5.2 Complex inner product spaces

中文

We now consider a complex vector space $V$, which in the first instance may be infinite dimensional. Vectors will continue to be denoted by lower case Roman letters such as $u$ and $v$, but complex scalars will be denoted by Greek letters such as $\alpha$, $\beta ,\dots$ . from the early par of the alphabet. The word inner product, or scalar product, on a complex vector space

$V$ will be reserved for a map $V\times V\to \mathbb{C}$ that assigns to every pair of vectors $u$, $v\in V:$ a complex scalar $⟨u\mid v⟩$ satisfying

(IP1) $⟨u\mid v⟩=\overline{⟨v\mid u⟩}$

(IP2) $⟨u\mid \alpha v+\beta w⟩=\alpha ⟨u\mid v⟩+\beta ⟨u\mid w⟩$ for all complex numbers $\alpha ,\beta$

(IP3) $⟨u|u⟩\ge 0$ and $⟨u\mid u⟩=0⟺u=0$

The condition (IP1) implies $⟨u\mid u⟩$ is always real, a necessary condition for (IP3) to make any sense. From (IP1) and (IP2)

$$\begin{array}{rl}& ⟨\alpha v+\beta w|u⟩=\overline{⟨u|\alpha v+\beta w⟩}\\ & =\overline{\alpha ⟨u|v⟩+\beta ⟨u|w⟩}\\ & \\ & =\overline{\alpha }\overline{⟨u|v⟩}+\overline{\beta }\overline{⟨w|v⟩},\\ & \end{array}$$

so that

$$⟨\alpha v+\beta w|u⟩=\overline{\alpha }⟨v|u⟩+\overline{\beta }⟨w|u⟩.\text{\hspace{1em}(5.10)}$$

This property is often described by saying that the inner product is antilinear with respect to the first argument.

A complex vector space with an inner product will simply be called an inner product space. If $V$ is finite dimensional it is often called a finite dimensional Hilbert space, but for infinite dimensional spaces the term Hilbert space only applies if the space is complete (see Chapter 13).

Mathematicians more commonly adopt a notation $(u,v)$ in place of our angular bracket notation, and demand linearity in thefirst argument, with antilinearity in the second. Our conventions follow that which is most popular with physicists and takes its origins in Dirac’s ‘bra’ and ‘ket’ terminology for quantum mechanics (see Chapter 14).

Example 5.4

On ${\mathbb{C}}^n$ set

$$⟨({\alpha }_1,\dots ,{\alpha }_n)|({\beta }_1,\dots ,{\beta }_n)⟩=\sum _{i=1}^n\overline{{\alpha }_i}{\beta }_i.$$

Conditions (IP1)–(IP3) are easily verified. We shall see directly that this is the archetypa finite dimensional inner product space. Every finite dimensional inner product space has a basis such that the inner product takes this form.

Example 5.5

A complex-valued function $\phi :[0,1]\to \mathbb{C}$ is said to be continuous if both the real and imaginary parts of the function $\phi (x)=f(x)+ig(x)$ ) are continuous. Let $\mathcal{C}[0,1]$ be the set of continuous complex-valued functions on the real line interval [0, 1], and define an inner product

$$⟨\phi |\psi ⟩={\int }_0^1\overline{\phi (x)}\psi (x)dx.$$

Conditions (IP1) and (IP2) are simple to prove, but in order to show (IP3) it is necessary to show that

$${\int }_0^1|f(x){|}^2+|g(x){|}^2\mathrm{d}x=0⟹f(x)=g(x)=0,\forall x\in [0,1].$$

If $f(a)\ne 0$ for some $0\le a\le 1$ then, by continuity, there exists an interval $[a-ϵ,a]$ or an interval $[a,a+ϵ]$ on which $\begin{array}{r}|f(x)|>\frac{1}{2}|f(a)|\end{array}$ . Then

$${\int }_0^1|\phi (x){|}^{2}dx>\frac{1}{2}ϵ|f(a){|}^2+{\int }_0^1|g(x){|}^{2}dx>0.$$

Hence $f(x)=0$ for all $x\in [0,1]$ . The proof that $g(x)=0$ is essentially identical.

