8.4 Isometries and unitary operators

A rotation of the plane moves almost every vector but changes no length and no angle. Section 6.3 met the matrices that behave like this, the unitary matrices, through a condition on their columns, $𝑈^{*} 𝑈 = 𝐼_{𝑛}$ . This section starts from the other end, from linear maps that preserve lengths, and shows that preserving lengths forces the rest: inner products are preserved, $𝑆^{*} 𝑆 = id$ , and the columns of every matrix in orthonormal bases are orthonormal. For operators on a finite-dimensional space these maps are the unitary operators of Table 8.1.1, the analogues of the complex numbers on the unit circle, and over $ℂ$ the spectral theorem describes them completely. Over $ℝ$ the description is the classical one, by rotations in perpendicular planes and reflections. We derive it in the second half of the section, together with the distance-preserving maps of $ℝ^{𝑛}$ that are not linear.

Throughout, $𝑉$ and $𝑊$ are inner product spaces over $𝔽$ , and inner products are linear in the first slot.

8.4.1Isometries

Definition 8.4.1 (Isometry). Let $𝑉$ and $𝑊$ be inner product spaces over $𝔽$ . A linear map $𝑆 \in L (𝑉, 𝑊)$ is an isometry if

‖ 𝑆 𝐯 ‖ = ‖ 𝐯 ‖ for every 𝐯 \in 𝑉 .

Since $𝑆$ is linear, $‖ 𝑆 𝐮 -$ $𝑆 𝐯 ‖ =$ $‖ 𝑆 (𝐮 - 𝐯) ‖ =$ $‖ 𝐮 -$ $𝐯 ‖$ , so an isometry preserves distances as well as lengths. In metric geometry “isometry” means any distance-preserving map, linear or not, and translations are the basic nonlinear examples. Our isometries are always linear; Section 8.4.6 shows that, in a real inner product space, the general distance-preserving maps differ from them only by a translation.

Example 8.4.2 (Isometries).

(a)Every unitary matrix $𝑈 \in 𝔽^{𝑛 \times 𝑛}$ gives an isometry $𝐱 \mapsto 𝑈 𝐱$ of $𝔽^{𝑛}$ (Exercise 6.3.37). In particular, the rotations $𝑅_{𝜃}$ and the reflections $𝑆_{𝜃}$ of Example 6.3.35 are isometries of $ℝ^{2}$ .
(b)More generally, let $𝐴 \in 𝔽^{𝑚 \times 𝑛}$ have orthonormal columns. Then $𝐴^{*} 𝐴 = 𝐼_{𝑛}$ (Theorem 6.3.29), so for every $𝐱 \in 𝔽^{𝑛}$ ,
${‖ 𝐴 𝐱 ‖}^{2} = {(𝐴 𝐱)}^{*} (𝐴 𝐱) = 𝐱^{*} 𝐴^{*} 𝐴 𝐱 = 𝐱^{*} 𝐱 = {‖ 𝐱 ‖}^{2} .$
For instance, $𝐴 = \frac{1}{3} (\begin{matrix} 1 & 2 \\ 2 & 1 \\ 2 & - 2 \end{matrix})$ maps $ℝ^{2}$ isometrically onto a plane in $ℝ^{3}$ ; it sends $(1, 1)$ to $(1, 1, 0)$ , and both have length $\sqrt{2}$ .
(c)If $𝑈$ is a subspace of $𝑉$ , the inclusion $𝑈 \to 𝑉$ , $𝐮 \mapsto 𝐮$ , is an isometry.
(d)Let $𝒪︀ = 𝐞_{1}, \dots, 𝐞_{𝑛}$ be an orthonormal basis of $𝑉$ . By Theorem 6.3.22(b), the coordinate map $𝐯 \mapsto {[𝐯]}_{𝒪︀}$ is an isometry from $𝑉$ to $𝔽^{𝑛}$ with the dot product, and so is its inverse $(𝑎_{1}, \dots, 𝑎_{𝑛}) \mapsto$ $𝑎_{1} 𝐞_{1} +$ $\dots +$ $𝑎_{𝑛} 𝐞_{𝑛}$ .

Non-example 8.4.3 (Maps that are not isometries).

The shear $(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ sends $𝐞_{2}$ to $(1, 1)$ , which has length $\sqrt{2}$ .
An orthogonal projection ${Pr}_{𝑈}$ with $𝑈 \neq 𝑉$ is not an isometry. For $𝐯 \notin 𝑈$ , the vector $𝐯 - {Pr}_{𝑈} 𝐯$ is nonzero and orthogonal to ${Pr}_{𝑈} 𝐯$ , so $‖ {Pr}_{𝑈} 𝐯 ‖ < ‖ 𝐯 ‖$ by Theorem 6.3.3.
The columns of $(\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix})$ are orthogonal but have length $\sqrt{2}$ . The matrix is $\sqrt{2}$ times an orthogonal matrix, so it multiplies every length by $\sqrt{2}$ . It preserves right angles without being an isometry (compare Exercise 8.4.34).

Intuition (Why lengths control angles). Over $ℝ$ , the law of cosines

{‖ 𝐮 - 𝐯 ‖}^{2} = {‖ 𝐮 ‖}^{2} - 2 ⟨ 𝐮, 𝐯 ⟩ + {‖ 𝐯 ‖}^{2}

computes the inner product of $𝐮$ and $𝐯$ from the three side lengths of the triangle with vertices $𝟎$ , $𝐮$ , $𝐯$ . A linear map $𝑆$ sends this triangle to the triangle with vertices $𝟎$ , $𝑆 𝐮$ , $𝑆 𝐯$ , whose third side is $𝑆 𝐮 -$ $𝑆 𝐯 =$ $𝑆 (𝐮 - 𝐯)$ . If $𝑆$ preserves lengths, the two triangles have the same side lengths, hence the same angle at $𝟎$ . Over $ℂ$ the polarization identity (Proposition 6.1.32) plays the role of the law of cosines. Linearity is used once, to know that the third side of the image is the image of the third side.

The main theorem lists the equivalent forms of the definition. Parts of it appeared as exercises in Section 6.6 (Exercise 6.1.51 and Exercise 6.6.40).

Theorem 8.4.4 (Characterizations of isometries). Let $𝑉$ and $𝑊$ be finite-dimensional inner product spaces with $𝑛 = dim 𝑉$ , and let $𝑆 \in L (𝑉, 𝑊)$ . The following are equivalent.

(I) $𝑆$ is an isometry.
(II) $𝑆$ preserves inner products: $⟨ 𝑆 𝐮, 𝑆 𝐯 ⟩ =$ $⟨ 𝐮, 𝐯 ⟩$ for all $𝐮, 𝐯 \in 𝑉$ .
(III) $𝑆^{*} 𝑆 = {id}_{𝑉}$ .
(IV) $𝑆$ maps every orthonormal list in $𝑉$ to an orthonormal list in $𝑊$ .
(V) $𝑆$ maps some orthonormal basis of $𝑉$ to an orthonormal list in $𝑊$ .
(VI)For some orthonormal bases $ℬ︀$ of $𝑉$ and $𝒞︀$ of $𝑊$ , the columns of $𝐴 = {[𝑆]}_{𝒞︀ ℬ︀}$ are orthonormal, that is, $𝐴^{*} 𝐴 = 𝐼_{𝑛}$ .
(VII)For all orthonormal bases $ℬ︀$ of $𝑉$ and $𝒞︀$ of $𝑊$ , the columns of ${[𝑆]}_{𝒞︀ ℬ︀}$ are orthonormal.

Proof (by polarization).standard

Uses. Proposition 6.1.32 (Polarization identities), Definition 6.6.8 (Adjoint), Proposition 6.1.37 (Inner products distinguish vectors), Theorem 6.4.20 (Orthonormal bases exist), Proposition 6.3.18 (Coordinates in an orthonormal basis), Corollary 6.3.6 (Pythagoras for orthogonal lists), Theorem 6.6.15 (Existence of adjoints), Theorem 2.4.2 (The matrix of a composite), Theorem 2.3.20 (Linear maps correspond to matrices), Theorem 6.3.29 (Orthonormal columns)

Techniques. Proving TFAE (Technique 2.7.1), Polarization (Technique 6.1.33), Move an operator across the inner product (Technique 6.6.20), Translate between maps and matrices (Technique 2.3.27)

Hint 1. For (I)

\Rightarrow

(II), express

⟨ 𝑆 𝐮, 𝑆 𝐯 ⟩

through lengths of the vectors

𝑆 𝐮 + 𝑐 𝑆 𝐯

, and use linearity to pull

𝑆

out.

Hint 2. For (II)

\Rightarrow

(III), rewrite

⟨ 𝑆 𝐮, 𝑆 𝐯 ⟩

with the adjoint as

⟨ 𝐮, 𝑆^{*} 𝑆 𝐯 ⟩

. For (V)

\Rightarrow

(I), expand

𝐯

in the orthonormal basis and apply Pythagoras to

𝐯

and to

𝑆 𝐯

Hint 3. For the matrix conditions,

{[𝑆^{*}]}_{ℬ︀ 𝒞︀} = 𝐴^{*}

in orthonormal bases, so

{[𝑆^{*} 𝑆]}_{ℬ︀ ℬ︀} = 𝐴^{*} 𝐴

Sketch. Polarization writes

⟨ 𝑆 𝐮, 𝑆 𝐯 ⟩

as a combination of

{‖ 𝑆 (𝐮 + 𝑐 𝐯) ‖}^{2} =

{‖ 𝐮 + 𝑐 𝐯 ‖}^{2}

, which gives (II). Then

⟨ 𝐮, 𝑆^{*} 𝑆 𝐯 ⟩ =

⟨ 𝐮, 𝐯 ⟩

for all

𝐮

forces

𝑆^{*} 𝑆 𝐯 = 𝐯

. From (III),

⟨ 𝑆 𝐞_{𝑗}, 𝑆 𝐞_{𝑖} ⟩ =

⟨ 𝐞_{𝑗}, 𝑆^{*} 𝑆 𝐞_{𝑖} ⟩ =

𝛿_{𝑖 𝑗}

. If

𝑆 𝐞_{1}, \dots, 𝑆 𝐞_{𝑛}

is orthonormal, Pythagoras gives

{‖ 𝑆 𝐯 ‖}^{2} =

\sum {| 𝑎_{𝑗} |}^{2} =

{‖ 𝐯 ‖}^{2}

. In orthonormal bases

𝑆^{*} 𝑆

has matrix

𝐴^{*} 𝐴

, so (III) holds iff

𝐴^{*} 𝐴 = 𝐼_{𝑛}

in one, or in all, such pairs of bases.

We prove (I) $\Rightarrow$ (II) $\Rightarrow$ (III) $\Rightarrow$ (IV) $\Rightarrow$ (V) $\Rightarrow$ (I), and then (III) $\Rightarrow$ (VII) $\Rightarrow$ (VI) $\Rightarrow$ (III).

(I) $\Rightarrow$ (II). Let $𝐮, 𝐯 \in 𝑉$ . For every scalar $𝑐$ , linearity gives $𝑆 𝐮 +$ $𝑐 𝑆 𝐯 =$ $𝑆 (𝐮 + 𝑐 𝐯)$ , so by (I)

‖ 𝑆 𝐮 + 𝑐 𝑆 𝐯 ‖ = ‖ 𝐮 + 𝑐 𝐯 ‖ .

If $𝔽 = ℝ$ , Proposition 6.1.32(a) with $𝑐 = \pm 1$ gives

⟨ 𝑆 𝐮, 𝑆 𝐯 ⟩ = \frac{1}{4} ({‖ 𝑆 𝐮 + 𝑆 𝐯 ‖}^{2} - {‖ 𝑆 𝐮 - 𝑆 𝐯 ‖}^{2}) = \frac{1}{4} ({‖ 𝐮 + 𝐯 ‖}^{2} - {‖ 𝐮 - 𝐯 ‖}^{2}) = ⟨ 𝐮, 𝐯 ⟩ .

If $𝔽 = ℂ$ , the same computation with the four values $𝑐 = 𝑖^{𝑘}$ in Proposition 6.1.32(b) gives $⟨ 𝑆 𝐮, 𝑆 𝐯 ⟩ =$ $⟨ 𝐮, 𝐯 ⟩$ .

(II) $\Rightarrow$ (III). Let $𝐯 \in 𝑉$ . By the definition of the adjoint (Definition 6.6.8) and (II), for every $𝐮 \in 𝑉$

⟨ 𝐮, 𝑆^{*} 𝑆 𝐯 ⟩ = ⟨ 𝑆 𝐮, 𝑆 𝐯 ⟩ = ⟨ 𝐮, 𝐯 ⟩ .

Conjugating both sides, $⟨ 𝑆^{*} 𝑆 𝐯, 𝐮 ⟩ =$ $⟨ 𝐯, 𝐮 ⟩$ for all $𝐮$ , so $𝑆^{*} 𝑆 𝐯 = 𝐯$ by Proposition 6.1.37. Hence $𝑆^{*} 𝑆 = {id}_{𝑉}$ .

(III) $\Rightarrow$ (IV). Let $𝐞_{1}, \dots, 𝐞_{𝑘}$ be orthonormal in $𝑉$ . Then

⟨ 𝑆 𝐞_{𝑗}, 𝑆 𝐞_{𝑖} ⟩ = ⟨ 𝐞_{𝑗}, 𝑆^{*} 𝑆 𝐞_{𝑖} ⟩ = ⟨ 𝐞_{𝑗}, 𝐞_{𝑖} ⟩ = 𝛿_{𝑖 𝑗},

so $𝑆 𝐞_{1}, \dots, 𝑆 𝐞_{𝑘}$ is orthonormal.

(IV) $\Rightarrow$ (V). $𝑉$ has an orthonormal basis (Theorem 6.4.20(a)), and by (IV) its image is orthonormal.

(V) $\Rightarrow$ (I). Let $𝐞_{1}, \dots, 𝐞_{𝑛}$ be an orthonormal basis of $𝑉$ with $𝑆 𝐞_{1}, \dots, 𝑆 𝐞_{𝑛}$ orthonormal, and let $𝐯 \in 𝑉$ . By Proposition 6.3.18, $𝐯 = \sum_{𝑗} 𝑎_{𝑗} 𝐞_{𝑗}$ with $𝑎_{𝑗} = ⟨ 𝐯, 𝐞_{𝑗} ⟩$ , so $𝑆 𝐯 =$ $\sum_{𝑗} 𝑎_{𝑗} 𝑆 𝐞_{𝑗}$ . Applying Corollary 6.3.6 to both orthonormal lists,

{‖ 𝑆 𝐯 ‖}^{2} = \sum_{𝑗 = 1}^{𝑛} {| 𝑎_{𝑗} |}^{2} {‖ 𝑆 𝐞_{𝑗} ‖}^{2} = \sum_{𝑗 = 1}^{𝑛} {| 𝑎_{𝑗} |}^{2} = \sum_{𝑗 = 1}^{𝑛} {| 𝑎_{𝑗} |}^{2} {‖ 𝐞_{𝑗} ‖}^{2} = {‖ 𝐯 ‖}^{2} .

(III) $\Rightarrow$ (VII). Let $ℬ︀, 𝒞︀$ be orthonormal bases of $𝑉$ and $𝑊$ , and $𝐴 = {[𝑆]}_{𝒞︀ ℬ︀}$ . By Theorem 6.6.15(b), ${[𝑆^{*}]}_{ℬ︀ 𝒞︀} = 𝐴^{*}$ , and by Theorem 2.4.2,

𝐴^{*} 𝐴 = {[𝑆^{*}]}_{ℬ︀ 𝒞︀} {[𝑆]}_{𝒞︀ ℬ︀} = {[𝑆^{*} 𝑆]}_{ℬ︀ ℬ︀} = {[{id}_{𝑉}]}_{ℬ︀ ℬ︀} = 𝐼_{𝑛} .

By Theorem 6.3.29, $𝐴^{*} 𝐴 = 𝐼_{𝑛}$ says exactly that the columns of $𝐴$ are orthonormal.

(VII) $\Rightarrow$ (VI). Orthonormal bases of $𝑉$ and $𝑊$ exist (Theorem 6.4.20(a)).

(VI) $\Rightarrow$ (III). The computation in (III) $\Rightarrow$ (VII), read backwards, gives ${[𝑆^{*} 𝑆]}_{ℬ︀ ℬ︀} =$ $𝐴^{*} 𝐴 =$ $𝐼_{𝑛} =$ ${[{id}_{𝑉}]}_{ℬ︀ ℬ︀}$ , and a linear map is determined by its matrix (Theorem 2.3.20). So $𝑆^{*} 𝑆 = {id}_{𝑉}$ . □

The polarization step is the only place where an identity for single vectors becomes one for pairs. There is a second route from lengths to $𝑆^{*} 𝑆$ that hides the pairs inside a result about self-adjoint operators.

Proof (through the quadratic form of $𝑆^{*} 𝑆 - {id}_{𝑉}$ ).standard

Uses. Proposition 6.6.18 (Properties of the adjoint), Definition 6.6.8 (Adjoint), Proposition 8.1.4 (Self-adjoint operators are determined by their diagonal values)

Techniques. Move an operator across the inner product (Technique 6.6.20)

Hint 1. Let

𝑃 = 𝑆^{*} 𝑆 - {id}_{𝑉}

. Is

𝑃

self-adjoint? Compute

⟨ 𝑃 𝐯, 𝐯 ⟩

Hint 2.

⟨ 𝑃 𝐯, 𝐯 ⟩ =

{‖ 𝑆 𝐯 ‖}^{2} -

{‖ 𝐯 ‖}^{2}

. Which result says that a self-adjoint operator with

⟨ 𝑃 𝐯, 𝐯 ⟩ = 0

for all

𝐯

is zero?

Sketch.

𝑃 = 𝑆^{*} 𝑆 - {id}_{𝑉}

is self-adjoint, and

⟨ 𝑃 𝐯, 𝐯 ⟩ =

{‖ 𝑆 𝐯 ‖}^{2} -

{‖ 𝐯 ‖}^{2}

. So (I) says that the quadratic form of

𝑃

vanishes, and Proposition 8.1.4 gives

𝑃 = 0

. Then (III)

\Rightarrow

(II) is one line, and the rest is as in the first proof.

We prove the cycle (I) $\Rightarrow$ (III) $\Rightarrow$ (II) $\Rightarrow$ (I); the remaining implications are as in the first proof.

Let $𝑃 = 𝑆^{*} 𝑆 - {id}_{𝑉}$ . By Proposition 6.6.18, ${(𝑆^{*} 𝑆)}^{*} =$ $𝑆^{*} {(𝑆^{*})}^{*} =$ $𝑆^{*} 𝑆$ , so $𝑃$ is self-adjoint. For $𝐯 \in 𝑉$ , the definition of the adjoint gives $⟨ 𝑆^{*} 𝑆 𝐯, 𝐯 ⟩ =$ $⟨ 𝑆 𝐯, 𝑆 𝐯 ⟩$ (apply Definition 6.6.8 to $𝑆^{*}$ , whose adjoint is $𝑆$ ), so

⟨ 𝑃 𝐯, 𝐯 ⟩ = ⟨ 𝑆 𝐯, 𝑆 𝐯 ⟩ - ⟨ 𝐯, 𝐯 ⟩ = {‖ 𝑆 𝐯 ‖}^{2} - {‖ 𝐯 ‖}^{2} .

If (I) holds, this vanishes for every $𝐯$ , and $𝑃 = 0$ by Proposition 8.1.4; that is (III). Conversely, if (III) holds, then for all $𝐮, 𝐯$ ,

⟨ 𝑆 𝐮, 𝑆 𝐯 ⟩ = ⟨ 𝐮, 𝑆^{*} 𝑆 𝐯 ⟩ = ⟨ 𝐮, 𝐯 ⟩,

which is (II), and taking $𝐮 = 𝐯$ in (II) gives (I). □

The first proof recovers inner products from lengths; the second recovers the operator $𝑆^{*} 𝑆 - {id}_{𝑉}$ from its values $⟨ 𝑃 𝐯, 𝐯 ⟩$ . They rest on the same fact (the proof of Proposition 8.1.4 is a polarization argument), but the second shows the shape of the condition. An isometry is a map whose “defect” $𝑆^{*} 𝑆 - {id}_{𝑉}$ has zero quadratic form, and Section 8.3 studies maps whose $𝑆^{*} 𝑆$ is merely compared with ${id}_{𝑉}$ .

Remark 8.4.5 (Without finite dimension). The implications (I)

\Rightarrow

(II)

\Rightarrow

(I) use neither bases nor adjoints, and neither does (II)

\Rightarrow

(IV), since

⟨ 𝑆 𝐞_{𝑗}, 𝑆 𝐞_{𝑖} ⟩ =

⟨ 𝐞_{𝑗}, 𝐞_{𝑖} ⟩ =

𝛿_{𝑖 𝑗}

for an orthonormal list; so they hold for a linear map between any two inner product spaces. So does (II)

\Rightarrow

(III), whenever

𝑆

has an adjoint. In particular, every isometry preserves inner products, orthogonality, and angles (Definition 6.2.16).

Corollary 8.4.6 (Properties of isometries). Let $𝑆 \in L (𝑉, 𝑊)$ be an isometry.

(a) $𝑆$ is injective, and it preserves inner products, distances, and orthogonality.
(b)If $𝑅 \in L (𝑊, 𝑋)$ is also an isometry, so is $𝑅 𝑆$ .
(c)If $𝑉$ and $𝑊$ are finite-dimensional, then $dim 𝑉 \leq dim 𝑊$ , and $𝑆 𝑆^{*}$ is the orthogonal projection ${Pr}_{im 𝑆}$ .
(d)If $dim 𝑉 = dim 𝑊 < \infty$ , then $𝑆$ is bijective, $𝑆^{- 1} = 𝑆^{*}$ , and $𝑆^{- 1}$ is an isometry.

Proof.routine

Uses. Remark 8.4.5 (Without finite dimension), Theorem 8.4.4 (Characterizations of isometries), Proposition 4.2.6 (Injective maps preserve dimension), Corollary 6.6.29 (Orthogonal projections are the self-adjoint idempotents), Proposition 6.6.18 (Properties of the adjoint), Theorem 4.2.11 (Rank–nullity)

Techniques. Test injectivity at $𝟎$ (Technique 2.6.12)

Hint 1. For injectivity, test at

𝟎

: what is

‖ 𝐯 ‖

𝑆 𝐯 = 𝟎

Hint 2. For (c), check that

𝑃 = 𝑆 𝑆^{*}

satisfies

𝑃^{2} = 𝑃 = 𝑃^{*}

, and compare

im 𝑃

with

im 𝑆

using

𝑆 = 𝑆 (𝑆^{*} 𝑆)

Hint 3. For (d),

𝑆

is injective between spaces of equal finite dimension, and

𝑆^{*}

is a left inverse.

Sketch. If

𝑆 𝐯 = 𝟎

then

‖ 𝐯 ‖ = ‖ 𝑆 𝐯 ‖ = 0

. Composites:

‖ 𝑅 𝑆 𝐯 ‖ =

‖ 𝑆 𝐯 ‖ =

‖ 𝐯 ‖

. Injectivity gives

dim 𝑉 =

dim im 𝑆 \leq

dim 𝑊

𝑃 = 𝑆 𝑆^{*}

has

𝑃^{2} =

𝑆 (𝑆^{*} 𝑆) 𝑆^{*} =

𝑃

and

𝑃^{*} = 𝑃

, and

im 𝑃 = im 𝑆

because

𝑆 = 𝑃 𝑆

. If

dim 𝑉 = dim 𝑊

𝑆

is bijective, and

𝑆^{*} =

𝑆^{*} (𝑆 𝑆^{- 1}) =

𝑆^{- 1}

(a) If $𝑆 𝐯 = 𝟎$ , then $‖ 𝐯 ‖ = ‖ 𝑆 𝐯 ‖ = 0$ , so $𝐯 = 𝟎$ ; hence $𝑆$ is injective. Inner products are preserved by Remark 8.4.5, distances because $𝑆 𝐮 -$ $𝑆 𝐯 =$ $𝑆 (𝐮 - 𝐯)$ , and orthogonality because it is a statement about inner products.

(b) $‖ 𝑅 𝑆 𝐯 ‖ =$ $‖ 𝑆 𝐯 ‖ =$ $‖ 𝐯 ‖$ for every $𝐯$ .

(c) Since $𝑆$ is injective, $dim 𝑉 =$ $dim im 𝑆 \leq$ $dim 𝑊$ (Proposition 4.2.6). Let $𝑃 = 𝑆 𝑆^{*}$ . By (III) of Theorem 8.4.4 and Proposition 6.6.18,

𝑃^{2} = 𝑆 (𝑆^{*} 𝑆) 𝑆^{*} = 𝑆 𝑆^{*} = 𝑃, 𝑃^{*} = {(𝑆^{*})}^{*} 𝑆^{*} = 𝑆 𝑆^{*} = 𝑃,

so $𝑃$ is the orthogonal projection onto $im 𝑃$ (Corollary 6.6.29). Clearly $im 𝑃 \subseteq im 𝑆$ , and $𝑆 = 𝑆 (𝑆^{*} 𝑆) = 𝑃 𝑆$ gives $im 𝑆 \subseteq im 𝑃$ . So $𝑃 = {Pr}_{im 𝑆}$ .

