of the definition.

\(\blacksquare\)

Let \(x \in B_r(p)\). Then \(d(x,p) \lt r\), and \(d(p,q) \lt r\) because \(q \in B_r(p)\). So \(d(x,q) \le \max\big(d(x,p), d(p,q)\big) \lt r\), and \(x \in B_r(q)\). Hence \(B_r(p) \subseteq B_r(q)\). Since \(d(q,p) \lt r\) as well, the same argument with \(p\) and \(q\) swapped gives \(B_r(q) \subseteq B_r(p)\). The proof for closed balls is identical, with \(\le\) in place of \(\lt\).

\(\blacksquare\)

Take balls \(B_{r_1}(x_1)\) and \(B_{r_2}(x_2)\) with a common point \(a\), and without loss of generality \(r_1 \le r_2\). We show \(B_{r_1}(x_1) \subseteq B_{r_2}(x_2)\). Let \(b \in B_{r_1}(x_1)\). Then \(d(b,a) \le \max\big(d(b,x_1), d(x_1,a)\big) \lt r_1 \le r_2\), and \(d(a,x_2) \lt r_2\). So

that is, \(b \in B_{r_2}(x_2)\).

\(\blacksquare\)

Suppose \(z \in B_r(x) \cap B_r(y)\). By Statement 1.16, \(z\) is a centre of both balls: \(B_r(x) = B_r(z) = B_r(y)\).

\(\blacksquare\)

Each such open ball lies inside \(B_r[a]\): if \(x \in B_r[a]\) and \(y \in B_r(x)\), then \(d(a,y) \le \max\big(d(a,x), d(x,y)\big) \le r\). Every point \(x\) of \(B_r[a]\) lies in one of the open balls, namely \(B_r(x)\) itself. And by Statement 1.18, any two of these open balls are either disjoint or identical. So they partition \(B_r[a]\).

\(\blacksquare\)

This is the homothety of Statement 3.4.2 with centre the origin, which is an isometry. So every side of the scaled triangle has the same hyperbolic length as the corresponding side of the original. The homothety is also a Möbius map of the form \(z \mapsto \lambda z\) with \(ad - bc = \lambda \gt 0\), so it is conformal and preserves angles between curves, which is why the angles agree too.

\(\blacksquare\)

Step 1. By homogeneity (Statement 3.4.6) there is an isometry sending \(A'\) to \(A\). The rotations about a point, which are Möbius maps of the form in Part 3.6, act transitively on the circle of points at a given distance from that point. So after a further isometry we may assume \(A' = A\) and \(B' = B\).

Step 2. The angle at \(A\) is equal, so \(C'\) lies on the geodesic through \(A\) and \(C\), possibly on the other side of \(AB\). If it lies on the other side, reflect in the geodesic \(AB\); reflections in geodesics are isometries (anti-linear maps of Part 3.6). After this, \(C'\) lies on the ray from \(A\) through \(C\). Likewise, the equal angle at \(B\) puts \(C'\) on the ray from \(B\) through \(C\).

Step 3. Two distinct geodesics meet in at most one point of \(\mathbb{H}^2\): two circles centred on the axis meet in points symmetric about the axis, so at most one of them lies in the upper half-plane. The geodesics \(AC\) and \(BC\) are distinct, and both contain \(C\) and \(C'\), so \(C' = C\).

\(\blacksquare\)

The triangle is the region \(R = \{(x,y) : -1 \le x \le 1,\ y \ge \sqrt{1 - x^2}\}\). For each fixed \(x\) in \((-1,1)\), the inner integral is

so

At \(-1\), the semicircle has a vertical tangent, the same direction as the side \(x = -1\), so the angle between them is \(0\). The same holds at \(1\). The vertex at infinity is an ideal point, and the two vertical sides are parallel, so that angle is \(0\) too.

\(\blacksquare\)

This is the Gauss–Bonnet theorem for the hyperbolic plane, which has curvature \(-1\). For a geodesic triangle, Gauss–Bonnet says that the area equals the total exterior turning \(\pi - (\alpha+\beta+\gamma)\), because the geodesic sides turn by nothing. The notes do not develop Gauss–Bonnet, so this is an outline, not a full proof here.

The formula is checked numerically in the simulation above: for random triangles, the Monte Carlo integral of \(dx\,dy/y^2\) over the triangle agreed with \(\pi - (\alpha+\beta+\gamma)\) to about three decimal places. Statement 3.8.1 is a special case, since all its angles are \(0\). Once the formula holds, \(\alpha+\beta+\gamma \lt \pi\) follows because the area is positive.

\(\blacksquare\)

If \(X\) is sent to \(X^{*}\), then \(X^{*}\) lies on the ray \(OX\) and \(OX \cdot OX^{*} = k^2\). The same equation, read the other way, says that \(X\) is the point on the ray \(OX^{*}\) with \(OX^{*} \cdot OX = k^2\). So \(X^{*}\) is sent back to \(X\), and \(f(f(X)) = X\). The points \(O\) and \(P_\infty\) are swapped and swapped back.

\(\blacksquare\)

Since \(OA \cdot OA^{*} = OB \cdot OB^{*} = k^2\), the converse of the power of a point theorem shows that \(A, B, B^{*}, A^{*}\) are concyclic. Equivalently, \(\dfrac{OA}{OB^{*}} = \dfrac{OB}{OA^{*}}\) and the triangles \(OAB\) and \(OB^{*}A^{*}\) share the angle at \(O\), so they are similar (SAS), which gives \(\angle OAB = \angle OB^{*}A^{*}\).

\(\blacksquare\)

By Statement 3.1.2, triangles \(OAB\) and \(OB'A'\) are similar, so

(If \(O\), \(A\), \(B\) are collinear, the formula follows directly from the definition.) In particular, inversion preserves the distance between \(A\) and \(B\) exactly when \(OA \cdot OB = k^2\).

\(\blacksquare\)

Write \(z = re^{i\theta}\) and let \(f(z)\) be its image. Then \(|z|\,|f(z)| = 1\), so \(|f(z)| = \tfrac1r\), and \(f(z)\) lies on the same ray, so \(\arg f(z) = \theta\). Hence

With \(z = x + iy\) and \(|z|^2 = x^2 + y^2\), this is \(\left(\dfrac{x}{x^2+y^2},\ \dfrac{y}{x^2+y^2}\right)\).

\(\blacksquare\)

Every point \(A \ne O\) of the line is sent to a point of the ray \(OA\), which lies on the same line, and \(O\) and \(P_\infty\) are swapped. So the line is sent into itself, and by Statement 3.1.1 onto itself. (It is not fixed pointwise: only the two points where it crosses \(\omega\) stay put.)

\(\blacksquare\)

Let \(l'\) be the image. Since \(P_\infty\) lies on \(l\), its image \(O\) lies on \(l'\). Let \(X\) be the foot of the perpendicular from \(O\) to \(l\), and \(A\) any other point of \(l\). By Statement 3.1.2, \(\angle OA'X' = \angle OXA = 90^\circ\). So every \(A'\) sees the segment \(OX'\) at a right angle, and lies on the circle with diameter \(OX'\).

\(\blacksquare\)

From the proof of Statement 3.1.6, \(OX'\) is a diameter of \(l'\), and \(X'\) lies on the ray \(OX\), which is perpendicular to \(l\). The centre of \(l'\) is the midpoint of \(OX'\), which lies on this line.

\(\blacksquare\)

By Statements 3.1.1 and 3.1.6, a circle through \(O\) is sent to a line not through \(O\), and a line not through \(O\) is sent to a circle through \(O\). By Statement 3.1.5, a line through \(O\) is sent to itself. The only case left is a circle \(\gamma\) not through \(O\), whose image is a circle not through \(O\).

Neither \(O\) nor \(P_\infty\) lies on \(\gamma\), so neither lies on its image \(\gamma'\). Let \(AB\) be the diameter of \(\gamma\) whose line passes through \(O\), and let \(C\) be any other point of \(\gamma\), so \(\angle ACB = 90^\circ\). Since \(OA \cdot OA' = OB \cdot OB' = OC \cdot OC' = k^2\), Statement 3.1.2 gives \(\triangle OCA \sim \triangle OA'C'\) and \(\triangle OCB \sim \triangle OB'C'\), so \(\angle OC'A' = \angle OAC\) and \(\angle OC'B' = \angle OBC\). If \(O\) lies outside \(\gamma\), with \(A\) between \(O\) and \(B\), then \(\angle OAC = 180^\circ - \angle CAB\), and therefore

So every \(C'\) lies on the circle with diameter \(A'B'\). (If \(O\) lies inside \(\gamma\), the same angle chase works with the two angles added instead of subtracted.) Hence \(\gamma'\) is a circle.

\(\blacksquare\)

Use the diameter \(AB\) of \(\gamma\) through \(O\) from the proof of Statement 3.1.8. Its image \(A'B'\) is a diameter of \(\gamma'\) lying on the same line through \(O\). Both centres are midpoints of diameters on that line, so they lie on the line through \(O\) and \(AB\).

\(\blacksquare\)

The points \(P\) and \(Q\) lie on \(\omega\), so they are fixed. Since \(OP \perp A'P\) and \(OQ \perp A'Q\), the quadrilateral \(A'POQ\) is cyclic, with \(OA'\) as a diameter. This circle passes through \(O\), so by Statement 3.1.6 (read backwards, using Statement 3.1.1) its image is a line. That line passes through the images of \(A'\), \(P\) and \(Q\), which are \(A\), \(P\) and \(Q\).

\(\blacksquare\)

Inversion is a bijection of the plane together with \(P_\infty\), so it preserves the number of common points. Two circles are tangent when they meet in exactly one point; a circle and a line likewise; and two lines meet at exactly one point (other than \(P_\infty\)) or are parallel, meeting only at \(P_\infty\). In particular, inverting about the point where two circles touch sends them to two parallel lines.

\(\blacksquare\)

Symmetry. Running a curve backwards does not change its length, so the curves from \(P\) to \(Q\) and from \(Q\) to \(P\) have the same lengths and the same infimum.

Triangle inequality. A curve from \(P\) to \(Q\) followed by a curve from \(Q\) to \(R\) is a curve from \(P\) to \(R\), so \(d(P,R) \le \ell(\gamma_1) + \ell(\gamma_2)\). Taking infima over \(\gamma_1\) and \(\gamma_2\) gives \(d(P,R) \le d(P,Q) + d(Q,R)\).

Positivity. Clearly \(d(P,P) = 0\). Suppose \(P = (x(a), y(a)) \ne Q\), and let \(\gamma\) be any curve from \(P\) to \(Q\). If \(\gamma\) never rises above the line \(y = 2y(a)\), then \(y(t) \le 2y(a)\) throughout, so

If \(\gamma\) does reach the line \(y = 2y(a)\), let \(t = r\) be the first time it does. Up to that time \(y(t) \le 2y(a)\), so

Either way \(\ell_{\text{hyp}}(\gamma) \ge \min\!\left(\tfrac12, \tfrac{d_{\text{euc}}(P,Q)}{2y(a)}\right)\), a positive number independent of \(\gamma\). So \(d_{\text{hyp}}(P,Q) \gt 0\).

\(\blacksquare\)

Parametrise the segment by \(x(t) = t\), \(y(t) = y\) for \(t\) in an interval of length \(e\). Then \(\sqrt{x'(t)^2 + y'(t)^2} = 1\), so

\(\blacksquare\)

This is the substitution argument of Statement 2.1.1, with the extra factor \(1/y\) carried along. If \(x(t) = X(\theta(t))\) and \(y(t) = Y(\theta(t))\), then \(\sqrt{x'^2 + y'^2} = \sqrt{X'^2 + Y'^2}\,\theta'(t)\) and \(y(t) = Y(\theta(t))\). The factor \(\theta'(t)\) is absorbed by the substitution \(\theta = \theta(t)\), so the integrals agree.

\(\blacksquare\)

Suppose \(y_1 \le y_2\). For any curve from \(P\) to \(Q\), \(|y'(t)| \le \sqrt{x'(t)^2 + y'(t)^2}\) and \(y(t) \gt 0\), so

Equality holds when \(x'(t) = 0\) and \(y' \ge 0\) throughout, that is, along the vertical segment. If \(y_1 \gt y_2\), swap \(P\) and \(Q\). Either way \(d_{\text{hyp}}(P,Q) = \left|\ln \tfrac{y_2}{y_1}\right|\).

\(\blacksquare\)

Since \(PQ\) is not vertical, its perpendicular bisector is not horizontal, so it meets the \(x\)-axis at some point \(R\). Then \(RP = RQ\), so the Euclidean circle \(C\) centred at \(R\) through \(P\) passes through \(Q\). Let \(\phi_1\) be the horizontal translation sending one of the two points where \(C\) meets the axis to the origin. Then \(\phi_1(C)\) is a circle through the origin, centred at some \((a, 0)\). In polar coordinates its equation is \(r = 2a\cos\theta\).

Now let \(\phi_2\) be inversion in the unit circle. It replaces \(r\) by \(\tfrac1r\), so it sends \(\phi_1(C)\) to the curve \(r = \dfrac{1}{2a\cos\theta}\), that is, \(x = r\cos\theta = \dfrac{1}{2a}\), a vertical line. So \(\phi_2 \circ \phi_1\), an isometry by Statements 3.4.1, 3.4.3 and 3.4.5, sends \(P\) and \(Q\) onto one vertical line.

\(\blacksquare\)

The vertical case is Statement 3.3.1. Otherwise, let \(\zeta = \phi_2 \circ \phi_1\) be the isometry of Statement 3.3.2, so \(P' = \zeta(P)\) and \(Q' = \zeta(Q)\) lie on a vertical line, and the shortest path between them is the vertical segment. An isometry and its inverse preserve lengths, so \(\zeta^{-1}\) sends that segment to a shortest path from \(P\) to \(Q\), and the image of the segment lies on \(\zeta^{-1}\) of the vertical line, which is the circle \(C\). So the shortest path is the arc of the semicircle centred at \(R\) through \(P\) and \(Q\).

\(\blacksquare\)

The vertical line \(x = 2\) passes through \(P\) and does not meet \(g\). So does every semicircle centred at \((c, 0)\) through \(P\) with \(c \gt \tfrac54\). Its radius is \(\rho = \sqrt{(2-c)^2 + 1}\), and its leftmost point \(c - \rho\) is positive exactly when \(c^2 \gt (c-2)^2 + 1\), that is, when \(c \gt \tfrac54\). Such a semicircle lies entirely in \(x \gt 0\), so it misses \(g\). Hence infinitely many geodesics through \(P\) miss \(g\), where the postulate allows exactly one.

\(\blacksquare\)

Each map sends \(\mathbb{H}^2\) to itself and does not change \(y(t)\). It changes \(x(t)\) by a constant or by a sign, so \(x'(t)^2\) and \(y'(t)\) are unchanged. Hence the integrand of \(\ell_{\text{hyp}}\) is unchanged, and so is the length of every curve.

\(\blacksquare\)

With \(y_0 = 0\), the image of \(\gamma\) is \(\big(\lambda(x(t) - x_0) + x_0,\ \lambda y(t)\big)\), with derivatives \(\lambda x'(t)\) and \(\lambda y'(t)\). So

(If \(y_0 \ne 0\), the denominator would be \(\lambda(y - y_0) + y_0\) and the \(\lambda\)'s would no longer cancel. That is why the centre must lie on the axis.)

