13.4 Bounded linear operators

中文

Let $V$ be any normed vector space. A linear operator $A:V\to V$ is said to be bounded if

$$\Vert Au\Vert \le K\Vert u\Vert$$

for some constant $K\ge 0$ and all $u\in V$

Theorem 13.11 · Boundedness and continuity of linear operators

A linear operator on a normed vector space is bounded if and only if it is continuous with respect to the norm topology.

Proof

If $A$ is bounded then it is continuous, for if $ϵ>0$ then for any pair of vectors $u$, $v$ such that $\Vert u-v\Vert <ϵ/K$

$$\Vert Au-Av\Vert =\Vert A(u-v)\Vert \le K\Vert u-v\Vert <ϵ.$$

Conversely, let $A$ be a continuous operator on $V$. If $A$ is not bounded, then for each $N>0$ there exists $u_N$ such that $\Vert A{u}_N\Vert \ge N\Vert u_N\Vert$ . Set

$$w_N=\frac{{\mathbf{u}}_N}{N\Vert {\mathbf{u}}_N\Vert },$$

so that

$$\Vert w_N\Vert =\frac{1}{N}\to 0.$$

Hence $w_N\to 0$ , but $\Vert A{w}_N\Vert \ge 1$ , so that $A{w}_n$ definitely does $not\to 0.$ , contradicting the assumption that $A$ is continuous. -

The operator norm of a bounded operator $A$ is defined as

$$\Vert A\Vert =\sup \{\Vert Au\Vert |\Vert u\Vert \le 1\}.$$

By Theorem 13.11, $A$ is continuous at $x=0$ . Hence there exists $ϵ>0$ such that $\Vert Ax\Vert \le 1$ for all $\Vert x\Vert \le ϵ$ . For any $u$ with $\Vert u\Vert \le 1$ let $v=ϵu$ so that $\Vert v\Vert \le ϵ$ and

$$\Vert Au\Vert =\frac{1}{ϵ}\Vert Av\Vert \le \frac{1}{ϵ}.$$

This shows $\Vert A\Vert$ always exists for a bounded operator.

Example 13.8

On ${\ell }^2$ define the two shift operators $S$ and $S^{\prime }$ by

$$S((x_1,x_2,x_3,\dots ))=(0,x_1,x_2,\dots )$$

and

$$S^{\prime }((x_1,x_2,x_3,\dots ))=(x_2,x_3,\dots ).$$

These operators are clearly linear, and satisfy

$$\Vert Sx\Vert =\Vert x\Vert \text{ and }\Vert S^{\prime }x\Vert \le \Vert x\Vert .$$

Hence the norm of the operator $S$ is 1, while $\Vert S^{\prime }\Vert$ is also 1 since equality holds for $x_1=0$

Example 13.9

Let $\alpha$ be any bounded measurable function on the Hilbert space $L^2(X)$ of square integrable functions on a measure space $X$. The multiplication operator $A_{\alpha }:L^2(X)\to L^2(X)$ defined by $A_{\alpha }(f)=\alpha f$ is a bounded linear operator, for $\alpha f$ is measurable for every $f\in L^2(X)$ , and it is square integrable since

$$|\alpha f{|}^2\le M^2|f{|}^2\text{ where }M=\underset{x\in X}{\sup }|\alpha (x)|.$$

The multiplication operator is well-defined on $L^2(X)$ , for if $f$ and $f^{\prime }$ are equal almost everywhere, $f\sim f^{\prime }$ , then $\alpha f\sim \alpha f^{\prime }$ ; thus there is no ambiguity in writing $A_{\alpha }f$ for $A_{\alpha }[f]$ Linearity is trivial, while boundedness follows from

$$\Vert A_{\alpha }f{\Vert }^2={\int }_X|\alpha f{|}^2\mathrm{d}\mu \le M^2{\int }_X|f{|}^2\mathrm{d}\mu =M^2\Vert f{\Vert }^2.$$

Exercise

If $A$ and $B$ are bounded linear operators on a normed vector space, show that $A+\lambda B$ and $AB$ are also bounded.

