8.1 Self-adjoint and normal operators

An operator with an orthonormal basis of eigenvectors is the best kind of operator to have on an inner product space. In that basis it is a diagonal matrix, coordinates are inner products (Proposition 6.3.18), and lengths are computed by Pythagoras. This chapter determines which operators have such a basis. The answer, proved in Section 8.2, depends on the field: over $ℝ$ they are the self-adjoint operators, and over $ℂ$ they are the normal operators, the ones that commute with their adjoint. This section studies the two classes before that theorem is available. Self-adjoint operators have real eigenvalues and mutually orthogonal eigenspaces, and on a nonzero finite-dimensional real space they always have an eigenvalue. Normal operators are exactly the operators $𝑇$ for which $𝑇$ and $𝑇^{*}$ stretch every vector by the same amount, and this one fact gives them orthogonal eigenspaces as well.

Throughout, $𝑉$ is an inner product space over $𝔽$ , where $𝔽$ is $ℝ$ or $ℂ$ , and we write $𝑇 𝐯$ for $𝑇 (𝐯)$ . Whenever $𝑇^{*}$ appears, $𝑇$ is assumed to have an adjoint; by Theorem 6.6.15 this is automatic when $𝑉$ is finite-dimensional. Results that need finite dimension say so.

8.1.1Self-adjoint operators and their eigenvalues

Recall that $𝑇 \in L (𝑉)$ is self-adjoint if $⟨ 𝑇 𝐯, 𝐰 ⟩ =$ $⟨ 𝐯, 𝑇 𝐰 ⟩$ for all $𝐯, 𝐰 \in 𝑉$ (Definition 6.6.26), and that in an orthonormal basis this says that the matrix of $𝑇$ is Hermitian (Corollary 6.6.28). Section 6.6.5 collected examples: orthogonal projections, the operators $𝑇^{*} 𝑇$ , and the maps $𝐱 \mapsto 𝐴 𝐱$ for real symmetric and complex Hermitian matrices $𝐴$ . Over $ℂ$ , the self-adjoint operators are the ones whose values $⟨ 𝑇 𝐯, 𝐯 ⟩$ are all real (Proposition 6.6.30). Their eigenvalues are real as well, and this is the first thing the defining identity gives when we feed it eigenvectors.

Intuition. If

𝑇 𝐯 = 𝜆 𝐯

, then

𝑇

acts on

𝐯

as the scalar

𝜆

, and the defining identity

⟨ 𝑇 𝐯, 𝐯 ⟩ =

⟨ 𝐯, 𝑇 𝐯 ⟩

moves this scalar from the first slot to the second. Pulled out of the first slot it is

𝜆

; pulled out of the second it is

𝜆

(Proposition 6.1.4). So

𝜆 = 𝜆

. With eigenvectors

𝐯

and

𝐰

for two different eigenvalues, the same identity, applied to the pair, gives

𝜆 ⟨ 𝐯, 𝐰 ⟩ =

𝜇 ⟨ 𝐯, 𝐰 ⟩

, and that forces

⟨ 𝐯, 𝐰 ⟩ = 0

Proposition 8.1.1 (Eigenvalues and eigenvectors of self-adjoint operators). Let $𝑇 \in L (𝑉)$ be self-adjoint.

(a)Every eigenvalue of $𝑇$ is real.
(b)If $𝑇 𝐯 = 𝜆 𝐯$ and $𝑇 𝐰 = 𝜇 𝐰$ with $𝜆 \neq 𝜇$ , then $⟨ 𝐯, 𝐰 ⟩ = 0$ . That is, eigenspaces of $𝑇$ for distinct eigenvalues are orthogonal: every vector of one is orthogonal to every vector of the other.

Proof (by moving $𝑇$ across the inner product).routine

Uses. Definition 6.6.26 (Self-adjoint operator), Definition 7.2.6 (Eigenvalue, eigenvector), Proposition 6.1.4 (Sesquilinearity), Proposition 6.1.27 (Basic properties of the norm), Notation 6.1.1 (Real and complex scalars)

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

Hint 1. Let

𝐯

be an eigenvector for

𝜆

. Compute

⟨ 𝑇 𝐯, 𝐯 ⟩

and

⟨ 𝐯, 𝑇 𝐯 ⟩

separately.

Hint 2. The first is

𝜆 {‖ 𝐯 ‖}^{2}

and the second is

𝜆 {‖ 𝐯 ‖}^{2}

, and

‖ 𝐯 ‖ \neq 0

Hint 3. For (b), compute both sides of

⟨ 𝑇 𝐯, 𝐰 ⟩ =

⟨ 𝐯, 𝑇 𝐰 ⟩

, and use (a) to drop the conjugate on

𝜇

Sketch. (a)

𝜆 {‖ 𝐯 ‖}^{2} =

⟨ 𝑇 𝐯, 𝐯 ⟩ =

⟨ 𝐯, 𝑇 𝐯 ⟩ =

𝜆 {‖ 𝐯 ‖}^{2}

, and

𝐯 \neq 𝟎

, so

𝜆 = 𝜆

. (b)

𝜆 ⟨ 𝐯, 𝐰 ⟩ =

⟨ 𝑇 𝐯, 𝐰 ⟩ =

⟨ 𝐯, 𝑇 𝐰 ⟩ =

𝜇 ⟨ 𝐯, 𝐰 ⟩ =

𝜇 ⟨ 𝐯, 𝐰 ⟩

, and

𝜆 \neq 𝜇

(a) Let $𝜆$ be an eigenvalue of $𝑇$ and $𝐯$ an eigenvector for it. Then

𝜆 {‖ 𝐯 ‖}^{2} = ⟨ 𝜆 𝐯, 𝐯 ⟩ = ⟨ 𝑇 𝐯, 𝐯 ⟩ = ⟨ 𝐯, 𝑇 𝐯 ⟩ = ⟨ 𝐯, 𝜆 𝐯 ⟩ = 𝜆 {‖ 𝐯 ‖}^{2},

where the middle equality is self-adjointness and the last uses Proposition 6.1.4. Since $𝐯 \neq 𝟎$ , ${‖ 𝐯 ‖}^{2} \neq 0$ (Proposition 6.1.27), so $𝜆 = 𝜆$ , which says that $𝜆 \in ℝ$ (Notation 6.1.1). Over $ℝ$ there was nothing to prove.

(b) By (a), $𝜇$ is real, so

𝜆 ⟨ 𝐯, 𝐰 ⟩ = ⟨ 𝑇 𝐯, 𝐰 ⟩ = ⟨ 𝐯, 𝑇 𝐰 ⟩ = 𝜇 ⟨ 𝐯, 𝐰 ⟩ = 𝜇 ⟨ 𝐯, 𝐰 ⟩ .

Hence $(𝜆 - 𝜇) ⟨ 𝐯, 𝐰 ⟩ = 0$ , and $𝜆 \neq 𝜇$ gives $⟨ 𝐯, 𝐰 ⟩ = 0$ . If $𝐯 \in 𝐸_{𝜆}$ and $𝐰 \in 𝐸_{𝜇}$ are arbitrary, then either one of them is $𝟎$ , and $⟨ 𝐯, 𝐰 ⟩ = 0$ trivially, or both are eigenvectors and the computation applies. So $𝐸_{𝜆}$ and $𝐸_{𝜇}$ are orthogonal. □

Proof (for (a), through the values $⟨ 𝑇 𝐯, 𝐯 ⟩$ ).routine

Uses. Definition 6.6.26 (Self-adjoint operator), Proposition 6.6.30 (Self-adjoint means real values), Proposition 6.1.27 (Basic properties of the norm)

Hint 1. For self-adjoint

𝑇

, every number

⟨ 𝑇 𝐯, 𝐯 ⟩

is real. Express an eigenvalue through such a number.

Hint 2. If

𝑇 𝐯 = 𝜆 𝐯

, then

⟨ 𝑇 𝐯, 𝐯 ⟩ = 𝜆 {‖ 𝐯 ‖}^{2}

Sketch.

⟨ 𝑇 𝐯, 𝐯 ⟩ =

⟨ 𝐯, 𝑇 𝐯 ⟩ =

⟨ 𝑇 𝐯, 𝐯 ⟩

is real for every

𝐯

. For an eigenvector,

𝜆 =

⟨ 𝑇 𝐯, 𝐯 ⟩ / {‖ 𝐯 ‖}^{2}

is a real number divided by a positive one.

For every $𝐯 \in 𝑉$ , self-adjointness and conjugate symmetry give $⟨ 𝑇 𝐯, 𝐯 ⟩ =$ $⟨ 𝐯, 𝑇 𝐯 ⟩ =$ $⟨ 𝑇 𝐯, 𝐯 ⟩$ , so $⟨ 𝑇 𝐯, 𝐯 ⟩ \in ℝ$ . (This is the easy direction of Proposition 6.6.30; it needs neither $ℂ$ nor finite dimension.) If $𝑇 𝐯 = 𝜆 𝐯$ with $𝐯 \neq 𝟎$ , then $⟨ 𝑇 𝐯, 𝐯 ⟩ = 𝜆 {‖ 𝐯 ‖}^{2}$ , so

𝜆 = \frac{⟨ 𝑇 𝐯, 𝐯 ⟩}{{‖ 𝐯 ‖}^{2}}

is a real number divided by a positive real number (Proposition 6.1.27), hence real. □

The second proof shows where eigenvalues live. The function $𝐯 \mapsto$ $⟨ 𝑇 𝐯, 𝐯 ⟩ / {‖ 𝐯 ‖}^{2}$ on nonzero vectors, the Rayleigh quotient of $𝑇$ , takes only real values, and at an eigenvector it equals the eigenvalue. So every eigenvalue of $𝑇$ lies between the infimum and the supremum of the Rayleigh quotient. Section 12.3 shows that in finite dimensions the smallest and largest eigenvalues are exactly the minimum and maximum of the Rayleigh quotient.

Example 8.1.2 (A Hermitian $2 \times 2$ matrix). Let $𝐴 = (\begin{matrix} 2 & 1 - 𝑖 \\ 1 + 𝑖 & 3 \end{matrix})$ , which is Hermitian (Example 6.6.27). Its characteristic polynomial is

𝑝_{𝐴} = (𝑥 - 2) (𝑥 - 3) - (1 - 𝑖) (1 + 𝑖) = 𝑥^{2} - 5 𝑥 + 4 = (𝑥 - 1) (𝑥 - 4),

so its eigenvalues are $1$ and $4$ , real although the entries of $𝐴$ are not. For $𝜆 = 1$ , the first row of $𝐴 -$ $𝐼_{2} =$ $(\begin{matrix} 1 & 1 - 𝑖 \\ 1 + 𝑖 & 2 \end{matrix})$ gives $𝑥_{1} + (1 - 𝑖) 𝑥_{2} = 0$ , and the second row is $1 + 𝑖$ times the first, so $𝐸_{1} =$ $span ((- 1 + 𝑖, 1))$ . For $𝜆 = 4$ , the first row of $𝐴 -$ $4 𝐼_{2} =$ $(\begin{matrix} - 2 & 1 - 𝑖 \\ 1 + 𝑖 & - 1 \end{matrix})$ gives $𝐸_{4} =$ $span ((1 - 𝑖, 2))$ . As predicted,

(- 1 + 𝑖, 1) \cdot (1 - 𝑖, 2) = (- 1 + 𝑖) (1 - 𝑖) + 2 = (- 1 + 𝑖) (1 + 𝑖) + 2 = - 2 + 2 = 0 .

The squared norms are $3$ and $6$ , so $𝐮_{1} = \frac{1}{\sqrt{3}} (- 1 + 𝑖, 1)$ , $𝐮_{2} = \frac{1}{\sqrt{6}} (1 - 𝑖, 2)$ is an orthonormal basis of $ℂ^{2}$ consisting of eigenvectors, and with the unitary matrix $𝑈$ whose columns are $𝐮_{1}$ and $𝐮_{2}$ , we get $𝑈^{*} 𝐴 𝑈 = diag (1, 4)$ .

Part (b) says nothing about two eigenvectors for the same eigenvalue: for the identity, every nonzero vector is an eigenvector. An orthonormal list of eigenvectors is obtained by choosing an orthonormal basis inside each eigenspace, with Gram–Schmidt (Algorithm 6.4.14), and Exercise 8.1.23 does this in an example.

Real self-adjoint operators have eigenvalues

Over $ℝ$ , part (a) of Proposition 8.1.1 has no content, since the eigenvalues of an operator on a real space are real by definition. The real question is whether there are any; the rotation by $𝜋 / 2$ has none (Example 7.2.18). Self-adjoint operators always do. The idea is to let a real symmetric matrix act on $ℂ^{𝑛}$ , where eigenvalues exist, and to apply part (a) there.

Theorem 8.1.3 (Self-adjoint operators have eigenvalues). Let

𝑉

be a finite-dimensional inner product space with

𝑉 \neq {𝟎}

, and let

𝑇 \in L (𝑉)

be self-adjoint. Then

𝑇

has an eigenvalue, and it is real. In particular, every real symmetric matrix

𝐴 \in ℝ^{𝑛 \times 𝑛}

with

𝑛 \geq 1

has a real eigenvalue.

Proof (by letting the matrix act on $ℂ^{𝑛}$ ).standard

Uses. Theorem 6.4.20 (Orthonormal bases exist), Corollary 6.6.28 (Self-adjointness in an orthonormal basis), Example 6.6.10 (The adjoint of a matrix), Theorem 7.3.27 (Existence of eigenvalues over $ℂ$ ), Proposition 8.1.1 (Eigenvalues and eigenvectors of self-adjoint operators), Theorem 7.3.6 (Eigenvalues are the roots of the characteristic polynomial), Definition 7.3.12 (Characteristic polynomial of an operator), Corollary 7.3.13 (Eigenvalues of an operator)

Techniques. Pass to orthonormal coordinates (Technique 6.7.2)

Hint 1. Over

ℂ

this is Theorem 7.3.27 together with Proposition 8.1.1(a). For

ℝ

, pass to the matrix

𝐴

𝑇

in an orthonormal basis.

Hint 2.

𝐴

is real and symmetric. Regard it as a complex matrix acting on

ℂ^{𝑛}

with the dot product. What kind of operator is

𝐱 \mapsto 𝐴 𝐱

there?

Hint 3. That operator is self-adjoint on

ℂ^{𝑛}

, so it has an eigenvalue

𝜆

, which is real. Now

𝜆

is a real root of

𝑝_{𝐴} = 𝑝_{𝑇}

Sketch. Choose an orthonormal basis

𝒪︀

; then

𝐴 = {[𝑇]}_{𝒪︀}

is Hermitian. On

ℂ^{𝑛}

with the dot product,

𝑇_{𝐴}

is self-adjoint since

𝐴^{*} = 𝐴

, so it has an eigenvalue

𝜆

(Theorem 7.3.27), which is real by Proposition 8.1.1(a). Hence

𝜆

is a root of

𝑝_{𝐴}

, which is the same polynomial whether

𝐴

is read as real or complex, and

𝑝_{𝐴} = 𝑝_{𝑇}

. So

𝜆 \in 𝔽

is a root of

𝑝_{𝑇}

, hence an eigenvalue of

𝑇

Let $𝑛 = dim 𝑉 \geq 1$ . By Theorem 6.4.20, $𝑉$ has an orthonormal basis $𝒪︀$ , and $𝐴 =$ ${[𝑇]}_{𝒪︀} \in$ $𝔽^{𝑛 \times 𝑛}$ satisfies $𝐴^{*} = 𝐴$ by Corollary 6.6.28. Regard $𝐴$ as an element of $ℂ^{𝑛 \times 𝑛}$ ; when $𝔽 = ℝ$ its entries are real and still $𝐴^{*} = 𝐴^{𝖳} = 𝐴$ . By Example 6.6.10, the operator $𝑇_{𝐴} : 𝐱 \mapsto 𝐴 𝐱$ on $ℂ^{𝑛}$ with the dot product has adjoint $𝑇_{𝐴^{*}} = 𝑇_{𝐴}$ , so it is self-adjoint. By Theorem 7.3.27, $𝑇_{𝐴}$ has an eigenvalue $𝜆 \in ℂ$ , and $𝜆 \in ℝ$ by Proposition 8.1.1(a). By Theorem 7.3.6, $𝑝_{𝐴} (𝜆) = 0$ .

The polynomial $𝑝_{𝐴} =$ $det (𝑥 𝐼_{𝑛} - 𝐴)$ is computed from the entries of $𝐴$ alone, so it is the same polynomial whether $𝐴$ is regarded as a real or a complex matrix, and it equals $𝑝_{𝑇}$ (Definition 7.3.12). So $𝜆$ is a root of $𝑝_{𝑇}$ lying in $ℝ \subseteq 𝔽$ , and by Corollary 7.3.13 it is an eigenvalue of $𝑇$ . It is real by Proposition 8.1.1(a). The last sentence of the theorem is the case $𝑉 = ℝ^{𝑛}$ with the dot product and $𝑇 = 𝑇_{𝐴}$ , which is self-adjoint by Example 6.6.10. □

The proof shows more. Every complex root of $𝑝_{𝐴}$ is an eigenvalue of the self-adjoint operator $𝑇_{𝐴}$ on $ℂ^{𝑛}$ , hence real, so by Corollary 7.1.35

𝑝_{𝐴} = (𝑥 - 𝜆_{1}) (𝑥 - 𝜆_{2}) \dots (𝑥 - 𝜆_{𝑛}) with all 𝜆_{𝑗} \in ℝ .

The characteristic polynomial of a Hermitian matrix, and in particular of a real symmetric matrix, splits over $ℝ$ . This is what the real spectral theorem in Section 8.2 needs.

The first proof leans on determinants and lets a real matrix act on $ℂ^{𝑛}$ . The second avoids determinants and never complexifies the operator, although the factorization over $ℝ$ that it uses still rests on the Fundamental Theorem of Algebra. Following the determinant-free proof of Theorem 7.3.27, it factors a polynomial that kills a vector. Over $ℝ$ some factors may be quadratics without real roots, and these are exactly where a real operator can fail to have eigenvalues; for a self-adjoint operator they turn out to be injective, hence harmless.

Proof (without determinants).hardBeyond 21-242

Uses. Theorem 7.3.27 (Existence of eigenvalues over $ℂ$ ), Proposition 8.1.1 (Eigenvalues and eigenvectors of self-adjoint operators), Proposition 7.2.35 (A basis of a Krylov subspace), Theorem 7.1.39 (Factorization over $ℝ$ ), Proposition 7.2.27 (Evaluating polynomials at an operator), Theorem 6.2.1 (Cauchy–Schwarz inequality), Theorem 7.2.9 (Eigenvalues and invertibility)

Techniques. Krylov sequences must become dependent (Technique 7.3.28), A sum of nonnegative terms (Technique 6.1.9)

Hint 1. Over

ℂ

, combine Theorem 7.3.27 with (a) of Proposition 8.1.1. Over

ℝ

, imitate the second proof of Theorem 7.3.27. What goes wrong over

ℝ

, and how could self-adjointness repair it?

Hint 2. Over

ℝ

, the polynomial

𝑝

with

𝑝 (𝑇) 𝐯 = 𝟎

factors into linear factors and quadratics

𝑥^{2} + 𝑏 𝑥 + 𝑐

with

𝑏^{2} < 4 𝑐

(Theorem 7.1.39). Show that each such quadratic gives an injective operator

𝑇^{2} + 𝑏 𝑇 + 𝑐 id

Hint 3. Compute

⟨ (𝑇^{2} + 𝑏 𝑇 + 𝑐 id) 𝐯, 𝐯 ⟩ =

{‖ 𝑇 𝐯 ‖}^{2} +

𝑏 ⟨ 𝑇 𝐯, 𝐯 ⟩ +

𝑐 {‖ 𝐯 ‖}^{2}

, bound the middle term with Cauchy–Schwarz, and complete the square.

Sketch. For

𝑏^{2} < 4 𝑐

and

𝐯 \neq 𝟎

⟨ (𝑇^{2} + 𝑏 𝑇 + 𝑐 id) 𝐯, 𝐯 ⟩ \geq

{(‖ 𝑇 𝐯 ‖ - | 𝑏 | ‖ 𝐯 ‖ / 2)}^{2} +

(𝑐 - 𝑏^{2} / 4) {‖ 𝐯 ‖}^{2} >

0

, so the quadratic factors are injective at

𝑇

. Take

𝐯 \neq 𝟎

and a monic

𝑝

of positive degree with

𝑝 (𝑇) 𝐯 = 𝟎

(Proposition 7.2.35). Factor

𝑝

over

ℝ

. Then

𝑝 (𝑇)

is the composite of its factors evaluated at

𝑇

, and it is not injective, so some linear factor

𝑇 - 𝜆 id

is not injective.

If $𝔽 = ℂ$ , then $𝑇$ has an eigenvalue by Theorem 7.3.27, and it is real by Proposition 8.1.1(a). So let $𝔽 = ℝ$ .

Quadratics without real roots are injective at $𝑇$ . Let $𝑏, 𝑐 \in ℝ$ with $𝑏^{2} < 4 𝑐$ , and let $𝑄 = 𝑇^{2} + 𝑏 𝑇 + 𝑐 id$ . For $𝐯 \in 𝑉$ , self-adjointness gives $⟨ 𝑇^{2} 𝐯, 𝐯 ⟩ =$ $⟨ 𝑇 𝐯, 𝑇 𝐯 ⟩ =$ ${‖ 𝑇 𝐯 ‖}^{2}$ , and Cauchy–Schwarz (Theorem 6.2.1) gives $𝑏 ⟨ 𝑇 𝐯, 𝐯 ⟩ \geq$ $- | 𝑏 | ‖ 𝑇 𝐯 ‖ ‖ 𝐯 ‖$ . Hence

\begin{aligned} ⟨ 𝑄 𝐯, 𝐯 ⟩ & \geq {‖ 𝑇 𝐯 ‖}^{2} - | 𝑏 | ‖ 𝑇 𝐯 ‖ ‖ 𝐯 ‖ + 𝑐 {‖ 𝐯 ‖}^{2} \\ = {(‖ 𝑇 𝐯 ‖ - \frac{| 𝑏 |}{2} ‖ 𝐯 ‖)}^{2} + (𝑐 - \frac{𝑏^{2}}{4}) {‖ 𝐯 ‖}^{2} . \end{aligned}

If $𝐯 \neq 𝟎$ , the last term is positive, so $⟨ 𝑄 𝐯, 𝐯 ⟩ > 0$ and $𝑄 𝐯 \neq 𝟎$ . So $𝑄$ is injective.