(d) By (a) and rank–nullity (Theorem 4.2.11), $dim im 𝑆 =$ $dim 𝑉 =$ $dim 𝑊$ , so $𝑆$ is also surjective. Then $𝑆^{*} =$ $𝑆^{*} (𝑆 𝑆^{- 1}) =$ $(𝑆^{*} 𝑆) 𝑆^{- 1} =$ $𝑆^{- 1}$ , and for $𝐰 \in 𝑊$ , $‖ 𝑆^{- 1} 𝐰 ‖ =$ $‖ 𝑆 (𝑆^{- 1} 𝐰) ‖ =$ $‖ 𝐰 ‖$ . □

For example, the matrix $𝐴 = \frac{1}{3} (\begin{matrix} 1 & 2 \\ 2 & 1 \\ 2 & - 2 \end{matrix})$ of Example 8.4.2(b) has

𝐴^{𝖳} 𝐴 = 𝐼_{2}, 𝐴 𝐴^{𝖳} = \frac{1}{9} (\begin{matrix} 5 & 4 & - 2 \\ 4 & 5 & 2 \\ - 2 & 2 & 8 \end{matrix}),

and $𝐴 𝐴^{𝖳}$ is the projection onto the plane $im 𝐴$ , whose unit normal is $\frac{1}{3} (2, - 2, 1)$ . Indeed $𝐴 𝐴^{𝖳} = 𝐼_{3} - 𝐧 𝐧^{𝖳}$ with $𝐧 = \frac{1}{3} (2, - 2, 1)$ . So $𝐴^{𝖳}$ is a left inverse of $𝐴$ but not a right inverse (compare Exercise 6.6.41).

The next corollary makes precise the remark in Section 6.3 that an $𝑛$ -dimensional inner product space “looks like” $𝔽^{𝑛}$ with the dot product.

Corollary 8.4.7 (Inner product spaces are classified by dimension). Let

𝑉

and

𝑊

be finite-dimensional inner product spaces over

𝔽

. There is a bijective isometry

𝑉 \to 𝑊

if and only if

dim 𝑉 = dim 𝑊

. In particular, for every orthonormal basis

𝒪︀

of an

𝑛

-dimensional

𝑉

, the coordinate map

𝐯 \mapsto {[𝐯]}_{𝒪︀}

is a bijective isometry

𝑉 \to 𝔽^{𝑛}

Proof.routine

Uses. Theorem 6.4.20 (Orthonormal bases exist), Theorem 2.2.1 (Linear maps are determined by a basis), Theorem 8.4.4 (Characterizations of isometries), Corollary 8.4.6 (Properties of isometries), Example 8.4.2 (Isometries)

Techniques. Construct a map by choosing it on a basis (Technique 2.2.4)

Hint 1. A bijective linear map preserves dimension. For the converse, which map should send what where?

Hint 2. Choose orthonormal bases of

𝑉

and

𝑊

, and define

𝑆

on the first basis by sending it to the second.

Sketch. A bijective linear map sends a basis to a basis, so the dimensions agree. Conversely, pick orthonormal bases

𝐞_{1}, \dots, 𝐞_{𝑛}

and

𝐟_{1}, \dots, 𝐟_{𝑛}

, and let

𝑆 𝐞_{𝑗} = 𝐟_{𝑗}

; by (V) of Theorem 8.4.4

𝑆

is an isometry, and it is bijective by Corollary 8.4.6(d).

𝑆 : 𝑉 \to 𝑊

is a bijective isometry, it is a bijective linear map, so it sends a basis of

𝑉

to a basis of

𝑊

and

dim 𝑉 = dim 𝑊

. Conversely, suppose

dim 𝑉 = dim 𝑊 = 𝑛

. By Theorem 6.4.20(a) there are orthonormal bases

𝐞_{1}, \dots, 𝐞_{𝑛}

𝑉

and

𝐟_{1}, \dots, 𝐟_{𝑛}

𝑊

, and by Theorem 2.2.1 there is a linear

𝑆 : 𝑉 \to 𝑊

with

𝑆 𝐞_{𝑗} = 𝐟_{𝑗}

for each

𝑗

. It maps an orthonormal basis to an orthonormal list, so it is an isometry by (V)

\Rightarrow

(I) of Theorem 8.4.4, and it is bijective by Corollary 8.4.6(d). The last sentence is Example 8.4.2(d): the coordinate map sends

𝐞_{𝑗}

to the standard basis vector

𝐞_{𝑗}

𝔽^{𝑛}

. □

So everything about lengths and angles in an $𝑛$ -dimensional inner product space can be computed in $𝔽^{𝑛}$ with the dot product, once an orthonormal basis is fixed. What does depend on the space is how particular operators sit inside it.

Exercise 8.4.8.routine Let $𝑉, 𝑊$ be finite-dimensional and $𝑆 \in L (𝑉, 𝑊)$ .

(a)Prove or disprove: if $‖ 𝑆 𝐞_{𝑗} ‖ = 1$ for every vector of some orthonormal basis $𝐞_{1}, \dots, 𝐞_{𝑛}$ of $𝑉$ , then $𝑆$ is an isometry.
(b)Prove or disprove: if $‖ 𝑆 𝐯 ‖ = 1$ for every unit vector $𝐯 \in 𝑉$ , then $𝑆$ is an isometry.

Solution.

Uses. Theorem 8.4.4 (Characterizations of isometries), Definition 8.4.1 (Isometry)

Hint 1. Condition (V) of Theorem 8.4.4 asks for the images to be orthonormal, not merely of length

1

. Try two basis vectors with the same image.

Hint 2. For (b), every nonzero vector is a positive multiple of a unit vector.

Sketch. (a) False:

𝑆 = (\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix})

sends

𝐞_{1}

and

𝐞_{2}

𝐞_{1}

, but

𝑆 (𝐞_{1} - 𝐞_{2}) = 𝟎

. (b) True:

‖ 𝑆 𝐯 ‖ =

‖ 𝐯 ‖ ‖ 𝑆 (𝐯 / ‖ 𝐯 ‖) ‖ =

‖ 𝐯 ‖

for

𝐯 \neq 𝟎

(a) False. Let $𝑆 \in L (ℝ^{2})$ have matrix $(\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix})$ . Then $𝑆 𝐞_{1} = 𝑆 𝐞_{2} = 𝐞_{1}$ , a unit vector, but $𝑆 (𝐞_{1} - 𝐞_{2}) = 𝟎$ while $‖ 𝐞_{1} - 𝐞_{2} ‖ = \sqrt{2}$ . The images of the basis have length $1$ but are not orthogonal to each other, and (V) of Theorem 8.4.4 needs both.

(b) True. Let $𝐯 \neq 𝟎$ . Then $𝐯 / ‖ 𝐯 ‖$ is a unit vector, so by hypothesis and homogeneity of the norm,

‖ 𝑆 𝐯 ‖ = ‖ 𝐯 ‖ \cdot ‖ 𝑆 (\frac{𝐯}{‖ 𝐯 ‖}) ‖ = ‖ 𝐯 ‖ .

For $𝐯 = 𝟎$ both sides are $0$ . So $𝑆$ is an isometry.

8.4.2Unitary operators

An isometry $𝑉 \to 𝑉$ is injective. In finite dimensions it is therefore invertible, and its inverse is again an isometry; in infinite dimensions it need not be (Example 8.4.12). The invertible ones get a name.

Definition 8.4.9 (Unitary operator). Let

𝑉

be an inner product space. An operator

𝑆 \in L (𝑉)

is unitary if it is an isometry and invertible. When

𝔽 = ℝ

, unitary operators are also called orthogonal operators.

In Table 8.1.1 the unitary row was described by $𝑇^{*} 𝑇 = 𝑇 𝑇^{*} = id$ . The two descriptions agree in every dimension.

Proposition 8.4.10 (Unitary means $𝑆^{*} = 𝑆^{- 1}$ ). Let $𝑉$ be an inner product space and $𝑆 \in L (𝑉)$ . Then $𝑆$ is unitary if and only if $𝑆$ has an adjoint and

𝑆^{*} 𝑆 = 𝑆 𝑆^{*} = {id}_{𝑉} .

In that case $𝑆^{*} = 𝑆^{- 1}$ , and $𝑆^{*}$ is unitary as well.

Proof.routine

Uses. Remark 8.4.5 (Without finite dimension), Definition 6.6.8 (Adjoint), Proposition 6.6.9 (Adjoints are unique and linear), Definition 2.5.12 (Invertible)

Techniques. Find the adjoint by isolating the input (Technique 6.6.11)

Hint 1. To show that

𝑆^{*}

exists, guess it: which map could satisfy

⟨ 𝑆 𝐯, 𝐰 ⟩ =

⟨ 𝐯, 𝑆^{*} 𝐰 ⟩

Hint 2. Write

𝐰 = 𝑆 (𝑆^{- 1} 𝐰)

and use that

𝑆

preserves inner products.

Sketch. If

𝑆

is unitary,

⟨ 𝑆 𝐯, 𝐰 ⟩ =

⟨ 𝑆 𝐯, 𝑆 𝑆^{- 1} 𝐰 ⟩ =

⟨ 𝐯, 𝑆^{- 1} 𝐰 ⟩

, so

𝑆^{- 1}

is the adjoint. Conversely

𝑆^{*} 𝑆 = 𝑆 𝑆^{*} = id

says that

𝑆

is invertible with inverse

𝑆^{*}

, and

{‖ 𝑆 𝐯 ‖}^{2} =

⟨ 𝐯, 𝑆^{*} 𝑆 𝐯 ⟩ =

{‖ 𝐯 ‖}^{2}

Suppose $𝑆$ is unitary. By Remark 8.4.5, $𝑆$ preserves inner products, so for all $𝐯, 𝐰 \in 𝑉$ ,

⟨ 𝑆 𝐯, 𝐰 ⟩ = ⟨ 𝑆 𝐯, 𝑆 (𝑆^{- 1} 𝐰) ⟩ = ⟨ 𝐯, 𝑆^{- 1} 𝐰 ⟩ .

So $𝑆^{- 1}$ is an adjoint of $𝑆$ (Definition 6.6.8), and by Proposition 6.6.9, $𝑆^{*} = 𝑆^{- 1}$ . Hence $𝑆^{*} 𝑆 = 𝑆 𝑆^{*} = {id}_{𝑉}$ .

Conversely, suppose $𝑆^{*} 𝑆 = 𝑆 𝑆^{*} = {id}_{𝑉}$ . Then $𝑆$ is invertible with $𝑆^{- 1} = 𝑆^{*}$ (Definition 2.5.12), and for every $𝐯$ ,

{‖ 𝑆 𝐯 ‖}^{2} = ⟨ 𝑆 𝐯, 𝑆 𝐯 ⟩ = ⟨ 𝐯, 𝑆^{*} 𝑆 𝐯 ⟩ = {‖ 𝐯 ‖}^{2} .

So $𝑆$ is unitary. Finally, $𝑆^{*} = 𝑆^{- 1}$ is invertible, and $‖ 𝑆^{- 1} 𝐯 ‖ =$ $‖ 𝑆 (𝑆^{- 1} 𝐯) ‖ =$ $‖ 𝐯 ‖$ , so $𝑆^{*}$ is unitary. □

Theorem 8.4.11 (Characterizations of unitary operators). Let $𝑉$ be a finite-dimensional inner product space and $𝑆 \in L (𝑉)$ . The following are equivalent.

(I) $𝑆$ is an isometry.
(II) $𝑆$ is unitary.
(III) $𝑆^{*} 𝑆 = {id}_{𝑉}$ .
(IV) $𝑆 𝑆^{*} = {id}_{𝑉}$ .
(V) $𝑆^{*}$ is an isometry.
(VI) $𝑆$ maps some orthonormal basis of $𝑉$ to an orthonormal basis of $𝑉$ .
(VII) $𝑆$ maps every orthonormal basis of $𝑉$ to an orthonormal basis of $𝑉$ .
(VIII) ${[𝑆]}_{𝒪︀}$ is a unitary matrix for some orthonormal basis $𝒪︀$ of $𝑉$ , or equivalently for every one.

Proof.routine

Uses. Theorem 8.4.4 (Characterizations of isometries), Corollary 8.4.6 (Properties of isometries), Proposition 8.4.10 (Unitary means $𝑆^{*} = 𝑆^{- 1}$ ), Remark 4.2.20 (One-sided inverses of operators), Proposition 6.6.18 (Properties of the adjoint), Corollary 6.3.14 (Counting orthogonal vectors), Definition 6.3.33 (Unitary and orthogonal matrices)

Techniques. Proving TFAE (Technique 2.7.1)

Hint 1. Most of the work is in Theorem 8.4.4 with

𝑊 = 𝑉

. What does finite dimension add?

Hint 2. An injective operator on a finite-dimensional space is invertible, and a one-sided inverse of such an operator is two-sided.

Hint 3. For (V), apply Theorem 8.4.4 to

𝑆^{*}

, whose adjoint is

𝑆

. For (VI) and (VII), an orthonormal list of length

dim 𝑉

is a basis.

Sketch. (I)

\Leftrightarrow

(III) is Theorem 8.4.4, and an isometry of a finite-dimensional space is invertible (Corollary 8.4.6(d)), which gives (II). Unitary operators satisfy (IV) by Proposition 8.4.10, and (IV)

\Rightarrow

(III) because one-sided inverses of operators are two-sided. (V) is (IV) by Theorem 8.4.4 applied to

𝑆^{*}

. An orthonormal list of length

𝑛

is an orthonormal basis, so (VI) and (VII) are (V) and (IV) of Theorem 8.4.4; and (VIII) is its (VI)–(VII) with

ℬ︀ = 𝒞︀ = 𝒪︀

We prove (I) $\Rightarrow$ (II) $\Rightarrow$ (IV) $\Rightarrow$ (III) $\Rightarrow$ (I), then (IV) $\Leftrightarrow$ (V), and finally (I) $\Leftrightarrow$ (VI) $\Leftrightarrow$ (VII) and (III) $\Leftrightarrow$ (VIII).

(I) $\Rightarrow$ (II). By Corollary 8.4.6(d) with $𝑊 = 𝑉$ , $𝑆$ is bijective, with $𝑆^{- 1} = 𝑆^{*}$ ; so $𝑆$ is an invertible isometry.

(II) $\Rightarrow$ (IV). Proposition 8.4.10.

(IV) $\Rightarrow$ (III). $𝑆^{*}$ is a right inverse of the operator $𝑆$ on the finite-dimensional space $𝑉$ , hence also a left inverse (Remark 4.2.20).

(III) $\Rightarrow$ (I). (III) $\Rightarrow$ (I) of Theorem 8.4.4.

(IV) $\Leftrightarrow$ (V). By Proposition 6.6.18, ${(𝑆^{*})}^{*} = 𝑆$ . So by (I) $\Leftrightarrow$ (III) of Theorem 8.4.4 applied to $𝑆^{*} \in L (𝑉)$ , $𝑆^{*}$ is an isometry if and only if $𝑆 𝑆^{*} = {id}_{𝑉}$ .

(I) $\Leftrightarrow$ (VI) $\Leftrightarrow$ (VII). Let $𝑛 = dim 𝑉$ . An orthonormal list of length $𝑛$ in $𝑉$ is a basis (Corollary 6.3.14). So (VI) says the same as (V) of Theorem 8.4.4, and (VII) says the same as its (IV) restricted to lists of length $𝑛$ . Thus (I) $\Rightarrow$ (VII) by (I) $\Rightarrow$ (IV) there, (VII) $\Rightarrow$ (VI) is clear, and (VI) $\Rightarrow$ (I) is (V) $\Rightarrow$ (I) there.

(III) $\Leftrightarrow$ (VIII). With $ℬ︀ = 𝒞︀ = 𝒪︀$ , (VI) and (VII) of Theorem 8.4.4 say that ${[𝑆]}_{𝒪︀}^{*} {[𝑆]}_{𝒪︀} =$ $𝐼_{𝑛}$ for some, respectively every, orthonormal basis $𝒪︀$ , which is the definition of a unitary matrix (Definition 6.3.33). Both are equivalent to (III). □

So for operators on finite-dimensional spaces “isometry”, “unitary”, and “unitary matrix in an orthonormal basis” mean the same thing, and we use them interchangeably. For a real matrix $𝑄$ , the operator $𝐱 \mapsto 𝑄 𝐱$ on $ℝ^{𝑛}$ with the dot product is orthogonal exactly when $𝑄$ is an orthogonal matrix.

Beyond 21-242 Infinite-dimensional spaces are not covered in 21-242.

Example 8.4.12 (The shift). On $𝔽 [𝑥]$ , define

⟨ 𝑎_{0} + 𝑎_{1} 𝑥 + \dots + 𝑎_{𝑑} 𝑥^{𝑑}, 𝑏_{0} + 𝑏_{1} 𝑥 + \dots + 𝑏_{𝑑} 𝑥^{𝑑} ⟩ = \sum_{𝑘 = 0}^{𝑑} 𝑎_{𝑘} 𝑏_{𝑘},

padding the shorter polynomial with zero coefficients. This is the dot product of coefficient vectors, so it is an inner product, and $1, 𝑥, 𝑥^{2}, \dots$ is an orthonormal list. Let $𝑀 \in L (𝔽 [𝑥])$ be multiplication by $𝑥$ , which shifts the coefficients one step to the right. Then $‖ 𝑥 𝑝 ‖ = ‖ 𝑝 ‖$ for every $𝑝$ , so $𝑀$ is an isometry, but it is not surjective: every $𝑥 𝑝$ has constant term $0$ , so $1 \notin im 𝑀$ . So $𝑀$ is an isometry of an inner product space into itself that is not unitary, and (I) $\Rightarrow$ (II) of Theorem 8.4.11 fails without finite dimension. Exercise 8.4.40 computes its adjoint, which is a left inverse but not a right inverse.

Isometries have two more invariants that we can read off right away.

Corollary 8.4.13 (Eigenvalues and determinant).

(a)If $𝑆 \in L (𝑉)$ is an isometry and $𝜆$ is an eigenvalue of $𝑆$ , then $| 𝜆 | = 1$ .
(b)If $𝑉$ is finite-dimensional and $𝑆 \in L (𝑉)$ is unitary, then $| det 𝑆 | = 1$ . In particular $det 𝑆 = \pm 1$ when $𝔽 = ℝ$ .

Proof (by norms and by $det (𝑈^{*}) = det 𝑈$ ).routine

Uses. Definition 7.2.6 (Eigenvalue, eigenvector), Definition 5.4.18 (Determinant of an operator), Theorem 8.4.11 (Characterizations of unitary operators), Theorem 5.4.14 (The transpose), Theorem 5.4.10 (Multiplicativity), Definition 5.3.6 (Determinant)

Hint 1. For (a), take norms in

𝑆 𝐯 = 𝜆 𝐯

Hint 2. For (b), pass to the unitary matrix

𝑈 = {[𝑆]}_{𝒪︀}

and take the determinant of

𝑈^{*} 𝑈 = 𝐼_{𝑛}

. You need

det (𝑈^{*}) = det 𝑈

Sketch. (a)

‖ 𝐯 ‖ =

‖ 𝑆 𝐯 ‖ =

| 𝜆 | ‖ 𝐯 ‖

with

𝐯 \neq 𝟎

. (b)

det 𝑆 = det 𝑈

for the unitary matrix

𝑈 = {[𝑆]}_{𝒪︀}

, and

1 =

det (𝑈^{*} 𝑈) =

det 𝑈 det 𝑈 =

{| det 𝑈 |}^{2}

, since conjugation commutes with the polynomial

det

and

det

is unchanged by transposing.

(a) Let $𝑆 𝐯 = 𝜆 𝐯$ with $𝐯 \neq 𝟎$ . Then $‖ 𝐯 ‖ =$ $‖ 𝑆 𝐯 ‖ =$ $‖ 𝜆 𝐯 ‖ =$ $| 𝜆 | ‖ 𝐯 ‖$ , and dividing by $‖ 𝐯 ‖ \neq 0$ gives $| 𝜆 | = 1$ .

(b) Let $𝒪︀$ be an orthonormal basis of $𝑉$ and $𝑈 = {[𝑆]}_{𝒪︀}$ , which is a unitary matrix by Theorem 8.4.11; $det 𝑆 = det 𝑈$ by Definition 5.4.18. The determinant is a polynomial in the entries with integer coefficients (Definition 5.3.6), and complex conjugation respects sums and products, so $det (𝑈) = det 𝑈$ , where $𝑈$ conjugates every entry. With Theorem 5.4.14, $det (𝑈^{*}) =$ $det (𝑈^{𝖳}) =$ $det (𝑈) =$ $det 𝑈$ . Now by Theorem 5.4.10,

1 = det 𝐼_{𝑛} = det (𝑈^{*} 𝑈) = det (𝑈^{*}) det 𝑈 = det 𝑈 det 𝑈 = {| det 𝑈 |}^{2} .

So $| det 𝑆 | = 1$ . Over $ℝ$ , $det 𝑆$ is a real number of absolute value $1$ . □

Proof (through the eigenvalues, over $ℂ$ ).routine

Uses. Theorem 7.5.22 (Trace and determinant from eigenvalues), Theorem 8.4.11 (Characterizations of unitary operators), Definition 5.4.18 (Determinant of an operator)

Hint 1. Over

ℂ

, the determinant is the product of the eigenvalues counted with multiplicity.

Hint 2. For a real orthogonal operator, regard its matrix as a complex unitary matrix.

Sketch. For (b):

det 𝑈 = 𝜆_{1} \dots 𝜆_{𝑛}

, and each

| 𝜆_{𝑗} | = 1

by (a). A real orthogonal matrix is also a complex unitary matrix, with the same determinant.

Part (a) is as in the first proof. For (b), let

𝑈 = {[𝑆]}_{𝒪︀}

for an orthonormal basis

𝒪︀

. Over

ℝ

𝑈

has real entries, and

𝑈^{*} 𝑈 = 𝐼_{𝑛}

also holds in

ℂ^{𝑛 \times 𝑛}

, so in either case we may regard

𝑈

as a complex unitary matrix with the same determinant. By Theorem 7.5.22,

det 𝑈 = 𝜆_{1} \dots 𝜆_{𝑛}

, where

𝜆_{1}, \dots, 𝜆_{𝑛}

are the complex eigenvalues of

𝑈

counted with multiplicity. Each is an eigenvalue of the isometry

𝐱 \mapsto 𝑈 𝐱

ℂ^{𝑛}

, so

| 𝜆_{𝑗} | = 1

by (a), and

| det 𝑈 | =

\prod_{𝑗} | 𝜆_{𝑗} | =

1

. □

For unitary operators, (a) is also Corollary 8.1.20(c), which was proved through normality. The norm argument here needs neither normality nor an adjoint, so it applies to every isometry, including the shift of Example 8.4.12 (which turns out to have no eigenvalues at all). Neither condition is sufficient for unitarity: the shear $(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ has determinant $1$ and the single eigenvalue $1$ , and it is not an isometry.

Invariant subspaces and the spectral form

For a normal operator, the orthogonal complement of an invariant subspace spanned by eigenvectors is invariant; this is what made the induction in the second proof of Theorem 8.2.5 work. For a unitary operator the same conclusion holds for every invariant subspace, with a proof that needs no eigenvectors. (By the spectral theorem it holds for normal operators on complex spaces too, but that is a consequence of the theorem, not a step towards it.)

Lemma 8.4.14 (Invariant complements). Let

𝑉

be finite-dimensional, let

𝑆 \in L (𝑉)

be unitary, and let

𝑈

be an

𝑆

-invariant subspace. Then

𝑆 (𝑈) = 𝑈

, the complement

𝑈^{⟂}

𝑆

-invariant, and

𝑆 ↾_{𝑈}

and

𝑆 ↾_{𝑈^{⟂}}

are unitary operators on

𝑈

and on

𝑈^{⟂}

Proof (directly).routine

Uses. Corollary 4.2.18 (Injective if and only if surjective), Corollary 8.4.6 (Properties of isometries), Theorem 8.4.11 (Characterizations of unitary operators), Definition 4.2.1 (Restriction)

Hint 1.

𝑆 ↾_{𝑈}

is an injective operator on the finite-dimensional space

𝑈

. So every

𝐮 \in 𝑈

𝑆 𝐮^{'}

for some

𝐮^{'} \in 𝑈

Hint 2. For

𝐰 \in 𝑈^{⟂}

, compute

⟨ 𝑆 𝐰, 𝑆 𝐮^{'} ⟩

Sketch.

𝑆 ↾_{𝑈}

is injective, hence onto

𝑈

. For

𝐰 \in 𝑈^{⟂}

and

𝐮 = 𝑆 𝐮^{'} \in 𝑈

⟨ 𝑆 𝐰, 𝐮 ⟩ =

⟨ 𝑆 𝐰, 𝑆 𝐮^{'} ⟩ =

⟨ 𝐰, 𝐮^{'} ⟩ =

0

. Both restrictions are isometries of finite-dimensional spaces into themselves, hence unitary.

Since $𝑆$ is injective, so is $𝑆 ↾_{𝑈} : 𝑈 \to 𝑈$ , and $𝑈$ is finite-dimensional, so $𝑆 ↾_{𝑈}$ is surjective (Corollary 4.2.18). That is, $𝑆 (𝑈) = 𝑈$ . Now let $𝐰 \in 𝑈^{⟂}$ and $𝐮 \in 𝑈$ . Write $𝐮 = 𝑆 𝐮^{'}$ with $𝐮^{'} \in 𝑈$ ; since $𝑆$ preserves inner products (Corollary 8.4.6(a)),

⟨ 𝑆 𝐰, 𝐮 ⟩ = ⟨ 𝑆 𝐰, 𝑆 𝐮^{'} ⟩ = ⟨ 𝐰, 𝐮^{'} ⟩ = 0 .

So $𝑆 𝐰 \in 𝑈^{⟂}$ , and $𝑈^{⟂}$ is $𝑆$ -invariant. The restrictions of $𝑆$ to $𝑈$ and to $𝑈^{⟂}$ preserve norms, so they are isometries of finite-dimensional spaces into themselves, hence unitary by Theorem 8.4.11. □

Proof (through the adjoint).routine

Uses. Proposition 8.4.10 (Unitary means $𝑆^{*} = 𝑆^{- 1}$ ), Exercise 6.6.38, Corollary 4.2.18 (Injective if and only if surjective), Theorem 8.4.11 (Characterizations of unitary operators)

Techniques. Move an operator across the inner product (Technique 6.6.20)

Hint 1.