A bounded operator $A:V\to V$ is said to be invertible if there exists a bounded operator $A^{-1}:V\to V$ such that $A{A}^{-1}=A^{-1}A=I\equiv {\mathrm{id}}_V.A^{-1}$ is called the inverse of $A$. It is clearly unique, for if $BA=CA$ then $B=BI=BA{A}^{-1}=CA{A}^{-1}=C$ . It is important that we specify $A^{-1}$ to be both a right and left inverse. For example, in ${\ell }^2$ , the shift operator $S$ defined in Example 13.8 has left inverse $S^{\prime }{}_{;}$ , since $S^{\prime }S=I$ , but it is not a right inverse for $S{S}^{\prime }(x_1,x_2,\dots )=(0,x_2,x_3,\dots )$ . Thus $S$ is not an invertible operator, despite the fact that it is injective and an isometry, $\Vert Sx\Vert =\Vert x\Vert$ . For a finite dimensional space these conditions would be enough to guarantee invertibility.

Theorem 13.12 · Neumann series and operator inversion

If $A$ is a bounded operator on a Banach space $V$, with $\Vert A\Vert <1$ , then the operator $I-A$ is invertible and

$$(I-A{)}^{-1}=\sum _{n=0}^{\infty }{A}^n.$$

Proof

Let $x$ be any vector in $V$. Since $\Vert A^{k}x\Vert \le \Vert A\Vert (A^{k-1}x)$ it follows by simple induction that $A^k$ is bounded and has norm $\Vert A^k\Vert \le (\Vert A\Vert {)}^k$ . The vectors $u_n=(I+A+$ $A^2+\cdots +A^n)x$ form a Cauchy sequence, since

$$\begin{array}{l}\Vert u_n-u_m\Vert =\Vert (A^{m+1}+\cdots +A^n)x\Vert \\ \le (\Vert A{\Vert }^{m+1}+\cdots +\Vert A{\Vert }^n)\Vert x\Vert \\ \le \frac{\Vert A{\Vert }^{m+1}}{1-\Vert A\Vert }\Vert x\Vert \\ \to 0\text{ as }m\to \infty .\end{array}$$

Since $V$ is a Banach space, $u_m\to u$ for some $u\in V$ , so there is a linear operator $T:V\to V$ such that $u=Tx$ . Furthermore, since $T-(I+A+\cdot \cdot \cdot +A^n)$ is a bounded linear operator, it follows that $T$ is bounded. Writing $T={\sum }_{k=1}^{\infty }{A}^k$ , in the sense that

$$\underset{m\to \infty }{\lim }\left(T-\sum _{k=1}^m{A}^k\right)x=0,$$

it is straightforward to verify that $(I-A)Tx=T(I-A)x=x$ , which shows that $T=$ $(I-A{)}^{-1}$ -

Adjoint operators

Let $A:\mathcal{H}\to \mathcal{H}$ be a bounded linear operator on a Hilbert space $\mathcal{H}.$ We define its adjoint to be the operator $A^{\ast }:\mathcal{H}\to \mathcal{H}$ that has the property

$$⟨u\mid Av⟩=⟨A^{\ast }u\mid v⟩\text{ for all }u,v\in \mathcal{H}.\text{\hspace{1em}(13.9)}$$

This operator is well-defined, linear and bounded.