Some linear factor is not injective. Fix $𝐯 \in 𝑉$ with $𝐯 \neq 𝟎$ . By Proposition 7.2.35 there is a monic $𝑝 \in ℝ [𝑥]$ of degree $𝑑 \geq 1$ with $𝑝 (𝑇) 𝐯 = 𝟎$ . By Theorem 7.1.39, and since $𝑝$ is monic,

𝑝 = (𝑥 - 𝜆_{1}) \dots (𝑥 - 𝜆_{𝑘}) (𝑥^{2} + 𝑏_{1} 𝑥 + 𝑐_{1}) \dots (𝑥^{2} + 𝑏_{𝑙} 𝑥 + 𝑐_{𝑙})

with $𝑏_{𝑗}^{2} < 4 𝑐_{𝑗}$ for each $𝑗$ . By Proposition 7.2.27(a),

𝑝 (𝑇) = (𝑇 - 𝜆_{1} id) \dots (𝑇 - 𝜆_{𝑘} id) 𝑄_{1} \dots 𝑄_{𝑙}, 𝑄_{𝑗} = 𝑇^{2} + 𝑏_{𝑗} 𝑇 + 𝑐_{𝑗} id .

Each $𝑄_{𝑗}$ is injective by the first part. If every $𝑇 - 𝜆_{𝑖} id$ were injective too (this includes the case $𝑘 = 0$ ), then $𝑝 (𝑇)$ would be a composite of injective maps, hence injective, contradicting $𝑝 (𝑇) 𝐯 = 𝟎$ with $𝐯 \neq 𝟎$ . So some $𝑇 - 𝜆_{𝑖} id$ is not injective, and $𝜆_{𝑖}$ is an eigenvalue of $𝑇$ by Theorem 7.2.9. □

For the rotation $𝐽 = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ the Krylov polynomial of $𝐞_{1}$ is $𝑥^{2} + 1$ , and $𝐽^{2} + id = 0$ is as far from injective as possible. The positivity estimate is exactly what a rotation violates: $⟨ 𝐽^{2} 𝐯, 𝐯 ⟩ = - {‖ 𝐯 ‖}^{2}$ , whereas for self-adjoint $𝑇$ the term $⟨ 𝑇^{2} 𝐯, 𝐯 ⟩ =$ ${‖ 𝑇 𝐯 ‖}^{2}$ is never negative.

8.1.2Testing self-adjoint operators on the diagonal

By Proposition 6.1.41, an operator $𝑇$ on a complex inner product space with $⟨ 𝑇 𝐯, 𝐯 ⟩ = 0$ for all $𝐯$ is $0$ , and over $ℝ$ this fails for the rotation $𝐽$ (Example 6.1.39). The rotation is far from self-adjoint, and for self-adjoint operators the real case behaves like the complex one. We will need this to characterize normal operators over both fields at once.

Intuition. Put

𝐯 = 𝐱 + 𝐲

into

⟨ 𝑇 𝐯, 𝐯 ⟩ = 0

. Over either field the result is

⟨ 𝑇 𝐱, 𝐲 ⟩ +

⟨ 𝑇 𝐲, 𝐱 ⟩ =

0

(Proposition 6.1.41, first proof), and for

𝐽

the two terms cancel without being

0

. If

𝑇

is self-adjoint, then

⟨ 𝑇 𝐲, 𝐱 ⟩ =

⟨ 𝐲, 𝑇 𝐱 ⟩

is the conjugate of

⟨ 𝑇 𝐱, 𝐲 ⟩

, so the sum is

2 Re ⟨ 𝑇 𝐱, 𝐲 ⟩

. Then

Re ⟨ 𝑇 𝐱, 𝐲 ⟩ = 0

for all

𝐲

, and the choice

𝐲 = 𝑇 𝐱

gives

{‖ 𝑇 𝐱 ‖}^{2} = 0

Proposition 8.1.4 (Self-adjoint operators are determined by their diagonal values). Let

𝑇 \in L (𝑉)

be self-adjoint, where

𝔽

ℝ

ℂ

. If

⟨ 𝑇 𝐯, 𝐯 ⟩ = 0

for every

𝐯 \in 𝑉

, then

𝑇 = 0

. Consequently, if

𝑆, 𝑇 \in L (𝑉)

are self-adjoint and

⟨ 𝑆 𝐯, 𝐯 ⟩ =

⟨ 𝑇 𝐯, 𝐯 ⟩

for every

𝐯 \in 𝑉

, then

𝑆 = 𝑇

Proof (by polarization).standard

Uses. Definition 6.6.26 (Self-adjoint operator), Proposition 6.1.4 (Sesquilinearity), Proposition 6.1.27 (Basic properties of the norm), Proposition 6.6.18 (Properties of the adjoint)

Techniques. Polarization (Technique 6.1.33), Show a vector is zero via its norm (Technique 6.1.36)

Hint 1. Expand

⟨ 𝑇 (𝐱 + 𝐲), 𝐱 + 𝐲 ⟩ -

⟨ 𝑇 (𝐱 - 𝐲), 𝐱 - 𝐲 ⟩

Hint 2. With self-adjointness, the result is

4 Re ⟨ 𝑇 𝐱, 𝐲 ⟩

. So

Re ⟨ 𝑇 𝐱, 𝐲 ⟩ = 0

for all

𝐱, 𝐲

Hint 3. Choose

𝐲 = 𝑇 𝐱

Sketch. Expanding, and using

⟨ 𝑇 𝐲, 𝐱 ⟩ =

⟨ 𝐲, 𝑇 𝐱 ⟩ =

⟨ 𝑇 𝐱, 𝐲 ⟩

, we get

0 =

⟨ 𝑇 (𝐱 + 𝐲), 𝐱 + 𝐲 ⟩ -

⟨ 𝑇 (𝐱 - 𝐲), 𝐱 - 𝐲 ⟩ =

4 Re ⟨ 𝑇 𝐱, 𝐲 ⟩

. With

𝐲 = 𝑇 𝐱

this is

4 {‖ 𝑇 𝐱 ‖}^{2}

. For the last claim, apply this to

𝑆 - 𝑇

, which is self-adjoint.

Let $𝐱, 𝐲 \in 𝑉$ . Expanding by linearity of $𝑇$ and sesquilinearity, the terms $⟨ 𝑇 𝐱, 𝐱 ⟩$ and $⟨ 𝑇 𝐲, 𝐲 ⟩$ cancel, and

⟨ 𝑇 (𝐱 + 𝐲), 𝐱 + 𝐲 ⟩ - ⟨ 𝑇 (𝐱 - 𝐲), 𝐱 - 𝐲 ⟩ = 2 ⟨ 𝑇 𝐱, 𝐲 ⟩ + 2 ⟨ 𝑇 𝐲, 𝐱 ⟩ .

Since $𝑇$ is self-adjoint, $⟨ 𝑇 𝐲, 𝐱 ⟩ =$ $⟨ 𝐲, 𝑇 𝐱 ⟩ =$ $𝑧$ , where $𝑧 = ⟨ 𝑇 𝐱, 𝐲 ⟩$ . So the right side is $2 (𝑧 + 𝑧) = 4 Re 𝑧$ , and the left side is $0$ by hypothesis. Hence $Re ⟨ 𝑇 𝐱, 𝐲 ⟩ = 0$ for all $𝐱, 𝐲 \in 𝑉$ . Taking $𝐲 = 𝑇 𝐱$ gives ${‖ 𝑇 𝐱 ‖}^{2} =$ $Re ⟨ 𝑇 𝐱, 𝑇 𝐱 ⟩ =$ $0$ , so $𝑇 𝐱 = 𝟎$ (Proposition 6.1.27). As $𝐱$ was arbitrary, $𝑇 = 0$ .

For the last statement, $𝑆 - 𝑇$ is self-adjoint, since ${(𝑆 - 𝑇)}^{*} =$ $𝑆^{*} -$ $𝑇^{*} =$ $𝑆 -$ $𝑇$ by (a) and (b) of Proposition 6.6.18, and $⟨ (𝑆 - 𝑇) 𝐯, 𝐯 ⟩ =$ $⟨ 𝑆 𝐯, 𝐯 ⟩ -$ $⟨ 𝑇 𝐯, 𝐯 ⟩ =$ $0$ for every $𝐯$ . So $𝑆 - 𝑇 = 0$ . □

Proof (with the test vector $𝐯 + 𝑇 𝐯$ ).standard

Uses. Definition 6.6.26 (Self-adjoint operator), Proposition 6.1.4 (Sesquilinearity), Proposition 6.1.27 (Basic properties of the norm)

Techniques. Plug in clever test vectors (Technique 6.1.40), Show a vector is zero via its norm (Technique 6.1.36)

Hint 1. We want

{‖ 𝑇 𝐯 ‖}^{2} =

⟨ 𝑇 𝐯, 𝑇 𝐯 ⟩ =

⟨ 𝑇^{2} 𝐯, 𝐯 ⟩

to vanish. Feed the hypothesis a vector built from

𝐯

and

𝑇 𝐯

Hint 2. Try

𝐯 + 𝑇 𝐯

. Two of the four terms in the expansion vanish by the hypothesis, applied at

𝐯

and at

𝑇 𝐯

Sketch.

⟨ 𝑇 (𝐯 + 𝑇 𝐯), 𝐯 + 𝑇 𝐯 ⟩ =

⟨ 𝑇 𝐯, 𝐯 ⟩ +

⟨ 𝑇 𝐯, 𝑇 𝐯 ⟩ +

⟨ 𝑇^{2} 𝐯, 𝐯 ⟩ +

⟨ 𝑇 (𝑇 𝐯), 𝑇 𝐯 ⟩

. The first and last terms vanish by hypothesis, and

⟨ 𝑇^{2} 𝐯, 𝐯 ⟩ =

⟨ 𝑇 𝐯, 𝑇 𝐯 ⟩

by self-adjointness, so

0 = 2 {‖ 𝑇 𝐯 ‖}^{2}

Let $𝐯 \in 𝑉$ and put $𝐰 = 𝑇 𝐯$ . Applying the hypothesis to $𝐯 + 𝐰$ and expanding,

0 = ⟨ 𝑇 𝐯, 𝐯 ⟩ + ⟨ 𝑇 𝐯, 𝐰 ⟩ + ⟨ 𝑇 𝐰, 𝐯 ⟩ + ⟨ 𝑇 𝐰, 𝐰 ⟩ .

The first and last terms are $0$ by hypothesis, applied at $𝐯$ and at $𝐰$ . The second is $⟨ 𝑇 𝐯, 𝑇 𝐯 ⟩ =$ ${‖ 𝑇 𝐯 ‖}^{2}$ , and the third is $⟨ 𝑇 (𝑇 𝐯), 𝐯 ⟩ =$ $⟨ 𝑇 𝐯, 𝑇 𝐯 ⟩ =$ ${‖ 𝑇 𝐯 ‖}^{2}$ because $𝑇$ is self-adjoint. So $2 {‖ 𝑇 𝐯 ‖}^{2} = 0$ , and $𝑇 𝐯 = 𝟎$ (Proposition 6.1.27). The last statement follows as in the first proof. □

The first proof is the polarization argument of Proposition 6.1.41, with self-adjointness supplying the second equation that the scalar $𝑖$ supplied there. The second proof chooses its test vector by looking at the goal. We want ${‖ 𝑇 𝐯 ‖}^{2}$ , which by self-adjointness is $⟨ 𝑇^{2} 𝐯, 𝐯 ⟩$ , and $𝐯 + 𝑇 𝐯$ is the simplest vector whose expansion contains that term.

Without self-adjointness, the real case is described by Exercise 6.1.52(b): an operator $𝑇$ on a real inner product space satisfies $⟨ 𝑇 𝐯, 𝐯 ⟩ = 0$ for all $𝐯$ exactly when $⟨ 𝑇 𝐯, 𝐰 ⟩ =$ $- ⟨ 𝐯, 𝑇 𝐰 ⟩$ for all $𝐯, 𝐰$ , that is, when $𝑇^{*} = - 𝑇$ . The rotation $𝐽$ is of this kind, and such operators have a name.

Definition 8.1.5 (Skew-adjoint operators, skew-Hermitian matrices). An operator

𝑇 \in L (𝑉)

is skew-adjoint if it has an adjoint and

𝑇^{*} = - 𝑇

. A matrix

𝐴 \in 𝔽^{𝑛 \times 𝑛}

is skew-Hermitian if

𝐴^{*} = - 𝐴

. A real skew-Hermitian matrix, one with

𝐴^{𝖳} = - 𝐴

, is skew-symmetric.

The matrices $(\begin{matrix} 0 & - 𝑏 \\ 𝑏 & 0 \end{matrix})$ with $𝑏 \in ℝ$ , among them $𝐽$ , are skew-symmetric, and every skew-symmetric matrix has zeros on its diagonal. A skew-Hermitian matrix has purely imaginary diagonal entries, as in $(\begin{matrix} 𝑖 & 1 + 𝑖 \\ - 1 + 𝑖 & 0 \end{matrix})$ . Differentiation on trigonometric polynomials is skew-adjoint (Exercise 6.6.39). Over $ℂ$ , $𝑇$ is skew-adjoint if and only if $𝑖 𝑇$ is self-adjoint, because ${(𝑖 𝑇)}^{*} = - 𝑖 𝑇^{*}$ (Proposition 6.6.18(b)). For every $𝑇$ , the operator $𝑇 - 𝑇^{*}$ is skew-adjoint.

Over $ℝ$ , every operator $𝑇$ splits as $𝑇 = 𝑆 + 𝐾$ with $𝑆 = (𝑇 + 𝑇^{*}) / 2$ self-adjoint and $𝐾 = (𝑇 - 𝑇^{*}) / 2$ skew-adjoint, and $⟨ 𝐾 𝐯, 𝐯 ⟩ = 0$ for every $𝐯$ . So $⟨ 𝑇 𝐯, 𝐯 ⟩ =$ $⟨ 𝑆 𝐯, 𝐯 ⟩$ , and by Proposition 8.1.4 these values determine $𝑆$ and nothing more. The values $⟨ 𝑇 𝐯, 𝐯 ⟩$ see exactly the self-adjoint part of a real operator.

Exercise 8.1.6.routine Let $𝐾 \in L (𝑉)$ be skew-adjoint and $𝑐 \in ℝ$ . Show that

{‖ (𝐾 - 𝑐 id) 𝐯 ‖}^{2} = {‖ 𝐾 𝐯 ‖}^{2} + 𝑐^{2} {‖ 𝐯 ‖}^{2} for every 𝐯 \in 𝑉 .

Deduce that $𝐾$ has no nonzero real eigenvalue, and that $id + 𝐾$ is injective.

Solution.

Uses. Definition 8.1.5 (Skew-adjoint operators, skew-Hermitian matrices), Definition 6.6.8 (Adjoint), Proposition 6.1.4 (Sesquilinearity), Proposition 6.1.27 (Basic properties of the norm)

Techniques. Move an operator across the inner product (Technique 6.6.20), Show a vector is zero via its norm (Technique 6.1.36)

Hint 1. Expand the left side. The cross terms are

- 𝑐 ⟨ 𝐾 𝐯, 𝐯 ⟩ -

𝑐 ⟨ 𝐯, 𝐾 𝐯 ⟩

, since

𝑐

is real.

Hint 2. Move

𝐾

across in the second cross term:

⟨ 𝐯, 𝐾 𝐯 ⟩ =

⟨ 𝐾^{*} 𝐯, 𝐯 ⟩ =

- ⟨ 𝐾 𝐯, 𝐯 ⟩

Sketch. The cross terms are

- 𝑐 (⟨ 𝐾 𝐯, 𝐯 ⟩ + ⟨ 𝐯, 𝐾 𝐯 ⟩) =

- 𝑐 (⟨ 𝐾 𝐯, 𝐯 ⟩ - ⟨ 𝐾 𝐯, 𝐯 ⟩) =

0

. If

𝐾 𝐯 = 𝑐 𝐯

with

𝑐 \neq 0

real and

𝐯 \neq 𝟎

, the identity gives

0 \geq 𝑐^{2} {‖ 𝐯 ‖}^{2} > 0

. For

id + 𝐾

, take

𝑐 = - 1

Since $𝑐$ is real, $𝑐 = 𝑐$ , and expanding gives

{‖ 𝐾 𝐯 - 𝑐 𝐯 ‖}^{2} = {‖ 𝐾 𝐯 ‖}^{2} - 𝑐 ⟨ 𝐾 𝐯, 𝐯 ⟩ - 𝑐 ⟨ 𝐯, 𝐾 𝐯 ⟩ + 𝑐^{2} {‖ 𝐯 ‖}^{2} .

By the conjugated form of the defining identity, $⟨ 𝐯, 𝐾 𝐯 ⟩ =$ $⟨ 𝐾^{*} 𝐯, 𝐯 ⟩ =$ $- ⟨ 𝐾 𝐯, 𝐯 ⟩$ , so the two middle terms cancel, which proves the identity.

If $𝐾 𝐯 = 𝑐 𝐯$ with $𝑐 \in ℝ$ , $𝑐 \neq 0$ , and $𝐯 \neq 𝟎$ , the identity gives $0 =$ ${‖ 𝐾 𝐯 ‖}^{2} +$ $𝑐^{2} {‖ 𝐯 ‖}^{2} \geq$ $𝑐^{2} {‖ 𝐯 ‖}^{2} >$ $0$ , a contradiction. So no nonzero real number is an eigenvalue of $𝐾$ . Finally, with $𝑐 = - 1$ , ${‖ (id + 𝐾) 𝐯 ‖}^{2} =$ ${‖ 𝐾 𝐯 ‖}^{2} +$ ${‖ 𝐯 ‖}^{2}$ , which is positive for $𝐯 \neq 𝟎$ . So $ker (id + 𝐾) = {𝟎}$ , and $id + 𝐾$ is injective.

8.1.3Normal operators

Suppose $𝑉$ is finite-dimensional and has an orthonormal basis $𝒪︀$ of eigenvectors of $𝑇$ . What does this force on $𝑇$ ? In the basis $𝒪︀$ , $𝑇$ is a diagonal matrix $𝐷$ , and $𝑇^{*}$ is $𝐷^{*}$ , which is diagonal too. Diagonal matrices commute, so $𝑇$ and $𝑇^{*}$ commute. Over $ℝ$ , moreover, $𝐷^{*} = 𝐷$ .

Proposition 8.1.7 (What an orthonormal eigenbasis forces). Let

𝑉

be finite-dimensional and

𝑇 \in L (𝑉)

. Suppose that

𝑉

has an orthonormal basis consisting of eigenvectors of

𝑇

. Then

𝑇^{*} 𝑇 = 𝑇 𝑇^{*}

. If moreover every eigenvalue of

𝑇

is real, which is automatic when

𝔽 = ℝ

, then

𝑇

is self-adjoint.

Proof.routine

Uses. Theorem 6.6.15 (Existence of adjoints), Definition 6.3.26 (Conjugate transpose), Theorem 2.4.2 (The matrix of a composite), Theorem 2.3.20 (Linear maps correspond to matrices)

Techniques. Pass to orthonormal coordinates (Technique 6.7.2), Translate between maps and matrices (Technique 2.3.27)

Hint 1. Write down the matrix

𝐷

𝑇

in the given basis, and use Theorem 6.6.15(b) for the matrix of

𝑇^{*}

Hint 2.

𝐷^{*}

is diagonal with entries

𝜆_{𝑗}

, and diagonal matrices commute.

Sketch.

{[𝑇]}_{𝒪︀} =

𝐷 =

diag (𝜆_{1}, \dots, 𝜆_{𝑛})

, so

{[𝑇^{*}]}_{𝒪︀} =

𝐷^{*} =

diag (𝜆_{1}, \dots, 𝜆_{𝑛})

. Then

{[𝑇^{*} 𝑇]}_{𝒪︀} =

𝐷^{*} 𝐷 =

𝐷 𝐷^{*} =

{[𝑇 𝑇^{*}]}_{𝒪︀}

, so

𝑇^{*} 𝑇 = 𝑇 𝑇^{*}

. If all

𝜆_{𝑗}

are real,

𝐷^{*} = 𝐷

and

𝑇^{*} = 𝑇

Let

𝒪︀ = 𝐮_{1}, \dots, 𝐮_{𝑛}

be the orthonormal basis, with

𝑇 𝐮_{𝑗} = 𝜆_{𝑗} 𝐮_{𝑗}

. Then

{[𝑇]}_{𝒪︀} =

𝐷 =

diag (𝜆_{1}, \dots, 𝜆_{𝑛})

, and by Theorem 6.6.15(b) and Definition 6.3.26,

{[𝑇^{*}]}_{𝒪︀} =

𝐷^{*} =

diag (𝜆_{1}, \dots, 𝜆_{𝑛})

. By Theorem 2.4.2,

{[𝑇^{*} 𝑇]}_{𝒪︀} = 𝐷^{*} 𝐷

and

{[𝑇 𝑇^{*}]}_{𝒪︀} = 𝐷 𝐷^{*}

, and both equal

diag ({| 𝜆_{1} |}^{2}, \dots, {| 𝜆_{𝑛} |}^{2})

. So

𝑇^{*} 𝑇 = 𝑇 𝑇^{*}

by Theorem 2.3.20. If every

𝜆_{𝑗}

is real, then

𝐷^{*} = 𝐷

, so

{[𝑇^{*}]}_{𝒪︀} = {[𝑇]}_{𝒪︀}

and

𝑇^{*} = 𝑇

. □

So orthonormal eigenbases occur only for operators that commute with their adjoints, and over $ℝ$ only for self-adjoint ones. The spectral theorems of Section 8.2 prove the converses in finite dimensions: over $ℂ$ the first condition suffices, and over $ℝ$ the second. Over $ℂ$ , self-adjointness would be too much to ask, since a diagonal matrix such as $diag (1, 𝑖)$ has an orthonormal eigenbasis without being Hermitian. The weaker condition gets a name.

Definition 8.1.8 (Normal operator, normal matrix). An operator $𝑇 \in L (𝑉)$ is normal if it has an adjoint and

𝑇^{*} 𝑇 = 𝑇 𝑇^{*} .

A matrix $𝐴 \in 𝔽^{𝑛 \times 𝑛}$ is normal if $𝐴^{*} 𝐴 = 𝐴 𝐴^{*}$ .

