University Mathematics — Year 1 · Bachelor Year 1

23Euclidean Spaces

Adding an inner product to a real vector space buys the geometric notions — lengths, angles, orthogonality, distances — and one theorem that towers over the chapter: every subspace admits an orthogonal projection, computable by Gram–Schmidt, realizing the shortest distance. The plane isometries close the chapter and the year’s geometry.

Throughout, $E$ is a real vector space.

23.1 Inner products

Definition 23.1

An inner product on $E$ is a map $\langle\cdot,\cdot\rangle \colon E \times E \to \R$ that is bilinear, symmetric, and positive definite ( $\langle x, x\rangle > 0$ for $x \neq 0$ ). A finite-dimensional space so equipped is a Euclidean space. The norm of $x$ is $\norm{x} = \sqrt{\langle x, x\rangle}$ , and $d(x, y) = \norm{x - y}$ .

Example 23.2

On $\R^n$ : the canonical product $\langle x, y\rangle = \sum x_i y_i$ . On $C(\intcc{a}{b})$ : $\langle f, g \rangle = \int_a^b fg$ (positive definiteness is Theorem 15.7 (4)). On $\R_n[X]$ : $\langle P, Q\rangle = \int_0^1 PQ$ , or $\sum_{i} P(x_i)Q(x_i)$ over $n+1$ distinct points.

Example 23.3 (The angle between two polynomials)

Once an inner product is chosen, any two nonzero vectors have an angle, via $\cos\theta = \frac{\langle x, y\rangle}{\norm x\,\norm y}$ (a legitimate cosine by Cauchy–Schwarz). For $X$ and $X^2$ in $\int_0^1$ :

\langle X, X^2\rangle = \frac14, \qquad \norm X = \frac1{\sqrt3}, \qquad \norm{X^2} = \frac1{\sqrt5}, \qquad \cos\theta = \frac{1/4}{1/\sqrt{15}} = \frac{\sqrt{15}}{4} \approx 0.968 :

an angle of about $14.5$ degrees — on $\intcc{0}{1}$ , the graphs of $x$ and $x^2$ are “nearly parallel” in the quadratic-mean sense, which is why removing that shared direction (Gram–Schmidt, below) leaves only the small correction $X^2 - X + \frac16$ .

Theorem 23.4 (Cauchy–Schwarz; norm properties)

For all $x, y \in E$ :

\abs{\langle x, y\rangle} \leq \norm x\, \norm y ,

with equality iff $x, y$ are proportional. Consequently $\norm\cdot$ satisfies the triangle inequality $\norm{x + y} \leq \norm x + \norm y$ (and $\norm{\lambda x} = \abs\lambda \norm x$ , $\norm x = 0 \iff x = 0$ ). Moreover:

\norm{x+y}^2 = \norm x^2 + 2\langle x, y\rangle + \norm y^2, \qquad \langle x, y \rangle = \tfrac14\bigl(\norm{x+y}^2 - \norm{x-y}^2\bigr) .

Proof. If $y = 0$ , everything is trivial. Otherwise the quadratic $t \mapsto \norm{x + ty}^2 = \norm x^2 + 2t\langle x, y\rangle + t^2 \norm y^2$ is $\geq 0$ for all $t$ : its discriminant is $\leq 0$ , which is Cauchy–Schwarz; equality means a double root $t_0$ , i.e. $x + t_0 y = 0$ (definiteness): proportionality. Triangle inequality: expand,

\norm{x + y}^2 = \norm x^2 + 2\langle x, y\rangle + \norm y^2 \leq \norm x^2 + 2\norm x\,\norm y + \norm y^2 = \bigl(\norm x + \norm y\bigr)^2 ,

the middle step being Cauchy–Schwarz; equality forces $\langle x, y\rangle = \norm x\norm y$ , the positive equality case, i.e. proportionality with nonnegative ratio — geometrically, the triangle degenerates only when the two vectors point the same way. The last two identities are direct expansions (the second, the polarization identity, recovers the product from the norm). ∎

23.2 Orthogonality

Definition 23.5

$x \perp y$ when $\langle x, y \rangle = 0$ . A family is orthogonal when its vectors are pairwise orthogonal, orthonormal when moreover each has norm $1$ . The orthogonal complement of a subspace $F$ is

F^{\perp} = \{x \in E : \forall y \in F,\ \langle x, y\rangle = 0\},

a subspace of $E$ .

Proposition 23.6

(Pythagoras) If $x \perp y$ then $\norm{x+y}^2 = \norm x^2 + \norm y^2$ . An orthogonal family of nonzero vectors is free. In an orthonormal basis $(e_1, \dots, e_n)$ , coordinates and products are

x = \sum_{i} \langle x, e_i\rangle\, e_i, \qquad \langle x, y \rangle = \sum_i \langle x, e_i\rangle \langle y, e_i\rangle, \qquad \norm x^2 = \sum_i \langle x, e_i\rangle^2 .

Proof. Pythagoras: expand. Freeness: take $\langle\,\cdot\,, x_j\rangle$ of a null combination: $\lambda_j \norm{x_j}^2 = 0$ . Coordinates: write $x = \sum \lambda_i e_i$ and take the product with $e_j$ : $\lambda_j = \langle x, e_j\rangle$ ; the two formulas follow by bilinearity. ∎

Example 23.7 (Orthonormal coordinates, with a Parseval check)

Expand $x = (1, 2, 3)$ in the orthonormal basis of Exercise 23.3,

e_1 = \tfrac{1}{\sqrt2}(1,1,0), \quad e_2 = \tfrac{1}{\sqrt6}(1,-1,2), \quad e_3 = \tfrac{1}{\sqrt3}(-1,1,1).

No system to solve — three inner products:

\langle x, e_1\rangle = \frac{3}{\sqrt2}, \qquad \langle x, e_2\rangle = \frac{1 - 2 + 6}{\sqrt6} = \frac{5}{\sqrt6}, \qquad \langle x, e_3\rangle = \frac{-1 + 2 + 3}{\sqrt3} = \frac{4}{\sqrt3}.

Certification by the norm formula of the proposition:

\frac{9}{2} + \frac{25}{6} + \frac{16}{3} = \frac{27 + 25 + 32}{6} = 14 = \norm{x}^2 = 1 + 4 + 9 .

This sum-of-squared-coordinates check (a finite Parseval identity) costs seconds and catches sign and normalization errors with near certainty — make it a habit whenever an orthonormal expansion is computed; its infinite-dimensional version, for the Fourier coefficients of Example 23.14, is a theorem of the Year 3 volume.

Theorem 23.8 (Gram–Schmidt)

Every Euclidean space has orthonormal bases. Explicitly, from any basis $(v_1, \dots, v_n)$ , the recipe

w_k = v_k - \sum_{i=1}^{k-1} \langle v_k, e_i\rangle\, e_i , \qquad e_k = \frac{w_k}{\norm{w_k}}

produces an orthonormal basis $(e_1, \dots, e_n)$ with $\operatorname{Vect}(e_1, \dots, e_k) = \operatorname{Vect}(v_1, \dots, v_k)$ for every $k$ .