Example 5.6

A complex-valued function on the real line, $\phi :\mathbb{R}\to \mathbb{C},$ , is said to be square integrable if $|\phi {|}^2$ is an integrable function on any closed interval of $R$ and ${\int }_{-\infty }^{\infty }|\phi (x){|}^2$ $\mathrm{d}x$ $<\infty$ . The set $L^2(\mathbb{R})$ of square integrable complex-valued functions on the real line is a complex vector space, for if $\alpha$ is a complex constant and $\phi$ and $\psi$ are any pair of square integrable functions, then

$${\int }_{-\infty }^{\infty }|\phi (x)+\alpha \psi (x){|}^{2}dx\le {\int }_{-\infty }^{\infty }|\phi (x){|}^{2}dx+|\alpha {|}^2{\int }_{-\infty }^{\infty }|\psi (x){|}^{2}dx<\infty .$$

On $L^2(\mathbb{R})$ define the inner product

$$⟨\phi |\psi ⟩={\int }_{-\infty }^{\infty }\overline{\phi (x)}\psi (x)dx.$$

This is well-defined for any pair of square integrable functions $\phi$ and $\psi$ for, after some algebraic manipulation, we find that

$$\overline{\phi }\psi =\frac{1}{2}(|\phi +\psi {|}^2-i|\phi +i\psi {|}^2-(1-i)(|\phi {|}^2+|\psi {|}^2)).$$

Hence the integral of the left-hand side is equal to a sum of integrals on the right-hand side, each of which has been shown to exist

The properties (IP1) and (IP2) are trivial to show but the proof of (IP3) along the lines given in Example 5.5 will not suffice here since we do not stipulate continuity for the functions in $L^2(\mathbb{R})$ . For example, the function $f(x)$ defined by $f(x)=0$ for all $x\ne 0$ and $f(0)=1$ is a positive non-zero function whose integral vanishes. The remedy is to identify any two real functions $f$ and $g$ having the property that $\begin{array}{r}{\int }_{-\infty }^{\infty }|f(x)-g(x){|}^2\mathrm{d}x=0\end{array}$ . Such a pair of functions will be said to be equal almost everywhere, and $L^2(\mathbb{R})$ ) must be interpreted as consisting ofequivalence classes of complex-valued functions whose real and imaginary parts are equal almost everywhere. A more complete discussion will be given in Chapter 13, Example 13.4.

Exercise

Show that the relation $f\equiv g⟺{\int }_{-\infty }^{\infty }|f(x)-g(x){|}^2\mathrm{d}x=0$ is an equivalence relation on $L^2(\mathbb{R})$ .

Norm of a vector

The norm of a vector $u$ in an inner product space, denoted $\Vert u\Vert$ , is defined to be the non negative real number

$$\Vert u\Vert =\sqrt{⟨u|u⟩}\ge 0.\text{\hspace{1em}(5.11)}$$

From (IP2) and Eq. (5.10) it follows immediately that

$$\Vert \alpha u\Vert =|\alpha |\Vert u\Vert .\text{\hspace{1em}(5.12)}$$

Theorem 5.4 · Cauchy–Schwarz inequality

(Cauchy–Schwarz inequality) For any pair of vectors $u$, $v$ in an inner product space

$$|⟨u|v⟩|\le \Vert u\Vert \Vert v\Vert .\text{\hspace{1em}(5.13)}$$

Proof

By (IP3), (IP2) and Eq. (5.10) we have for all $\lambda \in \mathbb{C}$

$$\begin{array}{l}0\le ⟨u+\lambda v|u+\lambda v⟩\\ =⟨u|u⟩+\lambda ⟨u|v⟩+\overline{\lambda }⟨v|u⟩+\lambda \overline{\lambda }⟨v|v⟩.\end{array}$$

Substituting the particular value

$$\lambda =-\frac{⟨v|u⟩}{⟨v|v⟩}$$

gives the inequality

$$\begin{array}{l}0\le ⟨u|u⟩-\frac{⟨v|u⟩⟨u|v⟩}{⟨v|v⟩}-\frac{\overline{⟨v|u⟩}⟨v|u⟩}{⟨v|v⟩}+\frac{|⟨v|u⟩{|}^2}{⟨v|v⟩}\\ =⟨u|u⟩-\frac{|⟨v|u⟩{|}^2}{⟨v|v⟩}.\end{array}$$

Hence, from (IP1),

$$|⟨u|v⟩{|}^2=|⟨v|u⟩{|}^2\ge ⟨u|u⟩⟨v|v⟩$$

and the desired result follows from Eq. (5.11) on taking the square roots of both sides of this inequality.