\(\blacksquare\)

It sends \(\mathbb{H}^2\) to itself, since the second coordinate keeps its sign. Let \(X(t) = \dfrac{x}{x^2+y^2}\) and \(Y(t) = \dfrac{y}{x^2+y^2}\). By the quotient rule, using \(\tfrac{d}{dt}(x^2 + y^2) = 2xx' + 2yy'\),

Squaring and adding, the cross terms cancel and \((y^2 - x^2)^2 + 4x^2y^2 = (x^2 + y^2)^2\), so \(X'^2 + Y'^2 = \dfrac{x'^2 + y'^2}{(x^2 + y^2)^2}\). Therefore

\(\blacksquare\)

Let the centre be \((x_0, 0)\) and the radius \(r\). The inversion is \(z \mapsto x_0 + \dfrac{r^2}{\overline{z - x_0}}\), the composite of four maps: translate horizontally by \(-x_0\), invert in the unit circle, apply the homothety with ratio \(r^2\) about the origin, and translate back by \(x_0\). Each step is an isometry by Statements 3.4.1, 3.4.2 and 3.4.3, and so is their composite by Statement 3.4.5.

\(\blacksquare\)

Let \(f\) and \(g\) be isometries and \(\gamma\) a curve. Since \(f\) preserves lengths, \(\ell_{\text{hyp}}(f(\gamma)) = \ell_{\text{hyp}}(\gamma)\). Applying the same fact to \(g\) and the curve \(f(\gamma)\) gives \(\ell_{\text{hyp}}(g(f(\gamma))) = \ell_{\text{hyp}}(f(\gamma)) = \ell_{\text{hyp}}(\gamma)\). So \(g \circ f\) preserves lengths, and hence distances.

\(\blacksquare\)

Let \(P = (x_1, y_1)\) and \(Q = (x_2, y_2)\). The homothety about the origin with ratio \(\tfrac{y_2}{y_1}\) sends \(P\) to \(\left(\tfrac{x_1 y_2}{y_1},\ y_2\right)\). The horizontal translation by \(x_2 - \tfrac{x_1 y_2}{y_1}\) then sends it to \(Q\). Both are isometries (Statements 3.4.1 and 3.4.2), and so is their composite.

\(\blacksquare\)

By Statement 3.3.1, \(B\) and \(C\) are the two points at distance \(l\) from \(A\) on the vertical line through \(A\). Let \(\Gamma\) be the circle with diameter \(BC\). Every point at distance \(l\) from \(A\) lies on some geodesic through \(A\), so it is enough to show that the two points at distance \(l\) from \(A\) along any other geodesic through \(A\) lie on \(\Gamma\).

Such a geodesic is a semicircle through \(A\) centred at a point \(F\) on the axis, with endpoints \(D\) and \(E\). Let \(G = (x_0, 0)\) be the foot of the vertical through \(A\). Since \(DE\) is a diameter, \(\angle DAE = 90^\circ\), and \(AG\) is the altitude of the right triangle \(DAE\), so

Invert in the circle with centre \(E\) and radius \(EA\). It is an isometry by Statement 3.4.4, since \(E\) lies on the axis, and it fixes \(A\). It sends the vertical line through \(A\) to the geodesic \(DE\), and it sends \(B\) and \(C\) to points \(B'\) and \(C'\) of \(DE\) at distance \(l\) from \(A\), lying on the rays \(EB\) and \(EC\).

Now \(BG \cdot GC = y_0 e^{-l} \cdot y_0 e^{l} = y_0^2 = AG^2 = GD \cdot GE\). This says that \(B\) is the orthocentre of the triangle \(DCE\): the triangles \(DGB\) and \(CGE\) are similar right triangles, so \(DB \perp CE\). The point \(C'\) lies on \(CE\) and on the semicircle with diameter \(DE\), so \(\angle DC'E = 90^\circ\), and \(C'\) is the foot of the perpendicular from \(D\) to \(CE\). Hence \(D\), \(B\), \(C'\) are collinear and \(\angle BC'C = 90^\circ\), so \(C'\) lies on the circle with diameter \(BC\). The same argument with \(B\) and \(C\) swapped puts \(B'\) on \(\Gamma\).

In coordinates, \(\Gamma\) has Euclidean centre \(\big(x_0,\ y_0\cosh l\big)\) and Euclidean radius \(y_0\sinh l\).

\(\blacksquare\)

The Euclidean centre is at height \(y_0\cosh l\), and \(\cosh l \gt 1\) for \(l \gt 0\), so it is strictly above \(A\), whose height is \(y_0\). Since the hyperbolic circle is a Euclidean circle, its Euclidean centre is a Euclidean notion; \(A\) is the centre only in the sense that it is the point the circle is defined around.

\(\blacksquare\)

Let \(d = d_{\text{hyp}}(P_0, P_1)\). By Statement 3.5.1, the hyperbolic circle \(\Gamma\) about \(P_1\) of radius \(d\) is a Euclidean circle centred on the \(y\)-axis, and \(P_0\) is its lowest point, so its tangent at \(P_0\) is horizontal.

If \(P_0\) is the point of \(g\) closest to \(P_1\), no point of \(g\) is inside \(\Gamma\), so \(g\) touches \(\Gamma\) at \(P_0\) and has a horizontal tangent there. The centre of \(g\) is on the \(x\)-axis directly below the point with horizontal tangent, so it is the origin, and the radius is \(y_0\).

Conversely, if \(g\) is the semicircle of radius \(y_0\) about the origin, every point of \(g\) has height at most \(y_0\), while every point of \(\Gamma\) other than \(P_0\) has height greater than \(y_0\). So \(g\) meets the closed disc bounded by \(\Gamma\) only at \(P_0\), and every other point of \(g\) is further from \(P_1\) than \(d\).

\(\blacksquare\)

Let \(r = \sqrt{y_1 y_2}\) and let \(\sigma(z) = r^2/\bar z\). This is inversion in the circle \(|z| = r\), so it is an isometry by Statement 3.4.4 (its centre is the origin, on the axis). Points of the circle are fixed by \(\sigma\), since \(|X| = r\) gives \(r^2/\bar X = X\). And \(\sigma(iy_1) = \dfrac{r^2}{-iy_1} = iy_2\), so \(\sigma\) swaps \(P\) and \(Q\).

So for any \(X\) on the circle, \(d(X,P) = d(\sigma X, \sigma P) = d(X,Q)\). The circle \(|z| = r\) is a geodesic by Statement 3.3.3, and it meets the \(y\)-axis at right angles at \((0,r)\), which is the hyperbolic midpoint of \(P\) and \(Q\) because \(\ln\tfrac{r}{y_1} = \ln\tfrac{y_2}{r}\).

Note. The simulation shows that no other points are equidistant from \(P\) and \(Q\). The converse is not proved in these notes, so treat it as an observed fact until you prove it.

\(\blacksquare\)

Multiplying numerator and denominator by \(c\bar z + d\),

which is positive whenever \(\operatorname{Im} z \gt 0\). Dividing all four coefficients by \(\sqrt{ad - bc}\) does not change the map, so we may assume \(ad - bc = 1\).

\(\blacksquare\)

Case \(a \ne 0\). Compose the horizontal translation by \(\tfrac{b}{a}\), inversion in the unit circle \(w \mapsto \tfrac{1}{\bar w}\), the homothety with ratio \(\tfrac{1}{a^2}\) about the origin, and the horizontal translation by \(\tfrac{c}{a}\):

Using \(1 + bc = ad\),

This is the anti-linear map. Composing it with one more inversion in the unit circle gives \(\dfrac{az + b}{cz + d}\), the linear map.

Case \(a = 0\). Then \(-bc = 1\), so \(c \ne 0\). The coefficients \((c,\ b + d,\ c,\ d)\) satisfy \(cd - (b+d)c = -bc = 1\) and have first entry \(c \ne 0\), so by the first case \(z \mapsto \dfrac{cz + b + d}{cz + d}\) is an isometry. Composing with the horizontal translation by \(-1\) gives

the linear map. One more inversion in the unit circle gives the anti-linear one.

The maps \(z \mapsto \dfrac{az+b}{cz+d}\) are called linear fractional, and \(z \mapsto \dfrac{c\bar z+d}{a\bar z+b}\) anti-linear fractional.

\(\blacksquare\)

Existence. Fix \(w \in W\). The map \(\varphi(v) = \langle Tv, w\rangle\) is linear in \(v\). Let \(e_1,\dots,e_n\) be an orthonormal basis of \(V\) and put \(u = \sum_j \overline{\varphi(e_j)}\, e_j\). Because the inner product is conjugate-linear in its second slot, \(\langle v, u\rangle = \sum_j \varphi(e_j)\langle v, e_j\rangle = \varphi\bigl(\sum_j \langle v,e_j\rangle e_j\bigr) = \varphi(v)\). Define \(T^{*}w = u\). Then \(\langle Tv, w\rangle = \langle v, T^{*}w\rangle\) for all \(v\).

Uniqueness. If \(u\) and \(u'\) both work for \(w\), then \(\langle v, u-u'\rangle = 0\) for every \(v\). Taking \(v = u-u'\) gives \(u = u'\).

Linearity. For scalars \(a,b\), \(\langle v, T^{*}(aw_1+bw_2)\rangle = \langle Tv, aw_1+bw_2\rangle = \bar a\langle Tv,w_1\rangle + \bar b\langle Tv,w_2\rangle = \langle v, aT^{*}w_1+bT^{*}w_2\rangle\) for all \(v\), so uniqueness gives linearity.

\(\blacksquare\)

Take \(w \in W\). If \(T^{*}w = 0\), then \(\langle Tv, w\rangle = \langle v, T^{*}w\rangle = 0\) for every \(v\), so \(w\) is orthogonal to \(\operatorname{range}T\). Conversely, if \(w \perp \operatorname{range}T\), then \(\langle v, T^{*}w\rangle = \langle Tv, w\rangle = 0\) for all \(v\); choosing \(v = T^{*}w\) gives \(\lVert T^{*}w\rVert^{2} = 0\). So \(\operatorname{null}T^{*} = (\operatorname{range}T)^{\perp}\).

For the second equation, apply the first to \(T^{*}\) in place of \(T\). Since \(T^{**} = T\) (both sides satisfy \(\langle T^{*}w, v\rangle = \langle w, Tv\rangle\)), this gives \(\operatorname{null}T = (\operatorname{range}T^{*})^{\perp}\). Taking orthogonal complements in the finite-dimensional space \(V\) gives \(\operatorname{range}T^{*} = (\operatorname{null}T)^{\perp}\).

\(\blacksquare\)

Since \(f_1,\dots,f_m\) is orthonormal, \(Te_k = \sum_j \langle Te_k, f_j\rangle f_j\). So the \((j,k)\) entry of the matrix of \(T\) is \(\langle Te_k, f_j\rangle\). Similarly the \((k,j)\) entry of the matrix of \(T^{*}\) is \(\langle T^{*}f_j, e_k\rangle\). By the defining equation this equals \(\langle f_j, Te_k\rangle\), which is the complex conjugate of \(\langle Te_k, f_j\rangle\). So the matrix of \(T^{*}\) is the conjugate transpose.

\(\blacksquare\)

For every \(v\), \(\lVert Tv\rVert^{2} = \langle Tv, Tv\rangle = \langle T^{*}Tv, v\rangle\) and \(\lVert T^{*}v\rVert^{2} = \langle TT^{*}v, v\rangle\). So \(\lVert Tv\rVert^{2} - \lVert T^{*}v\rVert^{2} = \langle Av, v\rangle\), where \(A = T^{*}T - TT^{*}\). The operator \(A\) is self-adjoint, because \((T^{*}T)^{*} = T^{*}T\) and \((TT^{*})^{*} = TT^{*}\).

So the statement reduces to: a self-adjoint \(A\) with \(\langle Av,v\rangle = 0\) for all \(v\) must be zero. For any \(u, w\), expanding gives \(0 = \langle A(u+w), u+w\rangle = \langle Au,u\rangle + \langle Aw,w\rangle + \langle Au,w\rangle + \langle Aw,u\rangle\). Here \(\langle Aw,u\rangle = \langle w, Au\rangle = \overline{\langle Au,w\rangle}\), so \(\operatorname{Re}\langle Au,w\rangle = 0\) for all \(u,w\). Replacing \(w\) by \(iw\) in the complex case gives \(\operatorname{Im}\langle Au,w\rangle = 0\) too, so \(\langle Au,w\rangle = 0\). (In the real case the first step already suffices.) Taking \(w = Au\) gives \(Au = 0\).

Hence \(T^{*}T = TT^{*}\) exactly when \(\lVert Tv\rVert = \lVert T^{*}v\rVert\) for all \(v\).

\(\blacksquare\)

If the list is a basis, it spans, so every vector has at least one representation. If a vector had two, subtracting them would give a nontrivial zero combination, contradicting independence. Hence exactly one.

Conversely, if every vector has exactly one representation, then the list spans (since representations exist) and the zero vector has only one representation, the trivial one. So the list is independent, and it is a basis.

\(\blacksquare\)

Among all spanning sublists of the given spanning list, choose one of smallest length. We claim this sublist is independent. If not, the dependence lemma (Statement 2.1.2) gives a vector in it lying in the span of the others. Removing that vector keeps the span, since the removed vector is a combination of what remains. This yields a shorter spanning sublist, contradicting minimality.

A spanning, independent sublist is a basis.

\(\blacksquare\)

Let \(w_1,\dots,w_m\) be a spanning list of \(V\), which exists by finite-dimensionality. Start with the independent list \(u_1,\dots,u_n\). Go through \(w_1,\dots,w_m\) in order, adjoining \(w_j\) to the current list exactly when \(w_j\) is not in its span.

The current list stays independent. If \(w_j\) is not in the span of the current list and \(a w_j + \sum b_i u_i = 0\) with \(a \ne 0\), then \(w_j\) is a combination of the rest, a contradiction; so \(a=0\), and then all \(b_i=0\) by independence of the current list. At the end every \(w_j\) lies in the span of the final list, so that list spans \(\operatorname{span}(w_1,\dots,w_m)=V\). It is therefore a basis extending the \(u_i\).

\(\blacksquare\)

Build up the independent list one vector at a time while keeping a spanning list of length \(m\). Start with the spanning list \(v_1,\dots,v_m\). Put \(u_1\) in front: the new list \(u_1,v_1,\dots,v_m\) is dependent, because \(u_1\) is a combination of the \(v_j\). Since \(u_1 \neq 0\) by independence, the dependence lemma removes some \(v_j\) and keeps the span. The list now has \(m\) vectors, \(u_1\) among them, and spans \(V\).

Repeat. At step \(k\), the list has the form \(u_1,\dots,u_{k-1}\) together with \(m-k+1\) of the \(v_j\), and it spans \(V\). Adding \(u_k\) in front gives a dependent list, so the dependence lemma removes a vector. That vector cannot be one of the \(u_i\), because a \(u_i\) in the span of earlier vectors would contradict independence. So a \(v_j\) is removed. If \(k-1=m\) were reached with \(u_{m+1}\) to add, \(u_{m+1}\) would lie in the span of \(u_1,\dots,u_m\), a contradiction. Hence \(n \le m\).

\(\blacksquare\)

Let \(B\) and \(B'\) be bases, of lengths \(m\) and \(n\). Since \(B\) spans and \(B'\) is independent, Statement 2.2.4 gives \(n \le m\). Exchanging the roles gives \(m \le n\). Hence \(m=n\), and \(\dim V\) is well defined.

\(\blacksquare\)

Build a list in \(U\) by starting empty and adding a vector of \(U\) that is not in the span of the list so far, as long as one exists. The list stays independent, and it cannot be longer than \(\dim V\): an independent list in \(V\) has length at most the length of a basis of \(V\), by Statement 2.2.4. So the process stops, with an independent list in \(U\) that spans \(U\). That list is a basis of \(U\), so \(U\) is finite-dimensional and \(\dim U \le \dim V\).