Proof

For fixed $u$, the map ${\phi }_u:v\mapsto ⟨u\mid Av⟩$ is clearly linear and continuous, on using Lemma 13.3. Hence ${\phi }_u$ is a linear functional, and by the Riesz representation theorem there exists a unique element $A\ast u\in \mathcal{H}$ such that

$$⟨A^{\ast }u\mid v⟩={\phi }_u(v)=⟨u\mid Av⟩.$$

The map $u\mapsto A\ast u$ is linear, since for an arbitrary vector $v$

$$\begin{array}{rl}⟨A^{\ast }(u+\lambda w)|v⟩& =⟨u+\lambda w|Av⟩\\ & =⟨u|Av⟩+\overline{\lambda }⟨w|Av⟩\\ & =⟨A^{\ast }u+\lambda A^{\ast }w|v⟩.\end{array}$$

To show that the linear operator $A^{\ast }$ is bounded, let $u$ be any vector,

$$\begin{array}{rl}& \Vert A^{\ast }u{\Vert }^2=|⟨A^{\ast }u|A^{\ast }u⟩|\\ & =|⟨u|A{A}^{\ast }u⟩|\\ & \le \Vert u\Vert \Vert A{A}^{\ast }u\Vert \\ & \le \Vert A\Vert \Vert u\Vert \Vert A^{\ast }u\Vert .\end{array}$$

Hence, either $A^{\ast }u=0$ or $\Vert A^{\ast }u\Vert \le \Vert A\Vert \Vert u\Vert$ . In either case $\Vert A^{\ast }u\Vert \le \Vert A\Vert \Vert u\Vert$

Theorem 13.13 · Algebraic properties of adjoint operators

The adjoint satisfies thefollowing properties:

(i) $(A+B{)}^{\ast }=A^{\ast }+B^{\ast },$

(ii) $(\lambda A{)}^{\ast }=\overline{\lambda }{A}^{\ast },$

(iii) $(AB{)}^{\ast }=B^{\ast }{A}^{\ast },$

(iv) $A^{\ast \ast }=A,$

(v) if $A$ is invertible then $(A^{-1}{)}^{\ast }=(A^{\ast }{)}^{-1}$

Proof

We provide proofs of (i) and (ii), leaving the others as exercises.

(i) For arbitrary $u,v\in \mathcal{H}$

$$\begin{array}{c}⟨(A+B{)}^{\ast }u|v⟩=⟨u|(A+B)v⟩=⟨u|Av+Bv⟩\\ =⟨u|Av⟩+⟨u|Bv⟩=⟨A^{\ast }u|v⟩+⟨B^{\ast }u|v⟩=⟨A^{\ast }u+B^{\ast }u|v⟩.\end{array}\text{\hspace{1em}(13.10)}$$

$\mathrm{As}⟨w\mid v⟩=⟨w^{\prime }\mid v⟩$ for all $v\in \mathcal{H}$ implies $w=w^{\prime },$ , we have

$$(A+B{)}^{\ast }u=A^{\ast }u+B^{\ast }u.$$

(ii) For any pair of vectors $u$, $v\in \mathcal{H}$

$$\begin{array}{c}⟨(\lambda A{)}^{\ast }u|v⟩=⟨u|\lambda Av⟩=\lambda ⟨u|Av⟩\\ =\lambda ⟨A^{\ast }u|v⟩=⟨\overline{\lambda }{A}^{\ast }u|v⟩.\end{array}\text{\hspace{1em}(13.11)}$$

The proofs of (iii)–(v) are on similar lines.

Example 13.10

The right shift operator $S$ on ${\ell }^2$ (see Example 13.8) induces the inner product

$$⟨x\mid Sy⟩=\overline{x_1}.0+\overline{x_2}{y}_1+\overline{x_3}{y}_2+\cdots =⟨S^{\prime }x\mid y⟩,$$

where $S^{\prime }$ is the left shift. Hence $S^{\ast }=S^{\prime }$ . Similarly $S^{\prime }{}^{\ast }=S_{;}$ , since

$$⟨x\mid S^{\prime }y⟩=\overline{x_1}{y}_2+\overline{x_2}{y}_3+\cdots =⟨S^{\prime }x\mid y⟩.$$