Corollary 5.5 · Equality in the Cauchy–Schwarz inequality

Equality in Eq. (5.13) can only result if $u$ and $v$ areproportional to each other,

$$|⟨u|v⟩|=\Vert u\Vert \Vert v\Vert ⟺u=\alpha v,\text{ for some }\alpha \in \mathbb{C}.$$

Proof

If $u=\alpha v$ then from Eq. (5.11) and Eq. (5.12) we have

$$|⟨u|v⟩|=|⟨\alpha v|v⟩|=|\alpha |⟨v|v⟩=\Vert \alpha v\Vert \Vert v\Vert =\Vert u\Vert \Vert v\Vert .$$

Conversely, $\mathrm{if}|⟨u|v⟩|=\Vert u\Vert \Vert v\Vert$ then,

$$|⟨u|v⟩{|}^2=\Vert u{\Vert }^2\Vert v{\Vert }^2$$

and reversing the steps in the proof of Lemma 5.4 with inequalities replaced by equalities gives

$$⟨u+\lambda v|u+\lambda v⟩=0\text{ where }\lambda =-\frac{⟨v|u⟩}{⟨v|v⟩}.$$

By (IP3) we conclude that $u=-\lambda v$ and the proposition follows with $\alpha =-\lambda$

Theorem 5.6 · Triangle inequality

(Triangle inequality) For any pair of vectors $u$ and $v$ in an inner product space,

$$\Vert u+v\Vert \le \Vert u\Vert +\Vert v\Vert .\text{\hspace{1em}(5.14)}$$

Proof

$$\begin{array}{l}(\Vert u+v\Vert {)}^2=⟨u+v|u+v⟩\\ =⟨u|u⟩+⟨v|v⟩+⟨v|u⟩+⟨u|v⟩\\ =\Vert u{\Vert }^2+\Vert v{\Vert }^2+2\mathrm{Re}(⟨u|v⟩)\\ \le \Vert u{\Vert }^2+\Vert v{\Vert }^2+2|⟨u|v⟩|\\ \le \Vert u{\Vert }^2+\Vert v{\Vert }^2+2\Vert u\Vert \Vert v\Vert \text{ by Eq. (5.13) }\\ =(\Vert u\Vert +\Vert v\Vert {)}^2.\end{array}$$

The triangle inequality Eq. (5.14) follows on taking square roots.

Orthonormal bases

Let $V$ be a finite dimensional inner product space with basis $e_1,e_2,\dots ,e_n$ . Define the components of the inner product with respect to this basis to be

$$h_{ij}=⟨e_i|e_j⟩=\overline{⟨e_j|e_i⟩}=\overline{h_{ji}}.\text{\hspace{1em}(5.15)}$$

The matrix of components $\mathsf{H}=[h_{ij}]$ is clearly hermitian,

$$\mathsf{H}={\mathsf{H}}^{†}\text{ where }{\mathsf{H}}^{†}={\overline{\mathsf{H}}}^T.$$

Under a change of basis Eq. (5.3) we have

$$⟨e_i|e_j⟩=\overline{A_i^k}⟨e_k^{\prime }|e_m^{\prime }⟩A_j^m$$

and the components of the inner product transform as

$$h_{ij}=\overline{A_i^k}{h}_{km}^{\prime }{A}_j^m.\text{\hspace{1em}(5.16)}$$

An identical argument can be used to express the primed components in terms of unprimed components,

$$h_{ij}^{\prime }=⟨e_i^{\prime }|e_j^{\prime }⟩={\overline{A^{\prime k}}}_i{h}_{km}{A}_j^{\prime m}.\text{\hspace{1em}(5.17)}$$

These equations have matrix equivalents,

$$\mathsf{H}={\mathsf{A}}^{†}{\mathsf{H}}^{\prime }\mathsf{A},{\mathsf{H}}^{\prime }={\mathsf{A}}^{\prime †}\mathsf{H}{\mathsf{A}}^{\prime },\text{\hspace{1em}(5.18)}$$

where ${\mathsf{A}}^{\prime }={\mathsf{A}}^{-1}$

Exercise

Show that the hermitian nature of the matrix $H$ is unchanged by a transformation Eq. (5.18).