\(\blacksquare\)

Let \(u_1,\dots,u_k\) be a basis of \(U\cap W\). Extend it, by Statement 2.2.3 applied inside \(U\), to a basis \(u_1,\dots,u_k,v_1,\dots,v_p\) of \(U\). Extend it likewise inside \(W\) to a basis \(u_1,\dots,u_k,w_1,\dots,w_q\) of \(W\). Then \(\dim U = k+p\), \(\dim W = k+q\), and \(\dim(U\cap W)=k\).

We claim \(u_1,\dots,u_k,v_1,\dots,v_p,w_1,\dots,w_q\) is a basis of \(U+W\). It spans, since it contains spanning lists of \(U\) and of \(W\). For independence, suppose \(\sum a_i u_i + \sum b_j v_j + \sum c_l w_l = 0\). Set \(x = \sum b_j v_j\). Then \(x \in U\), and \(x = -\sum a_i u_i - \sum c_l w_l\) lies in \(W\), since each \(u_i \in W\). So \(x \in U\cap W\), and \(x = \sum d_i u_i\) for some \(d_i\). Then \(\sum d_i u_i - \sum b_j v_j = 0\), and since \(u_1,\dots,u_k,v_1,\dots,v_p\) is independent, all \(b_j=0\) and \(d_i=0\). The original relation becomes \(\sum a_i u_i + \sum c_l w_l = 0\), and independence of the basis of \(W\) forces all \(a_i, c_l = 0\).

The basis has \(k+p+q\) vectors, so \(\dim(U+W) = k+p+q = (k+p)+(k+q)-k = \dim U+\dim W-\dim(U\cap W)\).

\(\blacksquare\)

Expand each argument using linearity: \(b(\sum_ix_ie_i,\sum_jy_je_j)=\sum_i x_i\,b(e_i,\sum_jy_je_j)=\sum_{i,j}x_iy_j\,b(e_i,e_j)\).

For the change of basis, \(b(f_k,f_l)=\sum_{i,j}P_{ik}P_{jl}\,b(e_i,e_j)=\sum_{i,j}P_{ik}A_{ij}P_{jl}=(P^{\top}AP)_{kl}\).

\(\blacksquare\)

Existence. Let \(e_1,\dots,e_n\) be an orthonormal basis. For fixed \(u\), the map \(v\mapsto b(u,v)\) is linear. Put \(w_u=\sum_jb(u,e_j)e_j\). Writing \(v=\sum_jv_je_j\), we get \(\langle w_u,v\rangle=\sum_jb(u,e_j)v_j=b(u,v)\).

Uniqueness and linearity. If two vectors \(w,w'\) both satisfy \(\langle w,v\rangle=b(u,v)=\langle w',v\rangle\) for all \(v\), then \(w-w'\) is orthogonal to everything, so \(w=w'\). Define \(Tu=w_u\). Then \(\alpha w_u+\beta w_{u'}\) satisfies the defining property for \(\alpha u+\beta u'\), because \(b\) is linear in its first slot. So \(T\) is linear.

Self-adjointness. \(\langle Tu,v\rangle=b(u,v)=b(v,u)=\langle Tv,u\rangle=\langle u,Tv\rangle\). Conversely, if \(T\) is self-adjoint, then \(b(u,v)=\langle Tu,v\rangle\) is bilinear and symmetric.

\(\blacksquare\)

Let \(T\) be the self-adjoint operator of 9.1.2, so \(q(v)=b(v,v)=\langle Tv,v\rangle\). By the spectral theorem for self-adjoint operators on a real inner product space (Part 7.2), there is an orthonormal basis \(e_1,\dots,e_n\) of eigenvectors, \(Te_j=\lambda_je_j\), with real \(\lambda_j\).

For \(v=\sum_jx_je_j\), \(\langle Tv,v\rangle=\sum_{j,k}x_jx_k\lambda_j\langle e_j,e_k\rangle=\sum_j\lambda_jx_j^2\), because the basis is orthonormal.

\(\blacksquare\)

If every \(\lambda_j\gt 0\) and \(v\ne 0\), then some \(x_k\neq 0\), so \(q(v)=\sum_j\lambda_jx_j^2\gt 0\).

Conversely, if \(\lambda_k\le 0\) for some \(k\), then \(q(e_k)=\lambda_k\le 0\), and \(e_k\neq 0\). So \(q\) is not positive definite.

\(\blacksquare\)

The coordinate identity. For every \(v \in V\) we have \([v]_B = P[v]_{B'}\). Indeed, write \(v = \sum_k x'_k b'_k\), where \(x'\) is the coordinate vector \([v]_{B'}\). Then \(v = \sum_k x'_k \sum_j P_{jk} b_j\), so the \(j\)th coordinate of \(v\) in \(B\) is \(\sum_k P_{jk}x'_k\), which is the \(j\)th entry of \(P[v]_{B'}\). Exchanging the roles of \(B\) and \(B'\), let \(Q\) be the matrix whose \(k\)th column is \([b_k]_{B'}\); the same argument gives \([v]_{B'} = Q[v]_B\) for every \(v\).

Substituting one identity into the other gives \([v]_B = PQ[v]_B\) for every \(v\). Taking \(v = b_j\), whose coordinate vector \([b_j]_B\) is the standard basis vector \(e_j\), gives \(PQe_j = e_j\) for every \(j\). Hence \(PQ = I\). Similarly \([v]_{B'} = QP[v]_{B'}\) for every \(v\), and taking \(v = b'_j\) gives \(QP = I\). So \(P\) is invertible with \(P^{-1} = Q\).

\(\blacksquare\)

Write \(A = [T]_B\) and \(A' = [T]_{B'}\). By definition \([Tv]_{B'} = A'[v]_{B'}\) for every \(v\). The coordinate identity from the proof of Statement 3.4.1 applies to the vector \(Tv\) and to \(v\), so

But \([v]_B = P[v]_{B'}\), so the left side is \(AP[v]_{B'}\). Therefore \((AP - PA')[v]_{B'} = 0\) for every \(v\). Taking \(v = b'_k\), so that \([v]_{B'} = e_k\), shows that the \(k\)th column of \(AP - PA'\) is zero. Hence \(AP = PA'\), and multiplying on the left by \(P^{-1}\) gives \(A' = P^{-1}AP\).

\(\blacksquare\)

Alternating forms are determined by one value. Let \(f\) be alternating, and \(e_1,\dots,e_n\) a basis. Expanding each argument in the basis gives \(f(v_1,\dots,v_n)=\sum a_{1i_1}\cdots a_{ni_n}f(e_{i_1},\dots,e_{i_n})\). Terms with a repeated index vanish. Replacing two arguments \(x\) and \(y\) by \(x+y\) in an alternating form gives \(f(\dots x\dots y\dots)+f(\dots y\dots x\dots)=0\), so swapping two arguments changes the sign. Therefore \(f(e_{\sigma(1)},\dots,e_{\sigma(n)})=\operatorname{sgn}(\sigma)f(e_1,\dots,e_n)\), and \(f\) is determined by \(f(e_1,\dots,e_n)\). So the space has dimension at most \(1\).

A nonzero alternating form exists. Let \(D(v_1,\dots,v_n)\) be the determinant (Leibniz formula) of the matrix whose \(j\)th column holds the coordinates of \(v_j\) in the basis. Each term of the Leibniz sum uses one entry from each column, so \(D\) is linear in each argument. If columns \(j\) and \(k\) are equal, pair each permutation \(\sigma\) with \(\sigma\circ(j\ k)\). These two terms have opposite signs and equal products, since the two columns agree. So the sum vanishes, and \(D\) is alternating. Also \(D(e_1,\dots,e_n)=1\).

\(\blacksquare\)

Take \(f=D\), the Leibniz form with respect to the given basis, so \(f(e_1,\dots,e_n)=1\). The \(j\)th column of \(A\) holds the coordinates of \(Te_j\), so \(f(Te_1,\dots,Te_n)=\det A\). By definition, \(f(Te_1,\dots,Te_n)=(\det T)\,f(e_1,\dots,e_n)=\det T\). Hence \(\det T=\det A\).

\(\blacksquare\)

Let \(f\neq 0\) be alternating. By definition, \(f(Sw_1,\dots,Sw_n)=\det S\cdot f(w_1,\dots,w_n)\) for all \(w_i\). Taking \(w_i=Tv_i\) and using the definition of \(\det T\), \(f(STv_1,\dots,STv_n)=\det S\cdot f(Tv_1,\dots,Tv_n)=\det S\cdot\det T\cdot f(v_1,\dots,v_n)\).

The left side is \(\det(ST)\,f(v_1,\dots,v_n)\) by definition. Taking \(v_i=e_i\) with \(f(e)\neq 0\) gives \(\det(ST)=\det S\det T\).

\(\blacksquare\)

If \(T\) is invertible. Then \(\det T\cdot\det T^{-1}=\det(I)=1\), since \(f(Iv_1,\dots,Iv_n)=f(v_1,\dots,v_n)\). So \(\det T\ne 0\).

If \(T\) is not invertible. In finite dimensions \(T\) is then not surjective, so \(\operatorname{range}T\) has dimension less than \(n\). The \(n\) vectors \(Te_1,\dots,Te_n\) span \(\operatorname{range}T\), so they are linearly dependent: some \(Te_j\) is a combination of the others. Expanding \(f(Te_1,\dots,Te_n)\) by linearity in slot \(j\) gives a sum of terms in which some vector appears twice, so \(f(Te_1,\dots,Te_n)=0\). Hence \(\det T\cdot f(e_1,\dots,e_n)=0\), and \(\det T=0\).

\(\blacksquare\)

Choose a basis in which \(T\) is upper triangular with diagonal entries \(\lambda_1,\dots,\lambda_n\), as in 8.2.4. In the Leibniz sum, a term \(\prod_i A_{\sigma(i),i}\) is nonzero only if \(\sigma(i)\le i\) for every \(i\), because the entries below the diagonal vanish. Then \(\sigma(1)=1\), \(\sigma(2)=2\), and so on, so only the identity survives. Hence \(\det A=\prod_k\lambda_k\).

By 9.2.2, \(\det T=\det A\). By 8.2.4(b), each eigenvalue \(\lambda\) occurs exactly \(\dim G(\lambda,T)\) times on the diagonal, so \(\prod_k\lambda_k=\prod_\lambda\lambda^{\dim G(\lambda,T)}\).

\(\blacksquare\)

(a) implies (b). If \(\mathcal M(T)\) is diagonal with respect to \(v_1,\dots,v_n\), then \(Tv_k=\lambda_kv_k\) for the diagonal entries \(\lambda_k\), so each \(v_k\) is an eigenvector and they form a basis.

(b) implies (a). If \(v_1,\dots,v_n\) are eigenvectors forming a basis, then \(Tv_k=\lambda_kv_k\) and the matrix is diagonal with entries \(\lambda_k\).

(b) implies (c). Each \(v_k\) lies in some eigenspace, and every vector of \(V\) is a sum of the \(v_k\). So \(V\) is contained in, and hence equal to, the sum of the eigenspaces.

(c) implies (b). If the eigenspaces span \(V\), their union is a list of eigenvectors that spans \(V\). A spanning list contains a basis, and every member of that basis is still an eigenvector.

\(\blacksquare\)

First, eigenvectors \(v_1,\dots,v_k\) for distinct eigenvalues \(\lambda_1,\dots,\lambda_k\) are linearly independent. Suppose not, and take a nontrivial relation \(\sum c_iv_i=0\) with the fewest nonzero coefficients. Relabel so that \(c_1,\dots,c_m\ne0\) and the rest vanish. Then \(m\ge2\), since each \(v_i\ne0\). Apply \(T-\lambda_mI\): the term with \(v_m\) disappears, leaving \(\sum_{i\lt m}c_i(\lambda_i-\lambda_m)v_i=0\). Its coefficient on \(v_1\) is \(c_1(\lambda_1-\lambda_m)\ne0\), so this is a nontrivial relation with fewer nonzero terms, a contradiction.

Now choose one eigenvector for each of the \(\dim V\) distinct eigenvalues. They are independent, and a list of \(\dim V\) independent vectors is a basis. By Statement 5.4.3, \(T\) is diagonalizable.

\(\blacksquare\)

Diagonalizable implies distinct roots. Let \(v_1,\dots,v_n\) be an eigenbasis with eigenvalues \(\mu_1,\dots,\mu_n\), and let \(\lambda_1,\dots,\lambda_k\) be the distinct values among the \(\mu_j\). Put \(q(z)=\prod_i(z-\lambda_i)\). The factors commute, and the factor with \(\lambda_i=\mu_j\) kills \(v_j\), so \(q(T)v_j=\prod_i(\mu_j-\lambda_i)v_j=0\) for every \(j\). Hence \(q(T)=0\). So \(m\) divides \(q\), and by unique factorization in the scalar field \(m\) is a product of some of the distinct factors \(z-\lambda_i\).

Distinct roots implies diagonalizable. Suppose \(m=\prod_{i=1}^k(z-\lambda_i)\) with distinct \(\lambda_i\). Let \(L_i(z)=\prod_{j\ne i}\dfrac{z-\lambda_j}{\lambda_i-\lambda_j}\). The polynomial \(\sum_iL_i-1\) has degree at most \(k-1\) and vanishes at the \(k\) distinct points \(\lambda_1,\dots,\lambda_k\), so it is zero. Hence \(\sum_iL_i(T)=I\). Next, \((z-\lambda_i)L_i(z)=c_i\,m(z)\) for a nonzero constant \(c_i\), so \((T-\lambda_iI)L_i(T)=c_i\,m(T)=0\). Thus each \(P_i=L_i(T)\) maps \(V\) into \(E_{\lambda_i}\). Every \(v\in V\) equals \(\sum_iP_iv\), so \(V=E_{\lambda_1}+\dots+E_{\lambda_k}\). By Statement 5.4.3, \(T\) is diagonalizable.

\(\blacksquare\)

Let \(\lambda\) be an eigenvalue with eigenvector \(x\ne0\), and choose \(j\) with \(|x_j|=\max_k|x_k|\); this maximum is positive. Row \(j\) of \(Ax=\lambda x\) reads \(\sum_ka_{jk}x_k=\lambda x_j\), so \((\lambda-a_{jj})x_j=\sum_{k\ne j}a_{jk}x_k\). Taking absolute values and using \(|x_k|\le|x_j|\), \[|\lambda-a_{jj}|\,|x_j|\le\sum_{k\ne j}|a_{jk}|\,|x_k|\le|x_j|\sum_{k\ne j}|a_{jk}|=|x_j|\,r_j.\] Dividing by \(|x_j|>0\) gives \(|\lambda-a_{jj}|\le r_j\), so \(\lambda\in D_j\).

\(\blacksquare\)

By Definition 5.1.1, \(\lambda\) is an eigenvalue exactly when there is \(v \ne 0\) with \(Tv = \lambda v\). This is the same as \(v \ne 0\) with \((T - \lambda I)v = 0\), that is, \(\operatorname{null}(T - \lambda I) \ne \{0\}\). By Statement 3.1.1, the null space is \(\{0\}\) exactly when the map is injective. So \(\lambda\) is an eigenvalue exactly when \(T - \lambda I\) is not injective.

\(\blacksquare\)

Suppose the statement fails, and choose a counterexample with \(m\) as small as possible: a list \(v_1,\dots,v_m\) of eigenvectors with distinct eigenvalues \(\lambda_1,\dots,\lambda_m\) that is linearly dependent. A single eigenvector is nonzero, so \(m \ge 2\). Take a dependence \(\sum_k a_kv_k = 0\), not all \(a_k\) zero.