𝑇

maps

𝑈

into

𝑈

if and only if

𝑇^{*}

maps

𝑈^{⟂}

into

𝑈^{⟂}

(Exercise 6.6.38). Apply this with

𝑇 = 𝑆^{*}

Hint 2. Why does

𝑆^{*} = 𝑆^{- 1}

map

𝑈

into

𝑈

Sketch.

𝑆 (𝑈) = 𝑈

as before, so

𝑆^{- 1} (𝑈) = 𝑈

, that is,

𝑈

is invariant under

𝑆^{*} = 𝑆^{- 1}

. By Exercise 6.6.38 applied to

𝑇 = 𝑆^{*}

𝑈^{⟂}

is invariant under

{(𝑆^{*})}^{*} = 𝑆

As in the first proof,

𝑆 (𝑈) = 𝑈

, so

𝑆^{- 1}

maps

𝑈

onto

𝑈

. By Proposition 8.4.10,

𝑆^{*} = 𝑆^{- 1}

, so

𝑈

𝑆^{*}

-invariant. By Exercise 6.6.38 applied to

𝑇 = 𝑆^{*}

, whose adjoint is

𝑆

, the complement

𝑈^{⟂}

𝑆

-invariant. The restrictions are unitary as in the first proof. □

The first proof uses only that $𝑆$ preserves inner products and maps $𝑈$ onto itself. The second places the lemma in the pattern of Section 6.6: invariance passes from $𝑇$ on $𝑈$ to $𝑇^{*}$ on $𝑈^{⟂}$ , and for unitary $𝑆$ the adjoint is the inverse, which has the same invariant subspaces. The lemma fails for isometries that are not onto: the shift $𝑀$ of Example 8.4.12 maps $𝑈 = span (𝑥, 𝑥^{2}, \dots)$ (the polynomials with constant term $0$ ) into itself, but $𝑈^{⟂} = span (1)$ is not $𝑀$ -invariant.

Over $ℂ$ every operator has an eigenvector, so the lemma splits off one eigenline at a time. Alternatively, unitary operators are normal, and the complex spectral theorem applies. Either way:

Theorem 8.4.15 (Spectral form of unitary operators). Let $𝑉$ be a finite-dimensional complex inner product space and $𝑆 \in L (𝑉)$ . The following are equivalent.

(I) $𝑆$ is unitary.
(II) $𝑆$ is normal and every eigenvalue of $𝑆$ has absolute value $1$ .
(III) $𝑉$ has an orthonormal basis consisting of eigenvectors of $𝑆$ , and every eigenvalue of $𝑆$ has absolute value $1$ .

In matrix form: $𝑈 \in ℂ^{𝑛 \times 𝑛}$ is unitary if and only if $𝑈 = 𝑊 𝐷 𝑊^{*}$ for some unitary $𝑊$ and some $𝐷 =$ $diag (𝑒^{𝑖 𝜃_{1}}, \dots, 𝑒^{𝑖 𝜃_{𝑛}})$ with $𝜃_{1}, \dots, 𝜃_{𝑛} \in ℝ$ .

Proof (from the complex spectral theorem).routine

Uses. Proposition 8.4.10 (Unitary means $𝑆^{*} = 𝑆^{- 1}$ ), Corollary 8.4.13 (Eigenvalues and determinant), Theorem 8.2.5 (Complex spectral theorem), Theorem 8.4.11 (Characterizations of unitary operators), Theorem 7.4.2 (Diagonalizable means an eigenbasis), Exercise 6.3.37, Example 6.3.35 (Unitary and orthogonal matrices)

Techniques. Proving TFAE (Technique 2.7.1)

Hint 1. Unitary operators satisfy

𝑆^{*} 𝑆 = 𝑆 𝑆^{*} = id

, so they are normal. Which theorem gives orthonormal eigenbases of normal operators?

Hint 2. For (III)

\Rightarrow

(I), show that

𝑆

maps the orthonormal eigenbasis to an orthonormal basis.

Sketch. (I)

\Rightarrow

(II):

𝑆^{*} 𝑆 = 𝑆 𝑆^{*} = id

, and eigenvalues of isometries have absolute value

1

. (II)

\Rightarrow

(III) is Theorem 8.2.5. (III)

\Rightarrow

(I):

⟨ 𝑆 𝐞_{𝑗}, 𝑆 𝐞_{𝑘} ⟩ =

𝜆_{𝑗} 𝜆_{𝑘} 𝛿_{𝑗 𝑘} =

𝛿_{𝑗 𝑘}

, so

𝑆

maps an orthonormal basis to one. For matrices, apply this to

𝐱 \mapsto 𝑈 𝐱

and put the eigenbasis into the columns of

𝑊

(I) $\Rightarrow$ (II). By Proposition 8.4.10, $𝑆^{*} 𝑆 = {id}_{𝑉} = 𝑆 𝑆^{*}$ , so $𝑆$ is normal, and its eigenvalues have absolute value $1$ by Corollary 8.4.13(a).

(II) $\Rightarrow$ (III). Theorem 8.2.5 gives an orthonormal eigenbasis, and the eigenvalues have absolute value $1$ by hypothesis.

(III) $\Rightarrow$ (I). Let $𝐞_{1}, \dots, 𝐞_{𝑛}$ be an orthonormal basis with $𝑆 𝐞_{𝑗} = 𝜆_{𝑗} 𝐞_{𝑗}$ and $| 𝜆_{𝑗} | = 1$ . Then

⟨ 𝑆 𝐞_{𝑗}, 𝑆 𝐞_{𝑘} ⟩ = 𝜆_{𝑗} 𝜆_{𝑘} ⟨ 𝐞_{𝑗}, 𝐞_{𝑘} ⟩ = 𝜆_{𝑗} 𝜆_{𝑘} 𝛿_{𝑗 𝑘} = {| 𝜆_{𝑗} |}^{2} 𝛿_{𝑗 𝑘} = 𝛿_{𝑗 𝑘},

so $𝑆$ maps an orthonormal basis to an orthonormal basis, and $𝑆$ is unitary by (VI) $\Rightarrow$ (II) of Theorem 8.4.11.

Matrices. If $𝑈$ is unitary, apply (I) $\Rightarrow$ (III) to $𝐱 \mapsto 𝑈 𝐱$ on $ℂ^{𝑛}$ with the dot product. Let $𝑊$ have the orthonormal eigenvectors as columns; it is unitary, and by Theorem 7.4.2, $𝑈 =$ $𝑊 𝐷 𝑊^{- 1} =$ $𝑊 𝐷 𝑊^{*}$ with $𝐷$ diagonal, whose entries have absolute value $1$ , so they are $𝑒^{𝑖 𝜃_{𝑗}}$ for some real $𝜃_{𝑗}$ . Conversely, $𝐷$ is unitary (Example 6.3.35(c)), and products of unitary matrices are unitary (Exercise 6.3.37), so $𝑊 𝐷 𝑊^{*}$ is unitary. □

Proof ((I) $\Rightarrow$ (III) by induction, without the spectral theorem).standard

Uses. Theorem 7.3.27 (Existence of eigenvalues over $ℂ$ ), Corollary 8.4.13 (Eigenvalues and determinant), Lemma 8.4.14 (Invariant complements), Corollary 6.4.25 (Dimension of the orthogonal complement), Theorem 6.4.24 (Orthogonal decomposition)

Techniques. Induction on dimension via an invariant subspace (Technique 7.5.12)

Hint 1. Over

ℂ

𝑆

has an eigenvector. Normalize it.

Hint 2. The eigenline is

𝑆

-invariant. Apply Lemma 8.4.14 and induct on

dim 𝑉

Sketch. Take a unit eigenvector

𝐞

;

span (𝐞)

is invariant, so its complement is invariant and

𝑆

restricted to it is unitary (Lemma 8.4.14). By induction the complement has an orthonormal eigenbasis; add

𝐞

. Eigenvalues have absolute value

1

by Corollary 8.4.13(a).

We prove (I)

\Rightarrow

(III) by induction on

𝑛 = dim 𝑉

; the other implications are as in the first proof. Every eigenvalue has absolute value

1

by Corollary 8.4.13(a), so we only need the orthonormal eigenbasis. If

𝑛 = 0

the empty list works. Let

𝑛 \geq 1

. By Theorem 7.3.27,

𝑆

has an eigenvector, and we scale it to a unit vector

𝐞

. The line

𝐿 = span (𝐞)

𝑆

-invariant, so by Lemma 8.4.14

𝐿^{⟂}

𝑆

-invariant and

𝑆 ↾_{𝐿^{⟂}}

is unitary. Since

dim 𝐿^{⟂} = 𝑛 - 1

(Corollary 6.4.25), the induction hypothesis gives an orthonormal basis

𝐞_{2}, \dots, 𝐞_{𝑛}

𝐿^{⟂}

consisting of eigenvectors of

𝑆 ↾_{𝐿^{⟂}}

, hence of

𝑆

. Since

𝑉 = 𝐿 \oplus 𝐿^{⟂}

(Theorem 6.4.24), the list

𝐞, 𝐞_{2}, \dots, 𝐞_{𝑛}

is an orthonormal basis of

𝑉

consisting of eigenvectors of

𝑆

. □

The first proof inherits everything from normal operators. The second shows why the unitary case is easier to prove than the normal one: the invariance of the complement comes from Lemma 8.4.14, which needs neither eigenvectors nor adjoint identities. Over $ℝ$ the second proof breaks at the first step, because a rotation of $ℝ^{2}$ has no eigenvector. Repairing it with two-dimensional invariant subspaces gives the normal form of Section 8.4.4.

Example 8.4.16 (Two unitary diagonalizations).

(a)The rotation $𝑅_{𝜃} =$ $(\begin{matrix} cos 𝜃 & - sin 𝜃 \\ sin 𝜃 & cos 𝜃 \end{matrix})$ with $𝜃 \notin 𝜋 ℤ$ has no real eigenvalue, but as a complex matrix
$𝑅_{𝜃} (\begin{matrix} 1 \\ - 𝑖 \end{matrix}) = (\begin{matrix} cos 𝜃 + 𝑖 sin 𝜃 \\ sin 𝜃 - 𝑖 cos 𝜃 \end{matrix}) = 𝑒^{𝑖 𝜃} (\begin{matrix} 1 \\ - 𝑖 \end{matrix}), 𝑅_{𝜃} (\begin{matrix} 1 \\ 𝑖 \end{matrix}) = 𝑒^{- 𝑖 𝜃} (\begin{matrix} 1 \\ 𝑖 \end{matrix}) .$
With $𝑊 = \frac{1}{\sqrt{2}} (\begin{matrix} 1 & 1 \\ - 𝑖 & 𝑖 \end{matrix})$ , which is unitary, $𝑅_{𝜃} =$ $𝑊 diag (𝑒^{𝑖 𝜃}, 𝑒^{- 𝑖 𝜃}) 𝑊^{*}$ . The same $𝑊$ works for every $𝜃$ , which reflects the fact that rotations of the plane commute with each other.
(b)Let $𝐶 \in ℝ^{𝑛 \times 𝑛}$ be the cyclic shift, $𝐶 𝐞_{𝑘} = 𝐞_{𝑘 + 1}$ , with indices $0, \dots, 𝑛 - 1$ read modulo $𝑛$ as in Definition 6.3.40. It is a permutation matrix, hence orthogonal, and ${(𝐶 𝐱)}_{𝑗} = 𝑥_{𝑗 - 1}$ . For the Fourier vector $𝐟_{𝑘} =$ ${(𝜔^{𝑗 𝑘})}_{𝑗 = 0}^{𝑛 - 1}$ ,
${(𝐶 𝐟_{𝑘})}_{𝑗} = 𝜔^{(𝑗 - 1) 𝑘} = 𝜔^{- 𝑘} {(𝐟_{𝑘})}_{𝑗},$
so $𝐶 𝐟_{𝑘} = 𝜔^{- 𝑘} 𝐟_{𝑘}$ . Since $Φ_{𝑛} / \sqrt{𝑛}$ is unitary (Proposition 6.3.41), $𝐶 = 𝑊 𝐷 𝑊^{*}$ with $𝑊 = Φ_{𝑛} / \sqrt{𝑛}$ and $𝐷 =$ $diag (1, 𝜔^{- 1}, \dots, 𝜔^{- (𝑛 - 1)})$ . The eigenvalues of $𝐶$ are all the $𝑛$ th roots of unity. For $𝑛 = 3$ , $𝐶$ is the rotation of $ℝ^{3}$ by $2 𝜋 / 3$ about the line through $(1, 1, 1)$ (Example 8.4.28).

Exercise 8.4.17.routine Find the matrix

𝑈 \in ℂ^{2 \times 2}

with

𝑈 (\begin{matrix} 1 \\ 1 \end{matrix}) = 𝑖 (\begin{matrix} 1 \\ 1 \end{matrix})

and

𝑈 (\begin{matrix} 1 \\ - 1 \end{matrix}) = - (\begin{matrix} 1 \\ - 1 \end{matrix})

. Show that

𝑈

is unitary, and compute

𝑈^{4}

and

𝑈^{*}

Solution.

Uses. Theorem 8.4.15 (Spectral form of unitary operators), Theorem 2.2.1 (Linear maps are determined by a basis), Proposition 6.3.34 (Characterizations of unitary matrices)

Hint 1. The vectors

(1, 1) / \sqrt{2}

and

(1, - 1) / \sqrt{2}

form an orthonormal basis. Put them in the columns of

𝑊

Hint 2.

𝑈 =

𝑊 diag (𝑖, - 1) 𝑊^{*}

. Powers and adjoints act on the diagonal factor.

Sketch.

𝑈

is determined by its values on a basis. With

𝑊 = \frac{1}{\sqrt{2}} (\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix}) = 𝑊^{*}

𝑈 =

𝑊 diag (𝑖, - 1) 𝑊 =

\frac{1}{2} (\begin{matrix} 𝑖 - 1 & 𝑖 + 1 \\ 𝑖 + 1 & 𝑖 - 1 \end{matrix})

. It is unitary by (III) of Theorem 8.4.15.

𝑈^{4} =

𝑊 diag (1, 1) 𝑊 =

𝐼_{2}

and

𝑈^{*} =

𝑊 diag (- 𝑖, - 1) 𝑊

An operator is determined by its values on the basis $(1, 1), (1, - 1)$ (Theorem 2.2.1), so there is exactly one such $𝑈$ . Let $𝑊 = \frac{1}{\sqrt{2}} (\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix})$ ; its columns are orthonormal and $𝑊 = 𝑊^{*} = 𝑊^{- 1}$ . The columns of $𝑊$ are eigenvectors for $𝑖$ and $- 1$ , so $𝑈 =$ $𝑊 diag (𝑖, - 1) 𝑊$ , and

𝑈 = \frac{1}{2} (\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix}) (\begin{matrix} 𝑖 & 0 \\ 0 & - 1 \end{matrix}) (\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix}) = \frac{1}{2} (\begin{matrix} 𝑖 & - 1 \\ 𝑖 & 1 \end{matrix}) (\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix}) = \frac{1}{2} (\begin{matrix} 𝑖 - 1 & 𝑖 + 1 \\ 𝑖 + 1 & 𝑖 - 1 \end{matrix}) .

It has an orthonormal eigenbasis with eigenvalues $𝑖$ and $- 1$ of absolute value $1$ , so it is unitary by (III) $\Rightarrow$ (I) of Theorem 8.4.15. Next, $𝑈^{4} =$ $𝑊 diag (𝑖^{4}, {(- 1)}^{4}) 𝑊 =$ $𝑊 𝑊 =$ $𝐼_{2}$ . Finally $𝑈^{*} =$ $𝑊^{*} {diag (𝑖, - 1)}^{*} 𝑊^{*} =$ $𝑊 diag (- 𝑖, - 1) 𝑊$ , which is

𝑈^{*} = \frac{1}{2} (\begin{matrix} - 𝑖 - 1 & - 𝑖 + 1 \\ - 𝑖 + 1 & - 𝑖 - 1 \end{matrix}),

the conjugate transpose of $𝑈$ , as it must be. As a check, $𝑈^{*} = 𝑈^{- 1} = 𝑈^{3}$ , since $𝑈^{4} = 𝐼_{2}$ .

Remark 8.4.18 (The unitary group). The identity is unitary, and composites and inverses of unitary operators are unitary, by Corollary 8.4.6(b) and Proposition 8.4.10. So the unitary operators on

𝑉

form a group under composition. For matrices, the unitary group

U (𝑛)

consists of the unitary

𝑛 \times 𝑛

matrices, the orthogonal group

O (𝑛)

of the real orthogonal ones, and the special orthogonal group

SO (𝑛)

of the orthogonal matrices with determinant

1

;

U (1)

is the unit circle in

ℂ

. Groups of matrices are the subject of Section 14.1. Under the analogy of Table 8.1.1,

U (1)

is the set of

𝑧

with

𝑧 𝑧 = 1

, and just as every complex number is

𝑧 = 𝑒^{𝑖 𝜃} | 𝑧 |

, every operator on a finite-dimensional space is a unitary operator times a positive one (Definition 8.3.1), the polar decomposition of Section 9.2.

Unitary operators can move any orthonormal list to any other of the same length. This is the precise form of the statement that an inner product space “looks the same” in all directions.

Proposition 8.4.19 (Moving orthonormal lists). Let $𝑉$ be finite-dimensional.

(a)If $𝐞_{1}, \dots, 𝐞_{𝑘}$ and $𝐟_{1}, \dots, 𝐟_{𝑘}$ are orthonormal lists in $𝑉$ , there is a unitary $𝑆 \in L (𝑉)$ with $𝑆 𝐞_{𝑗} = 𝐟_{𝑗}$ for $𝑗 = 1, \dots, 𝑘$ .
(b)If $𝑈$ is a subspace of $𝑉$ and $𝑆_{0} \in L (𝑈, 𝑉)$ is an isometry, there is a unitary $𝑆 \in L (𝑉)$ with $𝑆 ↾_{𝑈} = 𝑆_{0}$ .

Proof.routine

Uses. Theorem 6.4.20 (Orthonormal bases exist), Theorem 2.2.1 (Linear maps are determined by a basis), Theorem 8.4.11 (Characterizations of unitary operators), Theorem 8.4.4 (Characterizations of isometries)

Techniques. Construct a map by choosing it on a basis (Technique 2.2.4), Extend a basis of a subspace (Technique 4.1.18)

Hint 1. Extend both lists to orthonormal bases of

𝑉

. How long are the extensions?

Hint 2. Define

𝑆

on the first basis. For (b), apply (a) to an orthonormal basis of

𝑈

and its image.

Sketch. Both lists extend to orthonormal bases of length

𝑛 = dim 𝑉

; the linear map sending the first basis to the second, term by term, maps an orthonormal basis to one and is unitary. For (b),

𝑆_{0}

maps an orthonormal basis of

𝑈

to an orthonormal list; extend as in (a), and two linear maps agreeing on a basis of

𝑈

agree on

𝑈

(a) By Theorem 6.4.20(b), the lists extend to orthonormal bases $𝐞_{1}, \dots, 𝐞_{𝑛}$ and $𝐟_{1}, \dots, 𝐟_{𝑛}$ of $𝑉$ , both of length $𝑛 = dim 𝑉$ . By Theorem 2.2.1 there is $𝑆 \in L (𝑉)$ with $𝑆 𝐞_{𝑗} = 𝐟_{𝑗}$ for all $𝑗 \leq 𝑛$ . It maps an orthonormal basis to an orthonormal basis, so it is unitary by Theorem 8.4.11.

(b) Let $𝐞_{1}, \dots, 𝐞_{𝑘}$ be an orthonormal basis of $𝑈$ . By (IV) of Theorem 8.4.4, $𝑆_{0} 𝐞_{1}, \dots, 𝑆_{0} 𝐞_{𝑘}$ is orthonormal in $𝑉$ , so (a) gives a unitary $𝑆$ with $𝑆 𝐞_{𝑗} = 𝑆_{0} 𝐞_{𝑗}$ for all $𝑗$ . The linear maps $𝑆 ↾_{𝑈}$ and $𝑆_{0}$ agree on a basis of $𝑈$ , so they are equal by the uniqueness in Theorem 2.2.1. □

In particular, any unit vector can be moved to any other by a unitary operator. When a problem is unchanged by unitary operators (it involves only lengths and inner products), we may therefore assume that a given unit vector is $𝐞_{1}$ . Section 8.4.3 writes down an explicit operator that does this.

8.4.3Householder reflections

The simplest unitary operators other than scalars are reflections across hyperplanes.

Example 8.4.20 (Householder reflections). Let $𝐮 \in 𝔽^{𝑛}$ be a unit vector, and let

𝐻_{𝐮} = 𝐼_{𝑛} - 2 𝐮 𝐮^{*} .

Then $𝐻_{𝐮} 𝐱 =$ $𝐱 -$ $2 (𝐮^{*} 𝐱) 𝐮 =$ $𝐱 -$ $2 (𝐱 \cdot 𝐮) 𝐮$ . So $𝐻_{𝐮} 𝐮 = 𝐮 - 2 𝐮 = - 𝐮$ , and $𝐻_{𝐮} 𝐱 = 𝐱$ for every $𝐱 ⟂ 𝐮$ . In words, $𝐻_{𝐮}$ fixes the hyperplane $𝐮^{⟂}$ and reverses the normal direction: it is the reflection across $𝐮^{⟂}$ . By Theorem 6.4.31, $𝑃 = 𝐮 𝐮^{*}$ is the orthogonal projection onto $span (𝐮)$ , and

𝐻_{𝐮} = (𝐼_{𝑛} - 𝑃) - 𝑃,

which keeps the component of $𝐱$ in $𝐮^{⟂}$ and negates the component along $𝐮$ . This is the operator $- (2 𝑃 - 𝐼_{𝑛})$ of Exercise 8.1.30, so $𝐻_{𝐮}$ is Hermitian and unitary, and $𝐻_{𝐮}^{2} = 𝐼_{𝑛}$ . Directly: $𝐻_{𝐮}^{*} =$ $𝐼_{𝑛} -$ $2 {(𝐮 𝐮^{*})}^{*} =$ $𝐻_{𝐮}$ , and

𝐻_{𝐮}^{2} = 𝐼_{𝑛} - 4 𝐮 𝐮^{*} + 4 𝐮 (𝐮^{*} 𝐮) 𝐮^{*} = 𝐼_{𝑛} .

Extending $𝐮$ to an orthonormal basis $𝐮, 𝐮_{2}, \dots, 𝐮_{𝑛}$ of eigenvectors, with eigenvalues $- 1, 1, \dots, 1$ , gives $det 𝐻_{𝐮} = - 1$ (Theorem 7.5.22).

For instance, for $𝐮 = \frac{1}{\sqrt{3}} (1, - 1, - 1)$ ,

𝐻_{𝐮} = 𝐼_{3} - \frac{2}{3} (\begin{matrix} 1 & - 1 & - 1 \\ - 1 & 1 & 1 \\ - 1 & 1 & 1 \end{matrix}) = \frac{1}{3} (\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1 \end{matrix}),

the orthogonal matrix of Example 6.3.35(d). So that matrix is a reflection, across the plane $𝑥_{1} - 𝑥_{2} - 𝑥_{3} = 0$ .

The point of Householder reflections is that they can move a given vector to any other vector of the same length.

Proposition 8.4.21 (A reflection exchanging two vectors). Let $𝐱, 𝐲 \in 𝔽^{𝑛}$ with $𝐱 \neq 𝐲$ , $‖ 𝐱 ‖ = ‖ 𝐲 ‖$ , and $𝐱 \cdot 𝐲 \in ℝ$ . Let $𝐮 =$ $(𝐱 - 𝐲) / ‖ 𝐱 -$ $𝐲 ‖$ . Then

𝐻_{𝐮} 𝐱 = 𝐲 and 𝐻_{𝐮} 𝐲 = 𝐱 .

Proof (by computation).routine

Uses. Example 8.4.20 (Householder reflections), Proposition 6.1.29 (Expanding a norm), Proposition 6.1.4 (Sesquilinearity)

Hint 1.

𝐻_{𝐮} 𝐱 =

𝐱 -

2 (𝐱 \cdot 𝐮) 𝐮

. Write everything in terms of

𝐝 = 𝐱 - 𝐲

Hint 2. You need

2 (𝐱 \cdot 𝐝) = {‖ 𝐝 ‖}^{2}

. Expand both sides using

‖ 𝐱 ‖ = ‖ 𝐲 ‖

and

𝐱 \cdot 𝐲 = 𝐲 \cdot 𝐱

Sketch. With

𝐝 = 𝐱 - 𝐲

𝐻_{𝐮} 𝐱 =

𝐱 -

(2 (𝐱 \cdot 𝐝) / {‖ 𝐝 ‖}^{2}) 𝐝

. Both

2 (𝐱 \cdot 𝐝)

and

{‖ 𝐝 ‖}^{2}

equal

2 {‖ 𝐱 ‖}^{2} - 2 (𝐱 \cdot 𝐲)

, so the coefficient is

1

and

𝐻_{𝐮} 𝐱 = 𝐲

. Then

𝐻_{𝐮} 𝐲 = 𝐻_{𝐮}^{2} 𝐱 = 𝐱

Let $𝐝 = 𝐱 - 𝐲 \neq 𝟎$ , so $𝐮 = 𝐝 / ‖ 𝐝 ‖$ and

𝐻_{𝐮} 𝐱 = 𝐱 - 2 (𝐱 \cdot 𝐮) 𝐮 = 𝐱 - \frac{2 (𝐱 \cdot 𝐝)}{{‖ 𝐝 ‖}^{2}} 𝐝 .