Proof. Induction on $k$ . Assuming $(e_1, \dots, e_{k-1})$ orthonormal spanning $\operatorname{Vect}(v_1, \dots, v_{k-1})$ : the vector $w_k$ is orthogonal to each $e_j$ ( $j < k$ ) by construction ( $\langle w_k, e_j \rangle = \langle v_k, e_j\rangle - \langle v_k, e_j\rangle$ ), and $w_k \neq 0$ since $v_k \notin \operatorname{Vect}(v_1, \dots, v_{k-1})$ . Normalizing keeps orthogonality; the span statement holds since $e_k$ is a combination of $v_k$ and earlier $e_i$ ’s, invertibly. ∎

Example 23.9 (Gram–Schmidt on polynomials, in full)

Orthonormalize $(1, X, X^2)$ in $\R_2[X]$ with $\langle P, Q\rangle = \int_0^1 PQ$ . Step 1: $\norm{1}^2 = 1$ , so $e_1 = 1$ . Step 2: $w_2 = X - \langle X, 1\rangle\,1 = X - \frac12$ , and $\norm{w_2}^2 = \int_0^1\bigl(x - \frac12\bigr)^2 \dd x = \frac1{12}$ : $e_2 = \sqrt{12}\,\bigl(X - \frac12\bigr)$ . Step 3: $\langle X^2, e_1\rangle = \frac13$ and

\langle X^2, e_2\rangle = \sqrt{12}\int_0^1 x^2\Bigl(x - \frac12\Bigr)\dd x = \frac{\sqrt{12}}{12}, \qquad\text{so}\qquad w_3 = X^2 - \frac13 - \Bigl(X - \frac12\Bigr) = X^2 - X + \frac16 .

Its norm was computed in Exercise 23.9: $\norm{w_3}^2 = \frac1{180}$ , whence $e_3 = \sqrt{180}\,\bigl(X^2 - X + \frac16\bigr)$ . The polynomials $1$ , $X - \frac12$ , $X^2 - X + \frac16$ are, up to scale, the first Legendre polynomials of the interval $\intcc{0}{1}$ ; the construction continues one degree at a time, each new polynomial orthogonal to all its predecessors. Note how the algorithm recycles earlier work: the projection subtracted at step 3 is exactly the best affine approximation of $X^2$ found in Example 23.12 — Gram–Schmidt is iterated orthogonal projection.

Theorem 23.10 (Orthogonal projection)

Let $F$ be a subspace of the Euclidean space $E$ . Then

E = F \oplus F^{\perp},

and the associated projection $p_F$ onto $F$ (the orthogonal projection) is given, in any orthonormal basis $(e_1, \dots, e_k)$ of $F$ , by $p_F(x) = \sum_i \langle x, e_i\rangle e_i$ . It realizes the distance to $F$ : for all $y \in F$ ,

\norm{x - p_F(x)} \leq \norm{x - y},

with equality only for $y = p_F(x)$ ; one writes $d(x, F) = \norm{x - p_F(x)}$ .

Proof. Take an orthonormal basis $(e_i)_{i \leq k}$ of $F$ (Theorem 23.8 inside $F$ ) and set $\pi(x) = \sum \langle x, e_i\rangle e_i \in F$ . Then $x - \pi(x) \perp e_j$ for each $j$ (same cancellation as above), hence $x - \pi(x) \in F^\perp$ : $E = F + F^\perp$ . And $F \cap F^\perp = \{0\}$ : such a vector satisfies $\langle x, x\rangle = 0$ . So the sum is direct and $\pi = p_F$ .

Distance: for $y \in F$ , decompose $x - y = (x - p_F(x)) + (p_F(x) - y)$ , orthogonal pieces ( $F^\perp$ and $F$ ); Pythagoras:

\norm{x - y}^2 = \norm{x - p_F(x)}^2 + \norm{p_F(x) - y}^2 \geq \norm{x - p_F(x)}^2,

equality iff $y = p_F(x)$ . ∎

Example 23.11 (Projections never lengthen)

Applying Pythagoras to the split $x = p_F(x) + (x - p_F(x))$ :

\norm{p_F(x)}^2 = \norm x^2 - \norm{x - p_F(x)}^2 \leq \norm x^2 ,

with equality iff $x \in F$ . In an orthonormal basis $(e_1, \dots, e_k)$ of $F$ this reads $\sum_{i \leq k}\langle x, e_i\rangle^2 \leq \norm x^2$ (a Bessel inequality): however many orthonormal directions one measures, the squared coordinates never exceed the squared length — compare the exact equality of Example 23.7 when the family is a full basis. This one-line inequality is what makes Fourier coefficients summable in the Year 3 volume; here it already explains why adding more basis functions to a least squares fit can only decrease the residual.

Example 23.12 (Best quadratic approximation)

In $C(\intcc{0}{1})$ with $\langle f, g\rangle = \int_0^1 fg$ , the polynomial of degree $\leq 1$ closest to $f(x) = x^2$ in the associated (quadratic mean) distance is $p_F(f)$ where $F = \R_1[X]$ . Gram–Schmidt on $(1, X)$ : $e_1 = 1$ , $w_2 = X - \frac12$ , $\norm{w_2}^2 = \int_0^1 (x - \frac12)^2 = \frac{1}{12}$ , $e_2 = \sqrt{12}\,(X - \tfrac12)$ . Then

p_F(f) = \langle f, e_1\rangle e_1 + \langle f, e_2\rangle e_2 = \frac13 + 12\Bigl(\int_0^1 x^2\bigl(x - \tfrac12\bigr)\dd x\Bigr)\bigl(X - \tfrac12\bigr) = X - \frac{1}{6},

using $\int_0^1 x^2(x - \frac12)\dd x = \frac14 - \frac16 = \frac{1}{12}$ . The “least squares” idea in one line of linear algebra.

Method 23.13 (Three routes to a distance $d(x, F)$ )

Orthonormal basis of $F$ : then $p_F(x) = \sum_i \langle x, e_i\rangle e_i$ and, by Pythagoras,

$d(x, F)^2 = \norm{x}^2 - \norm{p_F(x)}^2 = \norm x^2 - \sum_i \langle x, e_i\rangle^2 ,$

often cheaper than computing $x - p_F(x)$ itself.
Normal equations: with any generating family of $F$ , solve $\langle x - p, v_j\rangle = 0$ for the coefficients of $p$ (Exercise 23.5) — no orthonormalization needed.
Through the complement: if $F^\perp$ is smaller than $F$ (e.g. $F$ a hyperplane, $F^\perp$ a line $\operatorname{Vect}(n)$ ), project onto $F^\perp$ instead:

$d(x, F) = \norm{p_{F^\perp}(x)} = \frac{\abs{\langle x, n\rangle}}{\norm n} ,$

which is the classical distance-to-a-plane formula (Exercise 25.8 uses it).

Route 3 is a special case of a general reflex: always project onto whichever of $F$ , $F^\perp$ has the smaller dimension.