Every \(a_j\) is nonzero. Otherwise \(\sum_{k \ne j} a_kv_k = 0\) would be a nontrivial dependence among \(m-1\) eigenvectors with distinct eigenvalues, contradicting the minimality of \(m\).

Apply \(T\) to the dependence: \(\sum_k a_k\lambda_kv_k = 0\). Subtract \(\lambda_m\) times the original dependence:

The list \(v_1,\dots,v_{m-1}\) is independent by minimality, so \(a_k(\lambda_k - \lambda_m) = 0\) for each \(k < m\). But \(a_k \ne 0\) and \(\lambda_k \ne \lambda_m\), a contradiction.

\(\blacksquare\)

Let \(\lambda_1,\dots,\lambda_m\) be distinct eigenvalues of \(T\), and pick an eigenvector \(v_k\) for each \(\lambda_k\). By Statement 5.1.2 the list \(v_1,\dots,v_m\) is linearly independent. An independent list in \(V\) has length at most \(\dim V\), so \(m \le \dim V\).

\(\blacksquare\)

Let \(n = \dim V \ge 1\) and pick \(v \ne 0\) in \(V\). The \(n+1\) vectors \(v, Tv, \dots, T^nv\) cannot be independent, since a list longer than \(\dim V\) is dependent. So there are scalars \(a_0,\dots,a_n\), not all zero, with \(\sum_k a_kT^kv = 0\).

Let \(m\) be the largest index with \(a_m \ne 0\). If \(m = 0\), then \(a_0v = 0\) with \(a_0 \ne 0\), forcing \(v = 0\), which is false. So \(m \ge 1\). Let \(p(z) = \sum_{k \le m} a_kz^k\). Then \(p\) is nonconstant and \(p(T)v = 0\).

By Statement 4.1.4, \(p(z) = c(z-\lambda_1)\cdots(z-\lambda_m)\) with \(c \ne 0\). Since polynomials in \(T\) multiply like the polynomials, \(p(T) = c\,(T-\lambda_1I)\cdots(T-\lambda_mI)\). Put \(S = (T-\lambda_1I)\cdots(T-\lambda_mI)\). Then \(Sv = c^{-1}p(T)v = 0\), and \(v \ne 0\), so \(S\) is not injective.

If every factor \(T - \lambda_jI\) were injective, then their composition \(S\) would be injective. So some \(T - \lambda_jI\) is not injective, and by Statement 5.1.1, \(\lambda_j\) is an eigenvalue of \(T\).

\(\blacksquare\)

Write \(S=T-\lambda I\). The union \(\bigcup_{j\ge1}\operatorname{null}S^j\) is \(G(\lambda,T)\). By 8.1.1 the null spaces increase with \(j\), and by 8.1.3 they are all equal to \(\operatorname{null}S^d\) for \(j\ge d\). So the union equals \(\operatorname{null}S^d\), which is a subspace. The eigenspace is \(\operatorname{null}S\subseteq\operatorname{null}S^d\).

\(\blacksquare\)

Splitting off one eigenspace. Let \(n=\dim V\ge 1\), let \(\lambda\) be an eigenvalue of \(T\) (one exists because \(V\) is a nonzero complex space; this is the fundamental theorem of algebra at work), and put \(S=T-\lambda I\), \(R=\operatorname{range}S^n\). Both \(\operatorname{null}S^n\) and \(R\) are \(T\)-invariant, since \(T\) commutes with \(S\).

They meet only in \(0\): if \(v=S^nw\) and \(S^nv=0\), then \(w\in\operatorname{null}S^{2n}=\operatorname{null}S^n\) by 8.1.3, so \(v=0\). By the fundamental theorem of linear maps, their dimensions add to \(n\). Hence \(V=G(\lambda,T)\oplus R\).

Induction. Induct on \(n\); the case \(n=0\) is trivial. Since \(G(\lambda,T)\ne 0\), the space \(R\) has dimension less than \(n\), so the induction hypothesis applies to \(T|_R\): \(R=\sum_\mu G(\mu,T|_R)\). Each \(G(\mu,T|_R)\) is contained in \(G(\mu,T)\), because the same vectors satisfy the same conditions. Also \(G(\lambda,T)\cap R=0\), so \(\lambda\) does not occur among the \(\mu\). Therefore \(V=G(\lambda,T)+\sum_\mu G(\mu,T)\subseteq\sum_{\text{all }\mu}G(\mu,T)\subseteq V\).

Directness. Suppose \(g_1+\cdots+g_m=0\) with \(g_i\in G(\mu_i,T)\) and the \(\mu_i\) distinct. The nonzero \(g_i\) are generalised eigenvectors for distinct eigenvalues, so by 8.1.6 they are independent. Since their sum is zero, there are none. Hence the sum is direct.

\(\blacksquare\)

Each \(G(\lambda_j,T)\) is \(T\)-invariant, since \(T\) commutes with \((T-\lambda_jI)^d\). Taking the concatenated basis, \(T\) sends each basis vector of the \(j\)th block into the span of that block, so the matrix is block diagonal.

On \(G_j=G(\lambda_j,T)\), let \(N_j=(T-\lambda_jI)|_{G_j}\). By 8.2.1, \(N_j^{d}=0\) on \(G_j\), so \(N_j\) is nilpotent, and the block of \(T\) is \(\lambda_jI+N_j\).

\(\blacksquare\)

Part (a). First, a nilpotent operator \(N\) on a space \(W\) has a basis with strictly upper-triangular matrix. Let \(W_j=\operatorname{null}N^j\), so \(W_0=0\) and \(W_d=W\) for \(d=\dim W\) by 8.1.4. Build a basis by adding, for \(j=1,2,\dots\), vectors that extend a basis of \(W_{j-1}\) to one of \(W_j\). Each new vector \(x\in W_j\) satisfies \(Nx\in W_{j-1}\), which is spanned by the vectors added earlier. So the column of \(N\) for \(x\) has nonzero entries only in earlier positions, and the matrix is strictly upper triangular.

Now apply this to each \(N_j=(T-\lambda_jI)|_{G_j}\) of 8.2.3. Each block \(\lambda_jI+N_j\) is then upper triangular with diagonal \(\lambda_j\). Concatenating the blocks gives an upper-triangular matrix for \(T\), since the blocks are block diagonal.

Part (b). Let \(v_1,\dots,v_n\) be any basis with upper-triangular matrix, and let \(U_k=\operatorname{span}(v_1,\dots,v_k)\). Each \(U_k\) is \(T\)-invariant and \(Tv_k\equiv\lambda_kv_k\) modulo \(U_{k-1}\). Put \(g_k(\mu)=\dim G(\mu,T|_{U_k})\), with \(g_0=0\).

Since \(G(\mu,T|_{U_{k-1}})=G(\mu,T|_{U_k})\cap U_{k-1}\), the difference \(g_k(\mu)-g_{k-1}(\mu)\) is the dimension of the image of \(G(\mu,T|_{U_k})\) in \(U_k/U_{k-1}\), which is one-dimensional, so the difference is at most \(1\). If \(\lambda_k\neq\mu\), take \(x=av_k+u\) in \(G(\mu,T|_{U_k})\) with \(u\in U_{k-1}\). Modulo \(U_{k-1}\), \((T-\mu I)^jx\equiv(\lambda_k-\mu)^jav_k\), and this must be \(0\), so \(a=0\) and \(x\in U_{k-1}\). Thus the difference is \(0\) when \(\lambda_k\ne\mu\).

Summing over \(k\), \(g_n(\mu)\le\#\{k:\lambda_k=\mu\}\) for every \(\mu\). By 8.2.2, \(\sum_\mu g_n(\mu)=\sum_\mu\dim G(\mu,T)=n\), and the total of the counts is also \(n\). A sum of inequalities with equal totals is an equality termwise, so \(g_n(\mu)=\#\{k:\lambda_k=\mu\}\).

\(\blacksquare\)

Fix \(j\), and let \(N=(T-\lambda_jI)|_{G_j}\), an operator on a space of dimension \(d_j\). It is nilpotent, so \(N^{d_j}=0\) by 8.1.4. Thus \((T-\lambda_jI)^{d_j}\) vanishes on \(G_j\).

Write \(p(z)=(z-\lambda_j)^{d_j}q_j(z)\), where \(q_j(z)=\prod_{i\neq j}(z-\lambda_i)^{d_i}\). Polynomials in \(T\) commute, so for \(v\in G_j\), \(p(T)v=q_j(T)(T-\lambda_jI)^{d_j}v=0\). Since \(V=\bigoplus_jG(\lambda_j,T)\) by 8.2.2, \(p(T)=0\).

\(\blacksquare\)

Expanding, \(\big\|\sum_ja_je_j\big\|^2=\sum_{j,k}a_j\overline{a_k}\langle e_j,e_k\rangle=\sum_{j,k}a_j\overline{a_k}\delta_{jk}=\sum_k|a_k|^2\).

For independence, suppose \(\sum_ka_ke_k=0\). Then \(\sum_k|a_k|^2=0\), so every \(a_k=0\).

\(\blacksquare\)

We induct on \(k\). Suppose \(e_1,\dots,e_{k-1}\) is orthonormal with the same span as \(v_1,\dots,v_{k-1}\). Independence of the \(v\)'s says \(v_k\notin\operatorname{span}(v_1,\dots,v_{k-1})=\operatorname{span}(e_1,\dots,e_{k-1})\), so \(w_k\ne0\) and \(e_k\) is defined. For \(i\lt k\), \[\langle w_k,e_i\rangle=\langle v_k,e_i\rangle-\sum_{j\lt k}\langle v_k,e_j\rangle\langle e_j,e_i\rangle=\langle v_k,e_i\rangle-\langle v_k,e_i\rangle=0,\] so \(e_k\) is orthogonal to the earlier vectors and has length \(1\). Thus \(e_1,\dots,e_k\) is orthonormal. Also \(e_k\in\operatorname{span}(v_1,\dots,v_k)\), and \(v_k=\|w_k\|e_k+\sum_{j\lt k}\langle v_k,e_j\rangle e_j\in\operatorname{span}(e_1,\dots,e_k)\). With the inductive hypothesis, the spans agree.

\(\blacksquare\)

Take any basis \(v_1,\dots,v_n\) of \(V\). It is linearly independent, so Statement 6.2.3 produces an orthonormal list \(e_1,\dots,e_n\). By 6.2.2 it is independent, so it is a list of \(n=\dim V\) independent vectors, hence a basis.

\(\blacksquare\)

By Statement 5.3.6, \(V\) has a basis \(v_1,\dots,v_n\) for which \(Tv_k\in\operatorname{span}(v_1,\dots,v_k)\) for all \(k\). Apply Gram–Schmidt (6.2.3) to get an orthonormal \(e_1,\dots,e_n\) with \(U_k:=\operatorname{span}(e_1,\dots,e_k)=\operatorname{span}(v_1,\dots,v_k)\). Then \(T\) maps each \(U_k\) into itself, because \(Tv_j\in\operatorname{span}(v_1,\dots,v_j)\subseteq U_k\) for \(j\le k\). In particular \(Te_k\in U_k=\operatorname{span}(e_1,\dots,e_k)\), and Statement 5.3.4 shows the matrix in the basis \(e_1,\dots,e_n\) is upper triangular.

\(\blacksquare\)

Existence. Take an orthonormal basis \(e_1,\dots,e_n\) (6.2.4) and put \(v=\sum_k\overline{\varphi(e_k)}\,e_k\). For \(u=\sum_ja_je_j\), linearity in \(u\) and 6.2.2-style orthonormality give \(\langle u,v\rangle=\sum_ja_j\,\varphi(e_j)=\varphi(u)\).

Uniqueness. If \(\langle u,v\rangle=\langle u,v'\rangle\) for all \(u\), then \(\langle u,v-v'\rangle=0\) for all \(u\). Taking \(u=v-v'\) gives \(v=v'\).

\(\blacksquare\)

(a) By (ii), \(\langle u,\lambda v\rangle=\overline{\langle\lambda v,u\rangle}=\overline{\lambda\langle v,u\rangle}=\overline\lambda\,\overline{\langle v,u\rangle}=\overline\lambda\langle u,v\rangle\), using (i), (ii) and the fact that conjugation is multiplicative.

(b) By (i), \(\langle 0,u\rangle=\langle 0+0,u\rangle=2\langle 0,u\rangle\), so \(\langle 0,u\rangle=0\).

(c) By (i), (a) and the definition of the norm, \(\|\lambda v\|^2=\lambda\overline\lambda\langle v,v\rangle=|\lambda|^2\|v\|^2\). Take square roots.

(d) This is (ii) in the real case, where conjugation does nothing.

\(\blacksquare\)

If \(v=0\), both sides are \(0\) and \(v=0\cdot u\), so the claim holds. Assume \(v\ne0\), so \(\|v\|^2>0\). Put \(c=\langle u,v\rangle/\|v\|^2\) and \(w=u-cv\). Expanding with (i), (ii) and 6.1.3(a), \[0\le\|w\|^2=\|u\|^2-\overline c\langle u,v\rangle-c\langle v,u\rangle+|c|^2\|v\|^2=\|u\|^2-\frac{|\langle u,v\rangle|^2}{\|v\|^2}.\] Multiplying by \(\|v\|^2\) gives \(|\langle u,v\rangle|^2\le\|u\|^2\|v\|^2\), and taking square roots proves the inequality.

For the equality case, equality holds exactly when \(\|w\|^2=0\), that is, when \(w=0\), i.e. \(u=cv\). Conversely, if \(u=\lambda v\) then \(|\langle\lambda v,v\rangle|=|\lambda|\|v\|^2=\|\lambda v\|\,\|v\|\).

\(\blacksquare\)

Expanding, \(\|u+v\|^2=\|u\|^2+\langle u,v\rangle+\langle v,u\rangle+\|v\|^2=\|u\|^2+2\operatorname{Re}\langle u,v\rangle+\|v\|^2\), since \(\langle v,u\rangle=\overline{\langle u,v\rangle}\). Because \(\operatorname{Re}z\le|z|\), Statement 6.1.4 gives \(2\operatorname{Re}\langle u,v\rangle\le2\|u\|\,\|v\|\). Hence \(\|u+v\|^2\le(\|u\|+\|v\|)^2\), and taking square roots finishes the proof.

\(\blacksquare\)

As in the proof of 6.1.5, \(\|u+v\|^2=\|u\|^2+2\operatorname{Re}\langle u,v\rangle+\|v\|^2\). If \(\langle u,v\rangle=0\), the middle term vanishes.

\(\blacksquare\)

Expanding each term, \(\|u+v\|^2=\|u\|^2+2\operatorname{Re}\langle u,v\rangle+\|v\|^2\) and \(\|u-v\|^2=\|u\|^2-2\operatorname{Re}\langle u,v\rangle+\|v\|^2\). Adding the two gives \(2\|u\|^2+2\|v\|^2\).

\(\blacksquare\)

Invertible implies bijective. Let \(S\) be an inverse of \(T\). If \(Tu = Tv\), then \(u = STu = STv = v\), so \(T\) is injective. For any \(w \in W\), the vector \(v = Sw\) satisfies \(Tv = TSw = w\), so \(T\) is surjective.