Example 13.11

Let $\alpha$ be a bounded measurable function on the Hilbert space $L^2(X)$ of square integrable functions on a measure space $X$, and $A_{\alpha }$ the multiplication operator defined in Example 13.9. For any pair of functions $f,$ , $g$ square integrable on $X,$ the equation $⟨A_{\alpha }^{\ast }f|g⟩=⟨f|A_{\alpha }g⟩$ reads

$${\int }_X\overline{A_{\alpha }^{\ast }f}g\mathrm{d}\mu ={\int }_X\overline{f}{A}_{\alpha }g\mathrm{d}\mu ={\int }_X\overline{f}\alpha g\mathrm{d}\mu .$$

Since $g$ is an arbitrary function from ${\mathcal{L}}^2(X)$ , we have $\overline{A_{\alpha }^{\ast }f}=\alpha \overline{f}a.e.$ .. and in terms of the equivalence classes of functions in $L^2(X)$ the adjoint operator reads

$$A_{\alpha }^{\ast }[f]=[\overline{\alpha }f].$$

The adjoint operator of a multiplication operator is the multiplication operator by the complex conjugate function.

We define the matrix element of the operator $A$ between the vectors $u$ and $v$ in $\mathcal{H}$ to be $⟨u\mid Av⟩$ . If the Hilbert space is separable and $e_i$ is an o.n. basis then, by Theorem 13.2, we may write

$$A{e}_j=\sum _i{a}_{ij}{e}_i\text{ where }{a}_{ij}=⟨e_i|A{e}_j⟩.$$

Thus the matrix elements of the operator between the basis vectors are identical with the components of the matrix of the operator with respect to this basis, $\mathsf{A}=[a_{ij}]$ . The adjoint operator has decomposition

$$A^{\ast }{e}_j=\sum _i{a}_{ij}^{\ast }{e}_i\text{ where }{a}_{ij}^{\ast }=⟨e_i|A^{\ast }{e}_j⟩.$$

The relation between the matrix elements $[a_{ij}^{\ast }]$ and $[a_{ij}]$ is determined by

$$a_{ij}^{\ast }=⟨e_i|A^{\ast }{e}_j⟩=⟨A{e}_i|e_j⟩=\overline{⟨e_j|A{e}_i⟩}=\overline{a_{ji}},$$

or, in matrix notation,

$${\mathsf{A}}^{\ast }\equiv [a_{ij}^{\ast }]=[\overline{a_{ji}}]=\overline{{\mathsf{A}}^T}={\mathsf{A}}^{†}.$$

In quantum mechanics it is common to use the conjugate transpose notation $A^{†}$ for the adjoint operator, but the equivalence with the complex adjoint matrix only holds for orthonorma bases.

Exercise

Show that in an o.n. basis $\begin{array}{r}Au={\sum }_i{u}_i^{\prime }{e}_i\end{array}$ where $u={\sum }_i{u}_i{e}_i$ and $\begin{array}{r}{u}_i^{\prime }={\sum }_j{a}_{ij}{u}_j\end{array}$

Hermitian operators

An operator $A$ is called hermitian if $A=A^{\ast }$ , so that

$$⟨u\mid Av⟩=⟨A^{\ast }u\mid v⟩=\overline{⟨v\mid A^{\ast }u⟩}=⟨Au\mid v⟩.$$

If $\mathcal{H}$ is separable and $e_1,e_2,\dots \mathrm{a}$ complete orthonormal set, then the matrix elements in this basis, $a_{ij}=⟨e_i|A{e}_j⟩$ , have the hermitian property

$$a_{ij}=\overline{a_{ji}}.$$

In other words, a bounded operator $A$ is hermitian if and only if its matrix with respect to any o.n. basis is hermitian,

$$\mathsf{A}=[a_{ij}]=\overline{{\mathsf{A}}^T}={\mathsf{A}}^{†}.$$

These operators are sometimes referred to as self-adjoint, but in line with modern usage we will use this term for a more general concept defined in Section 13.6.

Let $M$ be a closed subspace of $\mathcal{H}$ then, by Theorem 13.8, any $u\in \mathcal{H}$ has a unique decomposition

$$u=u^{\prime }+u^{\prime \prime }\text{ where }{u}^{\prime }\in M,u^{\prime \prime }\in M^{\perp }.$$

We define the projection operator $P_M:\mathcal{H}\to \mathcal{H}$ by $P_M(u)=u^{\prime }$ , which maps every vector of $\mathcal{H}$ onto its orthogonal projection in the subspace $M$.