For $𝐴 \in 𝔽^{𝑛 \times 𝑛}$ , the operator $𝑇_{𝐴}$ on $𝔽^{𝑛}$ with the dot product is normal exactly when $𝐴$ is, because ${(𝑇_{𝐴})}^{*} = 𝑇_{𝐴^{*}}$ (Example 6.6.10) and the matrix of a composite is the product of the matrices.

Example 8.1.9 (Normal operators).

(a)Self-adjoint and skew-adjoint operators are normal: if $𝑇^{*} = 𝑇$ , then $𝑇^{*} 𝑇 = 𝑇^{2} = 𝑇 𝑇^{*}$ , and if $𝑇^{*} = - 𝑇$ , then $𝑇^{*} 𝑇 = - 𝑇^{2} = 𝑇 𝑇^{*}$ .
(b)Unitary matrices are normal, since $𝑈^{*} 𝑈 = 𝐼_{𝑛} = 𝑈 𝑈^{*}$ by Proposition 6.3.34; in particular orthogonal matrices are normal. More generally, every operator with $𝑇^{*} 𝑇 = 𝑇 𝑇^{*} = id$ is normal. These are the unitary operators (Definition 8.4.9) of Section 8.4.
(c)For $𝑎, 𝑏 \in ℝ$ , the matrix $𝐴 =$ $(\begin{matrix} 𝑎 & - 𝑏 \\ 𝑏 & 𝑎 \end{matrix}) =$ $𝑎 𝐼_{2} +$ $𝑏 𝐽$ satisfies
$𝐴^{𝖳} 𝐴 = (\begin{matrix} 𝑎 & 𝑏 \\ - 𝑏 & 𝑎 \end{matrix}) (\begin{matrix} 𝑎 & - 𝑏 \\ 𝑏 & 𝑎 \end{matrix}) = (𝑎^{2} + 𝑏^{2}) 𝐼_{2} = 𝐴 𝐴^{𝖳} .$
When $𝑎 \neq 0$ , $𝑏 \neq 0$ , and $𝑎^{2} + 𝑏^{2} \neq 1$ , it is neither symmetric, skew-symmetric, nor orthogonal. Identifying $ℝ^{2}$ with $ℂ$ by $(𝑥, 𝑦) \leftrightarrow 𝑥 + 𝑖 𝑦$ , the map $𝐱 \mapsto 𝐴 𝐱$ is multiplication by $𝑎 + 𝑏 𝑖$ : it rotates by the argument of $𝑎 + 𝑏 𝑖$ and scales by $| 𝑎 + 𝑏 𝑖 |$ .
(d)Every diagonal matrix $𝐷$ is normal, as $𝐷^{*} 𝐷 = 𝐷 𝐷^{*}$ is diagonal with entries ${| 𝑑_{𝑗 𝑗} |}^{2}$ . For instance $diag (1, 2 𝑖)$ is normal but not Hermitian, not skew-Hermitian, and not unitary.

Non-example 8.1.10 (Operators that are not normal).

The nilpotent $𝑁 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ : here $𝑁^{*} 𝑁 = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})$ while $𝑁 𝑁^{*} = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ .
The shear $𝑆 = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ (Example 7.4.8), which is invertible: $𝑆^{𝖳} 𝑆 = (\begin{matrix} 1 & 1 \\ 1 & 2 \end{matrix})$ while $𝑆 𝑆^{𝖳} = (\begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix})$ .
$𝐵 = (\begin{matrix} 1 & 1 \\ 0 & 2 \end{matrix})$ , which is diagonalizable, with eigenvectors $(1, 0)$ for $1$ and $(1, 1)$ for $2$ : $𝐵^{𝖳} 𝐵 = (\begin{matrix} 1 & 1 \\ 1 & 5 \end{matrix})$ while $𝐵 𝐵^{𝖳} = (\begin{matrix} 2 & 2 \\ 2 & 4 \end{matrix})$ .
The complex symmetric matrix $𝐶 = (\begin{matrix} 1 & 𝑖 \\ 𝑖 & - 1 \end{matrix})$ of Exercise 6.6.34: $𝐶^{*} 𝐶 = (\begin{matrix} 2 & 2 𝑖 \\ - 2 𝑖 & 2 \end{matrix})$ while $𝐶 𝐶^{*} = (\begin{matrix} 2 & - 2 𝑖 \\ 2 𝑖 & 2 \end{matrix})$ .

None of these has an orthonormal eigenbasis, by Proposition 8.1.7. For $𝑁$ , $𝑆$ , and $𝐶$ there is no eigenbasis at all: each has a single eigenvalue ( $𝐶^{2} = 0$ , so $0$ is the only eigenvalue of $𝐶$ ) and is not a multiple of the identity. The matrix $𝐵$ does have an eigenbasis, but its eigenspaces are two lines that are not perpendicular, so no choice of eigenvectors is orthogonal.

Sums of normal operators need not be normal: $𝑁 =$ $(𝑁 + 𝑁^{*}) / 2 +$ $(𝑁 - 𝑁^{*}) / 2$ is a sum of a self-adjoint and a skew-adjoint operator (Exercise 8.1.24). So the normal operators do not form a subspace of $L (𝑉)$ , unlike the self-adjoint operators over $ℝ$ .

Corollary 8.1.11 (Normality in an orthonormal basis). Let

𝑉

be finite-dimensional, let

𝒪︀

be an orthonormal basis of

𝑉

, let

𝑇 \in L (𝑉)

, and let

𝐴 = {[𝑇]}_{𝒪︀}

. Then

𝑇

is normal, self-adjoint, skew-adjoint, or satisfies

𝑇^{*} 𝑇 = 𝑇 𝑇^{*} = id

if and only if

𝐴

is normal, Hermitian, skew-Hermitian, or unitary, respectively.

Proof.routine

Uses. 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), Proposition 6.3.34 (Characterizations of unitary matrices), Example 2.3.6 (The zero map and the identity map)

Techniques. Translate between maps and matrices (Technique 2.3.27)

Hint 1. Translate each operator identity into a matrix identity with

{[𝑇^{*}]}_{𝒪︀} = 𝐴^{*}

(Theorem 6.6.15(b)).

Hint 2. Two operators are equal if and only if their matrices in

𝒪︀

are equal.

Sketch.

{[𝑇^{*}]}_{𝒪︀} = 𝐴^{*}

{[𝑇^{*} 𝑇]}_{𝒪︀} = 𝐴^{*} 𝐴

{[𝑇 𝑇^{*}]}_{𝒪︀} = 𝐴 𝐴^{*}

{[- 𝑇]}_{𝒪︀} = - 𝐴

, and

{[{id}_{𝑉}]}_{𝒪︀} = 𝐼_{𝑛}

. Since

𝑇 \mapsto {[𝑇]}_{𝒪︀}

is injective, each identity among operators holds iff the same identity holds among the matrices.

By Theorem 6.6.15(b),

{[𝑇^{*}]}_{𝒪︀} = 𝐴^{*}

. By Theorem 2.4.2,

{[𝑇^{*} 𝑇]}_{𝒪︀} = 𝐴^{*} 𝐴

and

{[𝑇 𝑇^{*}]}_{𝒪︀} = 𝐴 𝐴^{*}

; also

{[- 𝑇]}_{𝒪︀} = - 𝐴

, since every entry of the matrix of

- 𝑇

is the negative of the corresponding entry of

𝐴

, and

{[{id}_{𝑉}]}_{𝒪︀} = 𝐼_{𝑛}

(Example 2.3.6). By Theorem 2.3.20, two operators are equal if and only if their matrices in

𝒪︀

are equal. So

𝑇^{*} 𝑇 = 𝑇 𝑇^{*}

iff

𝐴^{*} 𝐴 = 𝐴 𝐴^{*}

𝑇^{*} = 𝑇

iff

𝐴^{*} = 𝐴

𝑇^{*} = - 𝑇

iff

𝐴^{*} = - 𝐴

, and

𝑇^{*} 𝑇 = 𝑇 𝑇^{*} = {id}_{𝑉}

iff

𝐴^{*} 𝐴 = 𝐴 𝐴^{*} = 𝐼_{𝑛}

, which by Proposition 6.3.34 says that

𝐴

is unitary. □

Orthonormality matters here, as it did for Corollary 6.6.28. The matrix $𝐵$ of Non-example 8.1.10 is diagonal in the basis $(1, 0), (1, 1)$ , so its matrix in that basis is normal, but $𝑇_{𝐵}$ is not normal for the dot product.

Condition	Operator	Matrix	Eigenvalues
$𝑇^{*} = 𝑇$	self-adjoint	Hermitian (symmetric)	real
$𝑇^{*} = - 𝑇$	skew-adjoint	skew-Hermitian (skew-symmetric)	in $𝑖 ℝ$
$𝑇^{} 𝑇 = 𝑇 𝑇^{} = id$	unitary	unitary (orthogonal)	$\| 𝜆 \| = 1$
$𝑇^{} 𝑇 = 𝑇 𝑇^{}$	normal	normal	any

Table 8.1.1. Four classes of normal operators. The third column names the matrix of the operator in an orthonormal basis (Corollary 8.1.11), with the name used for real matrices in parentheses. The last column is Corollary 8.1.20. Under the analogy between

𝑇 \mapsto 𝑇^{*}

and

𝑧 \mapsto 𝑧

(Section 6.6.5), the rows correspond to the real numbers (

𝑧 = 𝑧

), the imaginary numbers (

𝑧 = - 𝑧

), the unit circle (

𝑧 𝑧 = 1

), and all of

ℂ

, where

𝑧 𝑧 = 𝑧 𝑧

always holds.

Exercise 8.1.12.routine Let

𝑇 \in L (𝑉)

be normal and

𝑐 \in 𝔽

. Show that

𝑇^{*}

𝑐 𝑇

, and

𝑇^{𝑘}

for

𝑘 \geq 1

are normal. Show also that if

𝑇

is invertible and

𝑇^{- 1}

has an adjoint, then

𝑇^{- 1}

is normal.

Solution.

Uses. Definition 8.1.8 (Normal operator, normal matrix), Proposition 6.6.18 (Properties of the adjoint)

Hint 1. Use Proposition 6.6.18 to write down each adjoint:

{(𝑇^{*})}^{*} = 𝑇

{(𝑐 𝑇)}^{*} = 𝑐 𝑇^{*}

{(𝑇^{𝑘})}^{*} = {(𝑇^{*})}^{𝑘}

{(𝑇^{- 1})}^{*} = {(𝑇^{*})}^{- 1}

Hint 2. For

𝑇^{𝑘}

: since

𝑇

and

𝑇^{*}

commute, any product of copies of

𝑇

and

𝑇^{*}

can be reordered. For

𝑇^{- 1}

, invert both sides of

𝑇^{*} 𝑇 = 𝑇 𝑇^{*}

Sketch.

{(𝑇^{*})}^{*} 𝑇^{*} =

𝑇 𝑇^{*} =

𝑇^{*} 𝑇 =

𝑇^{*} {(𝑇^{*})}^{*}

{(𝑐 𝑇)}^{*} (𝑐 𝑇) =

{| 𝑐 |}^{2} 𝑇^{*} 𝑇 =

{| 𝑐 |}^{2} 𝑇 𝑇^{*} =

(𝑐 𝑇) {(𝑐 𝑇)}^{*}

{(𝑇^{*})}^{𝑘} 𝑇^{𝑘} =

𝑇^{𝑘} {(𝑇^{*})}^{𝑘}

because

𝑇

and

𝑇^{*}

commute. Inverting

𝑇^{*} 𝑇 = 𝑇 𝑇^{*}

gives

𝑇^{- 1} {(𝑇^{*})}^{- 1} =

{(𝑇^{*})}^{- 1} 𝑇^{- 1}

, and

{(𝑇^{*})}^{- 1} = {(𝑇^{- 1})}^{*}

By Proposition 6.6.18(c), ${(𝑇^{*})}^{*} = 𝑇$ , so ${(𝑇^{*})}^{*} 𝑇^{*} =$ $𝑇 𝑇^{*} =$ $𝑇^{*} 𝑇 =$ $𝑇^{*} {(𝑇^{*})}^{*}$ , and $𝑇^{*}$ is normal. By (b) of the same proposition, ${(𝑐 𝑇)}^{*} = 𝑐 𝑇^{*}$ , so

{(𝑐 𝑇)}^{*} (𝑐 𝑇) = 𝑐 𝑐 𝑇^{*} 𝑇 = {| 𝑐 |}^{2} 𝑇 𝑇^{*} = (𝑐 𝑇) {(𝑐 𝑇)}^{*} .

By (d), applied $𝑘 - 1$ times, ${(𝑇^{𝑘})}^{*} = {(𝑇^{*})}^{𝑘}$ . Since $𝑇 𝑇^{*} = 𝑇^{*} 𝑇$ , we can move each factor $𝑇^{*}$ in ${(𝑇^{*})}^{𝑘} 𝑇^{𝑘}$ past each factor $𝑇$ , one swap at a time, so ${(𝑇^{*})}^{𝑘} 𝑇^{𝑘} =$ $𝑇^{𝑘} {(𝑇^{*})}^{𝑘}$ , and $𝑇^{𝑘}$ is normal.

Finally, suppose $𝑇$ is invertible and $𝑇^{- 1}$ has an adjoint. By Proposition 6.6.18(f), $𝑇^{*}$ is invertible with ${(𝑇^{*})}^{- 1} = {(𝑇^{- 1})}^{*}$ . Taking inverses of both sides of $𝑇^{*} 𝑇 = 𝑇 𝑇^{*}$ gives $𝑇^{- 1} {(𝑇^{*})}^{- 1} =$ ${(𝑇^{*})}^{- 1} 𝑇^{- 1}$ , that is, $𝑇^{- 1} {(𝑇^{- 1})}^{*} =$ ${(𝑇^{- 1})}^{*} 𝑇^{- 1}$ .

8.1.4Normal operators and their adjoints

The definition compares the operators $𝑇^{*} 𝑇$ and $𝑇 𝑇^{*}$ . Feeding both the same vector and taking an inner product with that vector turns this into a comparison of lengths.

Proposition 8.1.13 (Normal means $‖ 𝑇 𝐯 ‖ = ‖ 𝑇^{*} 𝐯 ‖$ ). Let $𝑇 \in L (𝑉)$ have an adjoint. Then $𝑇$ is normal if and only if

‖ 𝑇 𝐯 ‖ = ‖ 𝑇^{*} 𝐯 ‖ for every 𝐯 \in 𝑉 .

Proof (through the diagonal values of $𝑇^{*} 𝑇 - 𝑇 𝑇^{*}$ ).standard

Uses. Definition 6.6.8 (Adjoint), Proposition 6.6.18 (Properties of the adjoint), Proposition 8.1.4 (Self-adjoint operators are determined by their diagonal values), Definition 8.1.8 (Normal operator, normal matrix)

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

Hint 1. Write

{‖ 𝑇 𝐯 ‖}^{2}

and

{‖ 𝑇^{*} 𝐯 ‖}^{2}

in the form

⟨ 𝑋 𝐯, 𝐯 ⟩

for suitable operators

𝑋

Hint 2.

{‖ 𝑇 𝐯 ‖}^{2} -

{‖ 𝑇^{*} 𝐯 ‖}^{2} =

⟨ (𝑇^{*} 𝑇 - 𝑇 𝑇^{*}) 𝐯, 𝐯 ⟩

. Which proposition turns “this vanishes for every

𝐯

” into “

𝑇^{*} 𝑇 - 𝑇 𝑇^{*} = 0

”? Check its hypothesis.

Sketch. Moving operators across,

{‖ 𝑇 𝐯 ‖}^{2} -

{‖ 𝑇^{*} 𝐯 ‖}^{2} =

⟨ 𝑆 𝐯, 𝐯 ⟩

with

𝑆 = 𝑇^{*} 𝑇 - 𝑇 𝑇^{*}

. The operator

𝑆

is self-adjoint, so by Proposition 8.1.4,

𝑆 = 0

if and only if

⟨ 𝑆 𝐯, 𝐯 ⟩ = 0

for every

𝐯

For $𝐯 \in 𝑉$ , the defining identities of $𝑇^{*}$ and of ${(𝑇^{*})}^{*} = 𝑇$ (Proposition 6.6.18(c)) give

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

So ${‖ 𝑇 𝐯 ‖}^{2} -$ ${‖ 𝑇^{*} 𝐯 ‖}^{2} =$ $⟨ 𝑆 𝐯, 𝐯 ⟩$ , where $𝑆 = 𝑇^{*} 𝑇 - 𝑇 𝑇^{*}$ . If $𝑇$ is normal, then $𝑆 = 0$ and the norms agree. Conversely, $𝑆$ is self-adjoint: by (a)–(d) of Proposition 6.6.18,

𝑆^{*} = {(𝑇^{*} 𝑇)}^{*} - {(𝑇 𝑇^{*})}^{*} = 𝑇^{*} {(𝑇^{*})}^{*} - {(𝑇^{*})}^{*} 𝑇^{*} = 𝑇^{*} 𝑇 - 𝑇 𝑇^{*} = 𝑆 .

If $‖ 𝑇 𝐯 ‖ = ‖ 𝑇^{*} 𝐯 ‖$ for every $𝐯$ , then $⟨ 𝑆 𝐯, 𝐯 ⟩ = 0$ for every $𝐯$ , and Proposition 8.1.4 gives $𝑆 = 0$ . □

Proof (by polarization).standard

Uses. Proposition 6.1.32 (Polarization identities), Corollary 6.1.38 (Linear maps are determined by inner products), Definition 6.6.8 (Adjoint), Proposition 6.6.18 (Properties of the adjoint)

Techniques. Polarization (Technique 6.1.33)

Hint 1. Equal norms for all vectors should give equal inner products. Which identity recovers

⟨ 𝐱, 𝐲 ⟩

from norms?

Hint 2. Apply Proposition 6.1.32 to

⟨ 𝑇 𝐯, 𝑇 𝐰 ⟩

and to

⟨ 𝑇^{*} 𝐯, 𝑇^{*} 𝐰 ⟩

, using

𝑇 (𝐯 + 𝑐 𝐰) =

𝑇 𝐯 +

𝑐 𝑇 𝐰

Sketch. By polarization,

⟨ 𝑇 𝐯, 𝑇 𝐰 ⟩

is a fixed combination of the numbers

{‖ 𝑇 (𝐯 + 𝑐 𝐰) ‖}^{2}

with

𝑐 \in {\pm 1}

(over

ℝ

) or

𝑐 \in {\pm 1, \pm 𝑖}

(over

ℂ

), and the same combination of the numbers

{‖ 𝑇^{*} (𝐯 + 𝑐 𝐰) ‖}^{2}

gives

⟨ 𝑇^{*} 𝐯, 𝑇^{*} 𝐰 ⟩

. So

⟨ 𝑇^{*} 𝑇 𝐯, 𝐰 ⟩ =

⟨ 𝑇 𝑇^{*} 𝐯, 𝐰 ⟩

for all

𝐯, 𝐰

, and

𝑇^{*} 𝑇 = 𝑇 𝑇^{*}

If $𝑇$ is normal, then ${‖ 𝑇 𝐯 ‖}^{2} =$ $⟨ 𝑇^{*} 𝑇 𝐯, 𝐯 ⟩ =$ $⟨ 𝑇 𝑇^{*} 𝐯, 𝐯 ⟩ =$ ${‖ 𝑇^{*} 𝐯 ‖}^{2}$ for every $𝐯$ .

Conversely, suppose $‖ 𝑇 𝐱 ‖ = ‖ 𝑇^{*} 𝐱 ‖$ for every $𝐱 \in 𝑉$ , and let $𝐯, 𝐰 \in 𝑉$ . Over $ℂ$ , Proposition 6.1.32(b) and linearity of $𝑇$ and $𝑇^{*}$ give

⟨ 𝑇 𝐯, 𝑇 𝐰 ⟩ = \frac{1}{4} \sum_{𝑘 = 0}^{3} 𝑖^{𝑘} {‖ 𝑇 (𝐯 + 𝑖^{𝑘} 𝐰) ‖}^{2} = \frac{1}{4} \sum_{𝑘 = 0}^{3} 𝑖^{𝑘} {‖ 𝑇^{*} (𝐯 + 𝑖^{𝑘} 𝐰) ‖}^{2} = ⟨ 𝑇^{*} 𝐯, 𝑇^{*} 𝐰 ⟩,

and over $ℝ$ the same holds with Proposition 6.1.32(a), which uses only $𝐯 \pm 𝐰$ . Moving the operators across, the two ends of this chain are $⟨ 𝑇^{*} 𝑇 𝐯, 𝐰 ⟩$ and $⟨ 𝑇 𝑇^{*} 𝐯, 𝐰 ⟩$ . They agree for all $𝐯, 𝐰 \in 𝑉$ , so $𝑇^{*} 𝑇 = 𝑇 𝑇^{*}$ by Corollary 6.1.38. □

The first proof locates normality in a single self-adjoint operator, $𝑇^{*} 𝑇 - 𝑇 𝑇^{*}$ , whose diagonal values are the differences ${‖ 𝑇 𝐯 ‖}^{2} - {‖ 𝑇^{*} 𝐯 ‖}^{2}$ . The second proof produces something visibly stronger along the way: $𝑇$ and $𝑇^{*}$ give the same inner products, $⟨ 𝑇 𝐯, 𝑇 𝐰 ⟩ =$ $⟨ 𝑇^{*} 𝐯, 𝑇^{*} 𝐰 ⟩$ for all $𝐯, 𝐰$ .

Intuition.

𝑇

and

𝑇^{*}

need not be equal, but for a normal operator they stretch every vector by the same factor; they may send it in different directions. For

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

this fails as badly as possible:

𝑁 𝐞_{2} = 𝐞_{1}

has length

1

, while

𝑁^{*} 𝐞_{2} = 𝟎

. Everything else in this section follows from the equality of lengths. The first consequence is about kernels: a vector killed by

𝑇

is killed by

𝑇^{*}

Proposition 8.1.14 (Kernels and images of normal operators). Let $𝑇 \in L (𝑉)$ be normal.

(a) $ker 𝑇 = ker 𝑇^{*}$ .
(b)For every $𝜆 \in 𝔽$ , the operator $𝑇 - 𝜆 id$ is normal, and ${(𝑇 - 𝜆 id)}^{*} =$ $𝑇^{*} -$ $𝜆 id$ .
(c)If $𝑉$ is finite-dimensional, then $im 𝑇 =$ $im 𝑇^{*} =$ ${(ker 𝑇)}^{⟂}$ . So $𝑉 = ker 𝑇 \oplus im 𝑇$ , and $ker 𝑇$ and $im 𝑇$ are orthogonal.