Bijective implies invertible. Suppose \(T\) is injective and surjective. For each \(w \in W\) there is exactly one \(v \in V\) with \(Tv = w\). Call it \(S(w)\), which defines a function \(S \colon W \to V\). Linearity of \(S\) comes from uniqueness: if \(S(w_1) = v_1\) and \(S(w_2) = v_2\), then \(T(v_1+v_2) = w_1+w_2\) and \(T(\lambda v_1) = \lambda w_1\), so \(S(w_1+w_2) = v_1+v_2\) and \(S(\lambda w_1) = \lambda v_1\). By construction \(S(Tv) = v\) for every \(v\), since \(v\) is the unique preimage of \(Tv\), so \(ST = I_V\). Also \(T(Sw) = w\), so \(TS = I_W\). Hence \(T\) is invertible with inverse \(S\).

Uniqueness. If \(S\) and \(S'\) are both inverses, then \(S = S(TS') = (ST)S' = S'\). We write \(T^{-1}\) for the inverse.

\(\blacksquare\)

By Statement 3.3.1 it is enough to show that \(T\) is injective if and only if it is surjective. By the fundamental theorem of linear maps (Statement 3.1.2), \(\dim V = \dim\operatorname{null} T + \dim\operatorname{range} T\).

\(T\) is injective exactly when \(\operatorname{null} T = \{0\}\) (Statement 3.1.1), that is, when \(\dim\operatorname{null} T = 0\). By the theorem this happens exactly when \(\dim\operatorname{range} T = \dim V = \dim W\).

\(T\) is surjective exactly when \(\operatorname{range} T = W\). Since \(\operatorname{range} T\) is a subspace of \(W\), it equals \(W\) exactly when it has the same finite dimension as \(W\). That last fact holds because a basis of a subspace \(U \subseteq W\) with \(\dim U = \dim W\) is a linearly independent list of length \(\dim W\) in \(W\), hence a basis of \(W\), so \(U = W\).

Both conditions are therefore equivalent to \(\dim\operatorname{range} T = \dim W\), so they are equivalent to each other.

\(\blacksquare\)

Isomorphic implies same dimension. Let \(T \colon V \to W\) be an isomorphism, let \(n = \dim V\), and let \(e_1,\dots,e_n\) be a basis of \(V\). The list \(Te_1,\dots,Te_n\) spans \(W\): given \(w\in W\), surjectivity gives \(v\) with \(Tv = w\), and writing \(v = \sum a_je_j\) gives \(w = \sum a_jTe_j\). It is independent: if \(\sum c_jTe_j = 0\), then \(T(\sum c_je_j) = 0\), so \(\sum c_je_j = 0\) by injectivity, and all \(c_j = 0\). So it is a basis of \(W\), and \(\dim W = n\).

Same dimension implies isomorphic. Let \(n = \dim V = \dim W\), with bases \(e_1,\dots,e_n\) of \(V\) and \(f_1,\dots,f_n\) of \(W\). Define \(T\) by \(T(\sum a_je_j) = \sum a_jf_j\). This is well defined because the coefficients \(a_j\) are unique, and it is linear because adding or scaling vectors adds or scales their coefficients. It is injective: if \(\sum a_jf_j = 0\), then every \(a_j = 0\), so the original vector is \(0\). It is surjective: every \(w = \sum b_jf_j\) equals \(T(\sum b_je_j)\). By Statement 3.3.1, \(T\) is an isomorphism.

\(\blacksquare\)

Induct on \(n\). If \(n=0\) there is nothing to prove. Let \(n\ge 1\). Since \(N\) is nilpotent and \(W\neq 0\), \(N\) is not injective: if it were, \(N^m\) would be injective for every \(m\), which is impossible when \(N^m=0\).

Let \(W'=\operatorname{range}N\). It has dimension less than \(n\), is \(N\)-invariant, and \(N|_{W'}\) is nilpotent. By induction \(W'\) has a chain basis: vectors \(w_1,\dots,w_r\) and lengths \(m_1,\dots,m_r\) such that \(\{N^jw_i:0\le j\lt m_i\}\) is a basis of \(W'\) and \(N^{m_i}w_i=0\). The tops \(t_i=N^{m_i-1}w_i\) are independent and lie in \(\operatorname{null}N\). Extend them by vectors \(z_1,\dots,z_s\) to a basis of \(\operatorname{null}N\), so \(r+s=\dim\operatorname{null}N\).

Since each \(w_i\) is in the range of \(N\), choose \(x_i\) with \(Nx_i=w_i\). We claim that the vectors \(N^jx_i\) (for \(0\le j\le m_i\)) together with the \(z_l\) form a basis of \(W\). There are \(\sum_i(m_i+1)+s=\dim W'+\dim\operatorname{null}N=n\) of them, by rank–nullity, so it is enough to show independence.

Suppose \(\sum_i\sum_{j=0}^{m_i}a_{ij}N^jx_i+\sum_lb_lz_l=0\). Apply \(N\), using \(N^{j+1}x_i=N^jw_i\) and \(Nz_l=0\), and then use \(N^{m_i}w_i=0\). This gives \(\sum_i\sum_{j=0}^{m_i-1}a_{ij}N^jw_i=0\), so \(a_{ij}=0\) for \(j\lt m_i\). What remains is \(\sum_ia_{im_i}t_i+\sum_lb_lz_l=0\), and since these form a basis of \(\operatorname{null}N\), all coefficients vanish.

Reading the chain \(N^{m-1}w,\dots,Nw,w\) in that order, \(N\) sends each vector to the one before it and the first to \(0\), so the block is \(J(0,m)\).

\(\blacksquare\)

By 8.2.2, \(V=\bigoplus_jG(\lambda_j,T)\). By 8.2.3, the matrix of \(T\) in a basis adapted to this sum is block diagonal with blocks \(\lambda_jI+N_j\), where \(N_j\) is the nilpotent operator \((T-\lambda_jI)|_{G_j}\).

Apply 8.3.1 to each \(N_j\) to get a chain basis of \(G_j\) in which \(N_j\) is block diagonal with blocks \(J(0,m)\). In that basis \(T|_{G_j}=\lambda_jI+N_j\) is block diagonal with blocks \(\lambda_jI+J(0,m)=J(\lambda_j,m)\). Concatenating the bases over \(j\) gives the claimed basis.

\(\blacksquare\)

A polynomial lemma. We build polynomials \(p_m\) with \(p_m(0)=1\) and \(p_m(x)^2-(1+x)\) divisible by \(x^m\). Start with \(p_1=1\). Given \(p=p_m\) with \(p^2-(1+x)=x^mq(x)\), set \(p_{m+1}=p+cx^m\) with \(c=-q(0)/2\). Then \(p_{m+1}^2-(1+x)=x^m\bigl(q+2cp+c^2x^m\bigr)\), and the bracket vanishes at \(0\) because \(p(0)=1\) and \(q(0)+2c=0\). So \(p_{m+1}^2-(1+x)\) is divisible by \(x^{m+1}\).

Take \(m=\dim W\), so \(N^m=0\) by 8.1.4. Write \(p=p_m\) and \(p(x)^2-(1+x)=x^mq(x)\). Substituting \(N\) for \(x\) is a ring homomorphism, so \(p(N)^2-(I+N)=N^mq(N)=0\). Take \(R=p(N)\).

\(\blacksquare\)

Since \(T\) is invertible, \(0\) is not an eigenvalue. By 8.2.2, \(V=\bigoplus_jG_j\) with \(G_j=G(\lambda_j,T)\) and \(\lambda_j\ne 0\). On \(G_j\), write \(T=\lambda_jI+N_j\) with \(N_j\) nilpotent.

Choose a complex square root \(s_j\) of \(\lambda_j\), which exists in \(\mathbf{C}\). The operator \(N_j/\lambda_j\) is nilpotent, so by 8.3.3 there is \(R_j'\) on \(G_j\) with \(R_j'^2=I+N_j/\lambda_j\). Put \(R_j=s_jR_j'\). Then \(R_j^2=s_j^2(I+N_j/\lambda_j)=\lambda_jI+N_j=T|_{G_j}\).

Define \(R\) on \(V\) by \(R|_{G_j}=R_j\), which is well defined because \(V=\bigoplus_jG_j\). Then \(R^2\) and \(T\) agree on every \(G_j\), so \(R^2=T\).

\(\blacksquare\)

Part (a). Let \(A=(A_{ij})\) and \(B=(B_{ij})\). Then \(\operatorname{tr}(AB)=\sum_i\sum_jA_{ij}B_{ji}=\sum_j\sum_iB_{ji}A_{ij}=\operatorname{tr}(BA)\). Hence \(\operatorname{tr}(P^{-1}AP)=\operatorname{tr}((AP)P^{-1})=\operatorname{tr}A\). Matrices of \(T\) in two bases differ by such a conjugation, so \(\operatorname{tr}T\) does not depend on the basis.

Part (b). Use the upper-triangular basis of 8.2.4(a). Its diagonal entries are the eigenvalues, and by 8.2.4(b) each \(\lambda\) occurs \(\dim G(\lambda,T)\) times. So the trace, which is the sum of the diagonal, equals \(\sum_\lambda\lambda\dim G(\lambda,T)\).

\(\blacksquare\)

Let \(v \in V\). Write \(v = a_1v_1+\cdots+a_nv_n\), which is possible since the list spans. Linearity applied repeatedly, first to the sum and then to each scalar multiple, gives \(Tv = \sum a_iTv_i\). The same computation for \(S\) gives \(Sv = \sum a_iSv_i\). Since \(Tv_i = Sv_i\) for every \(i\), we get \(Tv = Sv\). As \(v\) was arbitrary, \(T=S\).

\(\blacksquare\)

Each \(v \in V\) has a unique representation \(v = \sum a_iv_i\) by Statement 2.2.1. Define \(Tv = \sum a_iw_i\). This is well defined because the representation is unique. It sends \(v_j\) to \(w_j\), since \(v_j\) has the representation with \(a_j=1\) and all others \(0\).

For linearity, let \(v = \sum a_iv_i\) and \(u = \sum b_iv_i\). Then \(u+v = \sum (a_i+b_i)v_i\) and \(\lambda v = \sum (\lambda a_i)v_i\). So \(T(u+v) = \sum(a_i+b_i)w_i = Tu+Tv\) and \(T(\lambda v) = \lambda Tv\). Uniqueness follows from Statement 3.1.1.

\(\blacksquare\)

Linearity gives \(T0 = T(0+0) = T0 + T0\), so \(T0 = 0\). Thus \(0\) lies in the null space. If \(Tu = Tv = 0\), then \(T(u+v) = 0\) and \(T(\lambda u) = 0\), so the null space is closed under the operations. For the range, \(0 = T0\) lies in it, and if \(Tu, Tv\) are in the range then \(Tu+Tv = T(u+v)\) and \(\lambda Tu = T(\lambda u)\) are too.

If \(T\) is injective and \(v \in \operatorname{null} T\), then \(Tv = 0 = T0\) gives \(v = 0\). Conversely, if \(\operatorname{null} T = \{0\}\) and \(Tu = Tv\), then \(T(u-v) = 0\), so \(u-v = 0\).

\(\blacksquare\)

Let \(u_1,\dots,u_p\) be a basis of \(U\), \(v_1,\dots,v_n\) of \(V\), and \(w_1,\dots,w_m\) of \(W\). Write \(\mathcal{M}(S) = B\) and \(\mathcal{M}(T) = A\), so \(Su_k = \sum_l B_{lk}v_l\) and \(Tv_l = \sum_j A_{jl}w_j\). Then, using linearity of \(T\) and interchanging a finite double sum,

The coefficient of \(w_j\) is \(\sum_l A_{jl}B_{lk} = (AB)_{jk}\), which is the \((j,k)\) entry of the product. So \(\mathcal{M}(T\circ S) = AB\).

\(\blacksquare\)

Let \(A\) be \(m\times n\), \(B\) be \(n\times p\), and \(C\) be \(p\times q\). The \((i,j)\) entry of \((AB)C\) is \(\sum_k (AB)_{ik}C_{kj} = \sum_k\sum_l A_{il}B_{lk}C_{kj}\). The \((i,j)\) entry of \(A(BC)\) is \(\sum_l A_{il}(BC)_{lj} = \sum_l A_{il}\sum_k B_{lk}C_{kj}\). Both are the same finite double sum over \(l\) and \(k\), so they agree.

For distributivity, \((A(B+C))_{ij} = \sum_k A_{ik}(B_{kj}+C_{kj}) = \sum_k A_{ik}B_{kj} + \sum_k A_{ik}C_{kj} = (AB+AC)_{ij}\).

\(\blacksquare\)

Row rank is at most column rank. Let \(r\) be the column rank of \(A\), an \(m\times n\) matrix. Choose \(r\) columns that form a basis of the column space, and let \(C\) be the \(m \times r\) matrix with those columns. Each column of \(A\) is a combination of the columns of \(C\), so \(A = CR\) for some \(r\times n\) matrix \(R\). Row \(i\) of \(A\) equals \(\sum_k C_{ik}\cdot(\text{row } k \text{ of } R)\), so every row of \(A\) lies in the span of the \(r\) rows of \(R\). Hence the row rank is at most \(r\).

Equality. Applying the inequality to \(A^{\mathsf T}\) gives \(\text{row rank}(A^{\mathsf T}) \le \text{column rank}(A^{\mathsf T})\). Since the rows of \(A^{\mathsf T}\) are the columns of \(A\), this reads \(\text{column rank}(A) \le \text{row rank}(A)\). Together with the first part, the two ranks are equal.

\(\blacksquare\)

Existence. Let \(n = \dim V\). The space \(\mathcal{L}(V)\) has dimension \(n^2\), as the matrix description of Part 3.2 shows. So the \(n^2+1\) operators \(I, T, \dots, T^{n^2}\) are linearly dependent, and some nonzero polynomial \(q\) has \(q(T) = 0\). Dividing \(q\) by its leading coefficient gives a monic annihilating polynomial. Among all monic annihilating polynomials choose one of least degree.

Uniqueness. Let \(\mu_1\) and \(\mu_2\) be monic annihilating polynomials of the same least degree \(d\). Their difference \(\mu_1 - \mu_2\) annihilates \(T\), because \((\mu_1 - \mu_2)(T) = \mu_1(T) - \mu_2(T) = 0\), and its degree is less than \(d\) since the leading terms cancel. By the Remark after Definition 5.2.1, \(\mu_1 - \mu_2 = 0\).

\(\blacksquare\)

Divide \(p\) by \(\mu\) using Statement 4.1.1: \(p = \mu s + r\) with \(\deg r < \deg \mu\). Then \(p(T) = \mu(T)s(T) + r(T) = r(T)\).

If \(\mu\) divides \(p\), then \(r = 0\), so \(p(T) = 0\). Conversely, if \(p(T) = 0\), then \(r(T) = 0\) with \(\deg r < \deg\mu\), so \(r = 0\) by the Remark after Definition 5.2.1. Hence \(p = \mu s\).

\(\blacksquare\)

Eigenvalue implies zero. Let \(Tv = \lambda v\) with \(v \ne 0\). Then \(T^kv = \lambda^kv\) for every \(k\), so \(p(T)v = p(\lambda)v\) for every polynomial \(p\). With \(p = \mu\), we get \(0 = \mu(T)v = \mu(\lambda)v\). Since \(v \ne 0\), \(\mu(\lambda) = 0\).

Zero implies eigenvalue. Suppose \(\mu(\lambda) = 0\). By Statement 4.1.2, \(\mu = (z-\lambda)s\) for a polynomial \(s\), which is monic of degree \(\deg \mu - 1\). By the Remark after Definition 5.2.1, \(s(T) \ne 0\), so there is \(w\) with \(s(T)w \ne 0\). Then \((T-\lambda I)\bigl(s(T)w\bigr) = \mu(T)w = 0\). So \(s(T)w\) is a nonzero eigenvector of \(T\) for \(\lambda\).