Example 23.14 (Trigonometric orthogonality: a Fourier preview)

On $C(\intcc{0}{2\pi})$ with $\langle f, g\rangle = \frac1\pi\int_0^{2\pi} fg$ , the family

\Bigl(\frac{1}{\sqrt2},\ \cos x,\ \sin x,\ \cos 2x,\ \sin 2x,\ \dots\Bigr)

is orthonormal: for instance $\langle \cos px, \cos qx\rangle = \frac1\pi\int_0^{2\pi}\cos px\cos qx\,\dd x = 0$ for $p \neq q$ (linearize the product into $\frac12[\cos(p{-}q)x + \cos(p{+}q)x]$ and integrate over full periods), while $\frac1\pi\int_0^{2\pi}\cos^2 px\,\dd x = 1$ . The orthogonal projection onto the span of the first $2N + 1$ of these functions therefore has coordinates $\langle f, e_i\rangle$ — integrals against cosines and sines. These are the Fourier coefficients of $f$ , and the projection is its best quadratic-mean trigonometric approximation; the Year 3 volume studies their convergence. Orthogonality is doing all the work: the formulas for the coefficients are Theorem 23.10 verbatim.

23.3 Isometries of the plane

Definition 23.15

An endomorphism $u$ of a Euclidean space is an isometry (or orthogonal map) when it preserves the norm: $\norm{u(x)} = \norm x$ for all $x$ — equivalently (polarization) it preserves the inner product; equivalently its matrix $A$ in an orthonormal basis satisfies $A^{\mathsf T} A = I$ . Isometries form a group, the orthogonal group $O(E)$ .

Example 23.16 (Recognizing an isometry at sight)

Is $A = \dfrac15\begin{pmatrix} 3 & -4\\ 4 & 3\end{pmatrix}$ orthogonal? Columns: norms $\frac15\sqrt{9 + 16} = 1$ and $\frac15\sqrt{16 + 9} = 1$ ; product $\frac1{25}(3\cdot(-4) + 4\cdot3) = 0$ . Yes — and $\det A = \frac{9 + 16}{25} = 1$ , so it is the rotation $R_\theta$ with $\cos\theta = \frac35$ , $\sin\theta = \frac45$ (the “ $3$ - $4$ - $5$ rotation”, whose angle is no remarkable fraction of $\pi$ ). By contrast $B = \frac{1}{\sqrt2}\begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix}$ has unit determinant-scaled look but non-unit first column ( $\frac1{\sqrt2}$ ): not orthogonal — determinant $\pm1$ alone certifies nothing, the columns must be checked.

Theorem 23.17 (Plane isometries)

In an orthonormal basis of a Euclidean plane, the matrices of isometries are exactly

R_\theta = \begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix} \quad (\text{rotation of angle } \theta,\ \det = 1),

S_\theta = \begin{pmatrix} \cos\theta & \sin\theta\\ \sin\theta & -\cos\theta \end{pmatrix} \quad (\det = -1),

the latter being the reflection in the line making angle $\frac\theta2$ with the first basis vector.

Proof. Let $A = \begin{pmatrix} a & c\\ b & d\end{pmatrix}$ with $A^{\mathsf T}A = I$ : columns are unit and orthogonal. The first column is $(\cos\theta, \sin\theta)$ for some $\theta$ ; the second, unit and orthogonal to it, is $\pm(-\sin\theta, \cos\theta)$ . The sign $+$ gives $R_\theta$ ; the sign $-$ gives $S_\theta$ . One checks $S_\theta^2 = I$ and that the vector $(\cos\frac\theta2, \sin\frac\theta2)$ is fixed while its orthogonal is reversed: a reflection. (And $R_\alpha R_\beta = R_{\alpha+\beta}$ : the rotation group is the angle group — compare Theorem 3.7.) ∎

Two reflections make a rotation: reflecting M = (2, 0.5) in the x-axis, then in the line y = x, lands at (-0.5, 2) — the image of M under the rotation of angle π2 about the origin, twice the angle π4 between the axes. The weekend problem turns this picture into the composition law of all plane isometries. — Two reflections make a rotation: reflecting $M = (2, 0.5)$ in the $x$ -axis, then in the line $y = x$ , lands at $(-0.5, 2)$ — the image of $M$ under the rotation of angle $\frac\pi2$ about the origin, *twice* the angle $\frac\pi4$ between the axes. The weekend problem turns this picture into the composition law of all plane isometries.

Remark 23.18 (Common pitfalls)

The projection formula needs an orthonormal basis: for a merely generating family $(v_i)$ of $F$ , the sum $\sum_i\langle x, v_i\rangle v_i$ is not $p_F(x)$ (test $F = \R^2$ , $v_1 = e_1$ , $v_2 = e_1 + e_2$ ); with a non-orthonormal family, solve the normal equations instead (Method 23.13 (2)). Orthogonal families must avoid $0$ to be free: the zero vector is orthogonal to everything, including itself — freeness in Proposition 23.6 requires nonzero vectors. $F^\perp$ depends on the inner product: in $\R_1[X]$ , the complement of $\operatorname{Vect}(X)$ for $\int_0^1 PQ$ is not the constants but $\operatorname{Vect}(1 - \frac32 X)$ — compute $\int_0^1 x(1 - \frac32 x) = \frac12 - \frac12 = 0$ ; “perpendicular” is meaningless until the product is named. Do not expand $\norm{x + y}$ linearly: the correct identity is $\norm{x+y}^2 = \norm x^2 + 2\langle x, y\rangle + \norm y^2$ ; the cross term vanishes only under orthogonality (Pythagoras), and the triangle inequality is an inequality. Sending unit vectors to unit vectors is not enough: $u(x, y) = (x + y,\ 0)$ maps both canonical basis vectors to the unit vector $(1, 0)$ , yet $\norm{u(1,1)} = 2 \neq \sqrt2$ : no isometry. The definition demands $\norm{u(x)} = \norm x$ for all $x$ ; in matrix terms $A^{\mathsf T}A = I$ , i.e. columns that are unit and pairwise orthogonal — both conditions, checked together.

Remark 23.19 (Where the inner product goes)

Orthogonal projection is the most applied theorem of the chapter: it underlies least squares (the weekend problem of Chapter 25 builds regression lines on it), Fourier coefficients (Example 23.14), and the normal equations of Exercise 23.5, which numerical analysis solves at scale. The classification $R_\theta / S_\theta$ is completed below: the weekend problem classifies all distance-preserving transformations of the plane, linear or not, and their finite groups — the mathematics of rosettes and regular polygons. In the Year 2 volume the inner product meets eigenvalue theory (symmetric matrices, quadratic forms); in Year 3, infinite-dimensional Euclidean geometry becomes Hilbert space theory.

Remark 23.20 (Perspectives inside Book 3)

Two bridges leave this chapter. Backward, to the linear algebra: the Gram matrix of Exercise 23.11 packages inner products into the determinant machinery of Chapter 22, and orthogonal projection is the special projector of Chapter 20 whose kernel is $F^\perp$ — all its algebra ( $p^2 = p$ , $s = 2p - \mathrm{id}$ ) applies verbatim, now with the bonus that $\norm{x - p(x)}$ is a distance. Forward, to analysis: Chapter 24 measures arc length with this chapter’s norm and classifies nothing without its isometries; Chapter 25 reads the gradient through Cauchy–Schwarz (steepest ascent) and closes the volume with least squares, which is Theorem 23.10 applied to a data vector in $\R^n$ . The inner product is the point where the book’s algebra and its analysis finally meet.

23.4 Exercises

Exercise 23.1 ★

In $\R^3$ canonical: compute $\langle u, v\rangle$ , $\norm u$ , $\norm v$ and the angle between $u = (1, 2, 2)$ and $v = (2, -2, 1)$ . Verify Cauchy–Schwarz numerically.