Proof.routine

Uses. Proposition 8.1.13 (Normal means $‖ 𝑇 𝐯 ‖ = ‖ 𝑇^{*} 𝐯 ‖$ ), Proposition 6.1.27 (Basic properties of the norm), Proposition 6.6.18 (Properties of the adjoint), Theorem 6.6.21 (Kernels and images of adjoints), Theorem 6.4.24 (Orthogonal decomposition)

Techniques. Show a vector is zero via its norm (Technique 6.1.36)

Hint 1. For (a),

𝑇 𝐯 = 𝟎

if and only if

‖ 𝑇 𝐯 ‖ = 0

Hint 2. For (b), compute both

{(𝑇 - 𝜆 id)}^{*} (𝑇 - 𝜆 id)

and

(𝑇 - 𝜆 id) {(𝑇 - 𝜆 id)}^{*}

, using

{(𝑇 - 𝜆 id)}^{*} =

𝑇^{*} -

𝜆 id

Hint 3. For (c), combine (a) with (c) and (d) of Theorem 6.6.21.

Sketch. (a)

‖ 𝑇 𝐯 ‖ = ‖ 𝑇^{*} 𝐯 ‖

, so one is

0

iff the other is. (b) Both products equal

𝑇^{*} 𝑇 -

𝜆 𝑇^{*} -

𝜆 𝑇 +

{| 𝜆 |}^{2} id

. (c)

im 𝑇 =

{(ker 𝑇^{*})}^{⟂} =

{(ker 𝑇)}^{⟂} =

im 𝑇^{*}

, and

𝑉 = ker 𝑇 \oplus {(ker 𝑇)}^{⟂}

(a) By Proposition 8.1.13, $‖ 𝑇 𝐯 ‖ = ‖ 𝑇^{*} 𝐯 ‖$ for every $𝐯$ , and a vector is $𝟎$ if and only if its norm is $0$ (Proposition 6.1.27). So $𝑇 𝐯 = 𝟎$ if and only if $𝑇^{*} 𝐯 = 𝟎$ .

(b) By (a), (b), and (e) of Proposition 6.6.18, $𝑇 - 𝜆 id$ has the adjoint $𝑇^{*} - 𝜆 id$ . Expanding,

\begin{aligned} (𝑇^{*} - 𝜆 id) (𝑇 - 𝜆 id) & = 𝑇^{*} 𝑇 - 𝜆 𝑇^{*} - 𝜆 𝑇 + {| 𝜆 |}^{2} id, \\ (𝑇 - 𝜆 id) (𝑇^{*} - 𝜆 id) & = 𝑇 𝑇^{*} - 𝜆 𝑇 - 𝜆 𝑇^{*} + {| 𝜆 |}^{2} id, \end{aligned}

and these are equal because $𝑇^{*} 𝑇 = 𝑇 𝑇^{*}$ .

(c) By (d) and (c) of Theorem 6.6.21, together with (a),

im 𝑇 = {(ker 𝑇^{*})}^{⟂} = {(ker 𝑇)}^{⟂} = im 𝑇^{*} .

By Theorem 6.4.24, applied to the finite-dimensional subspace $ker 𝑇$ , we get $𝑉 =$ $ker 𝑇 \oplus$ ${(ker 𝑇)}^{⟂} =$ $ker 𝑇 \oplus$ $im 𝑇$ , and the two summands are orthogonal by the definition of ${(ker 𝑇)}^{⟂}$ . □

Part (a) has a second proof that never mentions norms: by Proposition 6.6.23(a), applied to $𝑇$ and to $𝑇^{*}$ (whose adjoint is $𝑇$ ), $ker 𝑇 =$ $ker (𝑇^{*} 𝑇) =$ $ker (𝑇 𝑇^{*}) =$ $ker 𝑇^{*}$ .

Part (c) says that a normal operator splits a finite-dimensional $𝑉$ into two perpendicular pieces. It kills $ker 𝑇$ , and it maps $im 𝑇$ into itself injectively, since $im 𝑇 \cap ker 𝑇 = {𝟎}$ ; so on $im 𝑇$ it is a bijection. The nilpotent $𝑁$ shows what normality rules out: $ker 𝑁 =$ $im 𝑁 =$ $span (𝐞_{1})$ , so the image of $𝑁$ falls into its kernel. For a normal $𝑇$ with $𝑇^{2} = 0$ this would put $im 𝑇$ inside $ker 𝑇 \cap im 𝑇 = {𝟎}$ , so $𝑇 = 0$ , which is one of the three proofs in Example 6.7.3.

Part (b) is what lets us pass from kernels to eigenvectors. The move is common enough to name.

Technique 8.1.15 (Shift by a scalar). A statement about the kernel of an operator becomes a statement about eigenvectors once it is applied to $𝑇 - 𝜆 id$ , because $𝐸_{𝜆} (𝑇) =$ $ker (𝑇 - 𝜆 id)$ (Definition 7.2.13). So prove the statement for kernels, check that the hypotheses survive the shift $𝑇 \mapsto 𝑇 - 𝜆 id$ , and apply it to the shifted operator. Hypotheses that survive include being normal, being self-adjoint when $𝜆$ is real, commuting with a fixed operator, and being upper triangular in a fixed basis.

Reach for it when you want a fact about every eigenvalue and the case $𝜆 = 0$ is easier to see. The canonical example is Theorem 8.1.16, which is $ker 𝑇 = ker 𝑇^{*}$ applied to $𝑇 - 𝜆 id$ .

Used in 3 places and 1 practice exercise

Theorem 8.1.16, Theorem 8.1.18, Exercise 8.1.28; 1 practice exercise.

Theorem 8.1.16 (Eigenvectors of a normal operator and of its adjoint). Let $𝑇 \in L (𝑉)$ be normal, $𝐯 \in 𝑉$ , and $𝜆 \in 𝔽$ . Then

𝑇 𝐯 = 𝜆 𝐯 ⟺ 𝑇^{*} 𝐯 = 𝜆 𝐯 .

In particular $𝐸_{𝜆} (𝑇) =$ $𝐸_{𝜆} (𝑇^{*})$ , and $𝜆$ is an eigenvalue of $𝑇$ if and only if $𝜆$ is an eigenvalue of $𝑇^{*}$ , with the same eigenvectors.

Proof.routine

Uses. Proposition 8.1.14 (Kernels and images of normal operators), Definition 7.2.13 (Eigenspace)

Techniques. Shift by a scalar (Technique 8.1.15)

Hint 1. Rewrite

𝑇 𝐯 = 𝜆 𝐯

as a statement about a kernel.

Hint 2. Apply Proposition 8.1.14(a) to the normal operator

𝑇 - 𝜆 id

, whose adjoint is

𝑇^{*} - 𝜆 id

Sketch.

𝑇 𝐯 = 𝜆 𝐯

iff

𝐯 \in

ker (𝑇 - 𝜆 id) =

ker ({(𝑇 - 𝜆 id)}^{*}) =

ker (𝑇^{*} - 𝜆 id)

iff

𝑇^{*} 𝐯 = 𝜆 𝐯

, using (b) and then (a) of Proposition 8.1.14.

By Proposition 8.1.14(b), $𝑇 - 𝜆 id$ is normal with adjoint $𝑇^{*} - 𝜆 id$ . Applying Proposition 8.1.14(a) to it,

ker (𝑇 - 𝜆 id) = ker (𝑇^{*} - 𝜆 id) .

The left side is $𝐸_{𝜆} (𝑇)$ and the right side is $𝐸_{𝜆} (𝑇^{*})$ (Definition 7.2.13), which is the claim. □

Without normality, $𝑇$ and $𝑇^{*}$ usually have different eigenvectors. For $𝑁 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ , $𝑁 𝐞_{1} = 𝟎$ , but $𝑁^{*} 𝐞_{1} = 𝐞_{2}$ , and $𝐸_{0} (𝑁^{*}) =$ $span (𝐞_{2})$ . What survives in general is an orthogonality between the eigenvectors of $𝑇$ and those of $𝑇^{*}$ .

Lemma 8.1.17 (Eigenvectors of

𝑇

and of

𝑇^{*}

). Let

𝑇 \in L (𝑉)

have an adjoint. If

𝑇 𝐯 = 𝜆 𝐯

and

𝑇^{*} 𝐰 = 𝜈 𝐰

with

𝜆 \neq 𝜈

, then

⟨ 𝐯, 𝐰 ⟩ = 0

Proof.routine

Uses. Definition 6.6.8 (Adjoint), Proposition 6.1.4 (Sesquilinearity)

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

Hint. Compute

⟨ 𝑇 𝐯, 𝐰 ⟩ =

⟨ 𝐯, 𝑇^{*} 𝐰 ⟩

in two ways.

Sketch.

𝜆 ⟨ 𝐯, 𝐰 ⟩ =

⟨ 𝑇 𝐯, 𝐰 ⟩ =

⟨ 𝐯, 𝑇^{*} 𝐰 ⟩ =

𝜈 ⟨ 𝐯, 𝐰 ⟩

, and

𝜆 \neq 𝜈

By the defining identity of $𝑇^{*}$ and Proposition 6.1.4,

𝜆 ⟨ 𝐯, 𝐰 ⟩ = ⟨ 𝑇 𝐯, 𝐰 ⟩ = ⟨ 𝐯, 𝑇^{*} 𝐰 ⟩ = ⟨ 𝐯, 𝜈 𝐰 ⟩ = 𝜈 ⟨ 𝐯, 𝐰 ⟩ .

So $(𝜆 - 𝜈) ⟨ 𝐯, 𝐰 ⟩ = 0$ , and since $𝜆 \neq 𝜈$ , $⟨ 𝐯, 𝐰 ⟩ = 0$ . □

Figure 8.1.1. (a) The symmetric matrix

𝐴 = (\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix})

has the eigenspaces

𝐸_{1} (𝐴) =

span ((1, - 1))

and

𝐸_{3} (𝐴) =

span ((1, 1))

, which are perpendicular. (b)

𝐵 = (\begin{matrix} 1 & 1 \\ 0 & 2 \end{matrix})

has the eigenspaces

𝐸_{1} (𝐵) = span (𝐞_{1})

and

𝐸_{2} (𝐵) =

span ((1, 1))

(solid), at an angle of

𝜋 / 4

. The eigenspaces of

𝐵^{𝖳}

(dashed) are

𝐸_{1} (𝐵^{𝖳}) =

span ((1, - 1))

and

𝐸_{2} (𝐵^{𝖳}) =

span (𝐞_{2})

. Each eigenspace of

𝐵

is perpendicular to the eigenspace of

𝐵^{𝖳}

for the other eigenvalue (Lemma 8.1.17). For a normal matrix the solid and dashed lines coincide (Theorem 8.1.16), and the eigenspaces themselves become perpendicular.

Theorem 8.1.18 (Eigenspaces of a normal operator are orthogonal). Let

𝑇 \in L (𝑉)

be normal. If

𝑇 𝐯 = 𝜆 𝐯

and

𝑇 𝐰 = 𝜇 𝐰

with

𝜆 \neq 𝜇

, then

⟨ 𝐯, 𝐰 ⟩ = 0

. So eigenspaces of

𝑇

for distinct eigenvalues are orthogonal.

Proof (through the eigenvectors of $𝑇^{*}$ ).routine

Uses. Theorem 8.1.16 (Eigenvectors of a normal operator and of its adjoint), Lemma 8.1.17 (Eigenvectors of $𝑇$ and of $𝑇^{*}$ )

Hint 1. What does

𝑇^{*}

do to

𝐰

Hint 2.

𝑇^{*} 𝐰 = 𝜇 𝐰

. Apply Lemma 8.1.17 with

𝜈 = 𝜇

Sketch. By Theorem 8.1.16,

𝑇^{*} 𝐰 = 𝜇 𝐰

, and

𝜆 \neq 𝜇 = 𝜇

, so Lemma 8.1.17 applies.

By Theorem 8.1.16,

𝑇^{*} 𝐰 = 𝜇 𝐰

. Since

𝜆 \neq 𝜇 = 𝜇

, Lemma 8.1.17 with

𝜈 = 𝜇

gives

⟨ 𝐯, 𝐰 ⟩ = 0

. For arbitrary

𝐯 \in 𝐸_{𝜆} (𝑇)

and

𝐰 \in 𝐸_{𝜇} (𝑇)

, either one of them is

𝟎

or both are eigenvectors, so the eigenspaces are orthogonal. □

Proof (through kernels and images).standard

Uses. Proposition 8.1.14 (Kernels and images of normal operators), Theorem 6.6.21 (Kernels and images of adjoints), Definition 6.4.22 (Orthogonal complement)

Techniques. Shift by a scalar (Technique 8.1.15)

Hint 1. Let

𝑅 = 𝑇 - 𝜇 id

. Then

𝐰 \in ker 𝑅

. Show that

𝐯 \in im 𝑅

Hint 2.

𝑅 𝐯 = (𝜆 - 𝜇) 𝐯

with

𝜆 - 𝜇 \neq 0

. Now relate

ker 𝑅

im 𝑅

using that

𝑅

is normal.

Sketch. With

𝑅 = 𝑇 - 𝜇 id

𝐯 =

𝑅 (𝐯 / (𝜆 - 𝜇)) \in

im 𝑅

and

𝐰 \in

ker 𝑅 =

ker 𝑅^{*} =

{(im 𝑅)}^{⟂}

, by Proposition 8.1.14 and Theorem 6.6.21(a).

Let $𝑅 = 𝑇 - 𝜇 id$ , which is normal by Proposition 8.1.14(b). Since $𝑅 𝐯 = (𝜆 - 𝜇) 𝐯$ and $𝜆 - 𝜇 \neq 0$ , we have $𝐯 =$ $𝑅 ({(𝜆 - 𝜇)}^{- 1} 𝐯) \in$ $im 𝑅$ . On the other hand $𝐰 \in ker 𝑅$ , and by Proposition 8.1.14(a) and Theorem 6.6.21(a),

ker 𝑅 = ker 𝑅^{*} = {(im 𝑅)}^{⟂} .

So $𝐰$ is orthogonal to every vector of $im 𝑅$ , in particular to $𝐯$ . The statement about eigenspaces follows as in the first proof. □

The first proof reduces the theorem to a fact about single eigenvectors: $𝑇^{*}$ has the same eigenvectors as $𝑇$ , so the orthogonality of Lemma 8.1.17 applies to eigenvectors of $𝑇$ alone. The second proof is about subspaces and proves more. Every eigenvector $𝐰$ for $𝜇$ is orthogonal to all of $im (𝑇 - 𝜇 id)$ , which contains the eigenvectors for every other eigenvalue but in general much more. Part (b) of Proposition 8.1.1 is the special case of self-adjoint $𝑇$ .

Exercise 8.1.19.routine Let

𝑉

be a complex inner product space, let

𝑇 \in L (𝑉)

be normal, and let

𝐯, 𝐰 \in 𝑉

satisfy

𝑇 𝐯 = 2 𝐯

𝑇 𝐰 = 𝑖 𝐰

‖ 𝐯 ‖ = 1

, and

‖ 𝐰 ‖ = 3

. Compute

‖ 𝑇 (𝐯 + 𝐰) ‖

‖ 𝑇^{*} (𝐯 - 𝐰) ‖

, and

⟨ 𝑇 (𝐯 + 𝐰), 𝐯 - 𝐰 ⟩

Solution.

Uses. Theorem 8.1.18 (Eigenspaces of a normal operator are orthogonal), Theorem 8.1.16 (Eigenvectors of a normal operator and of its adjoint), Theorem 6.3.3 (Pythagorean theorem), Proposition 6.1.4 (Sesquilinearity)

Hint 1. The eigenvalues

2

and

𝑖

are distinct. What does that say about

⟨ 𝐯, 𝐰 ⟩

Hint 2.

𝐯 ⟂ 𝐰

, and

𝑇^{*} 𝐯 = 2 𝐯

𝑇^{*} 𝐰 = - 𝑖 𝐰

. Then use Pythagoras.

Sketch.

𝐯 ⟂ 𝐰

𝑇 (𝐯 + 𝐰) =

2 𝐯 +

𝑖 𝐰

has squared norm

4 + 9 = 13

𝑇^{*} (𝐯 - 𝐰) =

2 𝐯 +

𝑖 𝐰

too, with norm

\sqrt{13}

. Finally

⟨ 2 𝐯 + 𝑖 𝐰, 𝐯 - 𝐰 ⟩ =

2 -

9 𝑖

Since $2 \neq 𝑖$ , $⟨ 𝐯, 𝐰 ⟩ = 0$ by Theorem 8.1.18, and by Theorem 8.1.16, $𝑇^{*} 𝐯 = 2 𝐯$ and $𝑇^{*} 𝐰 = 𝑖 𝐰 = - 𝑖 𝐰$ .

First, $𝑇 (𝐯 + 𝐰) =$ $2 𝐯 +$ $𝑖 𝐰$ , a sum of orthogonal vectors, so by Theorem 6.3.3 its squared norm is $4 {‖ 𝐯 ‖}^{2} +$ ${| 𝑖 |}^{2} {‖ 𝐰 ‖}^{2} =$ $4 +$ $9 =$ $13$ , and $‖ 𝑇 (𝐯 + 𝐰) ‖ = \sqrt{13}$ .

Next, $𝑇^{*} (𝐯 - 𝐰) =$ $2 𝐯 -$ $(- 𝑖) 𝐰 =$ $2 𝐯 +$ $𝑖 𝐰$ , so $‖ 𝑇^{*} (𝐯 - 𝐰) ‖ = \sqrt{13}$ as well.

Finally, using $⟨ 𝐯, 𝐰 ⟩ = ⟨ 𝐰, 𝐯 ⟩ = 0$ and Proposition 6.1.4,

⟨ 2 𝐯 + 𝑖 𝐰, 𝐯 - 𝐰 ⟩ = 2 ⟨ 𝐯, 𝐯 ⟩ - 𝑖 ⟨ 𝐰, 𝐰 ⟩ = 2 - 9 𝑖 .

As a check, this also equals $⟨ 𝐯 + 𝐰, 𝑇^{*} (𝐯 - 𝐰) ⟩ =$ $⟨ 𝐯 + 𝐰, 2 𝐯 + 𝑖 𝐰 ⟩ =$ $2 +$ $𝑖 \cdot$ $9 =$ $2 -$ $9 𝑖$ .

Applied to the special classes of Example 8.1.9, Theorem 8.1.16 says what their eigenvalues can be.

Corollary 8.1.20 (Eigenvalues of self-adjoint, skew-adjoint, and unitary operators). Let $𝑇 \in L (𝑉)$ have an adjoint, and let $𝜆$ be an eigenvalue of $𝑇$ .

(a)If $𝑇^{*} = 𝑇$ , then $𝜆 \in ℝ$ .
(b)If $𝑇^{*} = - 𝑇$ , then $𝜆 = - 𝜆$ , that is, $𝜆 \in 𝑖 ℝ$ .
(c)If $𝑇^{*} 𝑇 = 𝑇 𝑇^{*} = id$ , then $| 𝜆 | = 1$ .

Proof.routine

Uses. Example 8.1.9 (Normal operators), Theorem 8.1.16 (Eigenvectors of a normal operator and of its adjoint), Proposition 1.2.35 (Zero products)

Hint 1. Each such

𝑇

is normal. Let

𝐯

be an eigenvector for

𝜆

. What is

𝑇^{*} 𝐯

Hint 2.

𝑇^{*} 𝐯 = 𝜆 𝐯

. Compare with what the hypothesis says about

𝑇^{*} 𝐯

, or for (c) compute

𝑇^{*} 𝑇 𝐯

Sketch. Each

𝑇

is normal, so

𝑇^{*} 𝐯 = 𝜆 𝐯

. (a)

𝜆 𝐯 = 𝑇 𝐯 = 𝜆 𝐯

. (b)

𝜆 𝐯 = - 𝜆 𝐯

. (c)

𝐯 = 𝑇^{*} 𝑇 𝐯 = 𝜆 𝜆 𝐯

In each case $𝑇$ is normal (Example 8.1.9). Let $𝐯$ be an eigenvector for $𝜆$ ; by Theorem 8.1.16, $𝑇^{*} 𝐯 = 𝜆 𝐯$ .

(a) $𝜆 𝐯 =$ $𝑇^{*} 𝐯 =$ $𝑇 𝐯 =$ $𝜆 𝐯$ , so $(𝜆 - 𝜆) 𝐯 = 𝟎$ , and $𝜆 = 𝜆$ since $𝐯 \neq 𝟎$ (Proposition 1.2.35).

(b) $𝜆 𝐯 =$ $𝑇^{*} 𝐯 =$ $- 𝑇 𝐯 =$ $- 𝜆 𝐯$ , so $𝜆 = - 𝜆$ in the same way.

(c) $𝐯 =$ $𝑇^{*} 𝑇 𝐯 =$ $𝑇^{*} (𝜆 𝐯) =$ $𝜆 𝜆 𝐯 =$ ${| 𝜆 |}^{2} 𝐯$ , so ${| 𝜆 |}^{2} = 1$ . □

Part (a) is a third proof of Proposition 8.1.1(a), and part (c) returns in Section 8.4. The converses fail without normality: $𝐵 = (\begin{matrix} 1 & 1 \\ 0 & 2 \end{matrix})$ has real eigenvalues and is not symmetric, and the shear $(\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ has only the eigenvalue $1$ and is not orthogonal. For normal operators on finite-dimensional complex spaces the converses hold, because by Section 8.2 such an operator is determined by its eigenvalues on an orthonormal eigenbasis.

8.1.5The Cartesian decomposition Beyond 21-242

The Cartesian decomposition is in LADR 7A but not in the lecture.

Every complex number is $𝑧 = 𝑥 + 𝑖 𝑦$ with $𝑥, 𝑦$ real, and then ${| 𝑧 |}^{2} = 𝑥^{2} + 𝑦^{2}$ . Operators split the same way: $𝑇 = 𝑆 + 𝐾$ with $𝑆 = (𝑇 + 𝑇^{*}) / 2$ self-adjoint and $𝐾 = (𝑇 - 𝑇^{*}) / 2$ skew-adjoint, and this is the only such splitting (over $ℂ$ this is Exercise 6.6.37, with $𝐾 = 𝑖 𝐵$ ). For numbers the two parts commute automatically. For operators they need not, and their commuting is exactly normality.