Two vectors $u$ and $v$ are said to be orthogonal if $⟨u|v⟩=0$ . A basis $e_1,e_2,\dots ,e_n$ is called an orthonormal basis if the vectors all have unit norm and are orthogonal to each

other,

$$⟨e_i|e_j⟩={\delta }_{ij}=\{\begin{array}{ll}1& \text{ if }i=j,\\ 0& \text{ if }i\ne j.\end{array}$$

Equivalently, a basis is orthonormal if the matrix of components of the inner product with respect to the basis is the unit matrix, $\mathsf{H}=\mathsf{I}$

Starting with an arbitrary basis $\{u_1,u_2,\dots ,u_n\}$ , it is always possible to construct an orthonormal basis by a process known as Schmidt orthonormalization, which closely mirrors the Gram–Schmidt process for Euclidean inner products, outlined in Example 5.2. Sequentially, the steps are:

1. Set $f_1=u_1,\mathrm{then}{e}_1=\frac{f_1}{\Vert f_1\Vert }.$

2. Set $f_2=u_2-⟨e_1|u_2⟩e_1$ , which is orthogonal to $e_1$ since

$$⟨e_1\mid f_2⟩=⟨e_1\mid u_2⟩-⟨e_1\mid u_2⟩\Vert e_1\Vert =0.$$

Normalize $f_2$ by setting $e_2=f_2/\Vert f_2\Vert$

3. Set $f_3=u_3-⟨e_1|u_3⟩e_1-⟨e_2|u_3⟩e_2$ , which is orthogonal to both $e_1$ and $e_2$ . Normalize to give $e_3=\frac{f_3}{\parallel f_3\parallel }$

4. Continue in this way until

$$f_n=u_n-\sum _{i=1}^{n-1}⟨e_i\mid u_n⟩e_i\text{ and }{e}_n=\frac{f_n}{\Vert f_n\Vert }.$$

Since each vector $e_i$ is a unit vector and is orthogonal to all the $e_j$ for $j<i$ defined by previous steps, they form an o.n. set. It is easily seen that any vector $v=v^i{u}_i$ of $V$ is a linear combination of the $e_i$ since each $u_j$ is a linear combination of $e_1,\dots ,e_j$ . Hence the vectors $\{e_i\}$ form a basis by Theorem 3.3 since they span $V$ and are $n$ in number.

With respect to an orthonormal basis the inner product of any pair of vectors $u=u^i{e}_i$ and $v=v^j{e}_j$ is given by

$$⟨u|v⟩=⟨u^i{e}_i|v^j{e}_j⟩=\overline{u^i}{v}^j⟨e_i|e_j⟩=\overline{u^i}{v}^j{\delta }_{ij}.$$

Hence

$$⟨u\mid v⟩\sum _{i=1}^n\overline{u^i}{v}^i=\overline{u^1}{v}^1+\overline{u^2}{v}^2+\cdots +\overline{u^n}{v}^n,$$

which is equivalent to the standard inner product defined on ${\mathbb{C}}^n$ in Example 5.4.

Example 5.7

Let an inner product have the following components in a basis $u_1,u_2,u_3;$

$$h_{11}=⟨u_1|u_1⟩=1h_{12}=⟨u_1|u_2⟩=0h_{13}=⟨u_1|u_3⟩=\frac{1}{2}(1+i)$$ $$h_{21}=⟨u_2|u_1⟩=0h_{22}=⟨u_2|u_2⟩=2h_{23}=⟨u_2|u_3⟩=0$$ $$h_{31}=⟨u_3|u_1⟩=\frac{1}{2}(1-i)h_{32}=⟨u_3|u_2⟩=0h_{33}=⟨u_3|u_3⟩=1.$$

Before proceeding it is important to realize that this inner product does in fact satisfy the positive definite condition (IP3). This would not be true, for example, if we had given $h_{13}=\overline{h_{31}}=1+i$ , for then the vector $v=u_1-\frac{1-i}{\sqrt{2}}{u}_3$ would have negative norm $⟨v|v⟩=2(1-\sqrt{2})<0$

In the above inner product, begin by setting $e_1=f_1=u_1$ . The next vector is

$$f_2=u_2-⟨e_1|u_2⟩e_1=u_2,e_2=\frac{u_2}{\Vert u_2\Vert }=\frac{1}{\sqrt{2}}{u}_2.$$