Solution

Solution of Exercise 23.1.

$\langle u, v\rangle = 2 - 4 + 2 = 0$ ; $\norm u = \norm v = 3$ . The vectors are orthogonal: the angle is $\frac\pi2$ . Cauchy–Schwarz: $\abs 0 \leq 9$ , comfortably.

Exercise 23.2 ★

Prove the parallelogram identity $\norm{x+y}^2 + \norm{x-y}^2 = 2\norm x^2 + 2\norm y^2$ in any Euclidean space, and use it to show that the sup norm on $\R^2$ , $\norm{(x,y)}_\infty = \max(\abs x, \abs y)$ , does not come from an inner product.

Solution

Solution of Exercise 23.2.

Expand both squared norms with the identity of Theorem 23.4 and add: the cross terms cancel.

Sup norm: take $x = (1, 0)$ , $y = (0, 1)$ . Then $\norm{x + y}_\infty = \norm{x - y}_\infty = 1$ and the identity would require $1 + 1 = 2(1) + 2(1) = 4$ : false. A norm violating the parallelogram identity comes from no inner product.

Exercise 23.3 ★

Apply Gram–Schmidt to $\bigl((1,1,0), (1,0,1), (0,1,1)\bigr)$ in canonical $\R^3$ .

Solution

Solution of Exercise 23.3.

$e_1 = \frac{1}{\sqrt 2}(1,1,0)$ .

$w_2 = (1,0,1) - \langle (1,0,1), e_1\rangle e_1 = (1,0,1) - \frac{1}{2}(1,1,0) = \bigl(\tfrac12, -\tfrac12, 1\bigr)$ ; $\norm{w_2} = \sqrt{\tfrac32}$ , so $e_2 = \frac{1}{\sqrt 6}(1, -1, 2)$ .

$w_3 = (0,1,1) - \langle\cdot, e_1\rangle e_1 - \langle\cdot, e_2\rangle e_2 = (0,1,1) - \frac12 (1,1,0) - \frac16(1,-1,2) = \bigl(-\tfrac23, \tfrac23, \tfrac23\bigr)$ ; normalizing, $e_3 = \frac{1}{\sqrt 3}(-1, 1, 1)$ .

(Check: pairwise products vanish; each has norm $1$ .)

Exercise 23.4 ★

In $\R^3$ , let $F = \operatorname{Vect}\bigl((1,1,1)\bigr)$ . Determine $F^\perp$ (equation and basis), the matrix of $p_F$ in the canonical basis, and $d\bigl((1, 2, 3), F\bigr)$ .

Solution

Solution of Exercise 23.4.

$F^\perp = \{(x,y,z) : x + y + z = 0\}$ , basis $\bigl((1,-1,0), (1,0,-1)\bigr)$ . With $e = \frac{1}{\sqrt3}(1,1,1)$ : $p_F(x) = \langle x, e\rangle e = \frac{x_1 + x_2 + x_3}{3}\,(1,1,1)$ , so

\operatorname{Mat}(p_F) = \frac13 \begin{pmatrix} 1&1&1\\ 1&1&1\\ 1&1&1\end{pmatrix}.

For $x = (1,2,3)$ : $p_F(x) = (2,2,2)$ and $d(x, F) = \norm{(1,2,3) - (2,2,2)} = \norm{(-1,0,1)} = \sqrt 2$ .

Exercise 23.5 ★★

(Normal equations) Let $F = \operatorname{Vect}\bigl((1,0,1),(0,1,1)\bigr) \subseteq \R^3$ and $x = (1, 1, 4)$ . Compute $p_F(x)$ by solving $\langle x - p, v \rangle = 0$ for the two generators ( $p = \alpha(1,0,1) + \beta(0,1,1)$ ), then $d(x, F)$ . Why is Gram–Schmidt unnecessary here?

Solution

Solution of Exercise 23.5.

Let $p = \alpha(1,0,1) + \beta(0,1,1) = (\alpha, \beta, \alpha + \beta)$ . Orthogonality of $x - p$ to the generators:

\langle x - p, (1,0,1)\rangle = (1 - \alpha) + (4 - \alpha - \beta) = 0, \qquad \langle x - p, (0,1,1)\rangle = (1 - \beta) + (4 - \alpha - \beta) = 0,

i.e. $2\alpha + \beta = 5$ and $\alpha + 2\beta = 5$ : $\alpha = \beta = \frac53$ . So $p_F(x) = \bigl(\frac53, \frac53, \frac{10}{3}\bigr)$ and

d(x, F) = \norm{x - p} = \norm{\bigl(-\tfrac23, -\tfrac23, \tfrac23\bigr)} = \frac{2}{\sqrt 3} .

Gram–Schmidt is unnecessary because the defining property of the projection — $x - p \perp F$ — is itself a linear system (“normal equations”) on the coefficients in any generating family.

Exercise 23.6 ★★

For $f$ continuous on $\intcc{0}{1}$ , prove

\Bigl(\int_0^1 f\Bigr)^{\!2} \leq \int_0^1 f^2 ,

with the case of equality, as an instance of Cauchy–Schwarz in $C(\intcc{0}{1})$ . Then prove $\bigl(\sum_{i=1}^n a_i\bigr)^2 \leq n \sum a_i^2$ for reals $a_i$ .

Solution

Solution of Exercise 23.6.

Cauchy–Schwarz with $g = 1$ :

\Bigl(\int_0^1 f \cdot 1\Bigr)^2 \leq \int_0^1 f^2 \cdot \int_0^1 1^2 = \int_0^1 f^2 ,

equality iff $f$ is proportional to $1$ , i.e. constant. Discrete version: in $\R^n$ with $x = (a_1, \dots, a_n)$ , $y = (1, \dots, 1)$ : $\bigl(\sum a_i\bigr)^2 \leq \norm x^2\, \norm y^2 = n \sum a_i^2$ , equality iff all $a_i$ equal.

Exercise 23.7 ★★

Prove that for every subspace $F$ of a Euclidean space: $(F^{\perp})^{\perp} = F$ and $\dim F^{\perp} = \dim E - \dim F$ .

Solution

Solution of Exercise 23.7.

By Theorem 23.10, $E = F \oplus F^\perp$ , so dimensions add: $\dim F^\perp = \dim E - \dim F$ . Inclusion $F \subseteq (F^\perp)^\perp$ is immediate (vectors of $F$ are orthogonal to everything in $F^\perp$ ). Dimensions: $\dim(F^\perp)^\perp = \dim E - \dim F^\perp = \dim F$ ; an inclusion with equal (finite) dimensions is an equality (Theorem 19.14).

Exercise 23.8 ★★

Identify the plane isometries of matrices

A = \frac{1}{\sqrt 2}\begin{pmatrix} 1 & -1\\ 1 & 1\end{pmatrix}, \qquad B = \frac{1}{5}\begin{pmatrix} 3 & 4\\ 4 & -3\end{pmatrix}

(type, angle or axis). Compute $A^8$ and $B^2$ without multiplying matrices.

Solution

Solution of Exercise 23.8.

$A$ : columns unit, determinant $+1$ : a rotation, with $\cos\theta = \frac{1}{\sqrt2}$ , $\sin\theta = \frac{1}{\sqrt2}$ : $\theta = \frac\pi4$ . Hence $A^8 = R_{8\pi/4} = R_{2\pi} = I$ .