Proposition 8.1.21 (Normal means commuting parts). Let $𝑇 \in L (𝑉)$ have an adjoint, and let $𝑆 = (𝑇 + 𝑇^{*}) / 2$ and $𝐾 = (𝑇 - 𝑇^{*}) / 2$ . Then $𝑆$ is self-adjoint, $𝐾$ is skew-adjoint, $𝑇 = 𝑆 + 𝐾$ , and

𝑇^{*} 𝑇 - 𝑇 𝑇^{*} = 2 (𝑆 𝐾 - 𝐾 𝑆) .

Hence $𝑇$ is normal if and only if $𝑆 𝐾 = 𝐾 𝑆$ . When $𝔽 = ℂ$ , writing $𝐾 = 𝑖 𝐵$ with $𝐵 = (𝑇 - 𝑇^{*}) / (2 𝑖)$ self-adjoint, $𝑇$ is normal if and only if $𝑆 𝐵 = 𝐵 𝑆$ .

Proof.routine

Uses. Proposition 6.6.18 (Properties of the adjoint), Definition 8.1.5 (Skew-adjoint operators, skew-Hermitian matrices)

Hint 1.

𝑇^{*} = 𝑆^{*} + 𝐾^{*} = 𝑆 - 𝐾

Hint 2. Expand

(𝑆 - 𝐾) (𝑆 + 𝐾) -

(𝑆 + 𝐾) (𝑆 - 𝐾)

Sketch.

𝑆^{*} = 𝑆

and

𝐾^{*} = - 𝐾

by Proposition 6.6.18, so

𝑇^{*} = 𝑆 - 𝐾

. Then

𝑇^{*} 𝑇 -

𝑇 𝑇^{*} =

(𝑆 - 𝐾) (𝑆 + 𝐾) -

(𝑆 + 𝐾) (𝑆 - 𝐾) =

2 (𝑆 𝐾 - 𝐾 𝑆)

. Over

ℂ

𝑆 𝐾 -

𝐾 𝑆 =

𝑖 (𝑆 𝐵 - 𝐵 𝑆)

By (a)–(c) of Proposition 6.6.18, $𝑆^{*} = (𝑇^{*} + 𝑇) / 2 = 𝑆$ and $𝐾^{*} =$ $(𝑇^{*} - 𝑇) / 2 =$ $- 𝐾$ , and clearly $𝑆 + 𝐾 = 𝑇$ . So $𝑇^{*} = 𝑆 - 𝐾$ , and

\begin{aligned} 𝑇^{*} 𝑇 - 𝑇 𝑇^{*} & = (𝑆 - 𝐾) (𝑆 + 𝐾) - (𝑆 + 𝐾) (𝑆 - 𝐾) \\ = (𝑆^{2} + 𝑆 𝐾 - 𝐾 𝑆 - 𝐾^{2}) - (𝑆^{2} - 𝑆 𝐾 + 𝐾 𝑆 - 𝐾^{2}) = 2 (𝑆 𝐾 - 𝐾 𝑆) . \end{aligned}

So $𝑇^{*} 𝑇 = 𝑇 𝑇^{*}$ if and only if $𝑆 𝐾 = 𝐾 𝑆$ . When $𝔽 = ℂ$ , $𝐵 = - 𝑖 𝐾$ satisfies $𝐵^{*} =$ $- 𝑖 𝐾^{*} =$ $𝑖 (- 𝐾) =$ $𝐵$ , so $𝐵$ is self-adjoint, $𝐾 = 𝑖 𝐵$ , and $𝑆 𝐾 -$ $𝐾 𝑆 =$ $𝑖 (𝑆 𝐵 - 𝐵 𝑆)$ , which vanishes if and only if $𝑆 𝐵 = 𝐵 𝑆$ . □

For $𝐵 = (\begin{matrix} 1 & 1 \\ 0 & 2 \end{matrix})$ , the parts are $𝑆 = (\begin{matrix} 1 & 1 / 2 \\ 1 / 2 & 2 \end{matrix})$ and $𝐾 = (\begin{matrix} 0 & 1 / 2 \\ - 1 / 2 & 0 \end{matrix})$ , and

2 (𝑆 𝐾 - 𝐾 𝑆) = 2 ((\begin{matrix} - 1 / 4 & 1 / 2 \\ - 1 & 1 / 4 \end{matrix}) - (\begin{matrix} 1 / 4 & 1 \\ - 1 / 2 & - 1 / 4 \end{matrix})) = (\begin{matrix} - 1 & - 1 \\ - 1 & 1 \end{matrix}) = 𝐵^{𝖳} 𝐵 - 𝐵 𝐵^{𝖳},

as the proposition predicts, so the parts of $𝐵$ do not commute.

Intuition. For a normal operator, $𝑇^{*} 𝑇 =$ $(𝑆 - 𝐾) (𝑆 + 𝐾) =$ $𝑆^{2} -$ $𝐾^{2}$ , and for every $𝐯$ ,

{‖ 𝑇 𝐯 ‖}^{2} = ⟨ 𝑆^{2} 𝐯, 𝐯 ⟩ - ⟨ 𝐾^{2} 𝐯, 𝐯 ⟩ = {‖ 𝑆 𝐯 ‖}^{2} + {‖ 𝐾 𝐯 ‖}^{2},

since $- ⟨ 𝐾^{2} 𝐯, 𝐯 ⟩ =$ $- ⟨ 𝐾 𝐯, 𝐾^{*} 𝐯 ⟩ =$ ${‖ 𝐾 𝐯 ‖}^{2}$ . This is ${| 𝑧 |}^{2} = 𝑥^{2} + 𝑦^{2}$ for operators, and Exercise 8.1.29 shows that it characterizes normal operators.

Exercises

Exercise 8.1.22.routine For each matrix, decide whether it is Hermitian, skew-Hermitian, unitary, or normal (any number of these may hold).

(a) $(\begin{matrix} 1 & 2 \\ - 2 & 1 \end{matrix}) \in ℝ^{2 \times 2}$ .
(b) $(\begin{matrix} 𝑖 & 1 \\ - 1 & 2 𝑖 \end{matrix})$ .
(c) $(\begin{matrix} 2 & 𝑖 \\ 𝑖 & 2 \end{matrix})$ .
(d) $(\begin{matrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}) \in ℝ^{3 \times 3}$ .
(e) $(\begin{matrix} 1 & 𝑖 \\ 0 & 1 \end{matrix})$ .

Solution.

Uses. Definition 6.3.26 (Conjugate transpose), Definition 8.1.5 (Skew-adjoint operators, skew-Hermitian matrices), Definition 6.3.33 (Unitary and orthogonal matrices), Definition 8.1.8 (Normal operator, normal matrix), Example 8.1.9 (Normal operators)

Hint 1. Compute

𝐴^{*}

first: it decides Hermitian and skew-Hermitian at once. A Hermitian or skew-Hermitian matrix is automatically normal.

Hint 2. For the rest, compute

𝐴^{*} 𝐴

and

𝐴 𝐴^{*}

. Unitary means

𝐴^{*} 𝐴 = 𝐼

Sketch. (a) Normal (

𝐴^{𝖳} 𝐴 = 𝐴 𝐴^{𝖳} = 5 𝐼

) and nothing else. (b)

𝐴^{*} = - 𝐴

: skew-Hermitian, hence normal; not unitary. (c)

𝐴^{*} 𝐴 = 𝐴 𝐴^{*} = 5 𝐼

: normal only. (d) A permutation matrix: orthogonal, hence unitary and normal. (e)

𝐴^{*} 𝐴 =

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

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

𝐴 𝐴^{*}

: none.

(a) $𝐴^{𝖳} = (\begin{matrix} 1 & - 2 \\ 2 & 1 \end{matrix})$ is neither $𝐴$ nor $- 𝐴$ . This is $𝑎 𝐼 + 𝑏 𝐽$ with $𝑎 = 1$ , $𝑏 = - 2$ , so by Example 8.1.9(c), $𝐴^{𝖳} 𝐴 =$ $𝐴 𝐴^{𝖳} =$ $5 𝐼_{2}$ . So $𝐴$ is normal, and not orthogonal since $5 𝐼_{2} \neq 𝐼_{2}$ .

(b) $𝐴^{*} =$ $(\begin{matrix} - 𝑖 & - 1 \\ 1 & - 2 𝑖 \end{matrix}) =$ $- 𝐴$ , so $𝐴$ is skew-Hermitian, hence normal. It is not Hermitian, since $𝐴 \neq 0$ . It is not unitary: $𝐴^{*} 𝐴 =$ $- 𝐴^{2} =$ $- (\begin{matrix} - 2 & 3 𝑖 \\ - 3 𝑖 & - 5 \end{matrix}) =$ $(\begin{matrix} 2 & - 3 𝑖 \\ 3 𝑖 & 5 \end{matrix}) \neq$ $𝐼_{2}$ .

(c) $𝐴^{*} = (\begin{matrix} 2 & - 𝑖 \\ - 𝑖 & 2 \end{matrix})$ , which is neither $𝐴$ nor $- 𝐴$ . Now

𝐴^{*} 𝐴 = (\begin{matrix} 2 & - 𝑖 \\ - 𝑖 & 2 \end{matrix}) (\begin{matrix} 2 & 𝑖 \\ 𝑖 & 2 \end{matrix}) = (\begin{matrix} 5 & 0 \\ 0 & 5 \end{matrix}), 𝐴 𝐴^{*} = (\begin{matrix} 2 & 𝑖 \\ 𝑖 & 2 \end{matrix}) (\begin{matrix} 2 & - 𝑖 \\ - 𝑖 & 2 \end{matrix}) = (\begin{matrix} 5 & 0 \\ 0 & 5 \end{matrix}),

so $𝐴$ is normal but not unitary. (Its parts in Proposition 8.1.21 are $𝑆 = 2 𝐼_{2}$ and $𝐾 = 𝑖 𝐵$ with $𝐵 = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ , and $𝑆$ commutes with $𝐵$ , as that proposition predicts.)

(d) The columns are $𝐞_{2}, 𝐞_{3}, 𝐞_{1}$ , an orthonormal basis, so $𝐴$ is orthogonal (Definition 6.3.33), hence unitary and normal. It is not symmetric, since $𝑎_{13} = 1 \neq 0 = 𝑎_{31}$ , and not skew-symmetric, since $𝑎_{13} = 1 \neq 0 = - 𝑎_{31}$ .

(e) $𝐴^{*} = (\begin{matrix} 1 & 0 \\ - 𝑖 & 1 \end{matrix})$ , and

𝐴^{*} 𝐴 = (\begin{matrix} 1 & 𝑖 \\ - 𝑖 & 2 \end{matrix}), 𝐴 𝐴^{*} = (\begin{matrix} 2 & 𝑖 \\ - 𝑖 & 1 \end{matrix}) .

These differ, so $𝐴$ is not normal, and therefore none of the other three.

Exercise 8.1.23.routine Let $𝐴 =$ $(\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}) \in$ $ℝ^{3 \times 3}$ .

(a)Find the eigenvalues and eigenspaces of $𝐴$ . (Write $𝐴 = 𝐼_{3} + 𝟏 𝟏^{𝖳}$ with $𝟏 = (1, 1, 1)$ .)
(b)Check that the eigenspaces are orthogonal, and find two eigenvectors for the same eigenvalue that are not orthogonal.
(c)Find an orthonormal basis of $ℝ^{3}$ consisting of eigenvectors of $𝐴$ .

Solution.

Uses. Definition 7.2.13 (Eigenspace), Corollary 7.2.23 (Eigenspaces are independent), Proposition 8.1.1 (Eigenvalues and eigenvectors of self-adjoint operators), Algorithm 6.4.14 (Gram–Schmidt)

Hint 1.

𝐴 𝐱 =

𝐱 +

(𝑥_{1} + 𝑥_{2} + 𝑥_{3}) 𝟏

. What happens for

𝐱 = 𝟏

, and for

𝐱

with

𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 0

Hint 2.

𝐸_{4} = span (𝟏)

and

𝐸_{1} =

{𝐱 | 𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 0}

. Their dimensions add up to

3

, so there are no other eigenvalues.

Hint 3. For (c), run Gram–Schmidt inside

𝐸_{1}

Sketch.

𝐴 𝟏 = 4 𝟏

, and

𝐴 𝐱 = 𝐱

when

𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 0

. So

𝐸_{4} = span (𝟏)

and

𝐸_{1}

is the plane

𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 0

;

𝐸_{1} ⟂ 𝐸_{4}

since that equation says

𝐱 \cdot 𝟏 = 0

. In

𝐸_{1}

(1, - 1, 0) \cdot

(1, 0, - 1) =

1

. Orthonormal eigenbasis:

(1, 1, 1) / \sqrt{3}

(1, - 1, 0) / \sqrt{2}

(1, 1, - 2) / \sqrt{6}

(a) For $𝐱 \in ℝ^{3}$ , $𝟏 𝟏^{𝖳} 𝐱 =$ $(𝑥_{1} + 𝑥_{2} + 𝑥_{3}) 𝟏$ , so $𝐴 𝐱 =$ $𝐱 +$ $(𝑥_{1} + 𝑥_{2} + 𝑥_{3}) 𝟏$ . Hence $(𝐴 - 𝐼_{3}) 𝐱 = 𝟎$ iff $𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 0$ , so $𝐸_{1} =$ ${𝐱 | 𝑥_{1} + 𝑥_{2} + 𝑥_{3} = 0}$ , a plane. And $(𝐴 - 4 𝐼_{3}) 𝐱 =$ $(𝑥_{1} + 𝑥_{2} + 𝑥_{3}) 𝟏 -$ $3 𝐱$ vanishes iff $𝐱$ is a multiple of $𝟏$ , say $𝐱 = 𝑐 𝟏$ , and then indeed $3 𝑐 𝟏 - 3 𝑐 𝟏 = 𝟎$ ; so $𝐸_{4} = span (𝟏)$ . The eigenspaces of distinct eigenvalues form a direct sum (Corollary 7.2.23), and $dim 𝐸_{1} + dim 𝐸_{4} = 3$ already, so there is no room for another eigenspace: the eigenvalues are $1$ and $4$ .

(b) For $𝐱 \in 𝐸_{1}$ , $𝐱 \cdot$ $𝟏 =$ $𝑥_{1} +$ $𝑥_{2} +$ $𝑥_{3} =$ $0$ , so $𝐸_{1} ⟂ 𝐸_{4}$ , as Proposition 8.1.1(b) predicts. Inside $𝐸_{1}$ , the eigenvectors $(1, - 1, 0)$ and $(1, 0, - 1)$ have dot product $1 \neq 0$ .

(c) Apply Gram–Schmidt (Algorithm 6.4.14) to the basis $(1, - 1, 0), (1, 0, - 1)$ of $𝐸_{1}$ : the second vector minus its projection onto the first is

(1, 0, - 1) - \frac{1}{2} (1, - 1, 0) = \frac{1}{2} (1, 1, - 2) .

Normalizing, and adding the normalized $𝟏$ from $𝐸_{4}$ , which is orthogonal to $𝐸_{1}$ by (b),

\frac{1}{\sqrt{3}} (1, 1, 1), \frac{1}{\sqrt{2}} (1, - 1, 0), \frac{1}{\sqrt{6}} (1, 1, - 2)

is an orthonormal basis of $ℝ^{3}$ consisting of eigenvectors of $𝐴$ .

Exercise 8.1.24.standard Let $𝑉$ be a finite-dimensional inner product space and $𝑆, 𝑇 \in L (𝑉)$ . Prove or disprove each statement.

(a)If $𝑆$ and $𝑇$ are normal, then $𝑆 + 𝑇$ is normal.
(b)If $𝑇^{2}$ is normal, then $𝑇$ is normal.
(c)If $ker 𝑇 = ker 𝑇^{*}$ , then $𝑇$ is normal.
(d)If $𝑇$ is normal, then $𝑇 + 𝑐 𝑇^{*}$ is normal for every $𝑐 \in 𝔽$ .
(e)If $𝔽 = ℝ$ , $𝑇$ is normal, and $⟨ 𝑇 𝐯, 𝐯 ⟩ = 0$ for every $𝐯 \in 𝑉$ , then $𝑇 = 0$ .

Solution.

Uses. Example 8.1.9 (Normal operators), Non-example 8.1.10 (Operators that are not normal), Proposition 6.6.18 (Properties of the adjoint), Example 6.1.39 (A rotation)

Hint 1. The nilpotent

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

is the source of counterexamples. Write it as a sum of a symmetric and a skew-symmetric matrix; compute

𝑁^{2}

Hint 2. For (c), look for an invertible operator that is not normal.

Hint 3. For (d), compute

{(𝑇 + 𝑐 𝑇^{*})}^{*} =

𝑇^{*} +

𝑐 𝑇

and expand both products.

Sketch. (a) False:

𝑁 =

(𝑁 + 𝑁^{𝖳}) / 2 +

(𝑁 - 𝑁^{𝖳}) / 2

. (b) False:

𝑁^{2} = 0

is normal. (c) False: the shear is invertible, so both kernels are

{𝟎}

. (d) True: the two products differ by

(1 - {| 𝑐 |}^{2}) (𝑇^{*} 𝑇 - 𝑇 𝑇^{*})

. (e) False: the rotation

𝐽

Let $𝑁 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ , acting on $𝔽^{2}$ with the dot product. It is not normal (Non-example 8.1.10).

(a) False. $𝑁 = 𝑆 + 𝐾$ with $𝑆 = \frac{1}{2} (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ symmetric and $𝐾 = \frac{1}{2} (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix})$ skew-symmetric. Both are normal (Example 8.1.9(a)), and their sum $𝑁$ is not.

(b) False. $𝑁^{2} = 0$ , which is normal, but $𝑁$ is not.

(c) False. The shear $𝑆 = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ is invertible, and so is $𝑆^{*} = 𝑆^{𝖳}$ , so $ker 𝑆 = ker 𝑆^{*} = {𝟎}$ ; but $𝑆$ is not normal (Non-example 8.1.10).

(d) True. By Proposition 6.6.18, ${(𝑇 + 𝑐 𝑇^{*})}^{*} =$ $𝑇^{*} +$ $𝑐 𝑇$ . Expanding,

\begin{aligned} (𝑇^{*} + 𝑐 𝑇) (𝑇 + 𝑐 𝑇^{*}) & = 𝑇^{*} 𝑇 + 𝑐 {(𝑇^{*})}^{2} + 𝑐 𝑇^{2} + {| 𝑐 |}^{2} 𝑇 𝑇^{*}, \\ (𝑇 + 𝑐 𝑇^{*}) (𝑇^{*} + 𝑐 𝑇) & = 𝑇 𝑇^{*} + 𝑐 𝑇^{2} + 𝑐 {(𝑇^{*})}^{2} + {| 𝑐 |}^{2} 𝑇^{*} 𝑇 . \end{aligned}

The difference is $(𝑇^{*} 𝑇 - 𝑇 𝑇^{*}) -$ ${| 𝑐 |}^{2} (𝑇^{*} 𝑇 - 𝑇 𝑇^{*}) =$ $(1 - {| 𝑐 |}^{2}) (𝑇^{*} 𝑇 - 𝑇 𝑇^{*})$ , which is $0$ when $𝑇$ is normal.

(e) False. The rotation $𝐽 = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ is orthogonal, hence normal, and $⟨ 𝐽 𝐱, 𝐱 ⟩ = 0$ for every $𝐱 \in ℝ^{2}$ (Example 6.1.39), but $𝐽 \neq 0$ . (Over $ℂ$ the statement is true for every operator, by Proposition 6.1.41.)

Exercise 8.1.25.standard Show that

𝐴 \in ℝ^{2 \times 2}

is normal if and only if either

𝐴

is symmetric or

𝐴 = (\begin{matrix} 𝑎 & - 𝑏 \\ 𝑏 & 𝑎 \end{matrix})

for some

𝑎, 𝑏 \in ℝ

. Describe the second family geometrically, and show that for

𝑏 \neq 0

it has no real eigenvalues.

Solution.

Uses. Definition 8.1.8 (Normal operator, normal matrix), Example 8.1.9 (Normal operators), Example 7.3.5 (Small cases)

Hint 1. Write

𝐴 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})

and compare

𝐴^{𝖳} 𝐴

with

𝐴 𝐴^{𝖳}

entry by entry.

Hint 2. The diagonal entries give

𝑏^{2} = 𝑐^{2}

. Treat

𝑐 = 𝑏

and

𝑐 = - 𝑏 \neq 𝑏

separately.

Sketch.

𝐴^{𝖳} 𝐴 =

(\begin{matrix} 𝑎^{2} + 𝑐^{2} & 𝑎 𝑏 + 𝑐 𝑑 \\ 𝑎 𝑏 + 𝑐 𝑑 & 𝑏^{2} + 𝑑^{2} \end{matrix})

𝐴 𝐴^{𝖳} =

(\begin{matrix} 𝑎^{2} + 𝑏^{2} & 𝑎 𝑐 + 𝑏 𝑑 \\ 𝑎 𝑐 + 𝑏 𝑑 & 𝑐^{2} + 𝑑^{2} \end{matrix})

. Normal iff

𝑏^{2} = 𝑐^{2}

and

𝑎 𝑏 +

𝑐 𝑑 =

𝑎 𝑐 +

𝑏 𝑑

. If

𝑐 = 𝑏

𝐴

is symmetric. If

𝑐 = - 𝑏 \neq 0

, the second equation gives

𝑏 (𝑎 - 𝑑) = 0

, so

𝑎 = 𝑑

. The second family is a rotation followed by a scaling, with characteristic polynomial

{(𝑥 - 𝑎)}^{2} + 𝑏^{2}

Let $𝐴 = (\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix})$ . Then

𝐴^{𝖳} 𝐴 = (\begin{matrix} 𝑎^{2} + 𝑐^{2} & 𝑎 𝑏 + 𝑐 𝑑 \\ 𝑎 𝑏 + 𝑐 𝑑 & 𝑏^{2} + 𝑑^{2} \end{matrix}), 𝐴 𝐴^{𝖳} = (\begin{matrix} 𝑎^{2} + 𝑏^{2} & 𝑎 𝑐 + 𝑏 𝑑 \\ 𝑎 𝑐 + 𝑏 𝑑 & 𝑐^{2} + 𝑑^{2} \end{matrix}) .