The last step is to set

$$f_3=u_3-⟨e_1|u_3⟩e_1-⟨e_2|u_3⟩e_2=u_3-\frac{1}{2}(1+i)u_1$$

which has norm squared

$$\begin{array}{c}(\Vert f_3\Vert {)}^2=⟨u_3|u_3⟩-\frac{1}{2}(1-i)⟨u_1|u_3⟩-\frac{1}{2}(1+i)⟨u_3|u_1⟩+\frac{1}{4}(1-i)(1+i)⟨u_1|u_1⟩\\ =1-\frac{1}{4}(1-i)(1+i)-\frac{1}{4}(1+i)(1-i)+\frac{1}{2}=\frac{1}{2}.\end{array}$$

Hence $e_3=f_3/\Vert f_3\Vert =\sqrt{2}(u_3-(1+i)u_1)$ completes the orthonormal basis.

The Schmidt orthonormalization procedure actually provides a good method for proving positive definiteness, since the process breaks down at some stage, producing a vector with non-positive norm if the inner product does not satisfy (IP3).

Exercise

Try to perform the Schmidt orthonormalization on the above inner product suggested with the change $h_{13}=\overline{h_{31}}=1+i$ , and watch it break down!

Unitary transformations

A linear operator $U:V\to V$ on an inner product space is said to be unitary if it preserves inner products,

$$⟨Uu|Uv⟩=⟨u|v⟩,\forall u,v\in V.\text{\hspace{1em}(5.19)}$$

Unitary operators clearly preserve the norm of any vector $v$,

$$\Vert Uv\Vert =\sqrt{⟨Uv|Uv⟩}=\sqrt{⟨v|v⟩}=\Vert v\Vert .$$

In fact it can be shown that a linear operator $U$ is unitary if and only if it is norm preserving (see Problem 5.7).

A unitary operator $U$ transforms any orthonormal basis $\{e_i\}$ into another o.n. basis $e_i^{\prime }=U{e}_i$ , since

$$⟨e_i^{\prime }|e_j^{\prime }⟩=⟨U{e}_i|U{e}_j⟩=⟨e_i|e_j⟩={\delta }_{ij}.\text{\hspace{1em}(5.20)}$$

The set $\{e_1^{\prime },\dots ,e_n^{\prime }\}$ is linearly independent, and is thus a basis, for if ${\alpha }^i{e}_i^{\prime }=0$ then $⟨e_j^{\prime }|{\alpha }^i{e}_i^{\prime }⟩={\alpha }^j=0$ . The map $U$ is onto since every vector $u=u^{\prime i}{e}_i^{\prime }=U(u^{\prime i}{e}_i)$ , and one-to-one since $Uv=0\Rightarrow v^i{e}_i^{\prime }=0\Rightarrow v=v^i{e}_i=0$ . Hence every unitary operator $U$ is invertible.

With respect to an orthonormal basis $\{e_i\}$ the components of the linear transformation $U$, defined by $U{e}_i=U_i^k{e}_k$ , form a unitary matrix $\mathsf{U}=[U_i^k]$ :

$$\begin{array}{c}{\delta }_{ij}=⟨U{e}_i|U{e}_j⟩=\overline{U_i^k}{U}_j^m⟨e_k|e_m⟩\\ =\overline{U_i^k}{U}_j^m{\delta }_{km}={\sum }_{k=1}^n\overline{U_i^k}{U}_j^k,\end{array}$$

or, in terms of matrices,

$$\mathsf{I}={\mathsf{U}}^{†}\mathsf{U}.$$

If $\left\{e_i\right\}$ and $\{e_j^{\prime }\}$ are any pair of orthonormal bases, then the linear operator $U$ defined by $e_i^{\prime }=U{e}_i$ is unitary since for any pair of vectors $u=u^i{e}_i$ and $v=v^j{e}_j$

$$\begin{array}{rl}⟨Uu|Uv⟩& =\overline{u^i}{v}^j⟨e_i^{\prime }|e_j^{\prime }⟩\\ & =\overline{u^i}{v}^j{\delta }_{ij}\\ & =\overline{u^i}{v}^j⟨e_i|e_j⟩=⟨u|v⟩.\end{array}$$

Thus all orthonormal bases are uniquely related by unitary transformations.