Theorem 13.14 · Orthogonal projections and idempotent Hermitian operators

For every subspace $M$, the projection operator $P_M$ is a bounded hermitian operator and satisfies $P_M^2=P_M$ (called an idempotent operator). Conversely any idempotent hermitian operator $P$ is a projection operator into some subspace.

Proof

1. $P_M$ is hermitian. For any two vectors from $u,v\in \mathcal{H}$

$$⟨u|P_{M}v⟩=⟨u|v^{\prime }⟩=⟨u^{\prime }+u^{\prime \prime }|v^{\prime }⟩=⟨u^{\prime }|v^{\prime }⟩$$

since $⟨u^{\prime \prime }|{\mathbf{v}}^{\prime }⟩=0$ . Similarly,

$$⟨P_{M}u|v⟩=⟨u^{\prime }|v⟩=⟨u^{\prime }|v^{\prime }+v^{\prime \prime }⟩=⟨u^{\prime }|v^{\prime }⟩.$$

Thus $P_M=P_M^{\ast }$

2. $P_M$ is bounded, for $\Vert P_{M}u{\Vert }^2\le \Vert u{\Vert }^2$ since

$$\Vert u{\Vert }^2=⟨u|u⟩=⟨u^{\prime }+u^{\prime \prime }|u^{\prime }+u^{\prime \prime }⟩=⟨u^{\prime }|u^{\prime }⟩+⟨u^{\prime \prime }|u^{\prime \prime }⟩\ge \Vert u^{\prime }{\Vert }^2.$$

3. $P_M$ is idempotent, for $P_M^{2}u=P_M{u}^{\prime }=u^{\prime }$ since $u^{\prime }\in M$ . Hence $P_M^2=P_M$

4. Suppose $P$ is hermitian and idempotent, $P^2=P$ . The operator $P$ is bounded and therefore continuous, for by the Cauchy–Schwarz inequality (5.13),

$$\Vert Pu{\Vert }^2=|⟨Pu|Pu⟩|=|⟨u|P^{2}u⟩|=|⟨u|Pu⟩|\le \Vert u\Vert \Vert Pu\Vert .$$

Hence either $\Vert Pu\Vert =0\mathrm{or}\Vert Pu\Vert \le \Vert u\Vert$

Let $M=\{u:|:u=Pu\}$ . This is obviously a vector subspace of $\mathcal{H}$. It is closed by continuity of $P$, for if $u_n\to u$ and $P{u}_n=u_n$ , then $P{u}_n\to Pu={\lim }_{n\to \infty }{u}_n=u$ . Thus $M$ is a subspace of $\mathcal{H}$. For any vector $v\in \mathcal{H}$ , set ${\mathbf{v}}^{\prime }=P\mathbf{v}$ and $v^{\prime \prime }=(I-P)v=v-v^{\prime }$ . Then $\mathbf{v}={\mathbf{v}}^{\prime }+{\mathbf{v}}^{\prime \prime }$ and $v^{\prime }\in M,v^{\prime \prime }\in M^{\perp }$ , for

$$P{v}^{\prime }=P(Pv)=P^{2}v=Pv=v^{\prime },$$

and for all $w\in M$

$$⟨v^{\prime \prime }|w⟩=⟨(I-P)v|w⟩=⟨v|w⟩-⟨Pv|w⟩=⟨v|w⟩-⟨v|Pw⟩=⟨v|w⟩-⟨v|w⟩=0.$$

Unitary operators

An operator $U:\mathcal{H}\to \mathcal{H}$ is called unitary if

$$⟨Uu\mid Uv⟩=⟨u\mid v⟩\text{ for all }u,v\in \mathcal{H}.$$

Since this implies $⟨U^{\ast }Uu\mid v⟩=⟨u\mid v⟩$ , an operator $U$ is unitary if and only if $U^{-1}=U^{\ast }$ Every unitary operator is isometric, $\Vert Uu\Vert =\Vert u\Vert$ for all $u\in \mathcal{H}-$ it preserves the distance $d(u,v)=\Vert u-v\Vert$ between any two vectors. Conversely, every isometric operator is unitary, for if $U$ is isometric then