\(\blacksquare\)

We prove by strong induction on odd \(n\) that every operator on a real \(n\)-dimensional space has an eigenvalue. For \(n = 1\), the space is spanned by one nonzero \(v\), and \(Tv\) lies in it, so \(Tv = \lambda v\) for some \(\lambda\). Now let \(n \ge 3\) be odd, let \(T\) act on \(V\) with \(\dim V = n\), and let \(\mu\) be the minimal polynomial of \(T\). Since \(V \ne 0\), \(\mu\) is not constant.

Case 1: \(\mu\) has a real zero \(\lambda\). Then \(\lambda\) is an eigenvalue by Statement 5.2.3.

Case 2: \(\mu\) has no real zero. By Statement 4.1.5, \(\mu\) is a product of real polynomials of degree \(1\) or \(2\). A factor of degree \(1\) would give a real zero of \(\mu\), so there is a real factor of degree \(2\). Divide it to make it monic, and call it \(q(z) = z^2 + \beta z + \gamma\). Then \(\mu = qs\) for a real polynomial \(s\) of smaller degree than \(\mu\). Any real zero of \(q\) would be a real zero of \(\mu\), so \(\beta^2 < 4\gamma\). Put \(\delta = \sqrt{\gamma - \beta^2/4} > 0\).

Let \(R = \operatorname{range} q(T)\) and \(U = \operatorname{null} q(T)\). Since \(q(T)\) commutes with \(T\), both \(R\) and \(U\) are invariant under \(T\).

(a) \(R \ne V\). Otherwise \(q(T)\) is surjective, so it is invertible by Statement 3.3.2. From \(q(T)s(T) = \mu(T) = 0\) we get \(s(T) = q(T)^{-1}\cdot 0 = 0\). But \(\deg s < \deg \mu\) and \(s \ne 0\), which contradicts the Remark after Definition 5.2.1.

(b) \(\dim U\) is even. On \(U\), define \(J = \delta^{-1}(T|_U + \tfrac{\beta}{2}I_U)\). Since \((T + \tfrac{\beta}{2}I)^2 = q(T) - \delta^2 I\) and \(q(T) = 0\) on \(U\), we get \(J^2 = -I_U\). Lemma: a real space \(W\) with an operator \(J\) satisfying \(J^2 = -I\) has even dimension. Proof of the lemma: let \(U_0\) be a \(J\)-invariant subspace of largest even dimension. Suppose \(U_0 \ne W\), and pick \(w \notin U_0\). Let \(P = \operatorname{span}(w, Jw)\). It is \(J\)-invariant, because \(J(Jw) = -w\). It has dimension \(2\): if \(Jw = cw\) with \(c\) real, then \(-w = J^2w = c^2w\), impossible for \(w \ne 0\). The subspace \(U_0 \cap P\) is \(J\)-invariant. It is not all of \(P\), since \(w \notin U_0\). If it were a line \(\operatorname{span}(x)\), then \(Jx = dx\) with \(d\) real, and \(-x = J^2x = d^2x\) would be impossible. So \(U_0 \cap P = \{0\}\), and \(U_0 + P\) is \(J\)-invariant of dimension \(\dim U_0 + 2\), which is even and larger, a contradiction. Hence \(U_0 = W\).

(c) Conclusion. By the fundamental theorem of linear maps (Statement 3.1.2), \(n = \dim U + \dim R\). Since \(n\) is odd and \(\dim U\) is even, \(\dim R\) is odd. In particular \(R \ne \{0\}\), and by (a) \(\dim R < n\). So \(T\) restricted to \(R\) is an operator on an odd-dimensional real space of smaller dimension. By the induction hypothesis it has an eigenvalue \(\lambda\) with eigenvector \(v \in R\), and then \(Tv = \lambda v\). So \(T\) has an eigenvalue.

\(\blacksquare\)

Let \(v\in\operatorname{null}T^k\), so \(T^kv=0\). Then \(T^{k+1}v=T(T^kv)=T(0)=0\), so \(v\in\operatorname{null}T^{k+1}\).

\(\blacksquare\)

It is enough to show that \(\operatorname{null}T^{j}=\operatorname{null}T^{j+1}\) implies \(\operatorname{null}T^{j+1}=\operatorname{null}T^{j+2}\), and then use induction on \(j\).

Suppose \(\operatorname{null}T^{j}=\operatorname{null}T^{j+1}\) and let \(v\in\operatorname{null}T^{j+2}\). Then \(T^{j+1}(Tv)=0\), so \(Tv\in\operatorname{null}T^{j+1}=\operatorname{null}T^{j}\). Hence \(T^{j}(Tv)=0\), that is \(T^{j+1}v=0\), so \(v\in\operatorname{null}T^{j+1}\). The reverse inclusion is 8.1.1.

Starting from \(j=m\) and repeating this gives \(\operatorname{null}T^{j}=\operatorname{null}T^{m}\) for all \(j\ge m\).

\(\blacksquare\)

Suppose \(\operatorname{null}T^j\neq\operatorname{null}T^{j+1}\) for every \(j=0,1,\dots,d\). By 8.1.1 the inclusions are then strict, and each strict inclusion of finite-dimensional subspaces raises the dimension by at least one. So \(\dim\operatorname{null}T^{d+1}\ge d+1\), which is impossible since \(\operatorname{null}T^{d+1}\subseteq V\) has dimension at most \(d\).

Hence there is a least \(k\le d\) with \(\operatorname{null}T^k=\operatorname{null}T^{k+1}\). By 8.1.2 the null spaces are equal from index \(k\) on, and since \(d\ge k\), this includes \(\operatorname{null}T^d\).

\(\blacksquare\)

Let \(d=\dim V\) and let \(m\) be such that \(T^m=0\). Then \(\operatorname{null}T^m=V\).

If \(m\le d\), then \(V=\operatorname{null}T^m\subseteq\operatorname{null}T^d\subseteq V\) by 8.1.1, so \(\operatorname{null}T^d=V\). If \(m\gt d\), then 8.1.3 gives \(\operatorname{null}T^d=\operatorname{null}T^m=V\). Either way \(T^dv=0\) for every \(v\), so \(T^d=0\).

\(\blacksquare\)

Since \(v\) is generalised for \(\mu\), there is a least \(k\ge 1\) with \((T-\mu I)^kv=0\). Put \(w=(T-\mu I)^{k-1}v\). By minimality \(w\neq 0\), and \((T-\mu I)w=0\), so \(Tw=\mu w\).

Since \(T-\lambda I\) and \(T-\mu I\) commute, \((T-\lambda I)^jw=(T-\mu I)^{k-1}(T-\lambda I)^jv=0\), where \(j\) is an exponent that kills \(v\) for \(\lambda\). But \((T-\lambda I)w=(\mu-\lambda)w\), so by induction \((T-\lambda I)^jw=(\mu-\lambda)^jw\). Therefore \((\mu-\lambda)^jw=0\), and since \(w\neq 0\), \(\mu=\lambda\).

\(\blacksquare\)

Induct on \(m\). The case \(m=1\) holds because \(v_1\neq 0\). Let \(m\ge 2\) and suppose \(\sum_i a_iv_i=0\) with not all \(a_i\) zero. Choose \(j_i\ge 1\) with \((T-\lambda_iI)^{j_i}v_i=0\), and let \(N\ge\max_i j_i\). Put \(S=(T-\lambda_mI)^N\).

Then \(Sv_m=0\). For \(i\lt m\), the vector \(w_i=Sv_i\) is killed by \((T-\lambda_iI)^{j_i}\), because \(S\) commutes with it. If \(w_i=0\), then \(v_i\) is killed by \((T-\lambda_mI)^N\), so \(v_i\) is a generalised eigenvector for \(\lambda_m\) as well as \(\lambda_i\), contradicting 8.1.5. So \(w_1,\dots,w_{m-1}\) are nonzero generalised eigenvectors for distinct eigenvalues, hence independent by the induction hypothesis.

Applying \(S\) to the relation gives \(\sum_{i\lt m}a_iw_i=0\). If all \(a_i\) with \(i\lt m\) vanished, then \(a_mv_m=0\) forces \(a_m=0\), a contradiction. So some \(a_i\) with \(i\lt m\) is nonzero, which contradicts the independence of the \(w_i\).

\(\blacksquare\)

Existence. Use strong induction on \(\deg p\). If \(p = 0\) or \(\deg p < \deg s\), take \(q = 0\) and \(r = p\). Otherwise let \(m = \deg p \ge n = \deg s\), with leading terms \(a z^m\) of \(p\) and \(b z^n\) of \(s\). The polynomial \(p_1 = p - (a/b)z^{m-n}s\) has its \(z^m\) terms cancelled, so \(p_1 = 0\) or \(\deg p_1 < m\). By induction \(p_1 = sq_1 + r\) with \(\deg r < n\), and then \(p = s\bigl((a/b)z^{m-n} + q_1\bigr) + r\).

Uniqueness. Suppose \(sq_1 + r_1 = sq_2 + r_2\) with both remainders of degree less than \(\deg s\). Then \(s(q_1 - q_2) = r_2 - r_1\). If \(q_1 \ne q_2\), the left side has degree at least \(\deg s\), while the right side has degree less than \(\deg s\) or is zero. This is a contradiction, so \(q_1 = q_2\) and then \(r_1 = r_2\).

\(\blacksquare\)

If \(p = (z-\lambda)q\), then \(p(\lambda) = 0\) because the first factor vanishes.

Conversely, divide \(p\) by \(z - \lambda\) using Statement 4.1.1: \(p = (z-\lambda)q + r\) with \(\deg r < 1\), so \(r\) is a constant \(c\). Evaluating at \(\lambda\) gives \(0 = p(\lambda) = 0\cdot q(\lambda) + c\), so \(c = 0\) and \(p = (z-\lambda)q\).

\(\blacksquare\)

Let \(\lambda_1,\dots,\lambda_k\) be distinct zeros of \(p\). We show by induction on \(k\) that \(p = (z-\lambda_1)\cdots(z-\lambda_k)\,r\) for some \(r \in \mathcal{P}(\mathbb{F})\). The case \(k = 1\) is Statement 4.1.2. For the inductive step, write \(p = (z-\lambda_1)\cdots(z-\lambda_{k-1})\,q\). Substituting \(\lambda_k\) gives \(0 = \prod_{i

Since \(p \ne 0\), the factor \(r\) is nonzero, and \(\deg p = k + \deg r \ge k\). Hence \(k \le m\).

\(\blacksquare\)

The first statement is the fundamental theorem of algebra. I cite it rather than prove it here; standard proofs use complex analysis (for instance Liouville's theorem applied to \(1/p\)) or topology.

For the consequence, induct on \(m = \deg p\). If \(m = 1\), then \(p = a_1z + a_0 = a_1(z - \lambda)\) with \(\lambda = -a_0/a_1\). If \(m \ge 2\), the theorem gives a zero \(\lambda\), so \(p = (z-\lambda)q\) by Statement 4.1.2. Here \(\deg q = m - 1 \ge 1\), and induction factors \(q\), which gives the factorisation of \(p\).

\(\blacksquare\)

Induct on \(\deg p\). Degrees \(1\) and \(2\) need nothing. Let \(p\) be real of degree at least \(3\).

Case 1: \(p\) has a real zero \(\lambda\). Divide \(p\) by \(z - \lambda\) inside real polynomials. The quotient \(q\) and remainder \(r\) are real, and \(r\) is the constant \(p(\lambda) = 0\). So \(p = (z-\lambda)q\) with \(q\) real of smaller degree, and induction applies to \(q\).

Case 2: \(p\) has no real zero. By Statement 4.1.4, \(p\) has a complex zero \(\alpha\), which is not real. Since \(p\) has real coefficients, \(p(\bar\alpha) = \overline{p(\alpha)} = 0\), so \(\bar\alpha\) is also a zero. Let \(q(z) = (z-\alpha)(z-\bar\alpha) = z^2 - 2(\operatorname{Re}\alpha)z + |\alpha|^2\), which has real coefficients. Dividing \(p\) by \(q\) within real polynomials gives \(p = qs + r\) with \(s\) real and \(r = uz + v\) with \(u, v\) real. Evaluating at \(\alpha\) gives \(u\alpha + v = 0\). Since \(\alpha\) is not real, \(u = 0\) and then \(v = 0\). So \(p = qs\), and \(s\) is real. If \(\deg s = 0\), then \(p\) is itself a real polynomial of degree \(2\) times a constant, which is a valid factor. Otherwise \(\deg s \ge 1\) and induction applies to \(s\).

\(\blacksquare\)

(a) implies (b). If \(Tv = \lambda v\) with \(v \neq 0\), then \(\lambda\lVert v\rVert^{2} = \langle Tv,v\rangle \geq 0\), so \(\lambda \geq 0\).

(b) implies (c). By Statement 7.2.3 (or 7.2.4), \(V\) has an orthonormal basis \(e_1,\dots,e_n\) with \(Te_j = \lambda_j e_j\) and every \(\lambda_j \geq 0\). Define \(R\) by \(Re_j = \sqrt{\lambda_j}\, e_j\). In this basis \(R\) has real diagonal entries, so \(R^{*} = R\), and \(R^{*}R e_j = \lambda_j e_j = Te_j\). Hence \(T = R^{*}R\).

(c) implies (a). \((R^{*}R)^{*} = R^{*}R^{**} = R^{*}R\), so \(R^{*}R\) is self-adjoint, and \(\langle R^{*}Rv,v\rangle = \lVert Rv\rVert^{2} \geq 0\).

\(\blacksquare\)

Existence. By Statement 7.3.1, the operator \(T\) has an orthonormal eigenbasis \(e_j\) with eigenvalues \(\lambda_j \geq 0\), and \(R\) defined by \(Re_j = \sqrt{\lambda_j}\,e_j\) is positive with \(R^{2} = T\).

Uniqueness. Let \(R\) be any positive operator with \(R^{2} = T\). Let \(\lambda\) be an eigenvalue of \(T\) with eigenspace \(E = \operatorname{null}(T - \lambda I)\). Since \(R\) commutes with \(R^{2} = T\), for \(v \in E\) we have \(T(Rv) = R(Tv) = \lambda Rv\), so \(R\) maps \(E\) into \(E\). The restriction \(R|_E\) is self-adjoint on \(E\). If \(Rv = \mu v\) with \(v \in E\), \(v \neq 0\), then \(\lambda v = Tv = R^{2}v = \mu^{2}v\), so \(\mu^{2} = \lambda\); also \(\mu \geq 0\) because \(R\) is positive. Hence \(\mu = \sqrt{\lambda}\). By the spectral theorem, \(E\) has an orthonormal basis of eigenvectors of \(R|_E\), and all of them have eigenvalue \(\sqrt{\lambda}\). So \(Rv = \sqrt{\lambda}\, v\) for every \(v \in E\).

The spectral theorem also gives \(V = \bigoplus_\lambda E_\lambda\) over the eigenvalues \(\lambda\) of \(T\). Since \(R\) acts as \(\sqrt{\lambda}\) on each \(E_\lambda\), it is determined by \(T\). So any two positive square roots agree.

\(\blacksquare\)

(a) implies (b). Polarisation: in a complex space \(\langle u,v\rangle = \frac14\sum_{k=1}^{4} i^{k}\lVert u + i^{k}v\rVert^{2}\), and in a real space \(\langle u,v\rangle = \frac14(\lVert u+v\rVert^{2} - \lVert u-v\rVert^{2})\). Since \(S\) is linear and preserves norms, it preserves each term, so it preserves \(\langle u,v\rangle\).