So $𝐴$ is normal if and only if $𝑐^{2} = 𝑏^{2}$ and $𝑎 𝑏 +$ $𝑐 𝑑 =$ $𝑎 𝑐 +$ $𝑏 𝑑$ . The first condition says $𝑐 = 𝑏$ or $𝑐 = - 𝑏$ .

If $𝑐 = 𝑏$ , then $𝐴$ is symmetric, and the second condition holds automatically. If $𝑐 = - 𝑏$ and $𝑏 \neq 0$ , the second condition reads $𝑎 𝑏 -$ $𝑏 𝑑 =$ $- 𝑎 𝑏 +$ $𝑏 𝑑$ , that is, $2 𝑏 (𝑎 - 𝑑) = 0$ , so $𝑎 = 𝑑$ , and $𝐴 = (\begin{matrix} 𝑎 & 𝑏 \\ - 𝑏 & 𝑎 \end{matrix})$ ; renaming $- 𝑏$ as $𝑏$ gives the stated form. Conversely, symmetric matrices are normal, and so are the matrices $(\begin{matrix} 𝑎 & - 𝑏 \\ 𝑏 & 𝑎 \end{matrix})$ by Example 8.1.9(c).

Writing $𝑎 +$ $𝑏 𝑖 =$ $𝑟 (cos 𝜃 + 𝑖 sin 𝜃)$ with $𝑟 = \sqrt{𝑎^{2} + 𝑏^{2}}$ , the matrix $(\begin{matrix} 𝑎 & - 𝑏 \\ 𝑏 & 𝑎 \end{matrix}) =$ $𝑟 (\begin{matrix} cos 𝜃 & - sin 𝜃 \\ sin 𝜃 & cos 𝜃 \end{matrix})$ is a rotation by $𝜃$ followed by scaling by $𝑟$ ; it is multiplication by $𝑎 + 𝑏 𝑖$ on $ℂ = ℝ^{2}$ . Its characteristic polynomial is ${(𝑥 - 𝑎)}^{2} + 𝑏^{2}$ (Example 7.3.5), which has no real root when $𝑏 \neq 0$ . So a real normal $2 \times 2$ matrix either is symmetric (and then, by Section 8.2, has an orthonormal eigenbasis), or is a scaled rotation, with no real eigenvalues at all (unless $𝑏 = 0$ , when it is $𝑎 𝐼_{2}$ ).

Exercise 8.1.26.standard Let

𝐱, 𝐲 \in ℂ^{𝑛}

be nonzero, and let

𝐴 = 𝐱 𝐲^{*} \in ℂ^{𝑛 \times 𝑛}

. Show that

𝐴

is normal if and only if

𝐲 = 𝑐 𝐱

for some

𝑐 \in ℂ

, and that in this case

𝐴

is Hermitian if and only if

𝑐 \in ℝ

Solution.

Uses. Proposition 6.3.27 (Properties of the conjugate transpose), Definition 8.1.8 (Normal operator, normal matrix), Proposition 2.4.14 (Algebra of matrices)

Techniques. Write a rank-one matrix as $𝐮 𝐯^{𝖳}$ (Technique 4.3.17)

Hint 1.

𝐴^{*} = 𝐲 𝐱^{*}

. Compute

𝐴^{*} 𝐴

and

𝐴 𝐴^{*}

, remembering that

𝐱^{*} 𝐱 = {‖ 𝐱 ‖}^{2}

is a scalar.

Hint 2.

𝐴^{*} 𝐴 = {‖ 𝐱 ‖}^{2} 𝐲 𝐲^{*}

and

𝐴 𝐴^{*} = {‖ 𝐲 ‖}^{2} 𝐱 𝐱^{*}

. If these are equal, apply both to

𝐲

Sketch.

𝐴^{*} 𝐴 = {‖ 𝐱 ‖}^{2} 𝐲 𝐲^{*}

and

𝐴 𝐴^{*} = {‖ 𝐲 ‖}^{2} 𝐱 𝐱^{*}

. If

𝐲 = 𝑐 𝐱

, both are

{| 𝑐 |}^{2} {‖ 𝐱 ‖}^{2} 𝐱 𝐱^{*}

. Conversely, applying both to

𝐲

gives

{‖ 𝐱 ‖}^{2} {‖ 𝐲 ‖}^{2} 𝐲 =

{‖ 𝐲 ‖}^{2} (𝐱^{*} 𝐲) 𝐱

, so

𝐲

is a multiple of

𝐱

. Then

𝐴 = 𝑐 𝐱 𝐱^{*}

and

𝐴^{*} = 𝑐 𝐱 𝐱^{*}

By Proposition 6.3.27, $𝐴^{*} =$ ${(𝐲^{*})}^{*} 𝐱^{*} =$ $𝐲 𝐱^{*}$ . Since $𝐱^{*} 𝐱 = {‖ 𝐱 ‖}^{2}$ and $𝐲^{*} 𝐲 = {‖ 𝐲 ‖}^{2}$ are $1 \times 1$ matrices, that is, scalars,

𝐴^{*} 𝐴 = 𝐲 (𝐱^{*} 𝐱) 𝐲^{*} = {‖ 𝐱 ‖}^{2} 𝐲 𝐲^{*}, 𝐴 𝐴^{*} = 𝐱 (𝐲^{*} 𝐲) 𝐱^{*} = {‖ 𝐲 ‖}^{2} 𝐱 𝐱^{*} .

If $𝐲 = 𝑐 𝐱$ , then $𝐲 𝐲^{*} = {| 𝑐 |}^{2} 𝐱 𝐱^{*}$ and ${‖ 𝐲 ‖}^{2} = {| 𝑐 |}^{2} {‖ 𝐱 ‖}^{2}$ , so both products equal ${| 𝑐 |}^{2} {‖ 𝐱 ‖}^{2} 𝐱 𝐱^{*}$ , and $𝐴$ is normal.

Conversely, suppose $𝐴^{*} 𝐴 = 𝐴 𝐴^{*}$ , and apply both sides to $𝐲$ :

{‖ 𝐱 ‖}^{2} {‖ 𝐲 ‖}^{2} 𝐲 = {‖ 𝐲 ‖}^{2} (𝐱^{*} 𝐲) 𝐱 .

Dividing by ${‖ 𝐱 ‖}^{2} {‖ 𝐲 ‖}^{2} \neq 0$ gives $𝐲 = 𝑐 𝐱$ with $𝑐 = (𝐱^{*} 𝐲) / {‖ 𝐱 ‖}^{2}$ .

In that case $𝐴 =$ $𝐱 {(𝑐 𝐱)}^{*} =$ $𝑐 𝐱 𝐱^{*}$ and $𝐴^{*} = 𝑐 𝐱 𝐱^{*}$ . Since $𝐱 𝐱^{*} \neq 0$ (its $(𝑗, 𝑗)$ entry is ${| 𝑥_{𝑗} |}^{2}$ , which is nonzero for some $𝑗$ ), $𝐴^{*} = 𝐴$ if and only if $𝑐 = 𝑐$ , that is, $𝑐 \in ℝ$ .

Exercise 8.1.27.standard Find the error in the following argument, and give an operator for which its conclusion fails. Then name a hypothesis on $𝑇$ under which the conclusion is true.

Claim. Let $𝑇 \in L (𝑉)$ have an adjoint. If $𝑇 𝐯 = 𝜆 𝐯$ , then $𝑇^{*} 𝐯 = 𝜆 𝐯$ .

Proof. We have $⟨ 𝑇^{*} 𝐯, 𝐯 ⟩ =$ $⟨ 𝐯, 𝑇 𝐯 ⟩ =$ $⟨ 𝐯, 𝜆 𝐯 ⟩ =$ $𝜆 ⟨ 𝐯, 𝐯 ⟩ =$ $⟨ 𝜆 𝐯, 𝐯 ⟩$ . Since $𝑇^{*} 𝐯$ and $𝜆 𝐯$ have the same inner product with $𝐯$ , they are equal.

Solution.

Uses. Proposition 6.1.37 (Inner products distinguish vectors), Non-example 8.1.10 (Operators that are not normal), Theorem 8.1.16 (Eigenvectors of a normal operator and of its adjoint)

Hint 1. Two vectors are equal if they have the same inner product with every vector (Proposition 6.1.37). How many vectors does the argument test against?

Hint 2. The argument shows only that

𝑇^{*} 𝐯 - 𝜆 𝐯 ⟂ 𝐯

. Try

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

with

𝐯 = 𝐞_{1}

Sketch. The computation is correct but shows only

⟨ 𝑇^{*} 𝐯 - 𝜆 𝐯, 𝐯 ⟩ = 0

; testing against the single vector

𝐯

does not determine

𝑇^{*} 𝐯

. For

𝑁

𝑁 𝐞_{1} = 0 𝐞_{1}

but

𝑁^{*} 𝐞_{1} = 𝐞_{2} \neq 𝟎

, and indeed

𝐞_{2} ⟂ 𝐞_{1}

. The claim holds for normal

𝑇

(Theorem 8.1.16).

The error. The displayed computation is correct, and it shows that $⟨ 𝑇^{*} 𝐯 - 𝜆 𝐯, 𝐯 ⟩ = 0$ , that is, $𝑇^{*} 𝐯 - 𝜆 𝐯$ is orthogonal to $𝐯$ . The last sentence is wrong. To conclude that two vectors are equal, they must have the same inner product with every vector of $𝑉$ (Proposition 6.1.37), and the argument tests only against the single vector $𝐯$ .

A counterexample. Let $𝑁 = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$ on $𝔽^{2}$ with the dot product, $𝐯 = 𝐞_{1}$ , and $𝜆 = 0$ . Then $𝑁 𝐞_{1} = 𝟎 = 0 𝐞_{1}$ , but $𝑁^{*} 𝐞_{1} =$ $𝑁^{𝖳} 𝐞_{1} =$ $𝐞_{2} \neq$ $𝟎$ . Consistently with what the argument does prove, $𝐞_{2} ⟂ 𝐞_{1}$ .

A correct version. If $𝑇$ is normal, the conclusion holds by Theorem 8.1.16. The missing ingredient is the equality of lengths $‖ (𝑇 - 𝜆 id) 𝐯 ‖ =$ $‖ (𝑇^{*} - 𝜆 id) 𝐯 ‖$ , which controls all of $𝑇^{*} 𝐯 - 𝜆 𝐯$ and not just its component along $𝐯$ .

Exercise 8.1.28.standard Let $𝑉$ be a complex inner product space, $𝑇 \in L (𝑉)$ self-adjoint, and $𝜆 = 𝑎 + 𝑏 𝑖$ with $𝑎, 𝑏 \in ℝ$ .

(a)Show that ${‖ (𝑇 - 𝜆 id) 𝐯 ‖}^{2} =$ ${‖ (𝑇 - 𝑎 id) 𝐯 ‖}^{2} +$ $𝑏^{2} {‖ 𝐯 ‖}^{2}$ for every $𝐯 \in 𝑉$ .
(b)Deduce once more that the eigenvalues of $𝑇$ are real.
(c)Suppose $𝑉$ is finite-dimensional and $𝑏 \neq 0$ . Show that $𝑇 - 𝜆 id$ is invertible and that $‖ {(𝑇 - 𝜆 id)}^{- 1} 𝐰 ‖ \leq$ $‖ 𝐰 ‖ / | 𝑏 |$ for every $𝐰 \in 𝑉$ .

Solution.

Uses. Definition 6.6.26 (Self-adjoint operator), Proposition 6.1.4 (Sesquilinearity), Proposition 6.1.27 (Basic properties of the norm), Corollary 4.2.18 (Injective if and only if surjective), Theorem 2.6.11 (Injectivity is decided at $𝟎$ )

Techniques. Move an operator across the inner product (Technique 6.6.20), Shift by a scalar (Technique 8.1.15)

Hint 1. Let

𝑅 = 𝑇 - 𝑎 id

, which is self-adjoint since

𝑎

is real. Then

𝑇 - 𝜆 id = 𝑅 - 𝑖 𝑏 id

; expand

{‖ 𝑅 𝐯 - 𝑖 𝑏 𝐯 ‖}^{2}

Hint 2. The cross terms are

𝑖 𝑏 ⟨ 𝑅 𝐯, 𝐯 ⟩ -

𝑖 𝑏 ⟨ 𝐯, 𝑅 𝐯 ⟩

, and

⟨ 𝐯, 𝑅 𝐯 ⟩ =

⟨ 𝑅 𝐯, 𝐯 ⟩

Hint 3. For (c), (a) gives

‖ (𝑇 - 𝜆 id) 𝐯 ‖ \geq

| 𝑏 | ‖ 𝐯 ‖

. Apply it to

𝐯 = {(𝑇 - 𝜆 id)}^{- 1} 𝐰

Sketch. With

𝑅 = 𝑇 - 𝑎 id

self-adjoint,

{‖ 𝑅 𝐯 - 𝑖 𝑏 𝐯 ‖}^{2} =

{‖ 𝑅 𝐯 ‖}^{2} +

𝑖 𝑏 ⟨ 𝑅 𝐯, 𝐯 ⟩ -

𝑖 𝑏 ⟨ 𝐯, 𝑅 𝐯 ⟩ +

𝑏^{2} {‖ 𝐯 ‖}^{2}

, and the middle terms cancel. If

𝑏 \neq 0

the right side is at least

𝑏^{2} {‖ 𝐯 ‖}^{2} > 0

for

𝐯 \neq 𝟎

, so

𝑇 - 𝜆 id

is injective, hence invertible in finite dimensions, and

‖ 𝐰 ‖ \geq

| 𝑏 | ‖ {(𝑇 - 𝜆 id)}^{- 1} 𝐰 ‖

(a) Let $𝑅 = 𝑇 - 𝑎 id$ . Since $𝑎$ is real, $𝑅^{*} = 𝑇^{*} - 𝑎 id = 𝑅$ , so $𝑅$ is self-adjoint. Now $𝑇 - 𝜆 id = 𝑅 - 𝑖 𝑏 id$ , and expanding with Proposition 6.1.4,

{‖ 𝑅 𝐯 - 𝑖 𝑏 𝐯 ‖}^{2} = {‖ 𝑅 𝐯 ‖}^{2} + - 𝑖 𝑏 ⟨ 𝑅 𝐯, 𝐯 ⟩ + (- 𝑖 𝑏) ⟨ 𝐯, 𝑅 𝐯 ⟩ + {| 𝑖 𝑏 |}^{2} {‖ 𝐯 ‖}^{2} .

Since $- 𝑖 𝑏 = 𝑖 𝑏$ and $⟨ 𝐯, 𝑅 𝐯 ⟩ =$ $⟨ 𝑅 𝐯, 𝐯 ⟩$ by self-adjointness, the two middle terms are $𝑖 𝑏 ⟨ 𝑅 𝐯, 𝐯 ⟩ -$ $𝑖 𝑏 ⟨ 𝑅 𝐯, 𝐯 ⟩ =$ $0$ . This leaves ${‖ 𝑅 𝐯 ‖}^{2} + 𝑏^{2} {‖ 𝐯 ‖}^{2}$ .

(b) If $𝑇 𝐯 = 𝜆 𝐯$ with $𝐯 \neq 𝟎$ , then (a) gives $0 =$ ${‖ (𝑇 - 𝑎 id) 𝐯 ‖}^{2} +$ $𝑏^{2} {‖ 𝐯 ‖}^{2} \geq$ $𝑏^{2} {‖ 𝐯 ‖}^{2}$ , so $𝑏 = 0$ and $𝜆 = 𝑎 \in ℝ$ .

(c) By (a), $‖ (𝑇 - 𝜆 id) 𝐯 ‖ \geq$ $| 𝑏 | ‖ 𝐯 ‖$ for every $𝐯$ . If $𝑏 \neq 0$ , this is positive for $𝐯 \neq 𝟎$ , so $ker (𝑇 - 𝜆 id) = {𝟎}$ and $𝑇 - 𝜆 id$ is injective (Theorem 2.6.11). An injective operator on a finite-dimensional space is surjective (Corollary 4.2.18), hence invertible. For $𝐰 \in 𝑉$ , put $𝐯 = {(𝑇 - 𝜆 id)}^{- 1} 𝐰$ ; then $‖ 𝐰 ‖ \geq | 𝑏 | ‖ 𝐯 ‖$ , which is the claimed bound.

Exercise 8.1.29.standardBeyond 21-242 Let $𝑇 \in L (𝑉)$ have an adjoint, and let $𝑆 = (𝑇 + 𝑇^{*}) / 2$ and $𝐾 = (𝑇 - 𝑇^{*}) / 2$ be its self-adjoint and skew-adjoint parts.

(a)Show that ${‖ 𝑇 𝐯 ‖}^{2} =$ ${‖ 𝑆 𝐯 ‖}^{2} +$ ${‖ 𝐾 𝐯 ‖}^{2} +$ $⟨ (𝑆 𝐾 - 𝐾 𝑆) 𝐯, 𝐯 ⟩$ for every $𝐯 \in 𝑉$ .
(b)Deduce that $𝑇$ is normal if and only if ${‖ 𝑇 𝐯 ‖}^{2} =$ ${‖ 𝑆 𝐯 ‖}^{2} +$ ${‖ 𝐾 𝐯 ‖}^{2}$ for every $𝐯 \in 𝑉$ .

Solution.

Uses. Proposition 8.1.21 (Normal means commuting parts), Proposition 6.6.18 (Properties of the 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.

{‖ 𝑇 𝐯 ‖}^{2} =

⟨ 𝑇^{*} 𝑇 𝐯, 𝐯 ⟩

and

𝑇^{*} = 𝑆 - 𝐾

. Expand

(𝑆 - 𝐾) (𝑆 + 𝐾)

Hint 2. For (b), the operator

𝑆 𝐾 - 𝐾 𝑆

is self-adjoint. Then use Proposition 8.1.4 and Proposition 8.1.21.

Sketch.

𝑇^{*} 𝑇 =

(𝑆 - 𝐾) (𝑆 + 𝐾) =

𝑆^{2} -

𝐾^{2} +

(𝑆 𝐾 - 𝐾 𝑆)

, and

⟨ 𝑆^{2} 𝐯, 𝐯 ⟩ =

{‖ 𝑆 𝐯 ‖}^{2}

- ⟨ 𝐾^{2} 𝐯, 𝐯 ⟩ =

{‖ 𝐾 𝐯 ‖}^{2}

. For (b),

{(𝑆 𝐾 - 𝐾 𝑆)}^{*} =

𝐾^{*} 𝑆^{*} -

𝑆^{*} 𝐾^{*} =

𝑆 𝐾 -

𝐾 𝑆

, so its diagonal values all vanish iff

𝑆 𝐾 = 𝐾 𝑆

, iff

𝑇

is normal.

(a) By Proposition 8.1.21, $𝑇^{*} = 𝑆 - 𝐾$ , so

𝑇^{*} 𝑇 = (𝑆 - 𝐾) (𝑆 + 𝐾) = 𝑆^{2} - 𝐾^{2} + (𝑆 𝐾 - 𝐾 𝑆) .

Now $⟨ 𝑆^{2} 𝐯, 𝐯 ⟩ =$ $⟨ 𝑆 𝐯, 𝑆^{*} 𝐯 ⟩ =$ ${‖ 𝑆 𝐯 ‖}^{2}$ and $- ⟨ 𝐾^{2} 𝐯, 𝐯 ⟩ =$ $- ⟨ 𝐾 𝐯, 𝐾^{*} 𝐯 ⟩ =$ $⟨ 𝐾 𝐯, 𝐾 𝐯 ⟩ =$ ${‖ 𝐾 𝐯 ‖}^{2}$ . Hence

{‖ 𝑇 𝐯 ‖}^{2} = ⟨ 𝑇^{*} 𝑇 𝐯, 𝐯 ⟩ = {‖ 𝑆 𝐯 ‖}^{2} + {‖ 𝐾 𝐯 ‖}^{2} + ⟨ (𝑆 𝐾 - 𝐾 𝑆) 𝐯, 𝐯 ⟩ .

(b) By Proposition 6.6.18, ${(𝑆 𝐾 - 𝐾 𝑆)}^{*} =$ $𝐾^{*} 𝑆^{*} -$ $𝑆^{*} 𝐾^{*} =$ $(- 𝐾) 𝑆 -$ $𝑆 (- 𝐾) =$ $𝑆 𝐾 -$ $𝐾 𝑆$ , so $𝑆 𝐾 - 𝐾 𝑆$ is self-adjoint. By (a), the identity ${‖ 𝑇 𝐯 ‖}^{2} =$ ${‖ 𝑆 𝐯 ‖}^{2} +$ ${‖ 𝐾 𝐯 ‖}^{2}$ holds for every $𝐯$ if and only if $⟨ (𝑆 𝐾 - 𝐾 𝑆) 𝐯, 𝐯 ⟩ =$ $0$ for every $𝐯$ , which by Proposition 8.1.4 holds if and only if $𝑆 𝐾 = 𝐾 𝑆$ . By Proposition 8.1.21 this says that $𝑇$ is normal.

Exercise 8.1.30.standard Let $𝑈 \in 𝔽^{𝑛 \times 𝑛}$ be both Hermitian and unitary.

(a)Show that $𝑈^{2} = 𝐼_{𝑛}$ .
(b)Show that $𝑃 = (𝐼_{𝑛} + 𝑈) / 2$ satisfies $𝑃^{2} = 𝑃 = 𝑃^{*}$ , so that $𝐱 \mapsto 𝑃 𝐱$ is an orthogonal projection of $𝔽^{𝑛}$ , and that $𝑈 = 2 𝑃 - 𝐼_{𝑛}$ . Describe $𝐱 \mapsto 𝑈 𝐱$ geometrically.
(c)Conversely, show that $2 𝑃 - 𝐼_{𝑛}$ is Hermitian and unitary whenever $𝑃^{2} = 𝑃 = 𝑃^{*}$ .
(d)Find $𝑈$ when $𝑃$ is the matrix of the orthogonal projection of $ℝ^{2}$ onto $span ((1, 1))$ .

Solution.

Uses. Definition 6.3.33 (Unitary and orthogonal matrices), Definition 6.3.26 (Conjugate transpose), Corollary 6.6.29 (Orthogonal projections are the self-adjoint idempotents), Proposition 6.4.29 (The orthogonal projection as a linear map), Theorem 6.4.31 (The matrix of an orthogonal projection)

Hint 1.

𝑈^{2} = 𝑈^{*} 𝑈

Hint 2.