$B$ : determinant $\frac{1}{25}(-9 - 16) = -1$ : a reflection $S_\theta$ with $\cos\theta = \frac35$ , $\sin\theta = \frac45$ ; its axis makes the angle $\frac\theta2$ with the $x$ -axis, i.e. the line directed by $\bigl(\cos\frac\theta2, \sin\frac\theta2\bigr)$ — concretely the axis is $\operatorname{Vect}\bigl((2, 1)\bigr)$ , since $B(2,1)^{\mathsf T} = \frac15(6+4, 8-3)^{\mathsf T} = (2,1)^{\mathsf T}$ . As a reflection, $B^2 = I$ .

Exercise 23.9 ★★★

(Minimum as a projection) Compute

\min_{(a, b) \in \R^2} \int_0^1 \bigl(x^2 - a - bx\bigr)^2 \dd x ,

using Example 23.12: the minimum is $\norm{f - p_F(f)}^2$ for $f = X^2$ , $F = \R_1[X]$ .

Solution

Solution of Exercise 23.9.

The quantity is $\norm{f - (a + bX)}^2$ in $C(\intcc{0}{1})$ with the integral inner product: minimal exactly at the orthogonal projection $a + bX = p_F(f) = X - \frac16$ (Example 23.12). The minimum is

\norm{f - p_F(f)}^2 = \int_0^1 \Bigl(x^2 - x + \frac16\Bigr)^{\!2} \dd x .

Expand: $\int_0^1 (x^2 - x + \frac16)^2 = \int_0^1 \bigl(x^4 - 2x^3 + \frac{4}{3}x^2 - \frac13 x + \frac{1}{36}\bigr)\dd x = \frac15 - \frac12 + \frac49 - \frac16 + \frac{1}{36} = \frac{1}{180}$ . So the minimum equals $\dfrac{1}{180}$ .

Exercise 23.10 ★★★

Let $u$ be an isometry of a Euclidean space $E$ . Prove that $\ker(u - \mathrm{id}) \perp \operatorname{im}(u - \mathrm{id})$ , and deduce $E = \ker(u - \mathrm{id}) \oplus \operatorname{im}(u - \mathrm{id})$ . (Compute $\langle x - u(x), y\rangle$ for $u(y) = y$ , using preservation of the product.)

Solution

Solution of Exercise 23.10.

Let $y \in \ker(u - \mathrm{id})$ (i.e. $u(y) = y$ ) and $x \in E$ . Then, using preservation of the inner product ( $\langle u(a), u(b)\rangle = \langle a, b\rangle$ ):

\langle x - u(x),\, y\rangle = \langle x, y\rangle - \langle u(x), y\rangle = \langle x, y\rangle - \langle u(x), u(y)\rangle = \langle x, y\rangle - \langle x, y\rangle = 0 :

every vector of $\operatorname{im}(u - \mathrm{id})$ is orthogonal to every fixed vector.

Therefore $\operatorname{im}(u - \mathrm{id}) \subseteq \ker(u - \mathrm{id})^{\perp}$ , and by rank–nullity plus Exercise 23.7, both have dimension $\dim E - \dim\ker(u - \mathrm{id})$ : they are equal. Then $E = \ker(u - \mathrm{id}) \oplus \ker(u - \mathrm{id})^\perp = \ker(u - \mathrm{id}) \oplus \operatorname{im}(u - \mathrm{id})$ .

Exercise 23.11 ★★

(Gram matrix) For vectors $v_1, \dots, v_k$ of a Euclidean space, let $G = \bigl(\langle v_i, v_j\rangle\bigr)_{1 \leq i, j \leq k}$ be their Gram matrix.

Prove that $(v_1, \dots, v_k)$ is free if and only if $G$ is invertible. (If $Gc = 0$ , compute $\norm{\sum_i c_i v_i}^2$ .)
Compute the Gram matrix of $(1, X, X^2)$ in $\R_2[X]$ with $\langle P, Q\rangle = \int_0^1 PQ$ , recognize the Hilbert matrix $H_3$ of the weekend problem of Chapter 22, and conclude freeness from $\det H_3 = \frac1{2160} \neq 0$ .

Solution

Solution of Exercise 23.11.

Suppose $Gc = 0$ for a column $c = (c_1, \dots, c_k)^{\mathsf T}$ . Then

$\Bigl\lVert\sum_i c_i v_i\Bigr\rVert^2 = \sum_{i,j} c_i c_j \langle v_i, v_j\rangle = c^{\mathsf T} G\, c = 0,$

so $\sum_i c_i v_i = 0$ . If the family is free this forces $c = 0$ : $G$ is invertible. Conversely, a nontrivial relation $\sum_j c_j v_j = 0$ gives, taking the product with each $v_i$ , the nontrivial relation $Gc = 0$ : $G$ singular. Freeness $\iff \det G \neq 0$ .
$\langle X^{i-1}, X^{j-1}\rangle = \int_0^1 x^{i+j-2}\dd x = \frac{1}{i+j-1}$ : the Gram matrix of $(1, X, X^2)$ is exactly the Hilbert matrix $H_3$ , whose determinant $\frac1{2160}$ was computed in the weekend problem of Chapter 22: nonzero, so the monomials are free — as expected, but now certified by a number.

Exercise 23.12 ★★★

Let $u$ be an endomorphism of a Euclidean space $E$ whose matrix $A$ in an orthonormal basis is both orthogonal ( $A^{\mathsf T}A = I$ ) and symmetric ( $A^{\mathsf T} = A$ ).

Show $A^2 = I$ , and deduce (via Theorem 20.15) that $E = \ker(u - \mathrm{id}) \oplus \ker(u + \mathrm{id})$ .
Show that the two subspaces are orthogonal, so that $u$ is the orthogonal symmetry with respect to $F = \ker(u - \mathrm{id})$ : reflection through $F$ . (For $u(x) = x$ and $u(y) = -y$ , compute $\langle x, y\rangle$ two ways.)
Classify the plane case: which matrices of Theorem 23.17 are symmetric, and what are the corresponding maps?

Solution

Solution of Exercise 23.12.

$A^2 = A A = A^{\mathsf T} A = I$ , so $u^2 = \mathrm{id}$ : $u$ is a symmetry, and Theorem 20.15 (2) gives $E = \ker(u - \mathrm{id}) \oplus \ker(u + \mathrm{id})$ .
Let $u(x) = x$ and $u(y) = -y$ . Since $u$ preserves the inner product,

$\langle x, y\rangle = \langle u(x), u(y)\rangle = \langle x, -y\rangle = -\langle x, y\rangle ,$

so $\langle x, y\rangle = 0$ : the two eigenspaces are orthogonal, $\ker(u + \mathrm{id}) = F^\perp$ for $F = \ker(u - \mathrm{id})$ , and $u$ is the orthogonal symmetry with respect to $F$ .
$R_\theta$ is symmetric iff $-\sin\theta = \sin\theta$ , i.e. $\theta \in \{0, \pi\}$ : the maps $\pm\mathrm{id}$ (identity and central symmetry). Every $S_\theta$ is symmetric: the line reflections. These are exactly the orthogonal symmetries of the plane, with $F$ equal to the whole plane, $\{0\}$ , or the reflection axis.