(b) implies (c). \(\langle (S^{*}S - I)u, v\rangle = \langle Su,Sv\rangle - \langle u,v\rangle = 0\) for all \(u,v\). Taking \(v = (S^{*}S-I)u\) gives \(S^{*}S = I\).

(c) implies (a). \(\lVert Sv\rVert^{2} = \langle S^{*}Sv,v\rangle = \langle v,v\rangle = \lVert v\rVert^{2}\).

(b) implies (d), because \(\langle Se_j, Se_k\rangle = \langle e_j,e_k\rangle\) for any orthonormal basis. (d) implies (a): if \(v = \sum_j\langle v,e_j\rangle e_j\), then \(\lVert Sv\rVert^{2} = \sum_j|\langle v,e_j\rangle|^{2} = \lVert v\rVert^{2}\) by orthonormality of the list \(Se_j\) and Parseval's identity for \(e_j\).

\(\blacksquare\)

For each \(u\in U\), the map \(v\mapsto\langle u,v\rangle\) is conjugate-linear, and the set where it vanishes is a subspace. So \(U^\perp\), the intersection of these sets, is a subspace. If \(x\in U\cap U^\perp\), then \(\langle x,x\rangle=0\), so \(x=0\).

For the sum, let \(e_1,\dots,e_m\) be an orthonormal basis of \(U\) (6.2.4 applied to \(U\)). Given \(v\in V\), set \(u=\sum_k\langle v,e_k\rangle e_k\in U\) and \(w=v-u\). Then \(\langle w,e_j\rangle=\langle v,e_j\rangle-\langle v,e_j\rangle=0\) for each \(j\), so for any \(x=\sum_jc_je_j\in U\), \(\langle w,x\rangle=\sum_j\overline{c_j}\langle w,e_j\rangle=0\). Hence \(w\in U^\perp\) and \(v=u+w\in U+U^\perp\). Together with \(U\cap U^\perp=\{0\}\), the sum is direct, and uniqueness of the decomposition follows.

\(\blacksquare\)

(a) If \(v=u+w\) and \(v'=u'+w'\) are the decompositions, then \(\alpha v+\beta v'=(\alpha u+\beta u')+(\alpha w+\beta w')\) with the two parts in \(U\) and \(U^\perp\). By uniqueness, \(P_U(\alpha v+\beta v')=\alpha P_Uv+\beta P_Uv'\).

(b) Every \(P_Uv\) lies in \(U\), and \(P_Uu=u\) for \(u\in U\) (take \(w=0\)), so the range is \(U\). If \(P_Uv=0\), then \(v=w\in U^\perp\); conversely \(P_Uw=0\) for \(w\in U^\perp\). So the null space is \(U^\perp\).

(c) Since \(P_Uv\in U\), we have \(P_U(P_Uv)=P_Uv\) by (b).

(d) Let \(u=\sum_k\langle v,e_k\rangle e_k\) and \(w=v-u\). The proof of 6.3.2 showed \(w\in U^\perp\) and \(u\in U\), so this is the decomposition and \(P_Uv=u\).

\(\blacksquare\)

Write \(v-u=(v-P_Uv)+(P_Uv-u)\). The first term lies in \(U^\perp\) by 6.3.2, and the second lies in \(U\) because both \(P_Uv\) and \(u\) do. So the two terms are orthogonal, and the Pythagorean theorem (6.1.6) gives \[\|v-u\|^2=\|v-P_Uv\|^2+\|P_Uv-u\|^2\ge\|v-P_Uv\|^2.\] This proves the inequality. Equality holds exactly when \(\|P_Uv-u\|=0\), that is, when \(u=P_Uv\).

\(\blacksquare\)

Existence. Let \(a_1, \dots, a_n\) be the columns of \(A\); they are linearly independent. Run Gram–Schmidt: set \(v_k = a_k - \sum_{j 0\), and \(R_{jk} = 0\) for \(j > k\). A square matrix with orthonormal columns is unitary.

Uniqueness. Suppose \(QR = Q'R'\) with both factorisations of the required kind. Then \(R\) is invertible, since \(A\) is, and \(U = Q'^{*}Q = R'R^{-1}\). The matrix \(U\) is unitary, and it is upper triangular with positive diagonal, because products and inverses of upper triangular matrices are upper triangular and the diagonal entries multiply. Now \(U^{-1} = U^{*}\) is lower triangular, since \(U\) is upper triangular, and \(U^{-1}\) is upper triangular. So \(U\) is diagonal, with diagonal entries of modulus \(1\) that are also positive. Hence \(U = I\), so \(Q = Q'\) and \(R = R'\).

\(\blacksquare\)

Existence. Since \(A\) is positive definite, its eigenvalues are real and positive: if \(Av = \lambda v\) with \(v \neq 0\), then \(\lambda\lVert v\rVert^{2} = \langle Av,v\rangle > 0\). By Statement 7.3.2 there is a positive square root \(B = \sqrt{A}\), with eigenvalues \(\sqrt{\lambda} > 0\) on the same eigenvectors. So \(B\) is invertible and \(B^{*}B = B^{2} = A\). By Statement 7.5.1, \(B = QR\) with \(Q\) unitary and \(R\) upper triangular with positive diagonal. Then \(A = B^{*}B = R^{*}Q^{*}QR = R^{*}R\).

Uniqueness. Suppose \(A = R^{*}R = S^{*}S\) with \(R\) and \(S\) upper triangular with positive diagonals. Both are invertible, and \(U = RS^{-1}\) satisfies \(U^{*}U = S^{-*}R^{*}RS^{-1} = S^{-*}S^{*}SS^{-1} = I\). So \(U\) is unitary. It is also upper triangular with positive diagonal, so \(U = I\) by the argument in the uniqueness part of Statement 7.5.1 (a unitary upper triangular matrix with positive diagonal is the identity). Hence \(R = S\).

\(\blacksquare\)

The span contains \(0\), since taking all \(a_i=0\) gives \(0\). It is closed under addition: \((\sum a_iv_i)+(\sum b_iv_i) = \sum(a_i+b_i)v_i\). It is closed under scaling: \(\lambda\sum a_iv_i = \sum(\lambda a_i)v_i\). Each \(v_j\) is in the span by taking \(a_j=1\) and all other coefficients \(0\).

If \(X\) is a subspace containing every \(v_j\), then each \(a_jv_j\) lies in \(X\) by scaling, and the finite sum \(\sum a_jv_j\) lies in \(X\) by repeated addition. So the span is contained in \(X\).

\(\blacksquare\)

Since the list is dependent, there are scalars \(a_1,\dots,a_m\), not all zero, with \(\sum a_iv_i = 0\). Let \(j\) be the largest index with \(a_j \neq 0\). Then \(a_jv_j = -\sum_{i

We have \(j \ge 2\): if \(j=1\), then \(a_1v_1 = 0\) with \(a_1 \neq 0\), which forces \(v_1 = 0\), contrary to assumption.

Removing \(v_j\) does not change the span: every combination that uses \(v_j\) can replace \(v_j\) by its expression in the earlier vectors.

\(\blacksquare\)

Suppose the list is independent, and \(w = \sum a_iv_i = \sum b_iv_i\) is a vector in the span. Subtracting gives \(\sum (a_i-b_i)v_i = 0\), so every \(a_i-b_i=0\). The representation is unique.

Conversely, suppose every vector in the span has exactly one representation. The zero vector lies in the span and has the representation with all coefficients \(0\). Uniqueness says no other representation of \(0\) exists, which is independence.

\(\blacksquare\)

Let \(Tv = \lambda v\) with \(v \neq 0\). Then \(\lambda\lVert v\rVert^{2} = \langle Tv, v\rangle\). Also \(\langle Tv, v\rangle = \langle v, Tv\rangle = \overline{\langle Tv,v\rangle}\) because \(T\) is self-adjoint. So \(\langle Tv,v\rangle\) is real, and dividing by \(\lVert v\rVert^{2} \neq 0\) shows \(\lambda \in \mathbb{R}\).

\(\blacksquare\)

Using self-adjointness and the realness of \(\beta\) from Statement 7.2.1, \(\alpha\langle u,v\rangle = \langle Tu, v\rangle = \langle u, Tv\rangle = \langle u, \beta v\rangle = \beta\langle u,v\rangle\). So \((\alpha - \beta)\langle u,v\rangle = 0\), and since \(\alpha \neq \beta\), \(\langle u,v\rangle = 0\).

\(\blacksquare\)

If \(V\) has such a basis \(e_1,\dots,e_n\), with \(Te_j = \lambda_j e_j\), then the matrix of \(T\) in this basis is the real diagonal matrix \(D = \operatorname{diag}(\lambda_1,\dots,\lambda_n)\). Because the basis is orthonormal, the matrix of \(T^{*}\) is the conjugate transpose \(D^{*} = D\) (Statement 7.1.3). So \(T^{*}\) and \(T\) have the same matrix, and \(T^{*} = T\).

If \(T\) is self-adjoint, we induct on \(\dim V\). The case \(\dim V = 0\) is empty. Let \(\dim V \geq 1\).

Step 1: there is an eigenvector. The unit sphere of a finite-dimensional real space is compact, so \(v \mapsto \langle Tv,v\rangle\) attains a maximum \(\lambda\) at some unit vector \(v_0\). Put \(S = T - \lambda I\). Then \(S\) is self-adjoint, \(\langle Sx,x\rangle \leq 0\) for every \(x\) (by scaling), and \(\langle Sv_0,v_0\rangle = 0\). For any \(w\) and real \(t\), \(0 \geq \langle S(v_0+tw), v_0+tw\rangle = 2t\operatorname{Re}\langle Sv_0,w\rangle + t^{2}\langle Sw,w\rangle\). Since this holds for all real \(t\), \(\operatorname{Re}\langle Sv_0,w\rangle = 0\). Replacing \(w\) by \(iw\) in the complex case gives \(\operatorname{Im}\langle Sv_0,w\rangle = 0\) too. Taking \(w = Sv_0\) shows \(Sv_0 = 0\), so \(Tv_0 = \lambda v_0\).

Step 2: split off the eigenvector. Let \(U = (\operatorname{span}v_0)^{\perp}\). For \(u \in U\), \(\langle Tu, v_0\rangle = \langle u, Tv_0\rangle = \lambda\langle u,v_0\rangle = 0\), so \(U\) is invariant under \(T\). The restriction \(T|_U\) is self-adjoint on the \((n-1)\)-dimensional space \(U\). By induction \(U\) has an orthonormal basis of eigenvectors, and adding \(v_0\) gives one for \(V\).

\(\blacksquare\)

If there is an orthonormal eigenbasis, the matrix of \(T\) in it is diagonal, \(D\). Diagonal matrices commute with their conjugate transposes, so \(DD^{*} = D^{*}D\), which means \(TT^{*} = T^{*}T\).

If \(T\) is normal, we induct on \(\dim V\). Let \(\dim V \geq 1\). By the fundamental theorem of algebra, \(T\) has an eigenvalue \(\lambda\). Let \(E = \operatorname{null}(T - \lambda I)\).

Eigenvectors of \(T\) are eigenvectors of \(T^{*}\). The operator \(T - \lambda I\) is normal: \((T-\lambda I)(T-\lambda I)^{*} = TT^{*} - \bar\lambda T - \lambda T^{*} + |\lambda|^{2}I\) and \((T-\lambda I)^{*}(T-\lambda I) = T^{*}T - \lambda T^{*} - \bar\lambda T + |\lambda|^{2}I\), which agree because \(TT^{*} = T^{*}T\). By Statement 7.1.4, \(\lVert (T-\lambda I)v\rVert = \lVert (T^{*}-\bar\lambda I)v\rVert\) for all \(v\). So for \(v \in E\), \(\lVert (T^{*}-\bar\lambda I)v\rVert = 0\), that is \(T^{*}v = \bar\lambda v\). Hence \(E\) is invariant under \(T^{*}\).

Splitting off \(E\). Let \(W = E^{\perp}\). For \(w \in W\) and \(v \in E\), \(\langle Tw, v\rangle = \langle w, T^{*}v\rangle = \langle w, \bar\lambda v\rangle = 0\), so \(W\) is \(T\)-invariant. For \(w \in W\) and \(v \in E\), \(\langle T^{*}w, v\rangle = \langle w, Tv\rangle = \lambda\langle w,v\rangle = 0\), so \(W\) is also \(T^{*}\)-invariant. Therefore \((T|_W)^{*} = T^{*}|_W\), and \((T|_W)(T|_W)^{*} = TT^{*}|_W\) equals \(T^{*}T|_W = (T|_W)^{*}(T|_W)\). So \(T|_W\) is normal on a space of smaller dimension, and by induction \(W\) has an orthonormal basis of eigenvectors. An orthonormal basis of \(E\) consists of eigenvectors with eigenvalue \(\lambda\). Together they form an orthonormal basis of \(V\).

\(\blacksquare\)

First, \(\operatorname{null}T^{*}T = \operatorname{null}T\). Indeed \(\langle T^{*}Tv, v\rangle = \lVert Tv\rVert^{2}\), so \(T^{*}Tv = 0\) forces \(Tv = 0\), and the converse is clear.

In the orthonormal eigenbasis of Definition 7.4.1, the eigenspace for eigenvalue \(0\) is spanned by the basis vectors with \(\lambda_j = 0\). So its dimension is the number of zero singular values. The number of positive singular values is therefore \(n - \dim\operatorname{null}T^{*}T = n - \dim\operatorname{null}T\), which equals \(\dim\operatorname{range}T\) by the fundamental theorem of linear maps (Part 3.1).

\(\blacksquare\)

Let \(m\) be the number of positive singular values, so \(m = \dim\operatorname{range}T\) by Statement 7.4.1. Order the eigenbasis \(e_1,\dots,e_n\) of \(T^{*}T\) so that \(T^{*}Te_j = s_j^{2}e_j\) with \(s_1,\dots,s_m > 0\) and \(s_j = 0\) for \(j > m\). For \(j > m\), \(\lVert Te_j\rVert^{2} = \langle T^{*}Te_j,e_j\rangle = 0\), so \(Te_j = 0\).

For \(j \leq m\) put \(f_j = s_j^{-1}Te_j\). Then \(\langle Te_j, Te_l\rangle = \langle T^{*}Te_j, e_l\rangle = s_j^{2}\delta_{jl}\), so \(f_1,\dots,f_m\) is orthonormal. For any \(v\), \(v = \sum_j\langle v,e_j\rangle e_j\), hence \(Tv = \sum_{j\leq m} \langle v,e_j\rangle Te_j = \sum_{j \leq m} s_j\langle v,e_j\rangle f_j\).

Finally extend \(e_1,\dots,e_m\) to an orthonormal basis of \(V\) and \(f_1,\dots,f_m\) to one of \(W\). This is possible because \(m \leq \min(n,p)\). Define \(s_j = 0\) for \(m < j \leq k\). The formula still holds, since the new terms have coefficient zero.

\(\blacksquare\)

Let \(R = \sqrt{T^{*}T}\), a positive operator. For every \(v\), \(\lVert Tv\rVert^{2} = \langle T^{*}Tv,v\rangle = \langle R^{2}v,v\rangle = \lVert Rv\rVert^{2}\). So \(\operatorname{null}T = \operatorname{null}R\), and \(\dim\operatorname{range}T = \dim\operatorname{range}R =: k\).

Define \(S_0 : \operatorname{range}R \to \operatorname{range}T\) by \(S_0(Rv) = Tv\). It is well defined: if \(Rv = Rw\), then \(\lVert T(v-w)\rVert = \lVert R(v-w)\rVert = 0\). It is linear and preserves norms, since \(\lVert S_0(Rv)\rVert = \lVert Tv\rVert = \lVert Rv\rVert\).