Since $𝐱 \cdot 𝐲$ is real, $𝐲 \cdot 𝐱 = 𝐱 \cdot 𝐲 = 𝐱 \cdot 𝐲$ . Using $‖ 𝐲 ‖ = ‖ 𝐱 ‖$ ,

\begin{aligned} 𝐱 \cdot 𝐝 & = {‖ 𝐱 ‖}^{2} - 𝐱 \cdot 𝐲, \\ {‖ 𝐝 ‖}^{2} & = {‖ 𝐱 ‖}^{2} - 𝐱 \cdot 𝐲 - 𝐲 \cdot 𝐱 + {‖ 𝐲 ‖}^{2} = 2 ({‖ 𝐱 ‖}^{2} - 𝐱 \cdot 𝐲) . \end{aligned}

So $2 (𝐱 \cdot 𝐝) = {‖ 𝐝 ‖}^{2}$ , and $𝐻_{𝐮} 𝐱 = 𝐱 - 𝐝 = 𝐲$ . Since $𝐻_{𝐮}^{2} = 𝐼_{𝑛}$ (Example 8.4.20), $𝐻_{𝐮} 𝐲 = 𝐻_{𝐮}^{2} 𝐱 = 𝐱$ . □

Proof (by decomposing along the mirror).routine

Uses. Example 8.4.20 (Householder reflections), Proposition 6.1.4 (Sesquilinearity)

Techniques. Orthogonal decomposition (Technique 6.2.4)

Hint 1. Write

𝐱 =

\frac{1}{2} (𝐱 + 𝐲) +

\frac{1}{2} (𝐱 - 𝐲)

. Where does each piece lie relative to the mirror

𝐮^{⟂}

Hint 2. Show that

𝐱 + 𝐲 ⟂ 𝐱 - 𝐲

; this is where both hypotheses are used.

Sketch.

(𝐱 + 𝐲) \cdot

(𝐱 - 𝐲) =

{‖ 𝐱 ‖}^{2} -

{‖ 𝐲 ‖}^{2} +

(𝐲 \cdot 𝐱 - 𝐱 \cdot 𝐲) =

0

, so

𝐱 + 𝐲

lies in the mirror and is fixed, while

𝐱 - 𝐲

is a multiple of

𝐮

and is negated. Hence

𝐻_{𝐮} 𝐱 =

\frac{1}{2} (𝐱 + 𝐲) -

\frac{1}{2} (𝐱 - 𝐲) =

𝐲

First,

(𝐱 + 𝐲) \cdot (𝐱 - 𝐲) = {‖ 𝐱 ‖}^{2} - 𝐱 \cdot 𝐲 + 𝐲 \cdot 𝐱 - {‖ 𝐲 ‖}^{2} = 0,

since $‖ 𝐱 ‖ = ‖ 𝐲 ‖$ and $𝐲 \cdot 𝐱 = 𝐱 \cdot 𝐲 = 𝐱 \cdot 𝐲$ . So $𝐱 + 𝐲 \in 𝐮^{⟂}$ , and $𝐻_{𝐮}$ fixes it, while $𝐱 - 𝐲 = ‖ 𝐱 - 𝐲 ‖ 𝐮$ is negated (Example 8.4.20). Hence

𝐻_{𝐮} 𝐱 = 𝐻_{𝐮} (\frac{1}{2} (𝐱 + 𝐲) + \frac{1}{2} (𝐱 - 𝐲)) = \frac{1}{2} (𝐱 + 𝐲) - \frac{1}{2} (𝐱 - 𝐲) = 𝐲,

and in the same way $𝐻_{𝐮} 𝐲 =$ $\frac{1}{2} (𝐱 + 𝐲) +$ $\frac{1}{2} (𝐱 - 𝐲) =$ $𝐱$ . □

The second proof is the picture in Figure 8.4.1: the mirror is the perpendicular bisector of the segment from $𝐱$ to $𝐲$ , which passes through $𝟎$ because $𝐱$ and $𝐲$ are equally long. The condition $𝐱 \cdot 𝐲 \in ℝ$ cannot be dropped over $ℂ$ : since $𝐻_{𝐮}$ is Hermitian, $(𝐻_{𝐮} 𝐱) \cdot 𝐱$ is always real (Proposition 6.6.30), so no Householder reflection maps $𝐱$ to a $𝐲$ with $𝐲 \cdot 𝐱 \notin ℝ$ . Multiplying $𝐲$ by a suitable scalar of absolute value $1$ first repairs this (Exercise).

Figure 8.4.1. The Householder reflection with

𝐱 = (3, 4)

and

𝐲 = (5, 0)

, both of length

5

. The unit normal is

𝐮 =

(𝐱 - 𝐲) / ‖ 𝐱 -

𝐲 ‖ =

\frac{1}{\sqrt{5}} (- 1, 2)

, and the mirror

𝐮^{⟂}

(dashed) is the line through

𝟎

and

𝐱 + 𝐲 = (8, 4)

, the perpendicular bisector of the dotted segment.

Take $𝐲 = - 𝑐 ‖ 𝐱 ‖ 𝐞_{1}$ , where $𝑐 = 𝑥_{1} / | 𝑥_{1} |$ (and $𝑐 = 1$ if $𝑥_{1} = 0$ ). Then $𝐱 \cdot$ $𝐲 =$ $- 𝑐 𝑥_{1} ‖ 𝐱 ‖ =$ $- | 𝑥_{1} | ‖ 𝐱 ‖$ is real, so unless $𝐱 = 𝐲$ already, one reflection turns $𝐱$ into $𝐲$ , whose entries after the first are $0$ . Applied to the first column of a matrix $𝐴 \in 𝔽^{𝑚 \times 𝑛}$ , and then to the first column of the lower right block, and so on (skipping a step when the column is already in the required form), this produces a unitary $𝑄 \in 𝔽^{𝑚 \times 𝑚}$ , a product of reflections, with $𝑅_{0} = 𝑄^{*} 𝐴$ upper triangular. This is a full QR factorization $𝐴 = 𝑄 𝑅_{0}$ ; when the columns of $𝐴$ are independent, keeping the first $𝑛$ columns of $𝑄$ and the top $𝑛 \times 𝑛$ block of $𝑅_{0}$ , and rescaling by scalars of absolute value $1$ to make the diagonal positive, gives the factorization of Theorem 6.4.38. This is how QR factorizations are computed in practice; Exercise carries it out for a $3 \times 2$ matrix.

Exercise 8.4.22.routine Find a Householder reflection

𝐻

with

𝐻 (2, 1, 2) = (3, 0, 0)

, and one with

𝐻 (2, 1, 2) =

(- 3, 0, 0)

. Check that the first is symmetric and orthogonal.

Solution.

Uses. Proposition 8.4.21 (A reflection exchanging two vectors), Example 8.4.20 (Householder reflections)

Hint 1. Both targets have length

3 = ‖ (2, 1, 2) ‖

, and all entries are real. Use Proposition 8.4.21 with

𝐮

parallel to

𝐱 - 𝐲

Hint 2.

𝐻_{𝐮} =

𝐼_{3} -

2 𝐮 𝐮^{𝖳} =

𝐼_{3} -

(2 / {‖ 𝐝 ‖}^{2}) 𝐝 𝐝^{𝖳}

for any nonzero

𝐝

parallel to

𝐮

Sketch. For

(3, 0, 0)

𝐝 = (- 1, 1, 2)

{‖ 𝐝 ‖}^{2} = 6

𝐻 =

\frac{1}{3} (\begin{matrix} 2 & 1 & 2 \\ 1 & 2 & - 2 \\ 2 & - 2 & - 1 \end{matrix})

. For

(- 3, 0, 0)

𝐝 = (5, 1, 2)

{‖ 𝐝 ‖}^{2} = 30

𝐻 =

\frac{1}{15} (\begin{matrix} - 10 & - 5 & - 10 \\ - 5 & 14 & - 2 \\ - 10 & - 2 & 11 \end{matrix})

Let $𝐱 = (2, 1, 2)$ , so $‖ 𝐱 ‖ = 3$ . Both targets have length $3$ and real dot products with $𝐱$ , so Proposition 8.4.21 applies with $𝐮 = 𝐝 / ‖ 𝐝 ‖$ , $𝐝 = 𝐱 - 𝐲$ , and $𝐻 =$ $𝐼_{3} -$ $(2 / {‖ 𝐝 ‖}^{2}) 𝐝 𝐝^{𝖳}$ .

For $𝐲 = (3, 0, 0)$ : $𝐝 = (- 1, 1, 2)$ and ${‖ 𝐝 ‖}^{2} = 6$ , so

𝐻 = 𝐼_{3} - \frac{1}{3} (\begin{matrix} 1 & - 1 & - 2 \\ - 1 & 1 & 2 \\ - 2 & 2 & 4 \end{matrix}) = \frac{1}{3} (\begin{matrix} 2 & 1 & 2 \\ 1 & 2 & - 2 \\ 2 & - 2 & - 1 \end{matrix}) .

Indeed

𝐻 𝐱 = \frac{1}{3} (4 + 1 + 4, 2 + 2 - 4, 4 - 2 - 2) = (3, 0, 0) .

The matrix is symmetric by inspection. Its columns $\frac{1}{3} (2, 1, 2)$ , $\frac{1}{3} (1, 2, - 2)$ , $\frac{1}{3} (2, - 2, - 1)$ have squared lengths $9 / 9 = 1$ and pairwise dot products $(2 + 2 - 4) / 9$ , $(4 - 2 - 2) / 9$ , $(2 - 4 + 2) / 9$ , all $0$ ; so $𝐻$ is orthogonal.

For $𝐲 = (- 3, 0, 0)$ : $𝐝 = (5, 1, 2)$ and ${‖ 𝐝 ‖}^{2} = 30$ , so

𝐻 = 𝐼_{3} - \frac{1}{15} (\begin{matrix} 25 & 5 & 10 \\ 5 & 1 & 2 \\ 10 & 2 & 4 \end{matrix}) = \frac{1}{15} (\begin{matrix} - 10 & - 5 & - 10 \\ - 5 & 14 & - 2 \\ - 10 & - 2 & 11 \end{matrix}),

and

𝐻 𝐱 = \frac{1}{15} (- 20 - 5 - 20, - 10 + 14 - 4, - 20 - 2 + 22) = (- 3, 0, 0) .

(In floating-point computation one chooses the sign of the target that makes $‖ 𝐝 ‖$ large, here the second, to avoid subtracting nearly equal numbers.)

8.4.4Real orthogonal operators Beyond 21-242

The normal form of real orthogonal operators, orientation, and rigid motions are not covered in lecture; see LADW Chapters 5 and 6 and LADR 7D.

Over $ℝ$ an orthogonal operator need not have an eigenvector: a rotation of the plane has none. So the spectral form of Theorem 8.4.15 has no direct real analogue. Instead, we will show that every orthogonal operator is assembled from rotations of perpendicular planes and from $\pm 1$ on perpendicular lines. We first list the orthogonal operators of the plane.

Lemma 8.4.23 (Orthogonal $2 \times 2$ matrices). Let $𝑄 \in ℝ^{2 \times 2}$ be orthogonal. Then there is exactly one $𝜃 \in [0, 2 𝜋)$ such that $𝑄$ is one of

𝑅_{𝜃} = (\begin{matrix} cos 𝜃 & - sin 𝜃 \\ sin 𝜃 & cos 𝜃 \end{matrix}) or 𝑆_{𝜃} = (\begin{matrix} cos 𝜃 & sin 𝜃 \\ sin 𝜃 & - cos 𝜃 \end{matrix}),

and $det 𝑅_{𝜃} = 1$ , $det 𝑆_{𝜃} = - 1$ . The matrix $𝑆_{𝜃}$ is the reflection across the line spanned by $𝐜 =$ $(cos (𝜃 / 2), sin (𝜃 / 2))$ : it fixes $𝐜$ and negates $𝐜^{'} =$ $(- sin (𝜃 / 2), cos (𝜃 / 2))$ , so it has the eigenvalues $1$ and $- 1$ .

Proof.routine

Uses. Definition 6.3.33 (Unitary and orthogonal matrices), Corollary 6.3.14 (Counting orthogonal vectors), Proposition 6.3.18 (Coordinates in an orthonormal basis), Definition 5.1.14 (The $2 \times 2$ determinant)

Techniques. Take an inner product to isolate a coefficient (Technique 6.3.12)

Hint 1. The first column is a unit vector, so it is

(cos 𝜃, sin 𝜃)

for exactly one

𝜃 \in [0, 2 𝜋)

Hint 2. The second column is a unit vector orthogonal to the first. Expand it in the orthonormal basis

(cos 𝜃, sin 𝜃),

(- sin 𝜃, cos 𝜃)

Sketch. The first column is

(cos 𝜃, sin 𝜃)

. The second column is orthogonal to it, so its coordinates in the orthonormal basis

(cos 𝜃, sin 𝜃),

(- sin 𝜃, cos 𝜃)

are

(0, 𝑡)

, and being a unit vector forces

𝑡 = \pm 1

. The sign is the determinant. For

𝑆_{𝜃}

, the angle-subtraction formulas give

𝑆_{𝜃} 𝐜 = 𝐜

and

𝑆_{𝜃} 𝐜^{'} = - 𝐜^{'}

The columns $(𝑎, 𝑏)$ and $(𝑐, 𝑑)$ of $𝑄$ are orthonormal (Definition 6.3.33). Since $𝑎^{2} + 𝑏^{2} = 1$ , there is exactly one $𝜃 \in [0, 2 𝜋)$ with $(𝑎, 𝑏) =$ $(cos 𝜃, sin 𝜃)$ . The vectors $(cos 𝜃, sin 𝜃)$ and $(- sin 𝜃, cos 𝜃)$ are orthonormal, hence an orthonormal basis of $ℝ^{2}$ (Corollary 6.3.14), so by Proposition 6.3.18

(𝑐, 𝑑) = (𝑐 cos 𝜃 + 𝑑 sin 𝜃) (cos 𝜃, sin 𝜃) + 𝑡 (- sin 𝜃, cos 𝜃), 𝑡 = - 𝑐 sin 𝜃 + 𝑑 cos 𝜃 .

The first coefficient is $(𝑐, 𝑑) \cdot (𝑎, 𝑏) = 0$ , and $1 = ‖ (𝑐, 𝑑) ‖ = | 𝑡 |$ , so $𝑡 = \pm 1$ . For $𝑡 = 1$ we get $𝑄 = 𝑅_{𝜃}$ and for $𝑡 = - 1$ we get $𝑄 = 𝑆_{𝜃}$ ; the determinants are ${cos}^{2} 𝜃 + {sin}^{2} 𝜃 = 1$ and $- {cos}^{2} 𝜃 - {sin}^{2} 𝜃 = - 1$ . Since $𝜃$ is determined by the first column and the sign by the determinant, the description is unique.

Write $𝜑 = 𝜃 / 2$ . By the angle-subtraction formulas,

\begin{aligned} 𝑆_{𝜃} 𝐜 & = (cos 𝜃 cos 𝜑 + sin 𝜃 sin 𝜑, sin 𝜃 cos 𝜑 - cos 𝜃 sin 𝜑) = (cos 𝜑, sin 𝜑) = 𝐜, \\ 𝑆_{𝜃} 𝐜^{'} & = (- cos 𝜃 sin 𝜑 + sin 𝜃 cos 𝜑, - sin 𝜃 sin 𝜑 - cos 𝜃 cos 𝜑) = (sin 𝜑, - cos 𝜑) = - 𝐜^{'} . \end{aligned}

□

So an orthogonal operator of the plane is a rotation if its determinant is $1$ and a reflection across a line if it is $- 1$ . Rotations by $0$ and $𝜋$ are $\pm 𝐼_{2}$ , the only orthogonal $2 \times 2$ matrices that are scalars. In higher dimensions we need one more lemma, the block version of “a triangular unitary matrix is diagonal” (Exercise 6.3.57(c)).

Lemma 8.4.24 (Block-triangular unitary matrices). Let

𝑅 \in 𝔽^{𝑛 \times 𝑛}

be unitary and block upper triangular, with square diagonal blocks

𝐵_{1}, \dots, 𝐵_{𝑚}

. Then all blocks above the diagonal are

0

, and each

𝐵_{𝑖}

is unitary.

Proof.standard

Uses. Proposition 6.3.34 (Characterizations of unitary matrices), Definition 6.3.33 (Unitary and orthogonal matrices), Definition 2.4.33 (Block matrix)

Techniques. A sum of nonnegative terms (Technique 6.1.9), Induction on matrix size (Technique 5.5.17)

Hint 1. Count the sum of

{| 𝑟_{𝑖 𝑗} |}^{2}

over the first block of columns, and then over the first block of rows.

Hint 2. Both the columns and the rows of a unitary matrix are orthonormal (Proposition 6.3.34). The first

𝑘

columns are zero outside

𝐵_{1}

Sketch. Let

𝐵_{1}

𝑘 \times 𝑘

. The first

𝑘

columns are orthonormal and vanish below

𝐵_{1}

, so the squared entries of

𝐵_{1}

add up to

𝑘

. The first

𝑘

rows are orthonormal, so the squared entries in those rows also add up to

𝑘

. Hence every entry to the right of

𝐵_{1}

0

. Then

𝑅 = (\begin{matrix} 𝐵_{1} & 0 \\ 0 & 𝑅^{'} \end{matrix})

with

𝐵_{1}

and

𝑅^{'}

unitary, and induction on

𝑚

finishes.

We induct on $𝑚$ . If $𝑚 = 1$ , then $𝑅 = 𝐵_{1}$ . Let $𝑚 \geq 2$ , let $𝐵_{1}$ be $𝑘 \times 𝑘$ , and write

𝑅 = (\begin{matrix} 𝐵_{1} & 𝑋 \\ 0 & 𝑅^{'} \end{matrix})

with $𝑋 \in 𝔽^{𝑘 \times (𝑛 - 𝑘)}$ and $𝑅^{'}$ block upper triangular with diagonal blocks $𝐵_{2}, \dots, 𝐵_{𝑚}$ . The first $𝑘$ columns of $𝑅$ are orthonormal and have zero entries below $𝐵_{1}$ , so adding their squared lengths gives $\sum {| 𝑏 |}^{2} = 𝑘$ , the sum over all entries $𝑏$ of $𝐵_{1}$ . By Proposition 6.3.34, the first $𝑘$ rows of $𝑅$ are orthonormal too, so the sum of ${| 𝑟 |}^{2}$ over all entries $𝑟$ of these rows is also $𝑘$ . These rows consist of the entries of $𝐵_{1}$ and of $𝑋$ , so

𝑘 = \sum_{𝑏 in 𝐵_{1}} {| 𝑏 |}^{2} + \sum_{𝑥 in 𝑋} {| 𝑥 |}^{2} = 𝑘 + \sum_{𝑥 in 𝑋} {| 𝑥 |}^{2} .

A sum of nonnegative terms vanishes only if each term does, so $𝑋 = 0$ . Now $𝑅 = (\begin{matrix} 𝐵_{1} & 0 \\ 0 & 𝑅^{'} \end{matrix})$ and

𝐼_{𝑛} = 𝑅^{*} 𝑅 = (\begin{matrix} 𝐵_{1}^{*} 𝐵_{1} & 0 \\ 0 & 𝑅^{' *} 𝑅^{'} \end{matrix}),

so $𝐵_{1}$ and $𝑅^{'}$ are unitary. By the induction hypothesis applied to $𝑅^{'}$ , its blocks above the diagonal vanish and $𝐵_{2}, \dots, 𝐵_{𝑚}$ are unitary. □

Theorem 8.4.25 (Normal form of orthogonal operators). Let $𝑉$ be a real inner product space with $1 \leq dim 𝑉 < \infty$ , and let $𝑆 \in L (𝑉)$ be orthogonal. Then $𝑉$ has an orthonormal basis $𝒪︀$ in which ${[𝑆]}_{𝒪︀}$ is block diagonal with diagonal blocks

𝑅_{𝜃_{1}}, \dots, 𝑅_{𝜃_{𝑘}}, 𝐼_{𝑝}, - 𝐼_{𝑞},

where $𝑘, 𝑝, 𝑞 \geq 0$ , $2 𝑘 + 𝑝 + 𝑞 = dim 𝑉$ , and $0 <$ $𝜃_{1} \leq$ $\dots \leq$ $𝜃_{𝑘} <$ $𝜋$ . Equivalently, every $𝑄 \in O (𝑛)$ can be written as $𝑄 = 𝑃 𝐵 𝑃^{𝖳}$ with $𝑃$ orthogonal and $𝐵$ block diagonal of this form.

Intuition. Read geometrically, the theorem says that

𝑉

splits into perpendicular planes, each rotated by its own angle, and perpendicular lines, each fixed or reversed. Over

ℂ

we could diagonalize; over

ℝ

the planes are what is left of pairs of conjugate eigenvalues

𝑒^{\pm 𝑖 𝜃_{𝑗}}

. The two proofs below are the two routes to real block forms we have seen: triangularize first and then use orthogonality to clear the upper part, or split off one small invariant subspace at a time.

Proof (from the real Schur form).hard

Uses. Theorem 7.5.27 (Real block upper-triangular form), Theorem 8.4.11 (Characterizations of unitary operators), Lemma 8.4.24 (Block-triangular unitary matrices), Lemma 8.4.23 (Orthogonal $2 \times 2$ matrices), Theorem 4.5.17 (Change of basis)

Techniques. Prove it for triangular matrices first (Technique 7.5.23)

Hint 1. Theorem 7.5.27(b) gives an orthonormal basis in which

{[𝑆]}_{𝒪︀}

is block upper triangular with blocks of size

1

and

2

. What does orthogonality of this matrix force?

Hint 2. By Lemma 8.4.24 the matrix is block diagonal with orthogonal blocks. Identify the

1 \times 1

and

2 \times 2

orthogonal blocks; recall that the

2 \times 2

blocks have no real eigenvalue.

Hint 3. To move

𝜃 \in (𝜋, 2 𝜋)

into

(0, 𝜋)

, replace one vector of the pair by its negative.

Sketch. The real Schur form gives an orthonormal

𝒪︀

with

𝑅 = {[𝑆]}_{𝒪︀}

block upper triangular, diagonal blocks of size

1

2

, the latter without real eigenvalues.

𝑅

is orthogonal, so it is block diagonal with orthogonal blocks. A

1 \times 1

block is

\pm 1

; a

2 \times 2

block without real eigenvalues is a rotation

𝑅_{𝜃}

with

𝜃 \notin {0, 𝜋}

, since reflections have eigenvalues

\pm 1

. Flipping one basis vector replaces

𝑅_{𝜃}

𝑅_{- 𝜃}

, and reordering the basis puts the blocks in order.

By Theorem 7.5.27(b), $𝑉$ has an orthonormal basis $𝒪︀$ such that $𝑅 = {[𝑆]}_{𝒪︀}$ is block upper triangular with diagonal blocks of size $1 \times 1$ or $2 \times 2$ , where no $2 \times 2$ diagonal block has a real eigenvalue. By Theorem 8.4.11, $𝑅$ is orthogonal, so by Lemma 8.4.24 it is block diagonal and each diagonal block is orthogonal.

A $1 \times 1$ orthogonal matrix $(𝑎)$ has $𝑎^{2} = 1$ , so it is $(1)$ or $(- 1)$ . A $2 \times 2$ diagonal block $𝐵$ is orthogonal without real eigenvalues. By Lemma 8.4.23, it is $𝑅_{𝜃}$ or $𝑆_{𝜃}$ for some $𝜃 \in [0, 2 𝜋)$ ; the reflection $𝑆_{𝜃}$ has the eigenvalue $1$ , and $𝑅_{0} = 𝐼_{2}$ , $𝑅_{𝜋} = - 𝐼_{2}$ have real eigenvalues, so $𝐵 = 𝑅_{𝜃}$ with $𝜃 \in$ $(0, 𝜋) \cup$ $(𝜋, 2 𝜋)$ . If $𝜃 \in (𝜋, 2 𝜋)$ , let $𝐚, 𝐛$ be the two vectors of $𝒪︀$ belonging to this block, and replace $𝐛$ by $- 𝐛$ . The new list is still orthonormal, and by Theorem 4.5.17 the block becomes

(\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}) 𝑅_{𝜃} (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}) = (\begin{matrix} cos 𝜃 & sin 𝜃 \\ - sin 𝜃 & cos 𝜃 \end{matrix}) = 𝑅_{2 𝜋 - 𝜃},

with $2 𝜋 - 𝜃 \in (0, 𝜋)$ . Finally, reordering $𝒪︀$ (keeping the two vectors of each rotation block together) permutes the diagonal blocks, so we can list the rotation blocks first, in increasing order of angle, then the blocks $(1)$ , then the blocks $(- 1)$ .

For the matrix form, apply this to $𝐱 \mapsto 𝑄 𝐱$ on $ℝ^{𝑛}$ with the dot product, and let $𝑃$ have the vectors of $𝒪︀$ as columns. Then $𝑃$ is orthogonal and $𝑄 =$ $𝑃 𝐵 𝑃^{- 1} =$ $𝑃 𝐵 𝑃^{𝖳}$ with $𝐵 = {[𝑆]}_{𝒪︀}$ . □

Proof (by induction on dimension).hard

Uses. Theorem 7.3.31 (Small invariant subspaces), Lemma 8.4.14 (Invariant complements), Lemma 8.4.23 (Orthogonal $2 \times 2$ matrices), Corollary 8.4.13 (Eigenvalues and determinant), Theorem 6.4.24 (Orthogonal decomposition), Remark 7.2.5 (Invariant subspaces and block matrices)

Techniques. Induction on dimension via an invariant subspace (Technique 7.5.12)

Hint 1. Theorem 7.3.31 gives an

𝑆

-invariant subspace

𝑈

of dimension

1

2

. What does Lemma 8.4.14 say about

𝑈^{⟂}

Hint 2. On

𝑈

𝑆

is orthogonal of dimension

1

2

. Use Lemma 8.4.23, and note that a reflection of a plane splits it into two eigenlines.