$$⟨U(u+v)|U(u+v)⟩-i⟨U(u+iv)|U(u+iv)⟩=⟨u+v|u+v⟩-i⟨u+iv|u+iv⟩.$$

Expanding both sides and using $⟨Uu|Uu⟩=⟨u|u⟩$ and $⟨Uv|Uv⟩=⟨v|v⟩$ , gives

$$2⟨Uu|Uv⟩=2⟨u|v⟩.$$

If $\{e_1,e_2,\dots \}$ is an orthonormal basis then so is

$$e_1^{\prime }=U{e}_1,e_2^{\prime }=U{e}_2,\dots ,$$

for

$$⟨e_i^{\prime }|e_j^{\prime }⟩=⟨U{e}_i|U{e}_j⟩=⟨U^{\ast }U{e}_i|e_j⟩=⟨e_i|e_j⟩={\delta }_{ij}.$$

Conversely for any pair of complete orthonormal sets $\{e_1,e_2,\dots \}$ and $\{e_1^{\prime },e_2^{\prime },\dots \}$ the operator defined by $U{e}_i=e_i^{\prime }$ is unitary, for if $u$ is any vector then, by Theorem 13.2,

$$u=\sum _i{u}_i{e}_i\text{ where }{u}_i=⟨e_i|u⟩.$$

Hence

$$Uu=\sum _i{u}_{i}U{e}_i=\sum _i{u}_i{e}_i^{\prime },$$

which gives

$$u_i=⟨e_i|u⟩=⟨e_i^{\prime }|Uu⟩.$$

Parseval’s identity Eq. (13.7) can be applied in the primed basis,

$$\begin{array}{l}⟨Uu|Uv⟩={\sum }_i⟨Uu|e_i^{\prime }⟩⟨e_i^{\prime }|Uv⟩\\ ={\sum }_i\overline{u_i}{v}_i\\ ={\sum }_i⟨u|e_i⟩⟨e_i|v⟩\\ =⟨u|v⟩,\end{array}$$

which shows that $U$ is a unitary operator.

Exercise

Show that if $U$ is a unitary operator then $\Vert U\Vert =1$

Exercise

Show that the multiplication operator $A_{\alpha }$ on $L^2(X)$ is unitary iff $|\alpha (x)|=1$ for all $x\in X$

Problems

Problem 13.14

The norm $\Vert \varphi \Vert$ of a bounded linear operator $\varphi :\mathcal{H}\to \mathbb{C}$ is defined as the greatest lower bound of all $M$ such that $|\varphi (u)|\le M\Vert u\Vert$ for all $u\in \mathcal{H}.\mathrm{If}\varphi (u)=⟨v|u⟩$ show that $\Vert \varphi \Vert =\Vert v\Vert$ Hence show that the bounded linear functional norm satisfies the parallelogram law

$$\Vert \varphi +\psi {\Vert }^2+\Vert \varphi -\psi {\Vert }^2=2\Vert \varphi {\Vert }^2+2\Vert \psi {\Vert }^2.$$

Problem 13.15

$\mathrm{If}\{e_n\}$ is a complete o.n. set in a Hilbert space $\mathcal{H}$, and ${\alpha }_n$ a bounded sequence of scalars, show that there exists a unique bounded operator $A$ such that $A{e}_n={\alpha }_n{e}_n$ . Find the norm of $A$.