The orthogonal complements \((\operatorname{range}R)^{\perp}\) and \((\operatorname{range}T)^{\perp}\) both have dimension \(n-k\). Choose an isometric isomorphism \(S_1\) between them, by sending an orthonormal basis to an orthonormal basis. Define \(S(a+b) = S_0a + S_1b\) for \(a \in \operatorname{range}R\) and \(b \in (\operatorname{range}R)^{\perp}\). Since \(S_0a \in \operatorname{range}T\) and \(S_1b \in (\operatorname{range}T)^{\perp}\), these are orthogonal, so \(\lVert S(a+b)\rVert^{2} = \lVert S_0a\rVert^{2} + \lVert S_1b\rVert^{2} = \lVert a\rVert^{2} + \lVert b\rVert^{2} = \lVert a+b\rVert^{2}\). So \(S\) is an isometry, and \(S(Rv) = S_0(Rv) = Tv\). That is \(T = SR\).

If \(T\) is invertible, \(R\) is invertible and \(S = TR^{-1}\). An isometry of a finite-dimensional space is injective and therefore invertible, and \(S^{*}S = I\) by Statement 7.3.3, so \(S\) is orthogonal or unitary.

\(\blacksquare\)

(a). Let \(v \in B\) and \(x_j = \langle v, e_j\rangle\). By Bessel's inequality \(\sum_{j\leq m}|x_j|^{2} \leq \lVert v\rVert^{2} \leq 1\). The image is \(Tv = \sum_{j\leq m} s_jx_jf_j\), so \(y_j = s_jx_j\) satisfies \(\sum (y_j/s_j)^{2} \leq 1\). Conversely, given such \(y\), put \(v = \sum_{j\leq m}(y_j/s_j)e_j\). Then \(\lVert v\rVert^{2} = \sum (y_j/s_j)^{2} \leq 1\) and \(Tv = \sum y_jf_j\).

(b). For \(\lVert v\rVert \leq 1\), by orthonormality of the \(f_j\) and Bessel's inequality, \(\lVert Tv\rVert^{2} = \sum_{j\leq m}s_j^{2}|x_j|^{2} \leq s_1^{2}\sum_{j\le m}|x_j|^{2} \leq s_1^{2}\). Equality holds at \(v = e_1\).

\(\blacksquare\)

The \(k\)th column of \(\mathcal M(T)\) holds the coefficients of \(Tv_k\) in the basis. The entries below the diagonal in that column are the coefficients of \(v_{k+1},\dots,v_n\). So those entries are all \(0\) exactly when \(Tv_k\) has no \(v_{k+1},\dots,v_n\) component, that is, when \(Tv_k\in\operatorname{span}(v_1,\dots,v_k)\). Requiring this for every column gives the statement, since the matrix is upper triangular exactly when every column has zeros below the diagonal.

\(\blacksquare\)

Let \(A\) be the matrix, so \(Tv_k=\lambda_kv_k+\sum_{i\lt k}A_{i,k}v_i\).

Every \(\lambda_k\) is an eigenvalue. Fix \(k\) and let \(S=T-\lambda_kI\). For \(j\lt k\), \(Sv_j\in\operatorname{span}(v_1,\dots,v_j)\subseteq\operatorname{span}(v_1,\dots,v_{k-1})\). For \(j=k\), the \(v_k\) term of \(Sv_k=\sum_{i\lt k}A_{i,k}v_i\) cancels, so \(Sv_k\in\operatorname{span}(v_1,\dots,v_{k-1})\). Hence \(S\) maps the \(k\)-dimensional space \(\operatorname{span}(v_1,\dots,v_k)\) into the \((k-1)\)-dimensional space \(\operatorname{span}(v_1,\dots,v_{k-1})\). By the fundamental theorem of linear maps (Part 3.1), \(S\) has a nonzero vector \(w\) in \(\operatorname{span}(v_1,\dots,v_k)\) with \(Sw=0\). So \(Tw=\lambda_kw\).

There are no others. Let \(Tw=\lambda w\) with \(w\ne0\), and write \(w=\sum_{j\le m}c_jv_j\) with \(c_m\ne0\). Since \(Tv_j\in\operatorname{span}(v_1,\dots,v_j)\), only the terms with \(j=m\) contribute to the \(v_m\) coefficient of \(Tw\), and that coefficient is \(c_mA_{m,m}=c_m\lambda_m\). Comparing with the \(v_m\) coefficient \(\lambda c_m\) of \(\lambda w\) gives \(c_m(\lambda_m-\lambda)=0\), so \(\lambda=\lambda_m\).

\(\blacksquare\)

We induct on \(n=\dim V\). If \(n=1\) there is nothing to prove. Let \(n\ge2\) and assume the result for smaller dimensions. Since \(V\) is a nonzero complex space, \(T\) has an eigenvalue \(\lambda\) (a consequence of the fundamental theorem of algebra). Put \(W=\operatorname{range}(T-\lambda I)\).

Three facts about \(W\). First, \(T-\lambda I\) is not injective, so \(\dim W

If \(W=\{0\}\), take any basis \(x_1,\dots,x_n\) of \(V\); the third fact gives \(Tx_k\in\operatorname{span}(x_1,\dots,x_k)\) for all \(k\), and Statement 5.3.4 finishes. Otherwise, the induction hypothesis applied to \(T|_W\) gives a basis \(w_1,\dots,w_m\) of \(W\) with \(Tw_k\in\operatorname{span}(w_1,\dots,w_k)\). Extend it to a basis \(w_1,\dots,w_m,x_1,\dots,x_r\) of \(V\). For each \(k\), \(Tx_k=\lambda x_k+(T-\lambda I)x_k\) and \((T-\lambda I)x_k\in W=\operatorname{span}(w_1,\dots,w_m)\), so \(Tx_k\in\operatorname{span}(w_1,\dots,w_m,x_k)\subseteq\operatorname{span}(w_1,\dots,w_m,x_1,\dots,x_k)\). Together with \(Tw_k\in\operatorname{span}(w_1,\dots,w_k)\), Statement 5.3.4 shows the matrix is upper triangular.

\(\blacksquare\)

If \(U\) is a subspace, it has a zero, so \(0 \in U\), and it is closed under the operations by definition. Conversely, suppose the three conditions hold. The additive identity of \(V\) lies in \(U\), and additive inverses exist in \(U\) because \((-1)v = -v\) lies in \(U\) for each \(v \in U\). The vector space axioms are identities that hold in \(V\), so they hold for elements of \(U\) too.

\(\blacksquare\)

Both \(0 \in U\) and \(0 \in W\), so \(0 = 0+0 \in U+W\). If \(u_1+w_1\) and \(u_2+w_2\) lie in \(U+W\), their sum is \((u_1+u_2)+(w_1+w_2)\), which lies in \(U+W\) since \(U\) and \(W\) are closed under addition. Scaling by \(\lambda\) gives \((\lambda u_1)+(\lambda w_1)\in U+W\). So \(U+W\) is a subspace. It contains \(U\) because \(u = u+0\), and \(W\) because \(w = 0+w\).

If a subspace \(X\) contains \(U\) and \(W\), then every \(u+w\) is a sum of two elements of \(X\), so it lies in \(X\). Hence \(U+W \subseteq X\).

\(\blacksquare\)

Suppose the sum is direct and \(0 = u_1+\cdots+u_m\). Also \(0 = 0+\cdots+0\) is a representation with each \(u_i=0\). By uniqueness, \(u_i = 0\) for all \(i\).

Conversely, suppose the only representation of \(0\) is the trivial one, and take \(v = u_1+\cdots+u_m = w_1+\cdots+w_m\) with \(u_i,w_i \in U_i\). Subtracting gives \(0 = (u_1-w_1)+\cdots+(u_m-w_m)\), with \(u_i-w_i \in U_i\). So each \(u_i-w_i = 0\), that is \(u_i = w_i\).

\(\blacksquare\)

Suppose the sum is direct and \(v \in U\cap W\). Then \(v = v + 0\) with \(v\in U\), \(0 \in W\), and also \(v = 0 + v\) with \(0 \in U\), \(v \in W\). Uniqueness gives \(v = 0\).

Conversely, suppose \(U \cap W = \{0\}\) and \(0 = u + w\) with \(u \in U\), \(w \in W\). Then \(u = -w\) lies in \(U\cap W\), so \(u = 0\) and \(w = 0\). By Statement 1.1.3, the sum is direct.

\(\blacksquare\)

For \(z\) in the plane, \(|z - i|^2 - |z + i|^2 = -4\operatorname{Im} z\), because \(|z \mp i|^2 = |z|^2 + 1 \mp 2\operatorname{Im} z\). So \(|c(z)| \lt 1\) exactly when \(\operatorname{Im} z \gt 0\), which shows that \(c\) maps \(\mathbb{H}^2\) into \(\mathbb{D}\).

To invert, solve \(w(z + i) = z - i\): \(z(w - 1) = -i(1 + w)\), so \(z = \dfrac{i(1+w)}{1-w}\). This is defined for \(w \ne 1\), and it has positive imaginary part whenever \(|w| \lt 1\), since \(\operatorname{Im} z = \operatorname{Re}\dfrac{1+w}{1-w} = \dfrac{1 - |w|^2}{|1-w|^2} \gt 0\). So \(c\) and \(c^{-1}\) are inverse bijections between \(\mathbb{H}^2\) and \(\mathbb{D}\). Finally \(c(i) = 0\).

\(\blacksquare\)

For real \(x\), \(|x - i|^2 = x^2 + 1 = |x + i|^2\), so \(|c(x)| = 1\). Also \(c(\infty) = 1\). Since \(c\) is a bijection and sends \(\mathbb{H}^2\) into the disc, its image of the boundary is exactly the circle.

\(\blacksquare\)

A geodesic is a vertical half-line or a semicircle centred on the axis, and it meets the real axis (or \(\infty\)) at right angles. The map \(c\) is holomorphic and injective, so it preserves angles between curves. It sends the real axis, including \(\infty\), into the unit circle (Statement 4.1.2). So the image of a geodesic meets the unit circle at the same angle as the geodesic meets the real axis, namely \(90^\circ\).

A circle or line that meets the unit circle at right angles is either a circle orthogonal to it or a diameter. The vertical line \(x = 0\) goes to the diameter \((-1,1)\), since \(c(iy) = \dfrac{y-1}{y+1}\) is real.

\(\blacksquare\)

Let \(z = c^{-1}(u) = i\zeta\) with \(\zeta = \dfrac{1+u}{1-u}\), and similarly \(w = i\eta\) with \(\eta = \dfrac{1+v}{1-v}\). Then \(\operatorname{Im} z = \operatorname{Re}\zeta = \dfrac{1-|u|^2}{|1-u|^2}\), and likewise for \(w\). Also \(\zeta - \eta = \dfrac{2(u-v)}{(1-u)(1-v)}\), so \(|z - w|^2 = \dfrac{4|u-v|^2}{|1-u|^2|1-v|^2}\).

The distance formula in \(\mathbb{H}^2\) gives \(\cosh d = 1 + \dfrac{|z-w|^2}{2\operatorname{Im}z\,\operatorname{Im}w}\). Substituting,

\(\blacksquare\)

\(|k(w)| \lt 1\) is equivalent to \(2|w| \lt 1 + |w|^2\), that is \((1 - |w|)^2 \gt 0\), which holds for \(w \ne 1\), and that inequality is the only constraint. For the inverse, note that \(|k| = \dfrac{2r}{1+r^2}\) with \(r = |w|\), so \(1 - |k|^2 = \left(\dfrac{1-r^2}{1+r^2}\right)^2\), and \(\sqrt{1-|k|^2} = \dfrac{1-r^2}{1+r^2}\). Substituting back gives \(w = \dfrac{k}{1+\sqrt{1-|k|^2}}\). Both maps preserve the direction of their argument.

\(\blacksquare\)

Let \(P = k(u)\) and \(Q = k(v)\). From \(1 - |k|^2 = \left(\dfrac{1-|u|^2}{1+|u|^2}\right)^2\),

The right-hand side of Statement 4.2.2 has numerator \((1-|u|^2)(1-|v|^2) + 2|u-v|^2 = 1 + |u|^2 + |v|^2 + |u|^2|v|^2 - 4u\cdot v\), which is the same expression. So the two formulas agree.

\(\blacksquare\)

Geodesics are chords, so a geodesic through \(P\) is determined by its direction \(\theta\). Two chords meet in the disc exactly when their lines meet at a point of the open disc. As \(\theta\) turns, the point of intersection with the line \(l\) moves along \(l\). The set of directions for which it lies in the disc is an open interval of directions, and its complement is a closed interval of directions. That complement contains infinitely many directions, each giving a different geodesic that misses \(l\). This is the failure of Euclid's postulate.

\(\blacksquare\)

Let \(r = |w|\). Then \(x_0^2 - x_1^2 - x_2^2 = \dfrac{(1+r^2)^2 - 4r^2}{(1-r^2)^2} = \dfrac{(1-r^2)^2}{(1-r^2)^2} = 1\), and \(x_0 \gt 0\) for \(r \lt 1\). Conversely, \(\dfrac{x_1 + ix_2}{1+x_0} = \dfrac{2w/(1-r^2)}{2/(1-r^2)} = w\), using \(1 + x_0 = \dfrac{2}{1-r^2}\). So the two maps are inverse to each other.

\(\blacksquare\)

Let \(x = x(u)\) and \(y = x(v)\). Then

by the calculation in Statement 4.3.2 and Statement 4.2.2.

\(\blacksquare\)

The plane through the origin containing \(O\) and the direction \((0, \cos\varphi, \sin\varphi)\) consists of points \((s, t\cos\varphi, t\sin\varphi)\). On the hyperboloid this means \(s^2 - t^2 = 1\), so \(s = \cosh\tau\) and \(t = \sinh\tau\) for a unique real \(\tau\). This gives the curve above. By Statement 4.4.2, \(\cosh d(O, X_\tau) = \langle O, X_\tau\rangle = \cosh\tau\), so \(d(O, X_\tau) = |\tau|\). Hence the curve is parametrised by arc length from \(O\).

\(\blacksquare\)

Since \(c\) is an isometry from \(\mathbb{H}^2\) to the disc model, the disc distance from \(0 = c(i)\) to \(w\) is \(d\). By Statement 4.2.2 with \(u = 0\), \(\cosh d = \dfrac{1+|w|^2}{1-|w|^2}\). Solving for \(|w|^2\) gives \(|w|^2 = \dfrac{\cosh d - 1}{\cosh d + 1}\), that is \(|w| = \tanh\dfrac{d}{2}\).

For the Klein model, \(|k| = \dfrac{2|w|}{1+|w|^2}\). With \(\tau = \tanh\tfrac{d}{2}\), this is \(\dfrac{2\tau}{1+\tau^2} = \tanh d\), by the double-angle identity for \(\tanh\). For the hyperboloid, \(x_0 = \dfrac{1+|w|^2}{1-|w|^2} = \cosh d\) by Statement 4.4.1.

\(\blacksquare\)

The first equality is the distance formula of Part 3.3 in the upper half-plane with \(z = i\). The second is Statement 4.2.2 with \(u = 0\) and \(v = w\). The third holds because \(1 - |k|^2 = \left(\dfrac{1-|w|^2}{1+|w|^2}\right)^2\), so \(\dfrac{1}{\sqrt{1-|k|^2}} = \dfrac{1+|w|^2}{1-|w|^2}\). The last equality is Statement 4.5.1.

\(\blacksquare\)