Sketch. Pick an invariant

𝑈

with

dim 𝑈 \in {1, 2}

. Then

𝑈^{⟂}

is invariant and both restrictions are orthogonal. On a line,

𝑆 = \pm id

. On a plane,

𝑆

is a rotation

𝑅_{𝜃}

in an orthonormal basis (normalize

𝜃

as in the first proof;

𝜃 \in {0, 𝜋}

gives

\pm 𝐼_{2}

) or a reflection, which is

diag (1, - 1)

in the basis of its eigenvectors. Apply induction to

𝑈^{⟂}

, concatenate, and reorder.

We induct on $𝑛 = dim 𝑉$ , proving the statement without the ordering of the blocks; reordering the basis as in the first proof then arranges them. By Theorem 7.3.31, $𝑆$ has an invariant subspace $𝑈$ with $dim 𝑈 \in {1, 2}$ . By Lemma 8.4.14, $𝑈^{⟂}$ is $𝑆$ -invariant and $𝑆 ↾_{𝑈}$ , $𝑆 ↾_{𝑈^{⟂}}$ are orthogonal.

The piece $𝑈$ . If $dim 𝑈 = 1$ , then $𝑈$ is spanned by an eigenvector, whose eigenvalue is real of absolute value $1$ (Corollary 8.4.13(a)), so $𝑆 ↾_{𝑈} = \pm {id}_{𝑈}$ , a block $(1)$ or $(- 1)$ in a unit basis vector of $𝑈$ . If $dim 𝑈 = 2$ , let $𝐚, 𝐛$ be an orthonormal basis of $𝑈$ ; the matrix of $𝑆 ↾_{𝑈}$ in it is orthogonal, so by Lemma 8.4.23 it is $𝑅_{𝜃}$ or $𝑆_{𝜃}$ . If it is $𝑆_{𝜃}$ , the unit eigenvectors $𝐜, 𝐜^{'}$ of the lemma (taken as combinations of $𝐚, 𝐛$ ) form an orthonormal basis of $𝑈$ in which the matrix is $diag (1, - 1)$ . If it is $𝑅_{𝜃}$ with $𝜃 \in {0, 𝜋}$ , it is $\pm 𝐼_{2}$ . Otherwise it is a rotation block, and replacing $𝐛$ by $- 𝐛$ if necessary makes $𝜃 \in (0, 𝜋)$ , as in the first proof.

The piece $𝑈^{⟂}$ . If $𝑈 = 𝑉$ we are done. Otherwise $1 \leq dim 𝑈^{⟂} < 𝑛$ , and the induction hypothesis gives an orthonormal basis of $𝑈^{⟂}$ in which $𝑆 ↾_{𝑈^{⟂}}$ has the required block form.

By Theorem 6.4.24, $𝑉 = 𝑈 \oplus 𝑈^{⟂}$ , so the two bases together form an orthonormal basis of $𝑉$ . Both summands are invariant, so in this basis $[𝑆]$ is block diagonal, with the blocks of $𝑆 ↾_{𝑈}$ followed by those of $𝑆 ↾_{𝑈^{⟂}}$ (Remark 7.2.5). □

The first proof spends its effort in the real Schur form and needs only bookkeeping afterwards; the second needs only a small invariant subspace and does the splitting by hand, using that invariant complements come for free for orthogonal operators. A third proof complexifies and pairs conjugate eigenvectors, as in Proposition 7.3.34; Exercise carries it out.

The numbers in the normal form are determined by $𝑆$ . The complex eigenvalues of ${[𝑆]}_{𝒪︀}$ , counted with multiplicity, are $𝑒^{\pm 𝑖 𝜃_{1}},$ $\dots,$ $𝑒^{\pm 𝑖 𝜃_{𝑘}}$ together with $1$ ( $𝑝$ times) and $- 1$ ( $𝑞$ times), by Proposition 7.3.15, since $𝑅_{𝜃}$ has characteristic polynomial $𝑥^{2} -$ $2 cos 𝜃 𝑥 +$ $1 =$ $(𝑥 - 𝑒^{𝑖 𝜃}) (𝑥 - 𝑒^{- 𝑖 𝜃})$ . In particular $det 𝑆 = {(- 1)}^{𝑞}$ (Proposition 5.3.17).

8.4.5Orientation and rotations of $ℝ^{3}$ Beyond 21-242

By Corollary 8.4.13(b), every orthogonal operator has determinant $1$ or $- 1$ . The sign records whether $𝑆$ preserves orientation: whether it maps a right-handed orthonormal basis to a right-handed one, or reverses the handedness as a mirror does. In the plane, Lemma 8.4.23 says the two kinds are the rotations and the reflections, and the “F” in the picture below shows the difference.

Definition 8.4.26 (Rotation). Let

𝑉

be a finite-dimensional real inner product space. An orthogonal operator

𝑆 \in L (𝑉)

is a rotation (or proper) if

det 𝑆 = 1

, and improper if

det 𝑆 = - 1

. The rotations of

ℝ^{𝑛}

form the group

SO (𝑛)

𝑆 = (\begin{matrix} 0.6 & 0.8 \\ 0.8 & - 0.6 \end{matrix}), det 𝑆 = - 1, 𝑆 (\begin{matrix} 2 \\ 1 \end{matrix}) = (\begin{matrix} 2 \\ 1 \end{matrix}), 𝑆 (\begin{matrix} - 1 \\ 2 \end{matrix}) = (\begin{matrix} 1 \\ - 2 \end{matrix}) .

The reflection

𝑆 =

(\begin{matrix} 0.6 & 0.8 \\ 0.8 & - 0.6 \end{matrix}) =

𝑆_{𝜃}

with

cos 𝜃 = 0.6

sin 𝜃 = 0.8

. It fixes the line through

(2, 1)

and reverses

(- 1, 2)

, and

det 𝑆 = - 1

: the “F” comes out mirrored. In the interactive version, drag

𝑆 𝐞_{2}

(- 0.8, 0.6)

to get the rotation

𝑅_{𝜃}

, with determinant

1

and an unmirrored “F”. Every other position of

𝑆 𝐞_{2}

leaves the orthogonal matrices.

The determinant is a continuous function of the entries that takes only the values $\pm 1$ on $O (𝑛)$ , so no path of orthogonal matrices leads from a rotation to a reflection. Conversely, by Theorem 8.4.25 every rotation is $𝑃 𝐵 𝑃^{𝖳}$ with an even number of blocks $- 1$ , which pair up into blocks $- 𝐼_{2} = 𝑅_{𝜋}$ ; letting all the angles shrink to $0$ moves $𝐵$ continuously to $𝐼_{𝑛}$ inside $SO (𝑛)$ . This is the precise sense in which the rotations are the orthogonal operators that can be “performed physically”.

In three dimensions a rotation has more structure than the word suggests: it always turns about a fixed axis.

Theorem 8.4.27 (Euler’s rotation theorem). Let $𝑉$ be a real inner product space with $dim 𝑉 = 3$ , and let $𝑆 \in L (𝑉)$ be a rotation. Then $𝑆$ has an axis: a unit vector $𝐚$ with $𝑆 𝐚 = 𝐚$ . Moreover, for every orthonormal basis $𝒪︀ = 𝐚, 𝐛, 𝐜$ of $𝑉$ starting with $𝐚$ ,

{[𝑆]}_{𝒪︀} = (\begin{matrix} 1 & 0 & 0 \\ 0 & cos 𝜃 & - sin 𝜃 \\ 0 & sin 𝜃 & cos 𝜃 \end{matrix})

for some $𝜃 \in [0, 2 𝜋)$ , and $tr 𝑆 = 1 + 2 cos 𝜃$ .

Intuition. We need

1

to be an eigenvalue. Three different reasons are available: a determinant identity that says

det (𝑆 - id)

equals its own negative, a count of complex eigenvalues on the unit circle with product

1

, and the normal form, in which an odd dimension leaves at least one

1 \times 1

block. Once an axis exists, the plane perpendicular to it is invariant, and Lemma 8.4.23 identifies what happens there.

Proof (by a determinant identity).standard

Uses. Theorem 8.4.11 (Characterizations of unitary operators), Theorem 5.4.10 (Multiplicativity), Theorem 5.4.14 (The transpose), Definition 5.1.1 (Determinant function), Theorem 7.2.9 (Eigenvalues and invertibility), Lemma 8.4.14 (Invariant complements), Lemma 8.4.23 (Orthogonal $2 \times 2$ matrices), Proposition 5.3.17 (Block-diagonal matrices)

Hint 1. Let

𝐴

be the matrix of

𝑆

in an orthonormal basis. Write

𝐴 - 𝐼_{3} = 𝐴 - 𝐴 𝐴^{𝖳}

and factor out

𝐴

Hint 2.

det (𝐴 - 𝐼_{3}) =

det 𝐴 \cdot

det (𝐼_{3} - 𝐴^{𝖳}) =

det (𝐼_{3} - 𝐴)

. Compare

det (𝐼_{3} - 𝐴)

with

det (𝐴 - 𝐼_{3})

in dimension

3

Hint 3. For the matrix, the plane

𝐚^{⟂}

is invariant. What is the determinant of

𝑆

on it?

Sketch.

det (𝐴 - 𝐼) =

det (𝐴 (𝐼 - 𝐴^{𝖳})) =

det 𝐴 det ({(𝐼 - 𝐴)}^{𝖳}) =

det (𝐼 - 𝐴) =

{(- 1)}^{3} det (𝐴 - 𝐼)

, so

det (𝐴 - 𝐼) = 0

and

1

is an eigenvalue. With a unit fixed vector

𝐚

𝐚^{⟂}

is invariant and

𝑆

is orthogonal on it with determinant

det 𝑆 / 1 = 1

, so it is a rotation

𝑅_{𝜃}

of that plane. The trace is

1 + 2 cos 𝜃

An axis. Let $𝐴 = {[𝑆]}_{ℬ︀}$ for some orthonormal basis $ℬ︀$ ; it is orthogonal with $det 𝐴 = det 𝑆 = 1$ . Using $𝐴 𝐴^{𝖳} = 𝐼_{3}$ , Theorem 5.4.10, and Theorem 5.4.14,

det (𝐴 - 𝐼_{3}) = det (𝐴 - 𝐴 𝐴^{𝖳}) = det (𝐴) det (𝐼_{3} - 𝐴^{𝖳}) = det ({(𝐼_{3} - 𝐴)}^{𝖳}) = det (𝐼_{3} - 𝐴) .

Since $𝐼_{3} -$ $𝐴 =$ $(- 1) (𝐴 - 𝐼_{3})$ and the determinant is multilinear in the three columns (Definition 5.1.1), $det (𝐼_{3} - 𝐴) =$ ${(- 1)}^{3} det (𝐴 - 𝐼_{3})$ . So $det (𝐴 - 𝐼_{3}) =$ $- det (𝐴 - 𝐼_{3})$ , that is, $det (𝐴 - 𝐼_{3}) = 0$ . Hence $1$ is an eigenvalue of $𝐴$ , and so of $𝑆$ (Theorem 7.2.9); a unit eigenvector $𝐚$ is an axis.

The matrix. Let $𝐚, 𝐛, 𝐜$ be an orthonormal basis. The line $span (𝐚)$ is $𝑆$ -invariant, so by Lemma 8.4.14 the plane $𝑃 =$ $span (𝐛, 𝐜) =$ $𝐚^{⟂}$ is invariant and $𝑆 ↾_{𝑃}$ is orthogonal. So ${[𝑆]}_{𝒪︀}$ is block diagonal with blocks $(1)$ and the orthogonal matrix $𝐵$ of $𝑆 ↾_{𝑃}$ in $𝐛, 𝐜$ , and $1 = det 𝑆 = 1 \cdot det 𝐵$ (Proposition 5.3.17). By Lemma 8.4.23, $𝐵 = 𝑅_{𝜃}$ for some $𝜃 \in [0, 2 𝜋)$ . Finally $tr 𝑆 = 1 + 2 cos 𝜃$ . □

Proof (by counting eigenvalues).standard

Uses. Theorem 7.1.32 (Fundamental Theorem of Algebra), Proposition B.1.4 (Roots of real polynomials come in conjugate pairs), Corollary 7.1.43 (Odd degree gives a real root), Theorem 7.5.22 (Trace and determinant from eigenvalues), Corollary 8.4.13 (Eigenvalues and determinant), Corollary 7.3.13 (Eigenvalues of an operator)

Techniques. Complexify, then take real and imaginary parts (Technique 7.3.35)

Hint 1. Regard the matrix

𝐴

𝑆

as a complex unitary matrix. Its three eigenvalues lie on the unit circle and multiply to

det 𝐴 = 1

Hint 2. The characteristic polynomial is real of degree

3

: its nonreal roots come in conjugate pairs, and it has a real root.

Sketch. The eigenvalues

𝜆_{1}, 𝜆_{2}, 𝜆_{3} \in ℂ

𝐴

have absolute value

1

and product

1

. If two are a conjugate pair, their product is

{| 𝜆 |}^{2} = 1

, so the third is

1

. If all are real, they are

\pm 1

with product

1

, so they are not all

- 1

. Either way

1

is a root of

𝑝_{𝑆}

, hence an eigenvalue of

𝑆

We show that $1$ is an eigenvalue; the rest is as in the first proof. Let $𝐴$ be the matrix of $𝑆$ in an orthonormal basis, regarded as a complex unitary matrix, and let $𝜆_{1}, 𝜆_{2}, 𝜆_{3}$ be its eigenvalues counted with multiplicity, the roots of $𝑝_{𝐴}$ (Theorem 7.1.32). Each satisfies $| 𝜆_{𝑗} | = 1$ (Corollary 8.4.13(a)), and $𝜆_{1} 𝜆_{2} 𝜆_{3} = det 𝐴 = 1$ (Theorem 7.5.22). Since $𝑝_{𝐴}$ is a real polynomial of odd degree, it has a real root (Corollary 7.1.43), say $𝜆_{3} \in {1, - 1}$ , and its nonreal roots come in conjugate pairs (Proposition B.1.4).

If $𝜆_{1} \notin ℝ$ , then $𝜆_{2} = 𝜆_{1}$ , so $𝜆_{1} 𝜆_{2} = {| 𝜆_{1} |}^{2} = 1$ and $𝜆_{3} = 1$ . If all three are real, each is $\pm 1$ , and their product is $1$ , so they are not all $- 1$ . In both cases $1$ is a root of $𝑝_{𝐴} = 𝑝_{𝑆}$ , hence an eigenvalue of $𝑆$ (Corollary 7.3.13). □

Proof (from the normal form).routine

Uses. Theorem 8.4.25 (Normal form of orthogonal operators), Proposition 5.3.17 (Block-diagonal matrices)

Hint 1. In the normal form,

2 𝑘 + 𝑝 + 𝑞 = 3

and

det 𝑆 = {(- 1)}^{𝑞}

Hint 2. The number

𝑞

of blocks

- 1

is even. Can

𝑝

0

Sketch.

det 𝑆 = {(- 1)}^{𝑞} = 1

forces

𝑞

even. If

𝑘 = 1

, then

𝑝 + 𝑞 = 1

, so

𝑞 = 0

and

𝑝 = 1

. If

𝑘 = 0

, then

𝑝 + 𝑞 = 3

with

𝑞 \in {0, 2}

, so

𝑝 \geq 1

. A block

(1)

is a fixed unit vector.

By Theorem 8.4.25,

{[𝑆]}_{𝒪︀}

is block diagonal with

𝑘

rotation blocks,

𝑝

blocks

(1)

, and

𝑞

blocks

(- 1)

, where

2 𝑘 + 𝑝 + 𝑞 = 3

. By Proposition 5.3.17,

det 𝑆 = {(- 1)}^{𝑞}

(each

𝑅_{𝜃}

has determinant

1

), so

𝑞

is even. If

𝑘 = 1

, then

𝑝 + 𝑞 = 1

forces

𝑞 = 0

and

𝑝 = 1

. If

𝑘 = 0

, then

𝑝 + 𝑞 = 3

with

𝑞 \in {0, 2}

, so

𝑝 \geq 1

. Either way some vector of

𝒪︀

is fixed by

𝑆

. The matrix and trace follow as in the first proof. □

The first proof is the shortest and shows what matters: in any odd dimension $𝑛$ , $det (𝐼_{𝑛} - 𝐴) =$ ${(- 1)}^{𝑛} det (𝐴 - 𝐼_{𝑛})$ , so every rotation of $ℝ^{𝑛}$ with $𝑛$ odd has a fixed vector (Exercise 8.4.42). In even dimensions this fails: $𝑅_{𝜃} \in SO (2)$ with $𝜃 \neq 0$ , and $(\begin{matrix} 𝑅_{𝜃} & 0 \\ 0 & 𝑅_{𝜑} \end{matrix}) \in$ $SO (4)$ with $𝜃, 𝜑 \neq 0$ , fix only $𝟎$ . The second proof is a computation with the spectrum, and the third reads the answer off the classification.

The trace formula identifies a rotation of $ℝ^{3}$ quickly. Its axis is a kernel vector of $𝑆 - id$ , and the angle $𝜃$ satisfies $cos 𝜃 = (tr 𝑆 - 1) / 2$ . The cosine determines $𝜃$ only up to sign; replacing $𝐛, 𝐜$ by $𝐜, 𝐛$ replaces $𝜃$ by $2 𝜋 - 𝜃$ , so the sense of the rotation depends on an orientation of the plane $𝐚^{⟂}$ .

Example 8.4.28 (Axis and angle). Let

𝑄 = \frac{1}{3} (\begin{matrix} 2 & - 1 & 2 \\ 2 & 2 & - 1 \\ - 1 & 2 & 2 \end{matrix}) .

Its columns have squared lengths $(4 + 4 + 1) / 9 = 1$ and pairwise dot products $(- 2 + 4 - 2) / 9$ , $(4 - 2 - 2) / 9$ , $(- 2 - 2 + 4) / 9$ , all $0$ , so $𝑄$ is orthogonal. Expanding along the first row, $det (3 𝑄) =$ $2 (4 + 2) +$ $(4 - 1) +$ $2 (4 + 2) =$ $27$ , so $det 𝑄 = 1$ and $𝑄$ is a rotation. Each row of $𝑄$ sums to $1$ , so $𝑄 (1, 1, 1) = (1, 1, 1)$ , and the axis is $𝐚 = \frac{1}{\sqrt{3}} (1, 1, 1)$ . Since $tr 𝑄 = 2 = 1 + 2 cos 𝜃$ , we get $cos 𝜃 = 1 / 2$ : $𝑄$ turns the plane $𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 0$ through $𝜋 / 3$ . For instance $𝑄 (1, - 1, 0) =$ $(1, 0, - 1)$ , and these two vectors of length $\sqrt{2}$ have dot product $1 = \sqrt{2} \cdot \sqrt{2} \cdot cos (𝜋 / 3)$ .

Squaring, $𝑄^{2} = (\begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix})$ , the cyclic shift $𝐶$ of Example 8.4.16(b) for $𝑛 = 3$ : it turns about the same axis through $2 𝜋 / 3$ , and its trace is $0 = 1 + 2 cos (2 𝜋 / 3)$ .

Since $SO (3)$ is a group, the composite of two rotations of $ℝ^{3}$ , about any two axes, is again a rotation about some axis, which is not obvious from the pictures (Exercise).

8.4.6Rigid motions Beyond 21-242

We now drop linearity and ask for maps that preserve distances.

Definition 8.4.29 (Rigid motion). Let $𝑉$ be a real inner product space. A function $𝑓 : 𝑉 \to 𝑉$ , not assumed linear, is a rigid motion if

‖ 𝑓 (𝐱) - 𝑓 (𝐲) ‖ = ‖ 𝐱 - 𝐲 ‖ for all 𝐱, 𝐲 \in 𝑉 .

Translations $𝐱 \mapsto 𝐱 + 𝐛$ and orthogonal operators are rigid motions, and so are their composites $𝐱 \mapsto 𝑆 𝐱 + 𝐛$ . These are affine maps (Definition 4.7.35) with orthogonal linear part. The theorem of this subsection is that there are no others. In Problem 6.8 the linearity of $𝑓 - 𝑓 (𝟎)$ was proved by polarization; here we give a different proof that uses distances only, through the observation that in an inner product space a point is determined by its distances to the two ends of a segment that it lies on.

Lemma 8.4.30 (Points between two points). Let $𝑉$ be an inner product space, let $𝐚, 𝐛, 𝐳 \in 𝑉$ , and let $𝑡 \in [0, 1]$ . If

‖ 𝐳 - 𝐚 ‖ = 𝑡 ‖ 𝐛 - 𝐚 ‖ and ‖ 𝐛 - 𝐳 ‖ = (1 - 𝑡) ‖ 𝐛 - 𝐚 ‖,

then $𝐳 = (1 - 𝑡) 𝐚 + 𝑡 𝐛$ .

Proof.standard

Uses. Theorem 6.2.11 (Triangle inequality), Proposition 6.1.27 (Basic properties of the norm)

Hint 1. Let

𝐱 = 𝐳 - 𝐚

and

𝐲 = 𝐛 - 𝐳

. Then

𝐱 + 𝐲 = 𝐛 - 𝐚

. What do the hypotheses say about

‖ 𝐱 ‖ + ‖ 𝐲 ‖

Hint 2. Equality holds in the triangle inequality, so one of

𝐱, 𝐲

is a nonnegative multiple of the other (Theorem 6.2.11(b)). Deduce that

𝐱

is a nonnegative multiple of

𝐛 - 𝐚

, and find the factor from

‖ 𝐱 ‖

Sketch. With

𝐱 = 𝐳 - 𝐚

𝐲 = 𝐛 - 𝐳

‖ 𝐱 ‖ + ‖ 𝐲 ‖ = ‖ 𝐱 + 𝐲 ‖

, so

𝐱

and

𝐲

point the same way, and then

𝐱 = 𝑠 (𝐛 - 𝐚)

with

𝑠 \geq 0

. Taking norms,

𝑠 = 𝑡

(if

𝐛 \neq 𝐚

; if

𝐛 = 𝐚

, then

𝐳 = 𝐚

directly).

If $𝐛 = 𝐚$ , the first hypothesis gives $‖ 𝐳 - 𝐚 ‖ = 0$ , so $𝐳 =$ $𝐚 =$ $(1 - 𝑡) 𝐚 +$ $𝑡 𝐛$ . Suppose $𝐛 \neq 𝐚$ , and let $𝐱 = 𝐳 - 𝐚$ , $𝐲 = 𝐛 - 𝐳$ , so that $𝐱 + 𝐲 = 𝐛 - 𝐚$ and

‖ 𝐱 ‖ + ‖ 𝐲 ‖ = 𝑡 ‖ 𝐛 - 𝐚 ‖ + (1 - 𝑡) ‖ 𝐛 - 𝐚 ‖ = ‖ 𝐱 + 𝐲 ‖ .

By Theorem 6.2.11(b), one of $𝐱, 𝐲$ is a nonnegative real multiple of the other. If $𝐱 = 𝑐 𝐲$ with $𝑐 \geq 0$ , then $𝐛 - 𝐚 = (𝑐 + 1) 𝐲$ , so $𝐱 =$ $𝑐 / (𝑐 + 1) (𝐛 - 𝐚)$ . If $𝐲 = 𝑐 𝐱$ with $𝑐 \geq 0$ , then $𝐛 - 𝐚 = (1 + 𝑐) 𝐱$ , so $𝐱 =$ $1 / (1 + 𝑐) (𝐛 - 𝐚)$ . Either way $𝐱 = 𝑠 (𝐛 - 𝐚)$ for some real $𝑠 \geq 0$ , and taking norms gives $𝑠 ‖ 𝐛 -$ $𝐚 ‖ =$ $‖ 𝐱 ‖ =$ $𝑡 ‖ 𝐛 -$ $𝐚 ‖$ , so $𝑠 = 𝑡$ . Hence $𝐳 =$ $𝐚 +$ $𝑡 (𝐛 - 𝐚) =$ $(1 - 𝑡) 𝐚 +$ $𝑡 𝐛$ . □

Theorem 8.4.31 (Distance-preserving maps are affine). Let $𝑉$ and $𝑊$ be real inner product spaces, and let $𝑓 : 𝑉 \to 𝑊$ satisfy $‖ 𝑓 (𝐱) -$ $𝑓 (𝐲) ‖ =$ $‖ 𝐱 -$ $𝐲 ‖$ for all $𝐱, 𝐲 \in 𝑉$ . Then $𝑆 (𝐱) =$ $𝑓 (𝐱) -$ $𝑓 (𝟎)$ defines an isometry $𝑆 \in L (𝑉, 𝑊)$ , so $𝑓 = 𝑆 + 𝑓 (𝟎)$ is affine.

In particular, if $𝑉$ is finite-dimensional, every rigid motion of $𝑉$ has the form $𝐱 \mapsto 𝑆 𝐱 + 𝐛$ with $𝑆$ orthogonal, and is bijective. The rigid motions of $ℝ^{𝑛}$ are the maps $𝐱 \mapsto 𝑄 𝐱 + 𝐛$ with $𝑄 \in O (𝑛)$ and $𝐛 \in ℝ^{𝑛}$ .