23.5 Problem: plane isometries and Leonardo’s theorem

Problem 23.1

An isometry of the plane is any map $f \colon \R^2 \to \R^2$ preserving distances: $\norm{f(x) - f(y)} = \norm{x - y}$ for all $x, y$ — no linearity assumed. This problem proves that such maps are exactly the translations, rotations, reflections and glide reflections (the classification of plane isometries), computes their compositions, and determines all their finite groups: Leonardo’s theorem, the mathematics behind rosette patterns. From Part II on we identify $\R^2$ with $\C$ (Chapter 3): the canonical inner product is $\langle z, w\rangle = \operatorname{Re}(z\conj w)$ and the norm is the modulus.

Part I — Every isometry is affine.

Check that translations $t_a(x) = x + a$ , linear isometries, and all their compositions are isometries, and that the isometries form a group under composition.
Let $f$ be an isometry with $f(0) = 0$ . Show that $f$ preserves norms, then — by polarization, Theorem 23.4 — that $\langle f(x), f(y)\rangle = \langle x, y\rangle$ for all $x, y$ .
Still with $f(0) = 0$ : expand $\norm{f(x + y) - f(x) - f(y)}^2$ and $\norm{f(\lambda x) - \lambda f(x)}^2$ using question 2, and conclude that $f$ is linear: $f \in O(\R^2)$ .
Deduce that every isometry $f$ writes uniquely as $f = t_a \circ g$ with $a = f(0)$ and $g$ a linear isometry (the linear part of $f$ ).
Call $f$ direct if $\det g = 1$ , indirect if $\det g = -1$ . Show that the linear part of a composition is the composition of the linear parts, and state the resulting sign rule (direct/indirect compose like $+1/-1$ ).

Part II — The four types. Via Theorem 23.17, the linear isometries of $\C$ are $z \mapsto az$ and $z \mapsto a\conj z$ with $\abs a = 1$ ; so every plane isometry is

f(z) = a z + b \quad (\text{direct}) \qquad\text{or}\qquad f(z) = a\conj z + b \quad (\text{indirect}), \qquad \abs a = 1 .

Verify the dictionary: $R_\theta$ is $z \mapsto \eu^{\iu\theta}z$ and $S_\theta$ is $z \mapsto \eu^{\iu\theta}\conj z$ (check both on $1$ and $\iu$ ).
(Direct case) Let $f(z) = az + b$ , $\abs a = 1$ . Show: if $a = 1$ , $f$ is a translation; if $a \neq 1$ , $f$ has the unique fixed point $z_0 = b/(1 - a)$ and $f(z) - z_0 = a(z - z_0)$ : a rotation of center $z_0$ and angle $\arg a$ .
(Indirect case) Let $f(z) = a\conj z + b$ , and $v = a\conj b + b$ . Show $f \circ f = t_v$ and $f \circ t_v = t_v \circ f$ . If $v = 0$ : show that the midpoint of $z$ and $f(z)$ is a fixed point, and that $f$ is a reflection in a line. If $v \neq 0$ : show that $r = t_{-v/2}\circ f$ is a reflection whose axis is parallel to $v$ , so that $f = t_{v/2} \circ r$ is a glide reflection. Conclude: every plane isometry is a translation, a rotation, a reflection or a glide reflection (the classification of plane isometries).
(Compositions) Show: the composition of rotations of angles $\alpha$ and $\beta$ is a rotation of angle $\alpha + \beta$ (a translation if $\alpha + \beta \in 2\pi\Z$ ); the composition of two reflections is a rotation of angle twice the angle between the axes (a translation if the axes are parallel).
Deduce that every plane isometry is a composition of at most three reflections.

Part III — Three identifications.

Classify completely $f(z) = \iu\conj z + 1 + \iu$ : type, axis, glide vector.
Let $f$ be the rotation of angle $\frac\pi2$ about $0$ and $g$ the rotation of angle $\frac\pi2$ about $1$ . Compute $g \circ f$ in the form $z \mapsto az + b$ and identify it (type, center, angle).
Let $r_1(z) = \conj z$ (reflection in the real axis) and $r_2(z) = \iu\conj z$ (reflection in the line $y = x$ ). Compute $r_2 \circ r_1$ and check question 9 on this example.

Part IV — Finite groups: Leonardo’s theorem. Let $G$ be a finite group of plane isometries.

Show that isometries preserve barycenters: if $\lambda_1 + \dots + \lambda_m = 1$ and $f = t_a \circ g$ ( $g$ linear), then $f\bigl(\sum_i \lambda_i x_i\bigr) = \sum_i \lambda_i f(x_i)$ .
Set $c = \frac{1}{\abs G}\sum_{g \in G} g(x_0)$ for any chosen $x_0$ . Show that every $h \in G$ fixes $c$ : a finite group of isometries has a common fixed point.
Deduce that, after conjugating by $t_{-c}$ , one may assume $G \subseteq O(\R^2)$ . Let $G^{+} = \{g \in G : \det g = 1\}$ ; show that either $G = G^{+}$ or $G^{+}$ has exactly index $2$ in $G$ (exhibit a bijection $G^+ \to G \setminus G^+$ ).
Show that a finite group of rotations about $c$ is cyclic: among its elements pick the rotation $R_{\theta_0}$ of smallest angle $\theta_0 \in \intoo{0}{2\pi}$ , and use euclidean division of angles to prove that it generates; conclude $\theta_0 = \frac{2\pi}n$ and $G^{+} = \{R_{\theta_0}^{\,k}\} \cong C_n$ .
Suppose $G \neq G^{+}$ and pick a reflection $s \in G$ . Show $G = G^{+} \cup sG^{+}$ , that $s r s = r^{-1}$ for every rotation $r \in G^{+}$ , and that all $n$ elements of $sG^{+}$ are reflections: $G$ is the dihedral group $D_n$ of order $2n$ .
Conclude (Leonardo’s theorem): every finite group of isometries of the plane is cyclic $C_n$ or dihedral $D_n$ .

Part V — Dividends, and synthesis.

Show directly that a finite group of isometries can contain no translation and no glide reflection other than the identity (consider the powers of such an element).
Let $P_n$ be the regular $n$ -gon with vertices the $n$ -th roots of unity ( $n \geq 3$ ). Show that its symmetry group is exactly $D_n$ : the $n$ rotations $z \mapsto \omega^k z$ and the $n$ reflections $z \mapsto \omega^k \conj z$ , $\omega = \eu^{2\iu\pi/n}$ , and no others.
Exhibit plane figures whose symmetry groups are $C_1$ , $D_1$ , $D_2$ and $C_3$ respectively.
List the eight elements of the symmetry group of the square with vertices $\pm1, \pm\iu$ as maps $z \mapsto \omega^k z$ or $z \mapsto \omega^k\conj z$ , and give the axis of each of the four reflections.
Let $f, g$ be rotations of the same angle $\theta \notin 2\pi\Z$ about distinct centers $c_1 \neq c_2$ . Compute $f\circ g - g\circ f$ pointwise and show $f\circ g \neq g\circ f$ ; show moreover that $(f \circ g)\circ(g\circ f)^{-1}$ is a nontrivial translation, so that any group containing $f$ and $g$ is infinite — a second explanation of the single center in Leonardo’s theorem.
Synthesis, in four sentences: which two structural results reduce arbitrary isometries to linear algebra (questions 3–4) and arbitrary finite groups to subgroups of $O(2)$ (question 15); what the complete list of plane isometries is and which invariants (direct/indirect, fixed points) separate the four types; why the composition rules of question 9 make reflections the generators of everything; and what Leonardo’s theorem adds on the finite scale. Name the two theorems proved in Parts II and IV.