In the language of Section 3.6 this is the active view, wherein vectors are ‘physically’ moved about in the inner product space by the unitary transformation. In the relatedpassive view, the change of basis is given by Eq. (5.3) – it is the components of vectors that are transformed, not the vectors themselves. If both bases are orthonormal the components of an inner product, given by Eq. (5.16), are $h_{ij}=h_{ij}^{\prime }={\delta }_{ij}$ , and setting $A_i^k=U_i^k$ in Eq. (5.18) implies the matrix $\mathsf{U}=[U_i^k]$ is unitary,

$$\mathsf{I}=\mathsf{H}={\mathsf{U}}^{†}{\mathsf{H}}^{\prime }\mathsf{U}={\mathsf{U}}^{†}\mathsf{IU}={\mathsf{U}}^{†}\mathsf{U}.$$

Thus, from both the active and passive viewpoint, orthonormal bases are related by unitary matrices.

Problems

Problem 5.6

Show that the norm defined by an inner product satisfies the parallelogram law

$$\Vert u+v{\Vert }^2+\Vert u-v{\Vert }^2=2\Vert u{\Vert }^2+2\Vert v{\Vert }^2.$$

Problem 5.7

On an inner product space show that

$$4⟨u|v⟩=\Vert u+v{\Vert }^2-\Vert u-v{\Vert }^2-i\Vert u+iv{\Vert }^2+i\Vert u-iv{\Vert }^2.$$

Hence show that a linear transformation $U:V\to V$ is unitary iff it is norm preserving,

$$⟨Uu|Uv⟩=⟨u|v⟩,\forall u,v\in V⟺\Vert Uv\Vert =\Vert v\Vert ,\forall v\in V.$$

Problem 5.8

Show that a pair of vectors $u$ and $v$ in a complex inner product space are orthogonal iff

$${\Vert \alpha u+\beta v\Vert }^2={\Vert \alpha u\Vert }^2+{\Vert \beta v\Vert }^2,\forall \alpha ,\beta \in \mathbb{C}.$$

Find a non-orthogonal pair of vectors $u$ and $v$ in a complex inner product space such that $\Vert u+v{\Vert }^2=$ $\Vert u{\Vert }^2+\Vert v{\Vert }^2$

Problem 5.9

Show that the formula

$$⟨\mathsf{A}|\mathsf{B}⟩=\mathrm{tr}({\mathsf{BA}}^{†})$$

defines an inner product on the vector space of $m\times n$ complex matrices $M(m,n)$

(a) Calculate $\Vert {\mathsf{I}}_n\Vert$ where ${\mathsf{I}}_n$ is the $n\times n$ identity matrix.

(b) What characterizes matrices orthogonal to ${\mathsf{I}}_n?$

(c) Show that all unitary $n\times n$ matrices $U$ have the same norm with respect to this inner product.

Problem 5.10

Let $S$ and $T$ be complex inner product spaces and let $U:S\to T$ be a linear map such that $\Vert Ux\Vert =\Vert x\Vert$ . Prove that

$$⟨Ux\mid Uy⟩=⟨x\mid y⟩\text{ for all }x,y\in S.$$

Problem 5.11

Let $V$ be a complex vector space with an ‘indefinite inner product’, defined as an inner product that satisfies (IP1), (IP2) but with (IP3) replaced by the non-singularity condition

(IP3’) $⟨u|v⟩=0$ for all $v\in V$ implies that $u=0$

(a) Show that similar results to Theorem 5.2 and Theorem 5.3 can be proved for such an indefinite inner product.

(b) If there are $p$ entries equal to $+1$ along the diagonal and $q$ entries equal to $-1$ , find the defining relations for the group of transformations $U(p,q)$ between orthonormal basis.

Problem 5.12

If $V$ is an inner product space, an operator $K:V\to V$ is called self-adjoint if

$$⟨u|Kv⟩=⟨Ku|v⟩$$

for any pair of vectors $u$, $v\in V$ . Let $\{e_i\}$ be an arbitrary basis, having $⟨e_i\mid e_j⟩=h_{ij}$ , and set $K{e}_k=$ $K_k^j{e}_j$ . Show that if $\mathsf{H}=[h_{ij}]$ and $\mathsf{K}=[K_j^k]$ then

$$\mathsf{HK}={\mathsf{K}}^{†}\mathsf{H}=(\mathsf{HK}{)}^{†}.$$

If $\left\{e_i\right\}$ is an orthonormal basis, show that $K$ is a hermitian matrix.