Proof (by betweenness).hard

Uses. Lemma 8.4.30 (Points between two points), Theorem 2.1.3 (Checking linearity), Definition 4.7.35 (Affine map), Theorem 8.4.11 (Characterizations of unitary operators), Proposition 2.3.17 (Every linear map $𝐹^{𝑛} \to 𝐹^{𝑚}$ is some $𝑇_{𝐴}$ )

Techniques. Choose a base point (Technique 4.7.4)

Hint 1. Use Lemma 8.4.30 to show that

𝑓 ((1 - 𝑡) 𝐚 + 𝑡 𝐛) =

(1 - 𝑡) 𝑓 (𝐚) +

𝑡 𝑓 (𝐛)

for

𝑡 \in [0, 1]

: the point on the left has the right distances to

𝑓 (𝐚)

and

𝑓 (𝐛)

Hint 2. Extend to all real

𝑡

: for

𝑡 > 1

, the point

𝐛

lies between

𝐚

and

(1 - 𝑡) 𝐚 + 𝑡 𝐛

Hint 3. Now

𝑆 (𝑡 𝐱) =

𝑓 ((1 - 𝑡) 𝟎 + 𝑡 𝐱) -

𝑓 (𝟎)

, and

𝐱 + 𝐲

is twice the midpoint of

𝐱

and

𝐲

Sketch. Distances from

𝑓 (𝐳)

𝑓 (𝐚)

and

𝑓 (𝐛)

equal those from

𝐳 = (1 - 𝑡) 𝐚 + 𝑡 𝐛

𝐚

and

𝐛

, so Lemma 8.4.30 forces

𝑓 (𝐳) =

(1 - 𝑡) 𝑓 (𝐚) +

𝑡 𝑓 (𝐛)

for

𝑡 \in [0, 1]

, and a relabelling gives all

𝑡 \in ℝ

. Then

𝑆 = 𝑓 - 𝑓 (𝟎)

satisfies

𝑆 (𝑡 𝐱) = 𝑡 𝑆 (𝐱)

and

𝑆 (𝐱 + 𝐲) =

2 𝑆 ((𝐱 + 𝐲) / 2) =

𝑆 (𝐱) +

𝑆 (𝐲)

, and

‖ 𝑆 𝐱 ‖ = ‖ 𝐱 ‖

. In finite dimension

𝑆

is orthogonal, hence bijective.

Step 1: segments go to segments. Let $𝐚, 𝐛 \in 𝑉$ , $𝑡 \in [0, 1]$ , and $𝐳 = (1 - 𝑡) 𝐚 + 𝑡 𝐛$ . Then $𝐳 - 𝐚 = 𝑡 (𝐛 - 𝐚)$ and $𝐛 -$ $𝐳 =$ $(1 - 𝑡) (𝐛 - 𝐚)$ , so, as $𝑓$ preserves distances,

‖ 𝑓 (𝐳) - 𝑓 (𝐚) ‖ = 𝑡 ‖ 𝑓 (𝐛) - 𝑓 (𝐚) ‖, ‖ 𝑓 (𝐛) - 𝑓 (𝐳) ‖ = (1 - 𝑡) ‖ 𝑓 (𝐛) - 𝑓 (𝐚) ‖ .

By Lemma 8.4.30 in $𝑊$ , $𝑓 (𝐳) =$ $(1 - 𝑡) 𝑓 (𝐚) +$ $𝑡 𝑓 (𝐛)$ .

Step 2: lines go to lines. We show that $𝑓 ((1 - 𝑡) 𝐚 + 𝑡 𝐛) =$ $(1 - 𝑡) 𝑓 (𝐚) +$ $𝑡 𝑓 (𝐛)$ for every $𝑡 \in ℝ$ . Let $𝑡 > 1$ and $𝐜 = (1 - 𝑡) 𝐚 + 𝑡 𝐛$ . With $𝑠 = 1 / 𝑡 \in (0, 1)$ , a direct check gives $𝐛 = (1 - 𝑠) 𝐚 + 𝑠 𝐜$ , so Step 1 gives $𝑓 (𝐛) =$ $(1 - 𝑠) 𝑓 (𝐚) +$ $𝑠 𝑓 (𝐜)$ . Solving for $𝑓 (𝐜)$ ,

𝑓 (𝐜) = 𝑡 𝑓 (𝐛) - (𝑡 - 1) 𝑓 (𝐚) = (1 - 𝑡) 𝑓 (𝐚) + 𝑡 𝑓 (𝐛) .

For $𝑡 < 0$ , write $(1 - 𝑡) 𝐚 +$ $𝑡 𝐛 =$ $(1 - 𝑡^{'}) 𝐛 +$ $𝑡^{'} 𝐚$ with $𝑡^{'} = 1 - 𝑡 > 1$ , and apply the case just proved with $𝐚$ and $𝐛$ exchanged.

Step 3: $𝑆$ is linear. Let $𝑆 (𝐱) =$ $𝑓 (𝐱) -$ $𝑓 (𝟎)$ . For $𝑡 \in ℝ$ , Step 2 with $𝐚 = 𝟎$ , $𝐛 = 𝐱$ gives

𝑆 (𝑡 𝐱) = 𝑓 ((1 - 𝑡) 𝟎 + 𝑡 𝐱) - 𝑓 (𝟎) = (1 - 𝑡) 𝑓 (𝟎) + 𝑡 𝑓 (𝐱) - 𝑓 (𝟎) = 𝑡 𝑆 (𝐱) .

For $𝐱, 𝐲 \in 𝑉$ , Step 1 with $𝑡 = 1 / 2$ gives $𝑓 (\frac{1}{2} 𝐱 + \frac{1}{2} 𝐲) =$ $\frac{1}{2} 𝑓 (𝐱) +$ $\frac{1}{2} 𝑓 (𝐲)$ , so, using homogeneity,

𝑆 (𝐱 + 𝐲) = 2 𝑆 (\frac{1}{2} (𝐱 + 𝐲)) = 2 (\frac{1}{2} 𝑓 (𝐱) + \frac{1}{2} 𝑓 (𝐲) - 𝑓 (𝟎)) = 𝑆 (𝐱) + 𝑆 (𝐲) .

So $𝑆$ is linear (Theorem 2.1.3), and $‖ 𝑆 𝐱 ‖ =$ $‖ 𝑓 (𝐱) -$ $𝑓 (𝟎) ‖ =$ $‖ 𝐱 -$ $𝟎 ‖ =$ $‖ 𝐱 ‖$ , so $𝑆$ is an isometry. Thus $𝑓 = 𝑆 + 𝑓 (𝟎)$ is affine (Definition 4.7.35).

Finite dimension. If $𝑊 = 𝑉$ is finite-dimensional, $𝑆$ is orthogonal by (I) $\Rightarrow$ (II) of Theorem 8.4.11, so it is bijective, and so is $𝑓 = 𝑆 + 𝐛$ , with inverse $𝐲 \mapsto 𝑆^{- 1} (𝐲 - 𝐛)$ . For $𝑉 = ℝ^{𝑛}$ , $𝑆$ is $𝐱 \mapsto 𝑄 𝐱$ for a matrix $𝑄$ (Proposition 2.3.17), which is orthogonal by Theorem 8.4.11. □

The proof used the inner product only through Lemma 8.4.30, that is, through the equality case of the triangle inequality: the “midpoint” of $𝐚$ and $𝐛$ , defined by distances alone, is unique. This is the idea behind the Mazur–Ulam theorem, which says that every surjective distance-preserving map between real normed spaces is affine. In a normed space such as $ℝ^{2}$ with ${‖ 𝐱 ‖}_{\infty} =$ $max (| 𝑥_{1} |, | 𝑥_{2} |)$ there are many points at distance $1$ from both $(0, 0)$ and $(2, 0)$ , and Exercise 8.4.45 shows that the conclusion of Theorem 8.4.31 then fails for maps that are not onto. Over $ℂ$ , Lemma 8.4.30 still holds and Steps 1–3 go through with real $𝑡$ , so a distance-preserving map of complex spaces is additive and real-homogeneous after subtracting $𝑓 (𝟎)$ ; complex conjugation on $ℂ^{𝑛}$ shows that it need not be complex-linear (Problem 6.8(d)).

In homogeneous coordinates (Section 4.7.6), the rigid motion $𝑓 (𝐱) = 𝑄 𝐱 + 𝐛$ of $ℝ^{𝑛}$ is the matrix

𝑀_{𝑓} = (\begin{matrix} 𝑄 & 𝐛 \\ 0 & 1 \end{matrix}) \in ℝ^{(𝑛 + 1) \times (𝑛 + 1)},

and by Theorem 4.7.42 composites and inverses correspond to products and inverses:

(\begin{matrix} 𝑄_{1} & 𝐛_{1} \\ 0 & 1 \end{matrix}) (\begin{matrix} 𝑄_{2} & 𝐛_{2} \\ 0 & 1 \end{matrix}) = (\begin{matrix} 𝑄_{1} 𝑄_{2} & 𝑄_{1} 𝐛_{2} + 𝐛_{1} \\ 0 & 1 \end{matrix}), {(\begin{matrix} 𝑄 & 𝐛 \\ 0 & 1 \end{matrix})}^{- 1} = (\begin{matrix} 𝑄^{𝖳} & - 𝑄^{𝖳} 𝐛 \\ 0 & 1 \end{matrix}) .

So the rigid motions of $ℝ^{𝑛}$ form a group, the Euclidean group. A rigid motion is proper if $det 𝑄 = 1$ ; these are the motions of a solid body in space, and in the plane they are the translations and the rotations about a point (Exercise 4.7.62).

Exercises

Exercise 8.4.32.routine Decide which of the following linear maps are isometries, and which of the isometries are unitary.

(a) $𝑇 : ℝ^{2} \to ℝ^{3}$ , $𝑇 (𝑥, 𝑦) =$ $\frac{1}{3} (𝑥 + 2 𝑦, 2 𝑥 + 𝑦, 2 𝑥 - 2 𝑦)$ .
(b) $𝑇 : ℂ^{2} \to ℂ^{2}$ , $𝑇 (𝑧, 𝑤) =$ $\frac{1}{\sqrt{2}} (𝑧 + 𝑖 𝑤, 𝑖 𝑧 + 𝑤)$ .
(c) $𝑇 : 𝑉 \to 𝑉$ , $(𝑇 𝑝) (𝑥) =$ $𝑝 (1 - 𝑥)$ , where $𝑉 = {ℝ [𝑥]}_{\leq 1}$ with $⟨ 𝑝, 𝑞 ⟩ =$ $𝑝 (0) 𝑞 (0) +$ $𝑝 (1) 𝑞 (1)$ .
(d) $𝑇 : 𝑉 \to 𝑉$ , $(𝑇 𝑝) (𝑥) =$ $𝑝 (𝑥 / 2)$ , with $𝑉$ as in (c).
(e) $𝑇 : ℂ^{2 \times 2} \to ℂ^{2 \times 2}$ , $𝑇 (𝐴) = 𝐴^{𝖳}$ , with the Frobenius inner product.

Solution.

Uses. Theorem 8.4.4 (Characterizations of isometries), Theorem 8.4.11 (Characterizations of unitary operators), Definition 8.4.1 (Isometry), Proposition 6.1.16 (Frobenius inner product)

Hint 1. For matrices, check whether the columns are orthonormal. For the polynomial maps, compute

{‖ 𝑇 𝑝 ‖}^{2}

from the definition of the inner product.

Hint 2. Finite-dimensional isometries

𝑉 \to 𝑉

are unitary; an isometry between spaces of different dimension is not even an operator.

Sketch. (a) Orthonormal columns

\frac{1}{3} (1, 2, 2)

\frac{1}{3} (2, 1, - 2)

: an isometry, not unitary. (b) Columns

\frac{1}{\sqrt{2}} (1, 𝑖)

\frac{1}{\sqrt{2}} (𝑖, 1)

are orthonormal: unitary. (c)

{‖ 𝑇 𝑝 ‖}^{2} =

{𝑝 (1)}^{2} +

{𝑝 (0)}^{2}

: unitary. (d)

𝑝 = 𝑥

has

{‖ 𝑇 𝑝 ‖}^{2} = 1 / 4 \neq 1

: no. (e)

{‖ 𝐴^{𝖳} ‖}^{2} =

\sum {| 𝑎_{𝑗 𝑖} |}^{2} =

{‖ 𝐴 ‖}^{2}

: unitary.

(a) The matrix of $𝑇$ is $\frac{1}{3} (\begin{matrix} 1 & 2 \\ 2 & 1 \\ 2 & - 2 \end{matrix})$ , with columns of squared length $(1 + 4 + 4) / 9 = 1$ and $(4 + 1 + 4) / 9 = 1$ and dot product $(2 + 2 - 4) / 9 = 0$ . So $𝑇$ is an isometry by Theorem 8.4.4 (with the standard bases). It maps $ℝ^{2}$ into $ℝ^{3}$ , so it is not an operator, and not unitary; its image is a plane.

(b) The matrix is $\frac{1}{\sqrt{2}} (\begin{matrix} 1 & 𝑖 \\ 𝑖 & 1 \end{matrix})$ . Its columns $\frac{1}{\sqrt{2}} (1, 𝑖)$ and $\frac{1}{\sqrt{2}} (𝑖, 1)$ have squared length $(1 + 1) / 2 = 1$ , and their dot product is $\frac{1}{2} (1 \cdot 𝑖 + 𝑖 \cdot 1) =$ $\frac{1}{2} (- 𝑖 + 𝑖) =$ $0$ . So $𝑇$ is an isometry of $ℂ^{2}$ , hence unitary (Theorem 8.4.11).

(c) $(𝑇 𝑝) (0) = 𝑝 (1)$ and $(𝑇 𝑝) (1) = 𝑝 (0)$ , so ${‖ 𝑇 𝑝 ‖}^{2} =$ ${𝑝 (1)}^{2} +$ ${𝑝 (0)}^{2} =$ ${‖ 𝑝 ‖}^{2}$ . So $𝑇$ is an isometry of the finite-dimensional space $𝑉$ , hence unitary.

(d) For $𝑝 = 𝑥$ , ${‖ 𝑝 ‖}^{2} = 0^{2} + 1^{2} = 1$ , while $𝑇 𝑝 = 𝑥 / 2$ has ${‖ 𝑇 𝑝 ‖}^{2} = 0 + 1 / 4$ . So $𝑇$ is not an isometry.

(e) $𝑇$ is linear, and by Proposition 6.1.16, ${‖ 𝐴^{𝖳} ‖}^{2} =$ $\sum_{𝑖, 𝑗} {| 𝑎_{𝑗 𝑖} |}^{2} =$ $\sum_{𝑖, 𝑗} {| 𝑎_{𝑖 𝑗} |}^{2} =$ ${‖ 𝐴 ‖}^{2}$ . So $𝑇$ is an isometry of a finite-dimensional space, hence unitary. (It also satisfies $𝑇^{2} = id$ , so $𝑇^{- 1} = 𝑇 = 𝑇^{*}$ : it is self-adjoint as well.)

Exercise 8.4.33.standard Prove or disprove each statement. Here $𝑉$ and $𝑊$ are finite-dimensional.

(a)If $𝑆 \in L (𝑉, 𝑊)$ is an isometry, then so is $𝑆^{*}$ .
(b)If $𝑉$ is complex and every eigenvalue of $𝑇 \in L (𝑉)$ has absolute value $1$ , then $𝑇$ is unitary.
(c)If $𝑉$ is complex, $𝑇 \in L (𝑉)$ is normal, and $‖ 𝑇 𝐯 ‖ = ‖ 𝐯 ‖$ for every eigenvector $𝐯$ of $𝑇$ , then $𝑇$ is unitary.
(d)Statement (c) with “complex” replaced by “real”.

Solution.

Uses. Corollary 8.4.6 (Properties of isometries), Theorem 8.4.15 (Spectral form of unitary operators), Theorem 8.2.5 (Complex spectral theorem), Example 8.1.9 (Normal operators)

Hint 1. (a) An isometry

ℝ \to ℝ^{2}

cannot have an injective adjoint. (b) Look for a non-normal operator with a single eigenvalue

1

Hint 2. (c) The spectral theorem supplies an orthonormal basis of eigenvectors. (d) A real normal operator may have no eigenvectors at all.

Sketch. (a) False:

𝑆 𝑥 = (𝑥, 0)

has

𝑆^{*} (𝑦_{1}, 𝑦_{2}) = 𝑦_{1}

, which kills

𝐞_{2}

. (b) False: the shear. (c) True: an orthonormal eigenbasis exists, and

‖ 𝑇 𝐞_{𝑗} ‖ = | 𝜆_{𝑗} | = 1

, so (III) of Theorem 8.4.15 holds. (d) False:

(\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix})

is normal, has no real eigenvalue, and is not orthogonal.

(a) False. Let $𝑆 : ℝ \to ℝ^{2}$ , $𝑆 𝑥 = (𝑥, 0)$ , an isometry. Its adjoint is $𝑆^{*} (𝑦_{1}, 𝑦_{2}) = 𝑦_{1}$ , since $⟨ 𝑆 𝑥, (𝑦_{1}, 𝑦_{2}) ⟩ =$ $𝑥 𝑦_{1}$ . Then $𝑆^{*} 𝐞_{2} = 0$ , so $𝑆^{*}$ is not injective and not an isometry (Corollary 8.4.6(a)). The statement is true when $dim 𝑉 = dim 𝑊$ , by Corollary 8.4.6(d).

(b) False. The shear $(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ on $ℂ^{2}$ has the single eigenvalue $1$ but maps $𝐞_{2}$ to $(1, 1)$ , of length $\sqrt{2}$ .

(c) True. By Theorem 8.2.5, $𝑉$ has an orthonormal basis $𝐞_{1}, \dots, 𝐞_{𝑛}$ with $𝑇 𝐞_{𝑗} = 𝜆_{𝑗} 𝐞_{𝑗}$ . By hypothesis $1 = ‖ 𝑇 𝐞_{𝑗} ‖ = | 𝜆_{𝑗} |$ , and every eigenvalue of $𝑇$ is one of the $𝜆_{𝑗}$ . So $𝑇$ is unitary by (III) $\Rightarrow$ (I) of Theorem 8.4.15.

(d) False. Let $𝑇 \in L (ℝ^{2})$ have matrix $𝐴 = (\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix})$ . It is normal, since $𝐴^{𝖳} 𝐴 =$ $4 𝐼_{2} =$ $𝐴 𝐴^{𝖳}$ (Example 8.1.9(c)). Its characteristic polynomial $𝑥^{2} + 4$ has no real root, so $𝑇$ has no eigenvectors and the hypothesis holds vacuously. But $‖ 𝑇 𝐞_{1} ‖ = 2$ , so $𝑇$ is not orthogonal. In (c), the eigenvectors of a complex normal operator span $𝑉$ ; over $ℝ$ they need not exist.

Exercise 8.4.34.standard Let

𝑉

and

𝑊

be finite-dimensional, and let

𝑇 \in L (𝑉, 𝑊)

preserve orthogonality:

⟨ 𝑇 𝐮, 𝑇 𝐯 ⟩ = 0

whenever

⟨ 𝐮, 𝐯 ⟩ = 0

. Show that either

𝑇 = 0

𝑇 = 𝑐 𝑆

for a real number

𝑐 > 0

and an isometry

𝑆

. (So the maps that preserve right angles are exactly the scalar multiples of isometries.)

Solution.

Uses. Theorem 8.4.4 (Characterizations of isometries), Theorem 6.4.20 (Orthonormal bases exist)

Techniques. Plug in clever test vectors (Technique 6.1.40)

Hint 1. Fix an orthonormal basis

𝐞_{1}, \dots, 𝐞_{𝑛}

. The images are orthogonal. You want them to have equal lengths.

Hint 2. For

𝑖 \neq 𝑗

, the vectors

𝐞_{𝑖} + 𝐞_{𝑗}

and

𝐞_{𝑖} - 𝐞_{𝑗}

are orthogonal. Apply the hypothesis to them.

Sketch. The images

𝑇 𝐞_{𝑖}

are pairwise orthogonal. From

(𝐞_{𝑖} + 𝐞_{𝑗}) ⟂ (𝐞_{𝑖} - 𝐞_{𝑗})

we get

0 =

⟨ 𝑇 𝐞_{𝑖} + 𝑇 𝐞_{𝑗}, 𝑇 𝐞_{𝑖} - 𝑇 𝐞_{𝑗} ⟩ =

{‖ 𝑇 𝐞_{𝑖} ‖}^{2} -

{‖ 𝑇 𝐞_{𝑗} ‖}^{2}

. So all

‖ 𝑇 𝐞_{𝑖} ‖

equal some

𝑐

; if

𝑐 > 0

𝑇 / 𝑐

maps an orthonormal basis to an orthonormal list.

If $𝑉 = {𝟎}$ , then $𝑇 = 0$ . Otherwise let $𝐞_{1}, \dots, 𝐞_{𝑛}$ be an orthonormal basis of $𝑉$ (Theorem 6.4.20). For $𝑖 \neq 𝑗$ , $⟨ 𝐞_{𝑖}, 𝐞_{𝑗} ⟩ = 0$ , so $⟨ 𝑇 𝐞_{𝑖}, 𝑇 𝐞_{𝑗} ⟩ = 0$ by hypothesis. Also

⟨ 𝐞_{𝑖} + 𝐞_{𝑗}, 𝐞_{𝑖} - 𝐞_{𝑗} ⟩ = {‖ 𝐞_{𝑖} ‖}^{2} - ⟨ 𝐞_{𝑖}, 𝐞_{𝑗} ⟩ + ⟨ 𝐞_{𝑗}, 𝐞_{𝑖} ⟩ - {‖ 𝐞_{𝑗} ‖}^{2} = 1 - 0 + 0 - 1 = 0,

so by hypothesis and the orthogonality of $𝑇 𝐞_{𝑖}$ and $𝑇 𝐞_{𝑗}$ ,

0 = ⟨ 𝑇 𝐞_{𝑖} + 𝑇 𝐞_{𝑗}, 𝑇 𝐞_{𝑖} - 𝑇 𝐞_{𝑗} ⟩ = {‖ 𝑇 𝐞_{𝑖} ‖}^{2} - {‖ 𝑇 𝐞_{𝑗} ‖}^{2} .

So $‖ 𝑇 𝐞_{𝑖} ‖ = 𝑐$ for one $𝑐 \geq 0$ and all $𝑖$ . If $𝑐 = 0$ , then $𝑇$ vanishes on a basis and $𝑇 = 0$ . If $𝑐 > 0$ , let $𝑆 = 𝑐^{- 1} 𝑇$ . Then $𝑆 𝐞_{1}, \dots, 𝑆 𝐞_{𝑛}$ are pairwise orthogonal unit vectors, so $𝑆$ is an isometry by (V) $\Rightarrow$ (I) of Theorem 8.4.4, and $𝑇 = 𝑐 𝑆$ . Conversely, $𝑐 𝑆$ preserves orthogonality since $⟨ 𝑐 𝑆 𝐮, 𝑐 𝑆 𝐯 ⟩ =$ $𝑐^{2} ⟨ 𝐮, 𝐯 ⟩$ .

Exercise 8.4.35.standard

(a)Show that an orthogonal matrix $𝑄 \in ℝ^{𝑛 \times 𝑛}$ with integer entries is a signed permutation matrix: each row and each column has exactly one nonzero entry, which is $1$ or $- 1$ .
(b)Count the orthogonal $𝑛 \times 𝑛$ matrices with integer entries, and count those with determinant $1$ (for $𝑛 \geq 1$ ).

Solution.

Uses. Definition 6.3.33 (Unitary and orthogonal matrices), Theorem 6.3.29 (Orthonormal columns), Definition 5.2.29 (Permutation matrix), Theorem 5.4.10 (Multiplicativity)

Techniques. A sum of nonnegative terms (Technique 6.1.9)

Hint 1. A column is a unit vector with integer entries. How can integers have squares adding up to

1

Hint 2. Orthogonal columns with one nonzero entry each must have it in different rows. For the determinant, multiplying the first row by

- 1

is a bijection between the two kinds.

Sketch. Each column has integer entries with

\sum 𝑞_{𝑖 𝑗}^{2} = 1

, so exactly one entry is

\pm 1

. Two columns with their nonzero entries in the same row would have dot product

\pm 1

, so the rows of these entries are distinct:

𝑄 = 𝐷 𝑃

with

𝑃

a permutation matrix and

𝐷

diagonal with entries

\pm 1

. Count:

𝑛!

permutations times

2^{𝑛}

signs. Negating the first row is a bijection between determinant

1

and

- 1

, so

2^{𝑛 - 1} 𝑛!

have determinant

1

(a) Let $𝐪_{𝑗}$ be the $𝑗$ th column. It is a unit vector (Theorem 6.3.29), so $\sum_{𝑖} 𝑞_{𝑖 𝑗}^{2} = 1$ with every $𝑞_{𝑖 𝑗} \in ℤ$ . A nonzero integer has square at least $1$ , so exactly one entry of $𝐪_{𝑗}$ is nonzero, and it is $\pm 1$ ; say it is in row $𝜎 (𝑗)$ . If $𝜎 (𝑗) = 𝜎 (𝑘)$ with $𝑗 \neq 𝑘$ , then $𝐪_{𝑗} \cdot 𝐪_{𝑘} = \pm 1 \neq 0$ , contradicting orthogonality. So $𝜎$ is injective, hence a permutation of ${1, \dots, 𝑛}$ , and every row also contains exactly one nonzero entry.

(b) Conversely, every signed permutation matrix has orthonormal columns (they are $\pm 𝐞_{𝜎 (𝑗)}$ for distinct $𝜎 (𝑗)$ ), so it is orthogonal with integer entries. Such a matrix is determined by the permutation $𝜎$ ( $𝑛!$ choices) and the signs of its $𝑛$ nonzero entries ( $2^{𝑛}$ choices), so there are $2^{𝑛} 𝑛!$ of them. Let $𝐷 = diag (- 1, 1, \dots, 1)$ . The map $𝑄 \mapsto 𝐷 𝑄$ negates the first row, so it maps signed permutation matrices to signed permutation matrices, it is its own inverse, and $det (𝐷 𝑄) = - det 𝑄$ (Theorem 5.4.10). So it exchanges those of determinant $1$ and those of determinant $- 1$ (the only possible values), and exactly half, $2^{𝑛 - 1} 𝑛!$ , have determinant $1$ .