𝑃^{2} =

(𝐼 + 2 𝑈 + 𝑈^{2}) / 4

. For the geometry, what does

𝑈

do to

im 𝑃

and to

ker 𝑃 = {(im 𝑃)}^{⟂}

Sketch. (a)

𝑈^{2} = 𝑈^{*} 𝑈 = 𝐼

. (b)

𝑃^{*} = 𝑃

and

𝑃^{2} =

(2 𝐼 + 2 𝑈) / 4 =

𝑃

𝑈 = 2 𝑃 - 𝐼

fixes

im 𝑃

and negates

{(im 𝑃)}^{⟂}

: it is the reflection across

im 𝑃

. (c)

{(2 𝑃 - 𝐼)}^{*} = 2 𝑃 - 𝐼

and

{(2 𝑃 - 𝐼)}^{2} =

4 𝑃^{2} -

4 𝑃 +

𝐼 =

𝐼

. (d)

𝑃 = \frac{1}{2} (\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix})

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

(a) $𝑈^{2} = 𝑈^{*} 𝑈 = 𝐼_{𝑛}$ , since $𝑈^{*} = 𝑈$ and $𝑈$ is unitary.

(b) $𝑃^{*} =$ $(𝐼_{𝑛} + 𝑈^{*}) / 2 =$ $𝑃$ , and by (a),

𝑃^{2} = \frac{𝐼_{𝑛} + 2 𝑈 + 𝑈^{2}}{4} = \frac{2 𝐼_{𝑛} + 2 𝑈}{4} = 𝑃 .

By Corollary 6.6.29, applied to $𝐱 \mapsto 𝑃 𝐱$ on $𝔽^{𝑛}$ with the dot product (Example 6.6.10), this map is the orthogonal projection onto $𝑊 = im 𝑃$ , and clearly $𝑈 = 2 𝑃 - 𝐼_{𝑛}$ . If $𝐱 \in 𝑊$ , then $𝑃 𝐱 = 𝐱$ and $𝑈 𝐱 = 𝐱$ ; if $𝐱 \in 𝑊^{⟂}$ , then $𝑃 𝐱 = 𝟎$ (Proposition 6.4.29) and $𝑈 𝐱 = - 𝐱$ . Since $𝔽^{𝑛} = 𝑊 \oplus 𝑊^{⟂}$ , $𝑈$ is the reflection across $𝑊$ : it keeps the component in $𝑊$ and reverses the component in $𝑊^{⟂}$ .

(c) If $𝑃^{2} = 𝑃 = 𝑃^{*}$ , then ${(2 𝑃 - 𝐼_{𝑛})}^{*} =$ $2 𝑃 -$ $𝐼_{𝑛}$ , and ${(2 𝑃 - 𝐼_{𝑛})}^{*} (2 𝑃 - 𝐼_{𝑛}) =$ $4 𝑃^{2} -$ $4 𝑃 +$ $𝐼_{𝑛} =$ $𝐼_{𝑛}$ . So $2 𝑃 - 𝐼_{𝑛}$ is Hermitian and unitary.

(d) With the unit vector $𝐮 = (1, 1) / \sqrt{2}$ , the projection matrix is $𝑃 =$ $𝐮 𝐮^{𝖳} =$ $\frac{1}{2} (\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix})$ (Theorem 6.4.31), so $𝑈 = 2 𝑃 - 𝐼_{2} = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ , the reflection across the line $𝑥_{1} = 𝑥_{2}$ , which swaps the coordinates.

Exercise 8.1.31.standard Theorem 8.1.3 says that every real symmetric matrix has a real eigenvalue. Prove or disprove the analogues over other fields.

(a)Every symmetric $𝐴 \in ℚ^{2 \times 2}$ has an eigenvalue in $ℚ$ .
(b)Every symmetric $𝐴 \in 𝔽_{2}^{2 \times 2}$ has an eigenvalue in $𝔽_{2}$ .
(c)Which steps of the first proof of Theorem 8.1.3 fail over $ℚ$ and over $𝔽_{2}$ ?

Solution.

Uses. Theorem 7.3.6 (Eigenvalues are the roots of the characteristic polynomial), Example 7.3.5 (Small cases), Proposition 8.1.1 (Eigenvalues and eigenvectors of self-adjoint operators)

Hint 1. The eigenvalues are the roots of

𝑥^{2} -

(tr 𝐴) 𝑥 +

det 𝐴

(Example 7.3.5). Choose

𝐴

so that this polynomial has no root in the field.

Hint 2. Over

ℚ

, try to get

𝑥^{2} - 2

. Over

𝔽_{2}

there are only two candidates,

0

and

1

; try to get

𝑥^{2} + 𝑥 + 1

Hint 3. For (c), the proof needs a field containing the complex roots of

𝑝_{𝐴}

, and an inner product to show that those roots lie in the original field.

Sketch. (a) False:

(\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix})

has

𝑝 = 𝑥^{2} - 2

. (b) False:

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

has

𝑝 = 𝑥^{2} + 𝑥 + 1

, with

𝑝 (0) = 𝑝 (1) = 1

. (c) Over

ℚ

the proof still produces a real eigenvalue,

\pm \sqrt{2}

, but the argument only shows that it is real, not rational. Over

𝔽_{2}

there is no inner product at all: the dot product has

(1, 1) \cdot (1, 1) = 0

, and the step

𝜆 {‖ 𝐯 ‖}^{2} =

𝜆 {‖ 𝐯 ‖}^{2} \Rightarrow

𝜆 =

𝜆

has no meaning.

(a) False. $𝐴 = (\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix})$ is symmetric with $𝑝_{𝐴} =$ $(𝑥 - 1) (𝑥 + 1) -$ $1 =$ $𝑥^{2} -$ $2$ (Example 7.3.5). Its roots $\pm \sqrt{2}$ are irrational, so by Theorem 7.3.6 $𝐴$ has no eigenvalue in $ℚ$ .

(b) False. $𝐴 = (\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}) \in 𝔽_{2}^{2 \times 2}$ is symmetric with $𝑝_{𝐴} =$ $𝑥 (𝑥 - 1) -$ $1 =$ $𝑥^{2} +$ $𝑥 +$ $1$ over $𝔽_{2}$ . Since $𝑝_{𝐴} (0) = 1$ and $𝑝_{𝐴} (1) = 1 + 1 + 1 = 1$ , it has no root in $𝔽_{2}$ , and $𝐴$ has no eigenvalue there.

(c) Over $ℚ$ , we may still regard $𝐴$ as a complex Hermitian matrix, and the proof shows that its eigenvalues are real. The step that fails is the last one: the proof concludes that a real root $𝜆$ of $𝑝_{𝑇}$ lies in the field and is therefore an eigenvalue (Corollary 7.3.13), and $ℝ \subseteq ℚ$ is false. Here the real roots $\pm \sqrt{2}$ are not rational. The theorem holds over $ℝ$ because $ℝ$ is the field in which the proof places the eigenvalue.

Over $𝔽_{2}$ the proof breaks at the start. There is no inner product on $𝔽_{2}^{2}$ in our sense, since positivity makes no sense in $𝔽_{2}$ , and the dot product is degenerate: $(1, 1) \cdot (1, 1) = 1 + 1 = 0$ . The key step of Proposition 8.1.1(a), dividing by ${‖ 𝐯 ‖}^{2} \neq 0$ and comparing $𝜆$ with $𝜆$ , has nothing to stand on. The roots of $𝑥^{2} + 𝑥 + 1$ do exist in a larger field with four elements, but nothing forces them back into $𝔽_{2}$ .

Exercise 8.1.32.hard Let

𝑉

be finite-dimensional and let

𝑇 \in L (𝑉)

be normal with

𝑇^{3} = 𝑇^{2}

. Show that

𝑇

is an orthogonal projection, that is,

𝑇 = {Pr}_{𝑊}

for some subspace

𝑊

𝑉

Solution.

Uses. Proposition 8.1.14 (Kernels and images of normal operators), Theorem 6.6.21 (Kernels and images of adjoints), Theorem 6.4.6 (The projection is characterized by orthogonality)

Techniques. Recognize a projection by its two conditions (Technique 6.4.7)

Hint 1. First show

𝑇^{2} = 𝑇

. The operator

𝑅 = 𝑇^{2} - 𝑇

satisfies

𝑇 𝑅 = 0

. Where does

im 𝑅

lie?

Hint 2.

im 𝑅 \subseteq ker 𝑇

because

𝑇 𝑅 = 0

, and

im 𝑅 \subseteq im 𝑇

because

𝑅 = 𝑇 (𝑇 - id)

. Use Proposition 8.1.14(c).

Hint 3. Once

𝑇^{2} = 𝑇

, check the two conditions characterizing

{Pr}_{im 𝑇} (𝐯)

Sketch.

𝑅 =

𝑇^{2} -

𝑇 =

𝑇 (𝑇 - id)

has

im 𝑅 \subseteq im 𝑇

, and

𝑇 𝑅 = 𝑇^{3} - 𝑇^{2} = 0

gives

im 𝑅 \subseteq ker 𝑇

. Since

ker 𝑇 \cap im 𝑇 = {𝟎}

for normal

𝑇

𝑅 = 0

. Then

𝑇 𝐯 \in im 𝑇

and

𝑇 (𝐯 - 𝑇 𝐯) = 𝟎

, so

𝐯 -

𝑇 𝐯 \in

ker 𝑇 =

{(im 𝑇)}^{⟂}

, and

𝑇 𝐯 = {Pr}_{im 𝑇} (𝐯)

Let $𝑅 = 𝑇^{2} - 𝑇$ . Since $𝑅 = 𝑇 (𝑇 - id)$ , every vector $𝑅 𝐯 =$ $𝑇 ((𝑇 - id) 𝐯)$ lies in $im 𝑇$ . Since $𝑇 𝑅 = 𝑇^{3} - 𝑇^{2} = 0$ , every vector $𝑅 𝐯$ lies in $ker 𝑇$ . By Proposition 8.1.14(c), $ker 𝑇$ and $im 𝑇$ are orthogonal, so their intersection is ${𝟎}$ (a vector orthogonal to itself is $𝟎$ ). Hence $𝑅 = 0$ , that is, $𝑇^{2} = 𝑇$ .

Let $𝑊 = im 𝑇$ and $𝐯 \in 𝑉$ . Then $𝑇 𝐯 \in 𝑊$ , and $𝑇 (𝐯 - 𝑇 𝐯) =$ $𝑇 𝐯 -$ $𝑇^{2} 𝐯 =$ $𝟎$ , so $𝐯 -$ $𝑇 𝐯 \in$ $ker 𝑇 =$ $ker 𝑇^{*} =$ ${(im 𝑇)}^{⟂} =$ $𝑊^{⟂}$ by Proposition 8.1.14(a) and Theorem 6.6.21(a). By Theorem 6.4.6, $𝑇 𝐯 = {Pr}_{𝑊} (𝐯)$ . So $𝑇 = {Pr}_{𝑊}$ .

Exercise 8.1.33.hard Let

𝐴 \in 𝔽^{𝑛 \times 𝑛}

satisfy

‖ 𝐴^{*} 𝐱 ‖ \leq ‖ 𝐴 𝐱 ‖

for every

𝐱 \in 𝔽^{𝑛}

(dot product). Show that

𝐴

is normal.

Solution.

Uses. Proposition 2.4.28 (Cyclicity of the trace), Theorem 6.4.20 (Orthonormal bases exist), Proposition 6.3.34 (Characterizations of unitary matrices), Proposition 8.1.4 (Self-adjoint operators are determined by their diagonal values), Example 6.6.10 (The adjoint of a matrix)

Techniques. Cyclicity of trace (Technique 2.4.29)

Hint 1. Let

𝑆 = 𝐴^{*} 𝐴 - 𝐴 𝐴^{*}

. Then

𝑆

is Hermitian and

𝐱^{*} 𝑆 𝐱 =

{‖ 𝐴 𝐱 ‖}^{2} -

{‖ 𝐴^{*} 𝐱 ‖}^{2} \geq

0

. What is

tr 𝑆

Hint 2.

tr 𝑆 = 0

by cyclicity of the trace. Fix a unit vector

𝐮

, extend it to an orthonormal basis, and let

𝑄

be the unitary matrix with these columns. Compute

tr (𝑄^{*} 𝑆 𝑄)

in two ways.

Hint 3. The diagonal entries of

𝑄^{*} 𝑆 𝑄

are the nonnegative numbers

𝐮_{𝑗}^{*} 𝑆 𝐮_{𝑗}

, and they add up to

0

Sketch.

𝑆 = 𝐴^{*} 𝐴 - 𝐴 𝐴^{*}

is Hermitian with

𝐱^{*} 𝑆 𝐱 =

{‖ 𝐴 𝐱 ‖}^{2} -

{‖ 𝐴^{*} 𝐱 ‖}^{2} \geq

0

and

tr 𝑆 = 0

. For a unit vector

𝐮 = 𝐮_{1}

completed to an orthonormal basis (columns of a unitary

𝑄

0 =

tr 𝑆 =

tr (𝑄^{*} 𝑆 𝑄) =

\sum_{𝑗} 𝐮_{𝑗}^{*} 𝑆 𝐮_{𝑗}

, a sum of nonnegative terms, so

𝐮^{*} 𝑆 𝐮 = 0

. By scaling,

𝐱^{*} 𝑆 𝐱 = 0

for all

𝐱

, and

𝑆 = 0

by Proposition 8.1.4.

Let $𝑆 = 𝐴^{*} 𝐴 - 𝐴 𝐴^{*}$ . Then $𝑆^{*} = 𝑆$ , and for $𝐱 \in 𝔽^{𝑛}$ ,

𝐱^{*} 𝑆 𝐱 = {(𝐴 𝐱)}^{*} (𝐴 𝐱) - {(𝐴^{*} 𝐱)}^{*} (𝐴^{*} 𝐱) = {‖ 𝐴 𝐱 ‖}^{2} - {‖ 𝐴^{*} 𝐱 ‖}^{2} \geq 0 .

By cyclicity of the trace (Proposition 2.4.28), $tr 𝑆 =$ $tr (𝐴^{*} 𝐴) -$ $tr (𝐴 𝐴^{*}) =$ $0$ .

Let $𝐮 \in 𝔽^{𝑛}$ be a unit vector. By Theorem 6.4.20(b), it extends to an orthonormal basis $𝐮 = 𝐮_{1}, 𝐮_{2}, \dots, 𝐮_{𝑛}$ of $𝔽^{𝑛}$ ; let $𝑄$ be the unitary matrix with these columns. Then $𝑄 𝑄^{*} = 𝐼_{𝑛}$ (Proposition 6.3.34), so by cyclicity

0 = tr 𝑆 = tr (𝑆 𝑄 𝑄^{*}) = tr (𝑄^{*} 𝑆 𝑄) = \sum_{𝑗 = 1}^{𝑛} 𝐮_{𝑗}^{*} 𝑆 𝐮_{𝑗} .

Each term is $\geq 0$ , so each is $0$ ; in particular $𝐮^{*} 𝑆 𝐮 = 0$ . Every nonzero $𝐱$ is a positive multiple of a unit vector, so $𝐱^{*} 𝑆 𝐱 = 0$ for all $𝐱$ . Now $𝐱^{*} 𝑆 𝐱 = ⟨ 𝑆 𝐱, 𝐱 ⟩$ for the dot product, and $𝐱 \mapsto 𝑆 𝐱$ is self-adjoint since $𝑆^{*} = 𝑆$ (Example 6.6.10). By Proposition 8.1.4, $𝑆 = 0$ , so $𝐴$ is normal.

Exercise 8.1.34.hard Let $𝐴 \in ℝ^{𝑛 \times 𝑛}$ satisfy $𝐴^{𝖳} = 𝐴^{2}$ .

(a)Show that $𝐴^{4} = 𝐴$ and that $𝐴$ is normal.
(b)Show that $𝐴^{3}$ is the matrix of an orthogonal projection.
(c)Find all such $𝐴$ when $𝑛 = 2$ .

Solution.

Uses. Proposition 2.4.23 (Transpose of a product), Definition 8.1.8 (Normal operator, normal matrix), Corollary 6.6.29 (Orthogonal projections are the self-adjoint idempotents), Exercise 8.1.25, Theorem 6.4.31 (The matrix of an orthogonal projection)

Hint 1. Transpose the equation

𝐴^{𝖳} = 𝐴^{2}

. For normality, compute

𝐴^{𝖳} 𝐴

and

𝐴 𝐴^{𝖳}

Hint 2.

𝐴^{3} = 𝐴^{𝖳} 𝐴

is symmetric. Compute

{(𝐴^{3})}^{2}

using

𝐴^{4} = 𝐴

Hint 3. For (c), use Exercise 8.1.25:

𝐴

is symmetric or a scaled rotation. In the symmetric case

𝐴 = 𝐴^{2}

. In the other case compare

𝐴^{𝖳}

and

𝐴^{2}

entry by entry.

Sketch. (a)

𝐴 =

{(𝐴^{2})}^{𝖳} =

{(𝐴^{𝖳})}^{2} =

𝐴^{4}

, and

𝐴^{𝖳} 𝐴 = 𝐴^{3} = 𝐴 𝐴^{𝖳}

. (b)

𝐴^{3} = 𝐴^{𝖳} 𝐴

is symmetric, and

{(𝐴^{3})}^{2} = 𝐴^{4} 𝐴^{2} = 𝐴^{3}

. (c) Symmetric:

𝐴 = 𝐴^{2}

, so

𝐴

is an orthogonal projection:

0

𝐼_{2}

, or

𝐮 𝐮^{𝖳}

with

‖ 𝐮 ‖ = 1

. Scaled rotation

(\begin{matrix} 𝑎 & - 𝑏 \\ 𝑏 & 𝑎 \end{matrix})

𝑏 \neq 0

𝐴^{𝖳} = 𝐴^{2}

gives

𝑎 = 𝑎^{2} - 𝑏^{2}

and

𝑏 = - 2 𝑎 𝑏

, so

𝑎 = - 1 / 2

𝑏 = \pm \sqrt{3} / 2

: the rotations by

\pm 2 𝜋 / 3

(a) Transposing $𝐴^{𝖳} = 𝐴^{2}$ and using Proposition 2.4.23, $𝐴 =$ ${(𝐴^{2})}^{𝖳} =$ ${(𝐴^{𝖳})}^{2} =$ ${(𝐴^{2})}^{2} =$ $𝐴^{4}$ . Moreover $𝐴^{𝖳} 𝐴 = 𝐴^{2} 𝐴 = 𝐴^{3}$ and $𝐴 𝐴^{𝖳} = 𝐴 𝐴^{2} = 𝐴^{3}$ , so $𝐴^{𝖳} 𝐴 = 𝐴 𝐴^{𝖳}$ , and $𝐴$ is normal.

(b) $𝐴^{3} = 𝐴^{𝖳} 𝐴$ is symmetric, since ${(𝐴^{𝖳} 𝐴)}^{𝖳} = 𝐴^{𝖳} 𝐴$ . And by (a), ${(𝐴^{3})}^{2} =$ $𝐴^{6} =$ $𝐴^{4} 𝐴^{2} =$ $𝐴 𝐴^{2} =$ $𝐴^{3}$ . So $𝑃 = 𝐴^{3}$ satisfies $𝑃^{2} = 𝑃 = 𝑃^{𝖳}$ , and $𝐱 \mapsto 𝑃 𝐱$ is an orthogonal projection by Corollary 6.6.29.

(c) By (a) and Exercise 8.1.25, either $𝐴$ is symmetric or $𝐴 = (\begin{matrix} 𝑎 & - 𝑏 \\ 𝑏 & 𝑎 \end{matrix})$ with $𝑏 \neq 0$ .

If $𝐴$ is symmetric, then $𝐴 = 𝐴^{𝖳} = 𝐴^{2}$ , so $𝐴^{2} = 𝐴 = 𝐴^{𝖳}$ and $𝐴$ is the matrix of an orthogonal projection ${Pr}_{𝑊}$ (Corollary 6.6.29). For $𝑊 = {𝟎}$ or $𝑊 = ℝ^{2}$ this gives $𝐴 = 0$ or $𝐴 = 𝐼_{2}$ . For a line $𝑊 = span (𝐮)$ with $‖ 𝐮 ‖ = 1$ , it gives $𝐴 = 𝐮 𝐮^{𝖳}$ (Theorem 6.4.31). All of these satisfy $𝐴^{𝖳} = 𝐴 = 𝐴^{2}$ .

If $𝐴 = (\begin{matrix} 𝑎 & - 𝑏 \\ 𝑏 & 𝑎 \end{matrix})$ with $𝑏 \neq 0$ , then

𝐴^{𝖳} = (\begin{matrix} 𝑎 & 𝑏 \\ - 𝑏 & 𝑎 \end{matrix}), 𝐴^{2} = (\begin{matrix} 𝑎^{2} - 𝑏^{2} & - 2 𝑎 𝑏 \\ 2 𝑎 𝑏 & 𝑎^{2} - 𝑏^{2} \end{matrix}) .

Equality requires $𝑎 = 𝑎^{2} - 𝑏^{2}$ and $𝑏 = - 2 𝑎 𝑏$ . As $𝑏 \neq 0$ , the second gives $𝑎 = - 1 / 2$ , and then the first gives $𝑏^{2} = 𝑎^{2} - 𝑎 = 3 / 4$ , so $𝑏 = \pm \sqrt{3} / 2$ . Conversely these values satisfy both equations. They give

𝐴 = (\begin{matrix} - 1 / 2 & \mp \sqrt{3} / 2 \\ \pm \sqrt{3} / 2 & - 1 / 2 \end{matrix}),

the rotations by $\pm 2 𝜋 / 3$ . This is consistent with (a): for a rotation $𝑅$ by $\pm 2 𝜋 / 3$ , $𝑅^{3} = 𝐼_{2}$ , so $𝑅^{𝖳} = 𝑅^{- 1} = 𝑅^{2}$ .

So for $𝑛 = 2$ the solutions are $0$ , $𝐼_{2}$ , the matrices $𝐮 𝐮^{𝖳}$ with $𝐮$ a unit vector, and the two rotations by $\pm 2 𝜋 / 3$ .

Exercise 8.1.35.hard Let $𝐾 \in 𝔽^{𝑛 \times 𝑛}$ be skew-Hermitian.