Solution

Solution of Problem 23.1.

1. $\norm{t_a(x) - t_a(y)} = \norm{x - y}$ ; a linear isometry preserves norms, hence distances ( $\norm{g(x) - g(y)} = \norm{g(x - y)} = \norm{x-y}$ ); and a composition of distance-preserving maps preserves distances. Every isometry is injective (distinct points stay at positive distance) and, by the classification below, bijective; identity and inverses are isometries: a group.

2. $\norm{f(x)} = \norm{f(x) - f(0)} = \norm{x - 0} = \norm x$ . Polarization:

\langle f(x), f(y)\rangle = \frac{\norm{f(x)}^2 + \norm{f(y)}^2 - \norm{f(x) - f(y)}^2}{2} = \frac{\norm x^2 + \norm y^2 - \norm{x - y}^2}{2} = \langle x, y\rangle .

3. Expanding with question 2 (each product of $f$ ’s equals the product of the arguments):

\begin{align*} \norm{f(x+y) - f(x) - f(y)}^2 &= \norm{x+y}^2 + \norm x^2 + \norm y^2\\ &\quad - 2\langle x+y, x\rangle - 2\langle x+y, y\rangle + 2\langle x, y\rangle = 0 , \end{align*}

as a direct check shows; similarly $\norm{f(\lambda x) - \lambda f(x)}^2 = \norm{\lambda x}^2 - 2\lambda\langle \lambda x, x\rangle + \lambda^2\norm x^2 = 0$ . So $f(x + y) = f(x) + f(y)$ and $f(\lambda x) = \lambda f(x)$ : $f$ is linear, and norm-preserving: $f \in O(\R^2)$ .

4. Set $a = f(0)$ and $g = t_{-a}\circ f$ : an isometry fixing $0$ , hence a linear isometry (question 3), and $f = t_a \circ g$ . Uniqueness: $t_a \circ g = t_{a'}\circ g'$ evaluated at $0$ gives $a = a'$ , then $g = g'$ .

5. $(t_a\circ g)\circ(t_{a'}\circ g') = t_{a + g(a')}\circ(g\circ g')$ , since $g\,t_{a'} = t_{g(a')}\,g$ for linear $g$ . So the linear part of a composition is $g \circ g'$ , and $\det(gg') = \det g \det g'$ : direct $\circ$ direct $=$ indirect $\circ$ indirect $=$ direct, direct $\circ$ indirect $=$ indirect — the sign rule of $\pm1$ .

6. $z \mapsto \eu^{\iu\theta}z$ sends $1$ to $(\cos\theta, \sin\theta)$ and $\iu$ to $\iu\eu^{\iu\theta} = (-\sin\theta, \cos\theta)$ : the columns of $R_\theta$ . And $z \mapsto \eu^{\iu\theta}\conj z$ sends $1$ to $(\cos\theta, \sin\theta)$ and $\iu$ to $-\iu\eu^{\iu\theta} = (\sin\theta, -\cos\theta)$ : the columns of $S_\theta$ .

7. $a = 1$ : $f = t_b$ . $a \neq 1$ : the fixed-point equation $az_0 + b = z_0$ has the unique solution $z_0 = b/(1 - a)$ , and then

f(z) - z_0 = az + b - (az_0 + b) = a(z - z_0):

in the frame centered at $z_0$ , $f$ is the multiplication by $a = \eu^{\iu\arg a}$ : the rotation of center $z_0$ and angle $\arg a$ .

8. $f(f(z)) = a\conj{(a\conj z + b)} + b = a\conj a\,z + a\conj b + b = z + v$ : $f^2 = t_v$ . Commutation: $f(z + v) = a\conj z + a\conj v + b$ and $a\conj v = a\conj{(a\conj b + b)} = b + a\conj b = v$ , so $f\circ t_v = t_v\circ f$ .

Case $v = 0$ : $f^2 = \mathrm{id}$ . For any $z$ , the midpoint $m = \frac{z + f(z)}2$ satisfies (isometries being affine, Part I) $f(m) = \frac{f(z) + f^2(z)}{2} = m$ : fixed points exist. Conjugating by the translation to a fixed point, $f$ becomes a linear indirect isometry, i.e. some $S_\theta$ (Theorem 23.17): a reflection in a line.

Case $v \neq 0$ : $r = t_{-v/2}\circ f$ is indirect and

r^2 = t_{-v/2}\,f\,t_{-v/2}\,f = t_{-v/2}\,t_{-v/2}\,f^2 = t_{-v}\,t_v = \mathrm{id}

(using the commutation), so $r$ is a reflection, and $f = t_{v/2}\circ r$ . Its axis is parallel to $v$ : $r$ commutes with $t_v$ (both $f$ and $t_{v/2}$ do), so $t_v$ maps the axis (the fixed line of $r$ ) to itself, forcing $v$ to direct it. Hence $f$ is a glide reflection with glide vector $v/2$ . Every isometry is now classified: translation or rotation (direct), reflection or glide reflection (indirect).

9. Rotations $f(z) = az + b$ , $g(z) = a'z + b'$ with $a = \eu^{\iu\alpha}$ , $a' = \eu^{\iu\beta}$ : $g\circ f(z) = a'az + (a'b + b')$ has linear coefficient $\eu^{\iu(\alpha+\beta)}$ : a rotation of angle $\alpha + \beta$ by question 7, or a translation when $\eu^{\iu(\alpha+\beta)} = 1$ . Reflections $r_i(z) = a_i\conj z + b_i$ with $a_i = \eu^{\iu\theta_i}$ (axis at angle $\theta_i/2$ ):

r_2\circ r_1(z) = a_2\conj{a_1}\,z + (a_2\conj{b_1} + b_2),

direct with angle $\theta_2 - \theta_1 = 2\bigl(\tfrac{\theta_2}2 - \tfrac{\theta_1}2\bigr)$ : twice the angle between the axes; parallel axes ( $\theta_1 = \theta_2$ ) give a translation.

10. A rotation of center $c$ and angle $\theta$ is the product of two reflections in lines through $c$ making angle $\theta/2$ (question 9, read backwards); a translation $t_v$ is the product of two reflections in parallel lines orthogonal to $v$ at distance $\norm v/2$ ; a reflection is one reflection; a glide reflection is a reflection composed with a translation, hence three. Maximum: three.

11. $a = \iu$ , $b = 1 + \iu$ : $v = a\conj b + b = \iu(1 - \iu) + 1 + \iu = (1 + \iu) + (1 + \iu) = 2 + 2\iu \neq 0$ : a glide reflection with glide vector $v/2 = 1 + \iu$ . The reflection $r = t_{-(1+\iu)}\circ f$ is $r(z) = \iu\conj z$ , whose axis is the line at angle $\frac12\arg\iu = \frac\pi4$ : the line $y = x$ . So $f$ is the glide reflection of axis $y = x$ and vector $(1, 1)$ .