Exercise 8.4.36.standard Let $𝑄 =$ $(𝑞_{𝑖 𝑗}) \in$ $ℝ^{𝑛 \times 𝑛}$ be orthogonal.

(a)Show that $| \sum_{𝑖, 𝑗} 𝑞_{𝑖 𝑗} | \leq 𝑛$ , and that the sum equals $𝑛$ if and only if every row of $𝑄$ sums to $1$ .
(b)Show that $\sum_{𝑖, 𝑗} | 𝑞_{𝑖 𝑗} | \leq$ $𝑛 \sqrt{𝑛}$ , with equality if and only if every $| 𝑞_{𝑖 𝑗} | = 1 / \sqrt{𝑛}$ . Give an example of equality for $𝑛 = 2$ .

Solution.

Uses. Theorem 6.2.1 (Cauchy–Schwarz inequality), Theorem 8.4.11 (Characterizations of unitary operators), Definition 6.3.33 (Unitary and orthogonal matrices), Proposition 6.3.34 (Characterizations of unitary matrices)

Techniques. Apply Cauchy–Schwarz to a well-chosen inner product (Technique 6.2.21)

Hint 1. The sum of all entries is

𝟏^{𝖳} 𝑄 𝟏 =

⟨ 𝑄 𝟏, 𝟏 ⟩

, where

𝟏 = (1, \dots, 1)

Hint 2. Apply Cauchy–Schwarz, and use

‖ 𝑄 𝟏 ‖ = ‖ 𝟏 ‖ = \sqrt{𝑛}

. For (b), apply Cauchy–Schwarz to each row, a unit vector, and the vector

𝟏

Sketch. (a)

\sum 𝑞_{𝑖 𝑗} = ⟨ 𝑄 𝟏, 𝟏 ⟩

, and

| ⟨ 𝑄 𝟏, 𝟏 ⟩ | \leq

‖ 𝑄 𝟏 ‖ ‖ 𝟏 ‖ =

𝑛

. Equality

= 𝑛

holds iff

𝑄 𝟏

is a nonnegative multiple of

𝟏

of the same length, i.e.

𝑄 𝟏 = 𝟏

. (b) For row

𝑖

\sum_{𝑗} | 𝑞_{𝑖 𝑗} | =

(| 𝑞_{𝑖 1} |, \dots) \cdot

𝟏 \leq

1 \cdot

\sqrt{𝑛}

, with equality iff all

| 𝑞_{𝑖 𝑗} |

are equal. Example:

\frac{1}{\sqrt{2}} (\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix})

Let $𝟏 = (1, \dots, 1) \in ℝ^{𝑛}$ , so $‖ 𝟏 ‖ = \sqrt{𝑛}$ , and $‖ 𝑄 𝟏 ‖ = \sqrt{𝑛}$ since $𝑄$ is orthogonal.

(a) The $𝑖$ th entry of $𝑄 𝟏$ is the $𝑖$ th row sum, so $\sum_{𝑖, 𝑗} 𝑞_{𝑖 𝑗} =$ $(𝑄 𝟏) \cdot$ $𝟏$ . By Cauchy–Schwarz (Theorem 6.2.1),

| \sum_{𝑖, 𝑗} 𝑞_{𝑖 𝑗} | = | (𝑄 𝟏) \cdot 𝟏 | \leq ‖ 𝑄 𝟏 ‖ ‖ 𝟏 ‖ = 𝑛 .

If every row sums to $1$ , then $𝑄 𝟏 = 𝟏$ and the sum is $𝟏 \cdot 𝟏 = 𝑛$ . Conversely, if the sum is $𝑛 = ‖ 𝑄 𝟏 ‖ ‖ 𝟏 ‖$ , then by Theorem 6.2.11(b) (the case of equality $⟨ 𝐱, 𝐲 ⟩ = ‖ 𝐱 ‖ ‖ 𝐲 ‖$ ), one of $𝑄 𝟏$ and $𝟏$ is a nonnegative multiple of the other. They have the same nonzero length, so $𝑄 𝟏 = 𝟏$ : every row sums to $1$ .

(b) Let $𝐫_{𝑖}$ be the $𝑖$ th row of $𝑄$ , a unit vector by Proposition 6.3.34, and let $𝐚_{𝑖} =$ $(| 𝑞_{𝑖 1} |, \dots, | 𝑞_{𝑖 𝑛} |)$ , which has the same length. By Cauchy–Schwarz,

\sum_{𝑗} | 𝑞_{𝑖 𝑗} | = 𝐚_{𝑖} \cdot 𝟏 \leq ‖ 𝐚_{𝑖} ‖ ‖ 𝟏 ‖ = \sqrt{𝑛},

and summing over the $𝑛$ rows gives $\sum_{𝑖, 𝑗} | 𝑞_{𝑖 𝑗} | \leq$ $𝑛 \sqrt{𝑛}$ . Equality holds if and only if it holds for every row, that is (as $𝐚_{𝑖} \neq 𝟎$ and both vectors have nonnegative entries), $𝐚_{𝑖}$ is a positive multiple of $𝟏$ for each $𝑖$ . Since $‖ 𝐚_{𝑖} ‖ = 1$ , this says $| 𝑞_{𝑖 𝑗} | = 1 / \sqrt{𝑛}$ for all $𝑖, 𝑗$ . For $𝑛 = 2$ , the matrix $\frac{1}{\sqrt{2}} (\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix})$ is orthogonal with all $| 𝑞_{𝑖 𝑗} | = 1 / \sqrt{2}$ , and its entries have absolute values adding up to $4 / \sqrt{2} = 2 \sqrt{2}$ .

Exercise 8.4.37.standard Let

𝑈 \in ℂ^{𝑛 \times 𝑛}

be unitary. Show that

| tr 𝑈 | \leq 𝑛

, with equality if and only if

𝑈 = 𝑐 𝐼_{𝑛}

for some

𝑐 \in ℂ

with

| 𝑐 | = 1

. Deduce that a real orthogonal

𝑄

with

tr 𝑄 = 𝑛

𝐼_{𝑛}

Solution.

Uses. Theorem 8.4.15 (Spectral form of unitary operators), Theorem 7.5.22 (Trace and determinant from eigenvalues), Theorem B.1.11 (Triangle inequality in $ℂ$ )

Hint 1. The trace is the sum of the eigenvalues, each on the unit circle.

Hint 2. For equality, the triangle inequality in

ℂ

forces all

𝜆_{𝑗}

to point in the same direction. Then use the spectral form.

Sketch.

tr 𝑈 = \sum 𝜆_{𝑗}

with

| 𝜆_{𝑗} | = 1

, so

| tr 𝑈 | \leq 𝑛

. Equality in the triangle inequality for complex numbers of equal modulus forces all

𝜆_{𝑗}

equal to one

𝑐

; then

𝑈 =

𝑊 (𝑐 𝐼_{𝑛}) 𝑊^{*} =

𝑐 𝐼_{𝑛}

. For real

𝑄

tr 𝑄 = 𝑛

gives

𝑐 = 1

By Theorem 8.4.15, $𝑈 = 𝑊 𝐷 𝑊^{*}$ with $𝑊$ unitary and $𝐷 =$ $diag (𝜆_{1}, \dots, 𝜆_{𝑛})$ , $| 𝜆_{𝑗} | = 1$ ; and $tr 𝑈 = 𝜆_{1} + \dots + 𝜆_{𝑛}$ (Theorem 7.5.22). By the triangle inequality in $ℂ$ (Theorem B.1.11),

| tr 𝑈 | \leq | 𝜆_{1} | + \dots + | 𝜆_{𝑛} | = 𝑛 .

If $𝑈 = 𝑐 𝐼_{𝑛}$ with $| 𝑐 | = 1$ , then $| tr 𝑈 | = 𝑛 | 𝑐 | = 𝑛$ . Conversely, if $| tr 𝑈 | = 𝑛$ , equality holds in the triangle inequality, so by its equality case all nonzero $𝜆_{𝑗}$ are positive real multiples of one another. They all have absolute value $1$ , so they are all equal to one $𝑐$ with $| 𝑐 | = 1$ . Then $𝐷 = 𝑐 𝐼_{𝑛}$ and $𝑈 =$ $𝑊 (𝑐 𝐼_{𝑛}) 𝑊^{*} =$ $𝑐 𝑊 𝑊^{*} =$ $𝑐 𝐼_{𝑛}$ .

If $𝑄$ is real orthogonal with $tr 𝑄 = 𝑛$ , it is unitary as a complex matrix, and $| tr 𝑄 | = 𝑛$ , so $𝑄 = 𝑐 𝐼_{𝑛}$ with $𝑛 𝑐 = tr 𝑄 = 𝑛$ , that is, $𝑄 = 𝐼_{𝑛}$ .

Exercise 8.4.38.hard Let $𝑉$ be a finite-dimensional complex inner product space and $𝑆 \in L (𝑉)$ unitary.

(a)Show that there is a self-adjoint $𝐴 \in L (𝑉)$ , with all eigenvalues in $[0, 2 𝜋)$ , such that $𝑆 = 𝑓 (𝐴)$ for $𝑓 (𝑡) = 𝑒^{𝑖 𝑡}$ (Definition 8.2.23).
(b)Show that $𝑆$ has a unitary square root: a unitary $𝑅$ with $𝑅^{2} = 𝑆$ .
(c)Suppose $dim 𝑉 = 𝑛$ and $𝑆$ has $𝑛$ distinct eigenvalues. Show that $𝑆$ has exactly $2^{𝑛}$ square roots $𝑅 \in L (𝑉)$ , and that all of them are unitary.

Solution.

Uses. Theorem 8.4.15 (Spectral form of unitary operators), Theorem 8.2.18 (Spectral decomposition), Definition 8.2.23 (Functions of an operator), Proposition 8.2.24 (Properties of $𝑓 (𝑇)$ ), Theorem 2.2.1 (Linear maps are determined by a basis), Definition 7.2.13 (Eigenspace)

Techniques. Apply a function to the eigenvalues (Technique 8.2.25)

Hint 1. Write

𝑆 = \sum_{𝑗} 𝜆_{𝑗} 𝑃_{𝑗}

with distinct

𝜆_{𝑗} = 𝑒^{𝑖 𝜃_{𝑗}}

𝜃_{𝑗} \in [0, 2 𝜋)

. Take

𝐴 = \sum_{𝑗} 𝜃_{𝑗} 𝑃_{𝑗}

Hint 2. For (b), take square roots of the eigenvalues:

𝑅 =

\sum_{𝑗} 𝑒^{𝑖 𝜃_{𝑗} / 2} 𝑃_{𝑗}

Hint 3. For (c), any square root

𝑅

commutes with

𝑆 = 𝑅^{2}

, so it maps each eigenline of

𝑆

into itself.

Sketch. (a)

𝐴 = \sum 𝜃_{𝑗} 𝑃_{𝑗}

is self-adjoint with the same eigenspaces, and

𝑓 (𝐴) =

\sum 𝑒^{𝑖 𝜃_{𝑗}} 𝑃_{𝑗} =

𝑆

. (b)

𝑅 =

\sum 𝑒^{𝑖 𝜃_{𝑗} / 2} 𝑃_{𝑗}

is unitary by Theorem 8.4.15 and

𝑅^{2} = 𝑆

by Proposition 8.2.24. (c)

𝑅 𝑆 = 𝑅^{3} = 𝑆 𝑅

, so

𝑅

preserves each eigenline

span (𝐞_{𝑗})

𝑅 𝐞_{𝑗} = 𝑟_{𝑗} 𝐞_{𝑗}

with

𝑟_{𝑗}^{2} = 𝜆_{𝑗}

, two choices each. Each of the

2^{𝑛}

choices defines a square root with an orthonormal eigenbasis and

| 𝑟_{𝑗} | = 1

, hence unitary.

(a) By Theorem 8.4.15 and Theorem 8.2.18, $𝑆 =$ $\sum_{𝑗 = 1}^{𝑘} 𝜆_{𝑗} 𝑃_{𝑗}$ , where $𝜆_{1}, \dots, 𝜆_{𝑘}$ are the distinct eigenvalues, $𝑃_{𝑗} = {Pr}_{𝐸_{𝜆_{𝑗}}}$ , and $| 𝜆_{𝑗} | = 1$ . Write $𝜆_{𝑗} = 𝑒^{𝑖 𝜃_{𝑗}}$ with $𝜃_{𝑗} \in [0, 2 𝜋)$ ; the $𝜃_{𝑗}$ are distinct since the $𝜆_{𝑗}$ are. Let $𝐴 = \sum_{𝑗} 𝜃_{𝑗} 𝑃_{𝑗}$ . By Proposition 8.2.24 (applied to $𝑆$ and the real-valued function $𝜆_{𝑗} \mapsto 𝜃_{𝑗}$ ), $𝐴$ is self-adjoint, its distinct eigenvalues are $𝜃_{1}, \dots, 𝜃_{𝑘}$ , and its eigenspaces are $𝐸_{𝜆_{𝑗}} (𝑆)$ . So by Definition 8.2.23 applied to $𝐴$ , $𝑓 (𝐴) =$ $\sum_{𝑗} 𝑒^{𝑖 𝜃_{𝑗}} 𝑃_{𝑗} =$ $𝑆$ .

(b) Let $𝑅 =$ $\sum_{𝑗} 𝑒^{𝑖 𝜃_{𝑗} / 2} 𝑃_{𝑗}$ , the function $𝜆_{𝑗} \mapsto 𝑒^{𝑖 𝜃_{𝑗} / 2}$ applied to $𝑆$ . By Proposition 8.2.24, $𝑅^{2} =$ $\sum_{𝑗} {(𝑒^{𝑖 𝜃_{𝑗} / 2})}^{2} 𝑃_{𝑗} =$ $𝑆$ , and $𝑅$ is normal with eigenvalues $𝑒^{𝑖 𝜃_{𝑗} / 2}$ of absolute value $1$ , so $𝑅$ is unitary by Theorem 8.4.15.

(c) Let $𝐞_{1}, \dots, 𝐞_{𝑛}$ be an orthonormal eigenbasis of $𝑆$ with $𝑆 𝐞_{𝑗} = 𝜆_{𝑗} 𝐞_{𝑗}$ and distinct $𝜆_{𝑗}$ . Suppose $𝑅^{2} = 𝑆$ . Then $𝑅 𝑆 = 𝑅^{3} = 𝑆 𝑅$ , so $𝑆 (𝑅 𝐞_{𝑗}) =$ $𝑅 𝑆 𝐞_{𝑗} =$ $𝜆_{𝑗} 𝑅 𝐞_{𝑗}$ . Thus $𝑅 𝐞_{𝑗} \in$ $𝐸_{𝜆_{𝑗}} (𝑆) =$ $span (𝐞_{𝑗})$ , which is one-dimensional because the eigenvalues are distinct; say $𝑅 𝐞_{𝑗} = 𝑟_{𝑗} 𝐞_{𝑗}$ . Then $𝜆_{𝑗} 𝐞_{𝑗} =$ $𝑅^{2} 𝐞_{𝑗} =$ $𝑟_{𝑗}^{2} 𝐞_{𝑗}$ , so $𝑟_{𝑗}$ is one of the two square roots of $𝜆_{𝑗}$ . Conversely, for each of the $2^{𝑛}$ choices of square roots $𝑟_{1}, \dots, 𝑟_{𝑛}$ , Theorem 2.2.1 gives exactly one $𝑅$ with $𝑅 𝐞_{𝑗} = 𝑟_{𝑗} 𝐞_{𝑗}$ , and $𝑅^{2} 𝐞_{𝑗} =$ $𝑟_{𝑗}^{2} 𝐞_{𝑗} =$ $𝑆 𝐞_{𝑗}$ for all $𝑗$ , so $𝑅^{2} = 𝑆$ . Different choices give different $𝑅$ , so there are exactly $2^{𝑛}$ square roots. Each has an orthonormal eigenbasis with eigenvalues $𝑟_{𝑗}$ , where $| 𝑟_{𝑗} | =$ ${| 𝜆_{𝑗} |}^{1 / 2} =$ $1$ , so each is unitary by Theorem 8.4.15.

Exercise 8.4.39.standard A student argues: “Let $𝑆$ be an orthogonal operator on a finite-dimensional real inner product space. Then $𝑆^{*} 𝑆 = 𝑆 𝑆^{*} = id$ , so $𝑆$ is normal, and by the spectral theorem $𝑉$ has an orthonormal basis of eigenvectors of $𝑆$ .”

(a)Find the error, and give a counterexample to the conclusion.
(b)Show that an orthogonal $𝑆$ has an orthonormal eigenbasis if and only if $𝑆^{2} = id$ , and describe such operators geometrically.

Solution.

Uses. Theorem 8.2.12 (Real spectral theorem), Theorem 8.2.5 (Complex spectral theorem), Proposition 8.4.10 (Unitary means $𝑆^{*} = 𝑆^{- 1}$ ), Corollary 8.4.13 (Eigenvalues and determinant), Proposition 8.2.2 (Necessary conditions)

Hint 1. Which spectral theorem applies over

ℝ

, and what is its hypothesis?

Hint 2. For (b), the real spectral theorem says: orthonormal eigenbasis iff self-adjoint. For orthogonal

𝑆

𝑆^{*} = 𝑆^{- 1}

Sketch. (a) Over

ℝ

the spectral theorem needs self-adjointness, not normality;

𝑅_{𝜋 / 2}

is orthogonal with no eigenvectors. (b)

𝑆

has an orthonormal eigenbasis iff

𝑆 = 𝑆^{*}

(Theorem 8.2.12) iff

𝑆 = 𝑆^{- 1}

iff

𝑆^{2} = id

. Such

𝑆

fixes

𝐸_{1}

and negates

𝐸_{- 1} = 𝐸_{1}^{⟂}

: it is the reflection through

𝐸_{1}

(a) The normal spectral theorem (Theorem 8.2.5) is a theorem about complex spaces. Over $ℝ$ the spectral theorem (Theorem 8.2.12) requires $𝑆$ to be self-adjoint, and an orthogonal operator need not be. The rotation $𝑅_{𝜋 / 2} = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ is orthogonal, but its characteristic polynomial $𝑥^{2} + 1$ has no real root, so it has no eigenvectors at all.

(b) By Theorem 8.2.12 (with Proposition 8.2.2 for the forward direction), $𝑆$ has an orthonormal eigenbasis if and only if $𝑆^{*} = 𝑆$ . Since $𝑆^{*} = 𝑆^{- 1}$ (Proposition 8.4.10), this says $𝑆 = 𝑆^{- 1}$ , that is, $𝑆^{2} = id$ . In that case the eigenvalues are real of absolute value $1$ (Corollary 8.4.13(a)), so $𝑉 = 𝐸_{1} \oplus 𝐸_{- 1}$ with the two eigenspaces orthogonal. So $𝑆$ fixes the subspace $𝐸_{1}$ and negates its orthogonal complement $𝐸_{- 1}$ ; it is the reflection through $𝐸_{1}$ , that is, $𝑆 = 2 {Pr}_{𝐸_{1}} - id$ (compare Exercise 8.1.30).

Exercise 8.4.40.standardBeyond 21-242 Let $𝑀$ be multiplication by $𝑥$ on $𝔽 [𝑥]$ , with the inner product of Example 8.4.12, and define $𝐿 \in L (𝔽 [𝑥])$ by

𝐿 (𝑎_{0} + 𝑎_{1} 𝑥 + \dots + 𝑎_{𝑑} 𝑥^{𝑑}) = 𝑎_{1} + 𝑎_{2} 𝑥 + \dots + 𝑎_{𝑑} 𝑥^{𝑑 - 1} .

(a)Show that $𝐿$ is the adjoint of $𝑀$ .
(b)Show that $𝐿 𝑀 = id$ and $𝑀 𝐿 = id - {Pr}_{span (1)}$ .
(c)Show that $𝑀$ has no eigenvalues, and that the only eigenvalue of $𝐿$ is $0$ .

Solution.

Uses. Example 8.4.12 (The shift), Definition 6.6.8 (Adjoint), Definition 7.2.6 (Eigenvalue, eigenvector), Definition 6.4.11 (Orthogonal projection)

Hint 1. Write

𝑝 = \sum 𝑎_{𝑘} 𝑥^{𝑘}

and

𝑞 = \sum 𝑏_{𝑘} 𝑥^{𝑘}

, and compute

⟨ 𝑥 𝑝, 𝑞 ⟩

and

⟨ 𝑝, 𝐿 𝑞 ⟩

coefficient by coefficient.

Hint 2. For the eigenvalues, compare degrees:

𝑀

raises the degree of a nonzero polynomial by

1

, and

𝐿

lowers it by

1

(or kills constants).

Sketch. (a)

⟨ 𝑀 𝑝, 𝑞 ⟩ =

\sum_{𝑘} 𝑎_{𝑘} 𝑏_{𝑘 + 1} =

⟨ 𝑝, 𝐿 𝑞 ⟩

. (b)

𝐿 (𝑥 𝑝) = 𝑝

;

𝑥 𝐿 (𝑞) = 𝑞 - 𝑏_{0}

, and

𝑏_{0} = {Pr}_{span (1)} 𝑞

since

1, 𝑥, 𝑥^{2}, \dots

is orthonormal. (c)

deg (𝑥 𝑝) = deg 𝑝 + 1

, so

𝑥 𝑝 = 𝜆 𝑝

is impossible for

𝑝 \neq 𝟎

. If

𝐿 𝑞 = 𝜆 𝑞

with

𝑞 \neq 𝟎

of degree

𝑑

: for

𝑑 \geq 1

deg 𝐿 𝑞 =

𝑑 -

1 <

𝑑

forces

𝜆 = 0

and

𝐿 𝑞 = 0

, a contradiction; so

𝑞

is a nonzero constant,

𝐿 𝑞 = 0

, and

𝜆 = 0

(a) Let $𝑝 = \sum_{𝑘} 𝑎_{𝑘} 𝑥^{𝑘}$ and $𝑞 = \sum_{𝑘} 𝑏_{𝑘} 𝑥^{𝑘}$ (finite sums, padded with zeros). Then $𝑀 𝑝 =$ $\sum_{𝑘} 𝑎_{𝑘} 𝑥^{𝑘 + 1}$ and $𝐿 𝑞 =$ $\sum_{𝑘} 𝑏_{𝑘 + 1} 𝑥^{𝑘}$ , so

⟨ 𝑀 𝑝, 𝑞 ⟩ = \sum_{𝑘 \geq 0} 𝑎_{𝑘} 𝑏_{𝑘 + 1} = ⟨ 𝑝, 𝐿 𝑞 ⟩ .

So $𝐿$ is the adjoint of $𝑀$ (Definition 6.6.8).

(b) $𝐿 (𝑀 𝑝) =$ $𝐿 (\sum_{𝑘} 𝑎_{𝑘} 𝑥^{𝑘 + 1}) =$ $\sum_{𝑘} 𝑎_{𝑘} 𝑥^{𝑘} =$ $𝑝$ , so $𝐿 𝑀 = id$ . For $𝑞$ as above, $𝑀 𝐿 𝑞 =$ $\sum_{𝑘 \geq 1} 𝑏_{𝑘} 𝑥^{𝑘} =$ $𝑞 -$ $𝑏_{0}$ . Since $1, 𝑥, 𝑥^{2}, \dots$ is orthonormal, $𝑏_{0} = ⟨ 𝑞, 1 ⟩$ , so $𝑏_{0} = {Pr}_{span (1)} 𝑞$ (Definition 6.4.11), and $𝑀 𝐿 = id - {Pr}_{span (1)}$ . In particular $𝑀 𝐿 \neq id$ , as $𝑀 𝐿 1 = 𝟎$ : $𝐿$ is a left inverse of $𝑀$ but not a right inverse.

(c) If $𝑝 \neq 𝟎$ , then $deg (𝑥 𝑝) = deg 𝑝 + 1$ , while $𝜆 𝑝$ is $𝟎$ or has degree $deg 𝑝$ . So $𝑀 𝑝 = 𝜆 𝑝$ never holds with $𝑝 \neq 𝟎$ , and $𝑀$ has no eigenvalue. For $𝐿$ : the constant $1$ satisfies $𝐿 1 = 𝟎 = 0 \cdot 1$ , so $0$ is an eigenvalue. Suppose $𝐿 𝑞 = 𝜆 𝑞$ with $𝑞 \neq 𝟎$ . If $deg 𝑞 = 𝑑 \geq 1$ , then $𝐿 𝑞$ is $𝟎$ or has degree $𝑑 - 1$ , so $𝜆 𝑞$ cannot have degree $𝑑$ ; hence $𝜆 𝑞 = 𝟎$ , $𝜆 = 0$ , and $𝐿 𝑞 = 𝟎$ , which forces $𝑞$ to be constant, contradicting $𝑑 \geq 1$ . So $𝑞$ is a nonzero constant, $𝐿 𝑞 = 𝟎$ , and $𝜆 = 0$ .

Exercise 8.4.41.routineBeyond 21-242 Show that

𝑄 = \frac{1}{9} (\begin{matrix} 1 & - 4 & 8 \\ 8 & 4 & 1 \\ - 4 & 7 & 4 \end{matrix})

is a rotation of $ℝ^{3}$ , and find its axis and its angle.

Solution.