(a)Show that $𝐼_{𝑛} + 𝐾$ and $𝐼_{𝑛} - 𝐾$ are invertible.
(b)Show that $𝑄 =$ $(𝐼_{𝑛} - 𝐾) {(𝐼_{𝑛} + 𝐾)}^{- 1}$ is unitary and that $- 1$ is not an eigenvalue of $𝑄$ .
(c)Compute $𝑄$ for $𝐾 = (\begin{matrix} 0 & - 𝑡 \\ 𝑡 & 0 \end{matrix})$ with $𝑡 \in ℝ$ , and identify it as a rotation.

Solution.

Uses. Exercise 8.1.6, Example 6.6.10 (The adjoint of a matrix), Corollary 3.5.25 (The invertible matrix theorem), Proposition 6.3.27 (Properties of the conjugate transpose), Definition 6.3.33 (Unitary and orthogonal matrices)

Techniques. Guess the inverse, then check both products (Technique 2.5.16)

Hint 1. For (a), apply Exercise 8.1.6 to

𝐱 \mapsto 𝐾 𝐱

and to

𝐱 \mapsto - 𝐾 𝐱

, both skew-adjoint for the dot product.

Hint 2. For (b), first show that

𝐼_{𝑛} - 𝐾

and

{(𝐼_{𝑛} + 𝐾)}^{- 1}

commute, using

(𝐼_{𝑛} - 𝐾) (𝐼_{𝑛} + 𝐾) =

𝐼_{𝑛} -

𝐾^{2} =

(𝐼_{𝑛} + 𝐾) (𝐼_{𝑛} - 𝐾)

. Then

𝑄^{*} =

{(𝐼_{𝑛} - 𝐾)}^{- 1} (𝐼_{𝑛} + 𝐾)

Hint 3. If

𝑄 𝐱 = - 𝐱

, write

𝐱 = (𝐼_{𝑛} + 𝐾) 𝐲

Sketch. (a)

{‖ (𝐼 + 𝐾) 𝐱 ‖}^{2} =

{‖ 𝐱 ‖}^{2} +

{‖ 𝐾 𝐱 ‖}^{2} >

0

for

𝐱 \neq 𝟎

, and the same for

- 𝐾

. (b)

𝑄^{*} =

{({(𝐼 + 𝐾)}^{- 1})}^{*} {(𝐼 - 𝐾)}^{*} =

{(𝐼 - 𝐾)}^{- 1} (𝐼 + 𝐾)

, and

𝑄^{*} 𝑄 =

{(𝐼 - 𝐾)}^{- 1} (𝐼 + 𝐾) (𝐼 - 𝐾) {(𝐼 + 𝐾)}^{- 1} =

𝐼

because

𝐼 + 𝐾

and

𝐼 - 𝐾

commute. If

𝑄 𝐱 = - 𝐱

with

𝐱 = (𝐼 + 𝐾) 𝐲

, then

(𝐼 - 𝐾) 𝐲 =

- (𝐼 + 𝐾) 𝐲

, so

𝐲 = 𝟎

. (c)

𝑄 =

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

, the rotation by

- 2 arctan 𝑡

(a) By Example 6.6.10, $𝐱 \mapsto 𝐾 𝐱$ on $𝔽^{𝑛}$ with the dot product has adjoint $𝐱 \mapsto 𝐾^{*} 𝐱 = - 𝐾 𝐱$ , so it is skew-adjoint, and so is $𝐱 \mapsto - 𝐾 𝐱$ . By Exercise 8.1.6, ${‖ (𝐼_{𝑛} + 𝐾) 𝐱 ‖}^{2} =$ ${‖ 𝐱 ‖}^{2} +$ ${‖ 𝐾 𝐱 ‖}^{2}$ , which is positive for $𝐱 \neq 𝟎$ , and likewise with $- 𝐾$ in place of $𝐾$ . So $𝐼_{𝑛} + 𝐾$ and $𝐼_{𝑛} - 𝐾$ have trivial null space and are invertible (Corollary 3.5.25).

(b) Since $(𝐼_{𝑛} - 𝐾) (𝐼_{𝑛} + 𝐾) =$ $𝐼_{𝑛} -$ $𝐾^{2} =$ $(𝐼_{𝑛} + 𝐾) (𝐼_{𝑛} - 𝐾)$ , multiplying on both sides by ${(𝐼_{𝑛} + 𝐾)}^{- 1}$ gives ${(𝐼_{𝑛} + 𝐾)}^{- 1} (𝐼_{𝑛} - 𝐾) =$ $(𝐼_{𝑛} - 𝐾) {(𝐼_{𝑛} + 𝐾)}^{- 1}$ . By Proposition 6.3.27, ${(𝐼_{𝑛} - 𝐾)}^{*} =$ $𝐼_{𝑛} +$ $𝐾$ and ${({(𝐼_{𝑛} + 𝐾)}^{- 1})}^{*} =$ ${({(𝐼_{𝑛} + 𝐾)}^{*})}^{- 1} =$ ${(𝐼_{𝑛} - 𝐾)}^{- 1}$ , so

𝑄^{*} = {({(𝐼_{𝑛} + 𝐾)}^{- 1})}^{*} {(𝐼_{𝑛} - 𝐾)}^{*} = {(𝐼_{𝑛} - 𝐾)}^{- 1} (𝐼_{𝑛} + 𝐾) .

Hence, using that $𝐼_{𝑛} + 𝐾$ and $𝐼_{𝑛} - 𝐾$ commute,

𝑄^{*} 𝑄 = {(𝐼_{𝑛} - 𝐾)}^{- 1} (𝐼_{𝑛} + 𝐾) (𝐼_{𝑛} - 𝐾) {(𝐼_{𝑛} + 𝐾)}^{- 1} = {(𝐼_{𝑛} - 𝐾)}^{- 1} (𝐼_{𝑛} - 𝐾) (𝐼_{𝑛} + 𝐾) {(𝐼_{𝑛} + 𝐾)}^{- 1} = 𝐼_{𝑛},

so $𝑄$ is unitary (Definition 6.3.33). Suppose $𝑄 𝐱 = - 𝐱$ , and let $𝐲 = {(𝐼_{𝑛} + 𝐾)}^{- 1} 𝐱$ , so that $𝐱 = (𝐼_{𝑛} + 𝐾) 𝐲$ and $𝑄 𝐱 = (𝐼_{𝑛} - 𝐾) 𝐲$ . Then $(𝐼_{𝑛} - 𝐾) 𝐲 =$ $- (𝐼_{𝑛} + 𝐾) 𝐲$ , so $2 𝐲 = 𝟎$ , hence $𝐲 = 𝟎$ and $𝐱 = 𝟎$ . So $- 1$ is not an eigenvalue of $𝑄$ .

(c) Here $𝐼_{2} + 𝐾 = (\begin{matrix} 1 & - 𝑡 \\ 𝑡 & 1 \end{matrix})$ with inverse $\frac{1}{1 + 𝑡^{2}} (\begin{matrix} 1 & 𝑡 \\ - 𝑡 & 1 \end{matrix})$ , and $𝐼_{2} - 𝐾 = (\begin{matrix} 1 & 𝑡 \\ - 𝑡 & 1 \end{matrix})$ . So

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

With $𝑡 = tan (𝜑 / 2)$ , the double-angle formulas give $cos 𝜑 =$ $(1 - 𝑡^{2}) / (1 + 𝑡^{2})$ and $sin 𝜑 = 2 𝑡 / (1 + 𝑡^{2})$ , so $𝑄 =$ $(\begin{matrix} cos 𝜑 & sin 𝜑 \\ - sin 𝜑 & cos 𝜑 \end{matrix})$ , the rotation by $- 𝜑 = - 2 arctan 𝑡$ . As $𝑡$ runs over $ℝ$ , $𝜑$ runs over $(- 𝜋, 𝜋)$ , so every rotation except the rotation by $𝜋$ , which is $- 𝐼_{2}$ , arises, in line with (b).

Exercise 8.1.36.standard Let

𝑉

be finite-dimensional and let

𝑇 \in L (𝑉)

be normal and diagonalizable. Show that

𝑉

has an orthonormal basis consisting of eigenvectors of

𝑇

. Deduce that a diagonalizable operator has an orthonormal eigenbasis if and only if it is normal.

Solution.

Uses. Theorem 7.4.2 (Diagonalizable means an eigenbasis), Theorem 6.4.20 (Orthonormal bases exist), Theorem 8.1.18 (Eigenspaces of a normal operator are orthogonal), Proposition 6.3.13 (Orthogonal lists are independent), Proposition 8.1.7 (What an orthonormal eigenbasis forces)

Techniques. Glue bases of the summands (Technique 4.4.14)

Hint 1. Choose an orthonormal basis of each eigenspace and put them together. Why is the result orthonormal?

Hint 2. Vectors from different eigenspaces are orthogonal by Theorem 8.1.18. Why does the combined list span

𝑉

Sketch. Let

𝜆_{1}, \dots, 𝜆_{𝑘}

be the distinct eigenvalues. Concatenate orthonormal bases of

𝐸_{𝜆_{1}}, \dots, 𝐸_{𝜆_{𝑘}}

. Within each piece the list is orthonormal, and across pieces orthogonal by Theorem 8.1.18. Its span contains every eigenspace, hence every vector of an eigenbasis, hence

𝑉

. The converse is Proposition 8.1.7.

Let $𝜆_{1}, \dots, 𝜆_{𝑘}$ be the distinct eigenvalues of $𝑇$ . By Theorem 6.4.20, each eigenspace $𝐸_{𝜆_{𝑗}}$ has an orthonormal basis ${𝒪︀}_{𝑗}$ . Let $𝒪︀$ be the concatenation of ${𝒪︀}_{1}, \dots, {𝒪︀}_{𝑘}$ . Two vectors of $𝒪︀$ from the same ${𝒪︀}_{𝑗}$ are orthogonal unit vectors, and two vectors from different eigenspaces are orthogonal by Theorem 8.1.18. So $𝒪︀$ is orthonormal, hence independent (Proposition 6.3.13), and it consists of eigenvectors.

It remains to show that $𝒪︀$ spans $𝑉$ . Since $𝑇$ is diagonalizable, $𝑉$ has a basis of eigenvectors (Theorem 7.4.2). Each of its vectors lies in some $𝐸_{𝜆_{𝑗}} =$ $span ({𝒪︀}_{𝑗}) \subseteq$ $span (𝒪︀)$ , so $𝑉 \subseteq span (𝒪︀)$ . Thus $𝒪︀$ is an orthonormal basis of eigenvectors.

For the last sentence: a normal diagonalizable operator has an orthonormal eigenbasis by what we just proved, and an operator with an orthonormal eigenbasis is normal by Proposition 8.1.7. (Section 8.2 shows that over $ℂ$ the hypothesis “diagonalizable” is automatic for normal operators.)

Exercise 8.1.37.hard Let $𝐴 \in ℂ^{𝑛 \times 𝑛}$ have eigenvalues $𝜆_{1}, \dots, 𝜆_{𝑛}$ , listed with algebraic multiplicity.

(a)Prove Schur’s inequality
$\sum_{𝑗 = 1}^{𝑛} {| 𝜆_{𝑗} |}^{2} \leq tr (𝐴^{*} 𝐴) = \sum_{𝑖, 𝑗} {| 𝑎_{𝑖 𝑗} |}^{2} .$
(b)Show that if equality holds, then $𝐴$ is normal.

Solution.

Uses. Theorem 7.5.18 (Schur’s theorem), Proposition 2.4.28 (Cyclicity of the trace), Proposition 6.1.16 (Frobenius inner product), Proposition 6.3.34 (Characterizations of unitary matrices), Proposition 8.1.7 (What an orthonormal eigenbasis forces)

Techniques. The trace of $𝐴^{*} 𝐴$ (Technique 6.1.18), Cyclicity of trace (Technique 2.4.29)

Hint 1. Triangularize

𝐴

unitarily with Schur’s theorem:

𝐴 = 𝑈 𝑅 𝑈^{*}

. How do

tr (𝐴^{*} 𝐴)

and

tr (𝑅^{*} 𝑅)

compare?

Hint 2.

tr (𝐴^{*} 𝐴) =

tr (𝑅^{*} 𝑅) =

\sum_{𝑖 \leq 𝑗} {| 𝑟_{𝑖 𝑗} |}^{2}

, and the diagonal of

𝑅

lists the eigenvalues.

Hint 3. Equality forces every entry of

𝑅

above the diagonal to vanish.

Sketch.

𝐴 = 𝑈 𝑅 𝑈^{*}

with

𝑅

upper triangular and diagonal

𝜆_{1}, \dots, 𝜆_{𝑛}

(Theorem 7.5.18). Then

tr (𝐴^{*} 𝐴) =

tr (𝑈 𝑅^{*} 𝑅 𝑈^{*}) =

tr (𝑅^{*} 𝑅) =

\sum_{𝑖 \leq 𝑗} {| 𝑟_{𝑖 𝑗} |}^{2} =

\sum_{𝑗} {| 𝜆_{𝑗} |}^{2} +

\sum_{𝑖 < 𝑗} {| 𝑟_{𝑖 𝑗} |}^{2}

. Equality means

𝑅

is diagonal, so

𝐴^{*} 𝐴 =

𝑈 𝑅^{*} 𝑅 𝑈^{*} =

𝑈 𝑅 𝑅^{*} 𝑈^{*} =

𝐴 𝐴^{*}

(a) By Theorem 7.5.18(b), $𝐴 = 𝑈 𝑅 𝑈^{*}$ with $𝑈$ unitary and $𝑅 = (𝑟_{𝑖 𝑗})$ upper triangular with diagonal entries $𝜆_{1}, \dots, 𝜆_{𝑛}$ . Then $𝐴^{*} 𝐴 =$ $𝑈 𝑅^{*} 𝑈^{*} 𝑈 𝑅 𝑈^{*} =$ $𝑈 𝑅^{*} 𝑅 𝑈^{*}$ , and by cyclicity of the trace (Proposition 2.4.28) and $𝑈^{*} 𝑈 = 𝐼_{𝑛}$ ,

tr (𝐴^{*} 𝐴) = tr (𝑅^{*} 𝑅 𝑈^{*} 𝑈) = tr (𝑅^{*} 𝑅) .

By Proposition 6.1.16, $tr (𝑅^{*} 𝑅) =$ $\sum_{𝑖, 𝑗} {| 𝑟_{𝑖 𝑗} |}^{2}$ , and since $𝑟_{𝑖 𝑗} = 0$ for $𝑖 > 𝑗$ ,

tr (𝐴^{*} 𝐴) = \sum_{𝑗 = 1}^{𝑛} {| 𝜆_{𝑗} |}^{2} + \sum_{𝑖 < 𝑗} {| 𝑟_{𝑖 𝑗} |}^{2} \geq \sum_{𝑗 = 1}^{𝑛} {| 𝜆_{𝑗} |}^{2} .

The formula $tr (𝐴^{*} 𝐴) =$ $\sum_{𝑖, 𝑗} {| 𝑎_{𝑖 𝑗} |}^{2}$ is Proposition 6.1.16 again.

(b) If equality holds, then $\sum_{𝑖 < 𝑗} {| 𝑟_{𝑖 𝑗} |}^{2} =$ $0$ , so $𝑟_{𝑖 𝑗} = 0$ for all $𝑖 < 𝑗$ and $𝑅$ is diagonal. Diagonal matrices commute with their conjugate transposes, so

𝐴^{*} 𝐴 = 𝑈 𝑅^{*} 𝑅 𝑈^{*} = 𝑈 𝑅 𝑅^{*} 𝑈^{*} = 𝐴 𝐴^{*} .

(Equivalently, the columns of $𝑈$ are an orthonormal eigenbasis of $𝐴$ , and Proposition 8.1.7 applies.) The converse, that normal matrices give equality, follows from the complex spectral theorem in Section 8.2.

Exercise 8.1.38.challenge Let

𝐴 \in ℂ^{𝑛 \times 𝑛}

be normal and let

𝑊 \subseteq ℂ^{𝑛}

be a subspace with

𝐴 𝑊 \subseteq 𝑊

. Show that

𝐴^{*} 𝑊 \subseteq 𝑊

, and deduce that

𝐴 𝑊^{⟂} \subseteq 𝑊^{⟂}

. (Let

𝑃

be the matrix of

{Pr}_{𝑊}

, so that

𝐴 𝑊 \subseteq 𝑊

says

𝑃 𝐴 𝑃 = 𝐴 𝑃

, and consider

𝑋 = 𝑃 𝐴 (𝐼_{𝑛} - 𝑃)

Solution.

Uses. Proposition 6.4.29 (The orthogonal projection as a linear map), Corollary 6.6.29 (Orthogonal projections are the self-adjoint idempotents), Proposition 2.4.28 (Cyclicity of the trace), Proposition 6.1.16 (Frobenius inner product), Exercise 6.6.38

Techniques. The trace of $𝐴^{*} 𝐴$ (Technique 6.1.18), Cyclicity of trace (Technique 2.4.29)

Hint 1.

𝑃^{2} = 𝑃 = 𝑃^{*}

, and

𝐴 𝑊 \subseteq 𝑊

means

𝑃 𝐴 𝑃 = 𝐴 𝑃

. Taking conjugate transposes,

𝑃 𝐴^{*} 𝑃 = 𝑃 𝐴^{*}

Hint 2. Write

𝑃 𝐴 𝐴^{*} 𝑃 =

𝑃 𝐴 (𝑃 + (𝐼 - 𝑃)) 𝐴^{*} 𝑃

, so

tr (𝑃 𝐴 𝐴^{*} 𝑃) =

tr (𝑃 𝐴 𝑃 𝐴^{*} 𝑃) +

tr (𝑋 𝑋^{*})

. Compute

tr (𝑃 𝐴 𝐴^{*} 𝑃)

another way, using

𝐴 𝐴^{*} = 𝐴^{*} 𝐴

and

𝐴 𝑃 = 𝑃 𝐴 𝑃

Hint 3. You should find

tr (𝑋 𝑋^{*}) = 0

, so

𝑋 = 0

. What does

𝑃 𝐴 (𝐼 - 𝑃) = 0

say about

𝐴^{*}

Sketch. With

𝑋 = 𝑃 𝐴 (𝐼 - 𝑃)

tr (𝑃 𝐴 𝐴^{*} 𝑃) =

tr (𝑃 𝐴 𝑃 𝐴^{*} 𝑃) +

tr (𝑋 𝑋^{*})

. On the other hand

tr (𝑃 𝐴 𝐴^{*} 𝑃) =

tr (𝑃 𝐴^{*} 𝐴 𝑃) =

tr ({(𝐴 𝑃)}^{*} (𝐴 𝑃)) =

tr ({(𝑃 𝐴 𝑃)}^{*} (𝑃 𝐴 𝑃)) =

tr (𝑃 𝐴^{*} 𝑃 𝐴 𝑃) =

tr (𝑃 𝐴 𝑃 𝐴^{*} 𝑃)

by cyclicity. So

tr (𝑋 𝑋^{*}) = 0

𝑋 = 0

, and

𝑋^{*} =

(𝐼 - 𝑃) 𝐴^{*} 𝑃 =

0

says

𝐴^{*} 𝑃 = 𝑃 𝐴^{*} 𝑃

, that is,

𝐴^{*} 𝑊 \subseteq 𝑊

. The last claim is Exercise 6.6.38 applied to

𝐴^{*}

Since $𝑃$ is the matrix of an orthogonal projection, $𝑃^{2} = 𝑃 = 𝑃^{*}$ (Corollary 6.6.29), and $im 𝑃 = 𝑊$ , with $𝑃 𝐰 = 𝐰$ for $𝐰 \in 𝑊$ (Proposition 6.4.29). The hypothesis $𝐴 𝑊 \subseteq 𝑊$ says that $𝐴 𝑃 𝐱 \in 𝑊$ for every $𝐱$ , that is, $𝑃 𝐴 𝑃 = 𝐴 𝑃$ .

Let $𝑋 = 𝑃 𝐴 (𝐼_{𝑛} - 𝑃)$ . Inserting $𝐼_{𝑛} = 𝑃 + (𝐼_{𝑛} - 𝑃)$ and using ${(𝐼_{𝑛} - 𝑃)}^{2} =$ $𝐼_{𝑛} -$ $𝑃 =$ ${(𝐼_{𝑛} - 𝑃)}^{*}$ ,

𝑃 𝐴 𝐴^{*} 𝑃 = 𝑃 𝐴 𝑃 𝐴^{*} 𝑃 + 𝑃 𝐴 (𝐼_{𝑛} - 𝑃) (𝐼_{𝑛} - 𝑃) 𝐴^{*} 𝑃 = (𝑃 𝐴 𝑃) {(𝑃 𝐴 𝑃)}^{*} + 𝑋 𝑋^{*} .

On the other hand, by normality, then $𝐴 𝑃 = 𝑃 𝐴 𝑃$ , then cyclicity of the trace (Proposition 2.4.28),

tr (𝑃 𝐴 𝐴^{*} 𝑃) = tr (𝑃 𝐴^{*} 𝐴 𝑃) = tr ({(𝐴 𝑃)}^{*} (𝐴 𝑃)) = tr ({(𝑃 𝐴 𝑃)}^{*} (𝑃 𝐴 𝑃)) = tr ((𝑃 𝐴 𝑃) {(𝑃 𝐴 𝑃)}^{*}) .

Comparing the traces of the two expressions for $𝑃 𝐴 𝐴^{*} 𝑃$ gives $tr (𝑋 𝑋^{*}) = 0$ . By Proposition 6.1.16, $tr (𝑋 𝑋^{*}) =$ $tr ({(𝑋^{*})}^{*} 𝑋^{*})$ is the sum of the squared moduli of the entries of $𝑋^{*}$ , so $𝑋^{*} = 0$ , that is, $(𝐼_{𝑛} - 𝑃) 𝐴^{*} 𝑃 = 0$ . So $𝐴^{*} 𝑃 = 𝑃 𝐴^{*} 𝑃$ , which says that $𝐴^{*} 𝐰 = 𝑃 𝐴^{*} 𝐰 \in 𝑊$ for every $𝐰 \in 𝑊$ : $𝐴^{*} 𝑊 \subseteq 𝑊$ .

Finally, by Exercise 6.6.38 applied to $𝐴^{*}$ (whose adjoint is $𝐴$ ), $𝐴^{*} 𝑊 \subseteq 𝑊$ implies $𝐴 𝑊^{⟂} \subseteq 𝑊^{⟂}$ .

More practice

The practice bank has 22 exercises on this section’s topics.