12. $f(z) = \iu z$ and $g(z) = 1 + \iu(z - 1) = \iu z + 1 - \iu$ , so

g\circ f(z) = \iu(\iu z) + 1 - \iu = -z + 1 - \iu :

linear coefficient $-1 = \eu^{\iu\pi}$ , a rotation of angle $\pi$ (a half-turn), of center $z_0 = \frac{1 - \iu}{1 - (-1)} = \frac{1 - \iu}{2}$ .

13. $r_2\circ r_1(z) = \iu\,\conj{\conj z} = \iu z$ : the rotation of angle $\frac\pi2$ about $0$ . The axes (the real axis, angle $0$ ; the line $y = x$ , angle $\frac\pi4$ ) meet at angle $\frac\pi4$ , and twice that is $\frac\pi2$ : question 9 confirmed.

14. With $f = t_a\circ g$ , $g$ linear, and $\sum_i \lambda_i = 1$ :

f\Bigl(\sum_i \lambda_i x_i\Bigr) = a + \sum_i \lambda_i\,g(x_i) = \sum_i \lambda_i\bigl(a + g(x_i)\bigr) = \sum_i \lambda_i f(x_i) .

15. For $h \in G$ , using question 14 (the coefficients $\frac1{\abs G}$ sum to $1$ ):

h(c) = \frac{1}{\abs G}\sum_{g \in G} h\bigl(g(x_0)\bigr) = \frac{1}{\abs G}\sum_{g' \in G} g'(x_0) = c ,

because $g \mapsto hg$ is a bijection of $G$ onto itself.

16. The conjugates $t_{-c}\,g\,t_{c}$ ( $g \in G$ ) form a group of isometries fixing $0$ , hence of linear isometries (question 3): a finite subgroup of $O(\R^2)$ . If some $s \in G$ is indirect, the map $g \mapsto sg$ sends $G^+$ injectively into $G\setminus G^+$ and $h \mapsto s^{-1}h$ is its inverse (sign rule of question 5): $\abs G = 2\,\abs{G^+}$ ; otherwise $G = G^+$ .

17. If $G^{+} = \{\mathrm{id}\}$ , it is $C_1$ . Otherwise write its elements as $R_\theta$ , $\theta \in \intco{0}{2\pi}$ , and let $\theta_0$ be the smallest positive angle occurring. For $R_\theta \in G^{+}$ , euclidean division $\theta = k\theta_0 + \rho$ with $0 \leq \rho < \theta_0$ gives $R_\rho = R_\theta R_{\theta_0}^{-k} \in G^{+}$ , so $\rho = 0$ by minimality: $G^{+} = \langle R_{\theta_0}\rangle$ . Dividing $2\pi$ by $\theta_0$ the same way shows $2\pi = n\theta_0$ : $G^{+} \cong C_n$ , generated by the rotation of angle $\frac{2\pi}n$ .

18. $G = G^{+} \cup sG^{+}$ by the counting of question 16. With $s(z) = a\conj z$ and $r(z) = \omega z$ (linear forms, after question 16; $\abs a = \abs\omega = 1$ ):

s r s(z) = a\,\conj{\omega\,a\conj z} = a\conj\omega\conj a\,z = \conj\omega\,z = r^{-1}(z) .

Each element $sr^k$ of $sG^{+}$ is indirect and satisfies $(sr^k)^2 = (sr^ks)\,r^k = r^{-k}r^k = \mathrm{id}$ : an involutive indirect isometry fixing $c$ , i.e. a reflection. So $G$ consists of $n$ rotations and $n$ reflections with the relations $r^n = s^2 = \mathrm{id}$ , $srs = r^{-1}$ : the dihedral group $D_n$ .

19. Combining: a finite group of plane isometries fixes a point $c$ (question 15), reduces to a finite subgroup of $O(2)$ (question 16), and is $C_n$ if it contains only rotations (question 17), $D_n$ otherwise (question 18): Leonardo’s theorem.

20. A translation $t_v \neq \mathrm{id}$ has powers $t_{kv}$ , all distinct ( $kv$ pairwise distinct for $v \neq 0$ ): infinite order. A glide reflection $f$ has $f^2 = t_v$ with $v \neq 0$ : again infinite order. Neither fits in a finite group — consistently with questions 15–19, which produced only rotations and reflections.

21. The $2n$ maps preserve the vertex set: $\omega^k \omega^j = \omega^{k+j}$ and $\omega^k\conj{\omega^j} = \omega^{k-j}$ ; being isometries, they preserve the polygon (segments go to segments). Conversely, a symmetry preserves the barycenter $0$ of the vertices (question 14), hence is linear, and permutes the vertices (they are the points of the polygon at maximal distance from $0$ ). A linear isometry sending the vertex $1$ to $\omega^k$ is $z \mapsto \omega^k z$ if direct, $z \mapsto \omega^k\conj z$ if indirect (its matrix is determined by one column and the sign): at most $2n$ symmetries, so exactly the $D_n$ above.

22. $C_1$ : a scalene triangle (no nontrivial symmetry). $D_1$ : an isoceles, non-equilateral triangle (one reflection). $D_2$ : a non-square rectangle (identity, the half-turn about the center, the two axis reflections). $C_3$ : a triskelion — three congruent bent arms attached at $120$ -degree intervals; the bend kills every reflection but leaves the rotations of order $3$ .

23. With $\omega = \iu$ : the rotations $z \mapsto z$ , $\iu z$ , $-z$ , $-\iu z$ (angles $0, \frac\pi2, \pi, \frac{3\pi}2$ ), and the reflections

z \mapsto \conj z \ (\text{axis } y = 0), \quad \iu\conj z \ (y = x), \quad -\conj z \ (x = 0), \quad -\iu\conj z \ (y = -x):

the two diagonals and the two medians of the square — the dihedral group $D_4$ , of order $8$ .

24. Write $a = \eu^{\iu\theta} \neq 1$ : $f(z) = az + c_1(1 - a)$ and $g(z) = az + c_2(1 - a)$ . Then

f\circ g(z) - g\circ f(z) = (1-a)\bigl(ac_2 + c_1 - ac_1 - c_2\bigr) = -(1 - a)^2(c_2 - c_1) \neq 0 ,

so $f g \neq g f$ . Both compositions have linear coefficient $a^2$ , so $(fg)\circ(gf)^{-1}$ has linear coefficient $1$ : it is the translation by $fg(z) - gf(z)$ , the nonzero constant above. A group containing $f$ and $g$ contains this translation and all its powers: it is infinite. Two rotation centers are one too many for a finite group — the geometric heart of Leonardo’s theorem.

25. (i) Questions 3–4 show every distance-preserving map is affine with orthogonal linear part, and question 15 pins every finite group to a fixed point: both results convert metric geometry into linear algebra at the origin. (ii) The complete list is translation, rotation, reflection, glide reflection; the determinant of the linear part separates direct from indirect, and the existence of fixed points separates the two types within each parity. (iii) By question 9, two reflections compose to any rotation or translation, so reflections generate the whole group — at most three suffice for any isometry. (iv) On the finite scale only two families survive, the cyclic and dihedral groups, which is why rosette ornaments come in exactly two kinds (with or without mirror axes). Part II proved the classification of plane isometries; Part IV proved Leonardo’s theorem.