Uses. Definition 8.4.26 (Rotation), Theorem 8.4.27 (Euler’s rotation theorem), Theorem 6.3.29 (Orthonormal columns)

Hint 1. Check that the columns are orthonormal and that

det 𝑄 = 1

Hint 2. The axis spans

ker (𝑄 - 𝐼_{3})

. The angle comes from

tr 𝑄 = 1 + 2 cos 𝜃

Sketch. The columns

\frac{1}{9} (1, 8, - 4)

\frac{1}{9} (- 4, 4, 7)

\frac{1}{9} (8, 1, 4)

are orthonormal and

det (9 𝑄) = 729

, so

det 𝑄 = 1

𝑄 (1, 2, 2) = (1, 2, 2)

, so the axis is

\frac{1}{3} (1, 2, 2)

, and

tr 𝑄 = 1 = 1 + 2 cos 𝜃

gives

𝜃 = 𝜋 / 2

The columns of $9 𝑄$ are $(1, 8, - 4)$ , $(- 4, 4, 7)$ , $(8, 1, 4)$ , each of squared length $81$ , with dot products $- 4 + 32 - 28 = 0$ , $8 + 8 - 16 = 0$ , and $- 32 + 4 + 28 = 0$ . So the columns of $𝑄$ are orthonormal and $𝑄$ is orthogonal. Expanding along the first row,

det (9 𝑄) = 1 (16 - 7) + 4 (32 + 4) + 8 (56 + 16) = 9 + 144 + 576 = 729 = 9^{3},

so $det 𝑄 = 1$ and $𝑄$ is a rotation (Definition 8.4.26).

For the axis, $9 𝑄 (1, 2, 2) =$ $(1 - 8 + 16, 8 + 8 + 2, - 4 + 14 + 8) =$ $(9, 18, 18)$ , so $𝑄 (1, 2, 2) = (1, 2, 2)$ , and the axis is $𝐚 = \frac{1}{3} (1, 2, 2)$ . (Since $𝑄 \neq 𝐼_{3}$ , the fixed space is exactly this line, by the matrix form in Theorem 8.4.27.) Finally $tr 𝑄 =$ $(1 + 4 + 4) / 9 =$ $1 =$ $1 +$ $2 cos 𝜃$ , so $cos 𝜃 = 0$ and $𝑄$ is a quarter turn about $𝐚$ .

Exercise 8.4.42.standardBeyond 21-242 Let $𝑄 \in O (𝑛)$ .

(a)Show that if $det 𝑄 = - 1$ , then $- 1$ is an eigenvalue of $𝑄$ .
(b)Show that if $det 𝑄 = 1$ and $𝑛$ is odd, then $1$ is an eigenvalue of $𝑄$ .
(c)Give a $𝑄 \in SO (4)$ with no real eigenvalue, and a $𝑄 \in O (2)$ with $det 𝑄 = - 1$ that does not have the eigenvalue $1$ , or show that none exists.

Solution.

Uses. Theorem 5.4.10 (Multiplicativity), Theorem 5.4.14 (The transpose), Definition 5.1.1 (Determinant function), Theorem 7.2.9 (Eigenvalues and invertibility), Lemma 8.4.23 (Orthogonal $2 \times 2$ matrices), Proposition 7.3.15 (Block-triangular matrices)

Hint 1. Imitate the first proof of Theorem 8.4.27: write

𝑄 + 𝐼_{𝑛} = 𝑄 + 𝑄 𝑄^{𝖳}

Hint 2. For (c), use rotation blocks. For the second part, recall what every element of

O (2)

with determinant

- 1

is.

Sketch. (a)

det (𝑄 + 𝐼) =

det (𝑄 (𝐼 + 𝑄^{𝖳})) =

det 𝑄 det (𝐼 + 𝑄) =

- det (𝑄 + 𝐼)

, so

det (𝑄 + 𝐼) = 0

. (b)

det (𝑄 - 𝐼) =

det 𝑄 det (𝐼 - 𝑄^{𝖳}) =

{(- 1)}^{𝑛} det (𝑄 - 𝐼) =

- det (𝑄 - 𝐼)

. (c)

diag (𝑅_{𝜋 / 2}, 𝑅_{𝜋 / 2})

has characteristic polynomial

{(𝑥^{2} + 1)}^{2}

. Every element of

O (2)

with determinant

- 1

is a reflection

𝑆_{𝜃}

, which has eigenvalue

1

: none exists.

(a) Using $𝑄 𝑄^{𝖳} = 𝐼_{𝑛}$ , Theorem 5.4.10, and Theorem 5.4.14,

det (𝑄 + 𝐼_{𝑛}) = det (𝑄 + 𝑄 𝑄^{𝖳}) = det 𝑄 \cdot det (𝐼_{𝑛} + 𝑄^{𝖳}) = det 𝑄 \cdot det (𝐼_{𝑛} + 𝑄) = - det (𝑄 + 𝐼_{𝑛}) .

So $det (𝑄 - (- 1) 𝐼_{𝑛}) =$ $0$ , and $- 1$ is an eigenvalue (Theorem 7.2.9).

(b) Similarly, $det (𝑄 - 𝐼_{𝑛}) =$ $det (𝑄 - 𝑄 𝑄^{𝖳}) =$ $det 𝑄 \cdot$ $det (𝐼_{𝑛} - 𝑄^{𝖳}) =$ $det (𝐼_{𝑛} - 𝑄)$ . By multilinearity in the $𝑛$ columns (Definition 5.1.1), $det (𝐼_{𝑛} - 𝑄) =$ ${(- 1)}^{𝑛} det (𝑄 - 𝐼_{𝑛}) =$ $- det (𝑄 - 𝐼_{𝑛})$ for odd $𝑛$ . So $det (𝑄 - 𝐼_{𝑛}) = 0$ , and $1$ is an eigenvalue.

(c) Let $𝑄 =$ $(\begin{matrix} 𝑅_{𝜋 / 2} & 0 \\ 0 & 𝑅_{𝜋 / 2} \end{matrix}) \in$ $ℝ^{4 \times 4}$ . Its columns are $𝐞_{2}, - 𝐞_{1}, 𝐞_{4}, - 𝐞_{3}$ , which are orthonormal, and $det 𝑄 = 1 \cdot 1 = 1$ . By Proposition 7.3.15, $𝑝_{𝑄} = {(𝑥^{2} + 1)}^{2}$ , which has no real root, so $𝑄$ has no real eigenvalue. For the second request, no such matrix exists: by Lemma 8.4.23, an element of $O (2)$ with determinant $- 1$ is a reflection $𝑆_{𝜃}$ , which has the eigenvalue $1$ .

Exercise 8.4.43.standard For $𝛼 \in ℝ$ , let $𝑆_{𝛼} =$ $(\begin{matrix} cos 𝛼 & sin 𝛼 \\ sin 𝛼 & - cos 𝛼 \end{matrix})$ , the reflection across the line at angle $𝛼 / 2$ (Example 6.3.35).

(a)Show that $𝑆_{𝛼} 𝑆_{𝛽} = 𝑅_{𝛼 - 𝛽}$ . In words: reflecting across a line $ℓ_{2}$ and then across a line $ℓ_{1}$ through $𝟎$ rotates the plane by twice the angle from $ℓ_{2}$ to $ℓ_{1}$ .
(b)Deduce that every rotation $𝑅_{𝜃}$ is a product of two reflections, and that the first of them (the one applied first) can be any reflection.

Solution.

Uses. Definition 2.4.1 (Matrix product)

Hint 1. Write

𝑆_{𝛼} = 𝑅_{𝛼} 𝐷

with

𝐷 = diag (1, - 1)

, and check that

𝐷 𝑅_{𝛽} 𝐷 = 𝑅_{- 𝛽}

Hint 2. For (b), given

𝜃

and

𝛽

, choose

𝛼

with

𝛼 - 𝛽 = 𝜃

Sketch.

𝑆_{𝛼} = 𝑅_{𝛼} 𝐷

, so

𝑆_{𝛼} 𝑆_{𝛽} =

𝑅_{𝛼} 𝐷 𝑅_{𝛽} 𝐷 =

𝑅_{𝛼} 𝑅_{- 𝛽} =

𝑅_{𝛼 - 𝛽}

. The mirrors are at angles

𝛼 / 2

and

𝛽 / 2

, which differ by half the rotation angle. Given

𝜃

and any

𝑆_{𝛽}

𝑅_{𝜃} = 𝑆_{𝜃 + 𝛽} 𝑆_{𝛽}

(a) Let $𝐷 = diag (1, - 1)$ . Multiplying out, $𝑅_{𝛼} 𝐷 =$ $(\begin{matrix} cos 𝛼 & sin 𝛼 \\ sin 𝛼 & - cos 𝛼 \end{matrix}) =$ $𝑆_{𝛼}$ , and

𝐷 𝑅_{𝛽} 𝐷 = (\begin{matrix} cos 𝛽 & sin 𝛽 \\ - sin 𝛽 & cos 𝛽 \end{matrix}) = 𝑅_{- 𝛽} .

So, since rotations compose by adding angles ( $𝑅_{𝛼} 𝑅_{𝛾} = 𝑅_{𝛼 + 𝛾}$ , by the addition formulas),

𝑆_{𝛼} 𝑆_{𝛽} = 𝑅_{𝛼} 𝐷 𝑅_{𝛽} 𝐷 = 𝑅_{𝛼} 𝑅_{- 𝛽} = 𝑅_{𝛼 - 𝛽} .

The mirror of $𝑆_{𝛼}$ is the line at angle $𝛼 / 2$ (Lemma 8.4.23, or directly), and that of $𝑆_{𝛽}$ is at angle $𝛽 / 2$ . The angle from the second mirror to the first is $(𝛼 - 𝛽) / 2$ , and the composite rotates by $𝛼 - 𝛽$ , twice that.

(b) Given $𝜃$ and any reflection $𝑆_{𝛽}$ , let $𝛼 = 𝜃 + 𝛽$ . By (a), $𝑆_{𝛼} 𝑆_{𝛽} = 𝑅_{𝜃}$ . Since every reflection of the plane through $𝟎$ is some $𝑆_{𝛽}$ , the reflection applied first can be chosen freely.

Exercise 8.4.44.challengeBeyond 21-242 Let $𝑄 \in O (𝑛)$ and $𝑟 = rank (𝑄 - 𝐼_{𝑛})$ . A reflection of $ℝ^{𝑛}$ means a Householder matrix $𝐻_{𝐮}$ .

(a)Show that if $𝑄 = 𝐻_{𝐮_{1}} \dots 𝐻_{𝐮_{𝑘}}$ , then $𝑘 \geq 𝑟$ and $det 𝑄 = {(- 1)}^{𝑘}$ .
(b)Show that $𝑄$ is a product of exactly $𝑟$ reflections. (This is the Cartan–Dieudonné theorem for $ℝ^{𝑛}$ , in its sharp form.)
(c)Deduce that every rotation of $ℝ^{3}$ other than $𝐼_{3}$ is a product of two reflections, and never of an odd number of reflections.

Solution.

Uses. Example 8.4.20 (Householder reflections), Proposition 8.4.21 (A reflection exchanging two vectors), Corollary 6.4.25 (Dimension of the orthogonal complement), Theorem 4.2.11 (Rank–nullity), Theorem 5.4.10 (Multiplicativity), Theorem 8.4.27 (Euler’s rotation theorem)

Techniques. Induction on dimension via an invariant subspace (Technique 7.5.12), Parity via an invariant (Technique 5.2.19)

Hint 1. (a) A vector orthogonal to every

𝐮_{𝑗}

is fixed by every factor. How large is the set of such vectors?

Hint 2. (b) Induct on

𝑟

. If

𝑄 \neq 𝐼_{𝑛}

, pick

𝐯

with

𝑄 𝐯 \neq 𝐯

, and let

𝐻

be the reflection with

𝐻 𝑄 𝐯 = 𝐯

(Proposition 8.4.21). Show that

𝐻 𝑄

fixes

𝐯

and every vector fixed by

𝑄

Hint 3. For the fixed vectors: if

𝑄 𝐟 = 𝐟

, then

⟨ 𝑄 𝐯 - 𝐯, 𝐟 ⟩ =

⟨ 𝑄 𝐯, 𝑄 𝐟 ⟩ -

⟨ 𝐯, 𝐟 ⟩ =

0

Sketch. (a)

𝑄

fixes

{𝐮_{1}, \dots, 𝐮_{𝑘}}^{⟂}

, of dimension at least

𝑛 - 𝑘

, so

𝑛 -

𝑟 =

dim ker (𝑄 - 𝐼) \geq

𝑛 -

𝑘

; and each factor has determinant

- 1

. (b) With

𝐮 \propto 𝑄 𝐯 - 𝐯

𝐻 𝑄

fixes

𝐯

and all of

𝐹 = ker (𝑄 - 𝐼)

(since

𝐹 ⟂ 𝐮

), so

rank (𝐻 𝑄 - 𝐼) \leq

𝑟 -

1

; by induction

𝐻 𝑄

is a product of at most

𝑟 - 1

reflections, so

𝑄 = 𝐻 (𝐻 𝑄)

is a product of at most

𝑟

, and by (a) of exactly

𝑟

. (c) For a rotation

𝑄 \neq 𝐼_{3}

ℝ^{3}

, the fixed space is the axis, so

𝑟 = 2

, and

{(- 1)}^{𝑘} = 1

forces

𝑘

even.

(a) Let $𝑋 =$ ${𝐮_{1}, \dots, 𝐮_{𝑘}}^{⟂} =$ ${span (𝐮_{1}, \dots, 𝐮_{𝑘})}^{⟂}$ . Each $𝐻_{𝐮_{𝑗}}$ fixes every vector orthogonal to $𝐮_{𝑗}$ (Example 8.4.20), so $𝑄 𝐱 = 𝐱$ for $𝐱 \in 𝑋$ , that is, $𝑋 \subseteq ker (𝑄 - 𝐼_{𝑛})$ . By Corollary 6.4.25, $dim 𝑋 =$ $𝑛 -$ $dim span (𝐮_{1}, \dots, 𝐮_{𝑘}) \geq$ $𝑛 -$ $𝑘$ . By rank–nullity (Theorem 4.2.11), $dim ker (𝑄 - 𝐼_{𝑛}) =$ $𝑛 -$ $𝑟$ , so $𝑛 - 𝑟 \geq 𝑛 - 𝑘$ , that is, $𝑘 \geq 𝑟$ . Each $𝐻_{𝐮_{𝑗}}$ has determinant $- 1$ (Example 8.4.20), so $det 𝑄 = {(- 1)}^{𝑘}$ by Theorem 5.4.10.

(b) By (a), no product of fewer than $𝑟$ reflections equals $𝑄$ , so it suffices to show that $𝑄$ is a product of at most $𝑟$ reflections. We use strong induction on $𝑟$ . If $𝑟 = 0$ , then $𝑄 = 𝐼_{𝑛}$ , the empty product. Let $𝑟 \geq 1$ , and let $𝐹 = ker (𝑄 - 𝐼_{𝑛})$ , of dimension $𝑛 - 𝑟 < 𝑛$ . Pick $𝐯 \notin 𝐹$ , so $𝑄 𝐯 \neq 𝐯$ . Since $‖ 𝑄 𝐯 ‖ = ‖ 𝐯 ‖$ and all entries are real, Proposition 8.4.21 gives a reflection $𝐻 = 𝐻_{𝐮}$ , $𝐮 =$ $(𝑄 𝐯 - 𝐯) / ‖ 𝑄 𝐯 -$ $𝐯 ‖$ , with $𝐻 𝑄 𝐯 = 𝐯$ . For $𝐟 \in 𝐹$ ,

⟨ 𝑄 𝐯 - 𝐯, 𝐟 ⟩ = ⟨ 𝑄 𝐯, 𝑄 𝐟 ⟩ - ⟨ 𝐯, 𝐟 ⟩ = 0,

because $𝑄 𝐟 = 𝐟$ and $𝑄$ preserves inner products. So $𝐟 ⟂ 𝐮$ , $𝐻 𝐟 = 𝐟$ , and $𝐻 𝑄 𝐟 = 𝐟$ . Hence $𝑄^{'} = 𝐻 𝑄 \in O (𝑛)$ fixes every vector of $𝐹 + span (𝐯)$ , which has dimension $𝑛 - 𝑟 + 1$ since $𝐯 \notin 𝐹$ . So $rank (𝑄^{'} - 𝐼_{𝑛}) \leq$ $𝑟 -$ $1$ , and by the induction hypothesis $𝑄^{'}$ is a product of $rank (𝑄^{'} - 𝐼_{𝑛}) \leq$ $𝑟 -$ $1$ reflections. Since $𝐻^{2} = 𝐼_{𝑛}$ , $𝑄 = 𝐻 𝑄^{'}$ is a product of at most $𝑟$ reflections.

(c) Let $𝑄 \in SO (3)$ , $𝑄 \neq 𝐼_{3}$ . By Theorem 8.4.27, in an orthonormal basis starting with an axis $𝑄$ has matrix $diag (1, 𝑅_{𝜃})$ with $𝑅_{𝜃} \neq 𝐼_{2}$ , that is, $𝜃 \neq 0$ . Then $𝑅_{𝜃} - 𝐼_{2}$ is invertible (its determinant is ${(cos 𝜃 - 1)}^{2} +$ ${sin}^{2} 𝜃 =$ $2 -$ $2 cos 𝜃 \neq$ $0$ ), so $ker (𝑄 - 𝐼_{3})$ is the axis line and $𝑟 = 2$ . By (b), $𝑄$ is a product of two reflections. If $𝑄 = 𝐻_{𝐮_{1}} \dots 𝐻_{𝐮_{𝑘}}$ , then $1 = det 𝑄 = {(- 1)}^{𝑘}$ by (a), so $𝑘$ is even.

Exercise 8.4.45.standardBeyond 21-242 On $ℝ^{2}$ , let ${‖ 𝐱 ‖}_{\infty} =$ $max (| 𝑥_{1} |, | 𝑥_{2} |)$ , and let $𝑓 : ℝ \to ℝ^{2}$ , $𝑓 (𝑡) = (𝑡, | 𝑡 |)$ .

(a)Show that ${‖ 𝑓 (𝑠) - 𝑓 (𝑡) ‖}_{\infty} =$ $| 𝑠 -$ $𝑡 |$ for all $𝑠, 𝑡 \in ℝ$ , but that $𝑓$ is not affine.
(b)Which step of the proof of Theorem 8.4.31 fails for ${‖ \cdot ‖}_{\infty}$ ? Find all points $𝐳 \in ℝ^{2}$ with ${‖ 𝐳 - 𝑓 (- 1) ‖}_{\infty} =$ ${‖ 𝐳 - 𝑓 (1) ‖}_{\infty} =$ $1$ .

Solution.

Uses. Theorem 8.4.31 (Distance-preserving maps are affine), Lemma 8.4.30 (Points between two points), Theorem 4.7.37 (Affine maps preserve affine combinations), Corollary B.1.12 (Two terms, and the reverse triangle inequality)

Hint 1.

| | 𝑠 | -

| 𝑡 | | \leq

| 𝑠 -

𝑡 |

(the reverse triangle inequality in

ℝ

). For non-affineness, look at the midpoint of

- 1

and

1

Hint 2. Lemma 8.4.30 uses equality in the triangle inequality. With the max norm, the “midpoint by distances” of

(- 1, 1)

and

(1, 1)

is not unique.

Sketch. (a)

{‖ 𝑓 (𝑠) - 𝑓 (𝑡) ‖}_{\infty} =

max (| 𝑠 - 𝑡 |, | | 𝑠 | - | 𝑡 | |) =

| 𝑠 -

𝑡 |

. But

𝑓 (0) = (0, 0)

is not the midpoint

(0, 1)

𝑓 (- 1) = (- 1, 1)

and

𝑓 (1) = (1, 1)

. (b) Lemma 8.4.30 fails: the points at max-distance

1

from both

(- 1, 1)

and

(1, 1)

are

(0, 𝑦)

with

0 \leq 𝑦 \leq 2

, a whole segment, and

𝑓 (0)

is one of them.

(a) For $𝑠, 𝑡 \in ℝ$ , $𝑓 (𝑠) -$ $𝑓 (𝑡) =$ $(𝑠 - 𝑡, | 𝑠 | - | 𝑡 |)$ , and $| | 𝑠 | -$ $| 𝑡 | | \leq$ $| 𝑠 -$ $𝑡 |$ by the reverse triangle inequality for real numbers (Corollary B.1.12). So ${‖ 𝑓 (𝑠) - 𝑓 (𝑡) ‖}_{\infty} =$ $| 𝑠 -$ $𝑡 |$ . If $𝑓$ were affine, it would preserve affine combinations (Theorem 4.7.37), in particular midpoints: $𝑓 (0) =$ $\frac{1}{2} (𝑓 (- 1) + 𝑓 (1))$ . But $𝑓 (0) = (0, 0)$ , while $\frac{1}{2} ((- 1, 1) + (1, 1)) =$ $(0, 1)$ .

(b) Step 1 of the proof, which rests on Lemma 8.4.30, fails: in an inner product space the point $𝐳$ is determined by its distances to $𝐚$ and $𝐛$ , but not for ${‖ \cdot ‖}_{\infty}$ . Let $𝐳 = (𝑥, 𝑦)$ . The conditions are

max (| 𝑥 + 1 |, | 𝑦 - 1 |) = 1 and max (| 𝑥 - 1 |, | 𝑦 - 1 |) = 1 .

Both force $| 𝑥 + 1 | \leq 1$ and $| 𝑥 - 1 | \leq 1$ , so $𝑥 = 0$ ; then $| 𝑥 \pm 1 | = 1$ , and both conditions hold exactly when $| 𝑦 - 1 | \leq 1$ . So the solutions are the points $(0, 𝑦)$ with $0 \leq 𝑦 \leq 2$ , a whole segment; the midpoint $(0, 1)$ is one of them, and $𝑓 (0) = (0, 0)$ is another. The max norm has no strict equality case in the triangle inequality: ${‖ (1, 1) ‖}_{\infty} +$ ${‖ (1, - 1) ‖}_{\infty} =$ $2 =$ ${‖ (2, 0) ‖}_{\infty}$ , although neither vector is a multiple of the other.

Exercise 8.4.46.standardBeyond 21-242 Let

𝑓, 𝑔

be rigid motions of

ℝ^{𝑛}

, and let

𝐩_{0}, \dots, 𝐩_{𝑛}

be affinely independent (Definition 4.7.18) with

𝑓 (𝐩_{𝑖}) = 𝑔 (𝐩_{𝑖})

for all

𝑖

. Show that

𝑓 = 𝑔

. Deduce that a rigid motion of

ℝ^{2}

fixing three non-collinear points is the identity.

Solution.

Uses. Theorem 8.4.31 (Distance-preserving maps are affine), Proposition 4.7.20 (Affine independence through differences), Theorem 2.2.1 (Linear maps are determined by a basis)

Techniques. Choose a base point (Technique 4.7.4)

Hint 1. By Theorem 8.4.31,

𝑓 = 𝑄 𝐱 + 𝐛

and

𝑔 = 𝑄^{'} 𝐱 + 𝐛^{'}

. Measure positions from

𝐩_{0}

Hint 2.

𝑓 (𝐩_{𝑖}) -

𝑓 (𝐩_{0}) =

𝑄 (𝐩_{𝑖} - 𝐩_{0})

. The differences

𝐩_{𝑖} - 𝐩_{0}

form a basis.

Sketch. Write

𝑓 (𝐱) = 𝑄 𝐱 + 𝐛

and

𝑔 (𝐱) = 𝑄^{'} 𝐱 + 𝐛^{'}

. Then

𝑄 (𝐩_{𝑖} - 𝐩_{0}) =

𝑓 (𝐩_{𝑖}) -

𝑓 (𝐩_{0}) =

𝑔 (𝐩_{𝑖}) -

𝑔 (𝐩_{0}) =

𝑄^{'} (𝐩_{𝑖} - 𝐩_{0})

, and these

𝑛

differences are a basis, so

𝑄 = 𝑄^{'}

; then

𝐛 =

𝑓 (𝐩_{0}) -

𝑄 𝐩_{0} =

𝐛^{'}

By Theorem 8.4.31, $𝑓 (𝐱) = 𝑄 𝐱 + 𝐛$ and $𝑔 (𝐱) = 𝑄^{'} 𝐱 + 𝐛^{'}$ with $𝑄, 𝑄^{'}$ orthogonal. For each $𝑖$ ,

𝑄 (𝐩_{𝑖} - 𝐩_{0}) = 𝑓 (𝐩_{𝑖}) - 𝑓 (𝐩_{0}) = 𝑔 (𝐩_{𝑖}) - 𝑔 (𝐩_{0}) = 𝑄^{'} (𝐩_{𝑖} - 𝐩_{0}) .

By Proposition 4.7.20, the $𝑛$ vectors $𝐩_{1} -$ $𝐩_{0},$ $\dots,$ $𝐩_{𝑛} -$ $𝐩_{0}$ are linearly independent, hence a basis of $ℝ^{𝑛}$ . The linear maps $𝐱 \mapsto 𝑄 𝐱$ and $𝐱 \mapsto 𝑄^{'} 𝐱$ agree on this basis, so $𝑄 = 𝑄^{'}$ (Theorem 2.2.1). Then $𝐛 =$ $𝑓 (𝐩_{0}) -$ $𝑄 𝐩_{0} =$ $𝑔 (𝐩_{0}) -$ $𝑄^{'} 𝐩_{0} =$ $𝐛^{'}$ , so $𝑓 = 𝑔$ .

In $ℝ^{2}$ , three points are affinely independent exactly when they are not collinear. Applying the first part with $𝑔 = id$ shows that a rigid motion fixing three non-collinear points is the identity. (Two points do not suffice: the reflection across the line through them fixes both.)

More practice

The practice bank has 15 exercises on this section’s topics.

8.4.1Isometries

8.4.2Unitary operators

Invariant subspaces and the spectral form

8.4.3Householder reflections

8.4.4Real orthogonal operators Beyond 21-242

8.4.5Orientation and rotations of ℝ3 Beyond 21-242

8.4.6Rigid motions Beyond 21-242

Exercises

More practice

8.4.5Orientation and rotations of $ℝ^{3}$ Beyond 21-242