Problem 13.16

For bounded linear operators $A$, $B$ on a normed vector space $V$ show that

$$\Vert \lambda A\Vert =|\lambda |\Vert A\Vert ,\Vert A+B\Vert \le \Vert A\Vert +\Vert B\Vert ,\Vert AB\Vert \le \Vert A\Vert \Vert B\Vert .$$

Hence show that $A$ is a genuine norm on the set of bounded linear operators on $V$.

Problem 13.17

Prove properties (iii)–(v) of Theorem 13.13. Show tha $\Vert A^{\ast }\Vert =\Vert A\Vert$

Problem 13.18

Let $A$ be a bounded operator on a Hilbert space $\mathcal{H}$ with a one-dimensional range.

(a) Show that there exist vectors $u$, $v$ such that $Ax=⟨v|x⟩u$ for all $x\in \mathcal{H}$

(b) Show that $A^2=\lambda A$ for some scalar $\lambda$, and that $\Vert A\Vert =\Vert u\Vert \Vert v\Vert$

(c) Prove that $A$ is hermitian, $A^{\ast }=A$ , if and only if there exists a real number $a$ such that $v=au$

Problem 13.19

For every bounded operator $A$ on a Hilbert space $\mathcal{H}$ show that the exponentia operator

$${\mathrm{e}}^A=\sum _{n=0}^{\infty }\frac{A^n}{n!}$$

is well-defined and bounded on $\mathcal{H}$. Show that

(a) ${\mathrm{e}}^0=I$

(b) For all positive integers $n$, $({\mathbf{e}}^{\mathbf{A}}{)}^n={\mathbf{e}}^{n\mathbf{A}}$

(c) ${\mathrm{e}}^A$ is invertible for all bounded operators $A$ (even if $A$ is not invertible) and ${\mathfrak{e}}^{-A}=({\mathfrak{e}}^A{)}^{-1}$

(d) If $A$ and $B$ are commuting operators then ${\mathbf{e}}^{\mathbf{A}+\mathbf{B}}={\mathbf{e}}^{\mathbf{A}}{\mathbf{e}}^{\mathbf{B}}$

(e) If $A$ is hermitian then ${\mathrm{e}}^{iA}$ is unitary.

Problem 13.20

Show that the sum of two projection operators $P_M+P_N$ is a projection operator iff $P_M{P}_N=0$ . Show that this condition is equivalent to $M\perp N$

Problem 13.21

Verify that the operator on three-dimensional Hilbert space $\mathcal{H}$, having matrix repre sentation in an o.n. basis

$$\left(\begin{array}{ccc}\frac{1}{2}& 0& \frac{i}{2}\\ 0& 1& 0\\ -\frac{i}{2}& 0& \frac{1}{2}\end{array}\right)$$

is a projection operator, and find a basis of the subspace it projects onto.

Problem 13.22

Let $\omega ={\mathrm{e}}^{2\pi i/3}$ . Show that $1+\omega +{\omega }^2=0$

(a) In Hilbert space of three dimensions let $V$ be the subspace spanned by the vectors $(1,\omega ,{\omega }^2)$ and $(1,{\omega }^2,\omega )$ . Find the vector $u_0$ in this subspace that is closest to the vector $u=(1,-1,1)$

(b) Verify that $\mathbf{u}-{\mathbf{u}}_0$ is orthogonal to $V$.

(c) Find the matrix representing the projection operator $P_V$ into the subspace $V$.

Problem 13.23

An operator $A$ is called normal if it is bounded and commutes with its adjoint, $A^{\ast }A=A{A}^{\ast }$ . Show that the operator

$$A\psi (x)=c\psi (x)+i{\int }_a^{b}K(x,y)\psi (y)\mathrm{d}y$$

on $L^2([a,b])$ , where $c$ is a real number and $K(x,y)=\overline{K(y,x)}$ , is normal.

(a) Show that an operator $A$ is normal if and only if $\Vert Au\Vert =\Vert A^{\ast }u\Vert$ for all vectors $u\in \mathcal{H}$

(b) Show that if $A$ and $B$ are commuting normal operators, $AB$ and $A+\lambda B$ are normal for all $\lambda \in \mathbb{C}$