Wichita State University Logo

Math 511: Linear Algebra

7.3 Hermitian Matrices

7.3.1 Complex Valued Vectors and Matrices

$\require{color}$ $\definecolor{brightblue}{rgb}{.267, .298, .812}$ $\definecolor{darkblue}{rgb}{0.0, 0.0, 1.0}$ $\definecolor{palepink}{rgb}{1, .73, .8}$ $\definecolor{softmagenta}{rgb}{.99,.34,.86}$ $\def\ihat{\mathbf{\hat{\unicode{x0131}}}}$ $\def\jhat{\mathbf{\hat{\unicode{x0237}}}}$ $\def\khat{\mathbf{\hat{k}}}$ $\def\tombstone{\unicode{x220E}}$

Up until this point, we have primarily discussed vectors and matrices in $\mathbb{R}^n$, with occasional reference to the possibility of working with complex values. It is now time to address complex values directly. For vectors, we will consider $\mathbb{C}^n$ where vectors $\mathbf{z}\in\mathbb{C}^n$ have the form

$$ \mathbf{z} = \begin{bmatrix} z_1 \\ z_2 \\ \vdots \\ z_n \end{bmatrix} = \begin{bmatrix} x_1 + y_1 i \\ x_2 + y_2 i \\ \vdots \\ x_n + y_n i \end{bmatrix} $$
with $x_j,y_j\in\mathbb{R}$ for all $j=1,2,\ldots,n$ and $i = \sqrt{-1}$ is the imaginary unit. The scalar field for complex-valued vectors will be $\mathbb{C}$.

Complex Inner Products

If $\alpha = a + bi$ is a complex-valued scalar, its length is given by

$$ \left|\alpha\right| = \sqrt{\bar{\alpha}\alpha} = \sqrt{a^2 + b^2} $$
where $\bar{\alpha} = a - bi$. The length of a vector

$$ \mathbf{z} = \begin{bmatrix} z_1 \\ z_2 \\ \vdots \\ z_n \end{bmatrix} $$
is given by

$$ \begin{align*} \left\| \mathbf{z} \right\| &= \left(|z_1|^2 + |z_2|^2 + \ldots + |z_n|^2\right)^\frac{1}{2} \\ \\ &= \left(\bar{z}_1z_1 + \bar{z}_2 z_2 + \ldots + \bar{z}_n z_n\right)^\frac{1}{2} \\ \\ &= \left(\bar{\mathbf{z}}^T \mathbf{z}\right)^\frac{1}{2} \end{align*} $$
For convenience, we define the symbol $\mathbf{z}^H$ to be the conjugate transpose or Hermitian transpose of the vector $\mathbf{z}$, so

$$ \bar{\mathbf{z}}^T = \mathbf{z}^H \qquad\qquad\qquad \|\mathbf{z}\| = \left(\mathbf{z}^H\mathbf{z}\right)^\frac{1}{2} $$

Definition of Complex Inner Product

Let $V$ be a vector space over the complex numbers. An inner product on $V$ is a operation that assigns to each pair of vectors $\mathbf{z},\mathbf{w}\in V$ a complex number $\langle \mathbf{z},\mathbf{w}\rangle$ satisfying the conditions
I. $\langle \mathbf{z},\mathbf{z} \rangle \geq 0$, and $\langle \mathbf{z},\mathbf{z} \rangle = 0$ if and only if $\,\mathbf{z} = \mathbf{0}$.
II. $\langle \mathbf{z},\mathbf{w} \rangle = \overline{\langle \mathbf{w},\mathbf{z} \rangle}$ for all $\,\mathbf{z},\mathbf{w}\in V$.
III. $\langle \alpha\mathbf{z} + \beta\mathbf{w},\mathbf{u} \rangle = \alpha\langle\mathbf{z},\mathbf{u}\rangle + \beta\langle\mathbf{w},\mathbf{u}\rangle$.

Note that condition II states a complex inner product is conjugate symmetric instead of merely symmetric. It will require some simple changes, but taking this fact into account will allows use our previous theorems on real inner product spaces for complex inner product spaces. It is particularly useful for us to recover the theorems leading to Parseval's Identity , so lets restate those with an complex valued inner product:

Theorem 5.3.2

The Scalar Projection of a Coordinate Vector onto an Orthonormal Basis Vector is the Coordinate

Let $\left\{\mathbf{u}_1,\mathbf{u}_2,\ldots,\mathbf{u}_n\right\}$ be an orthonormal basis for a complex inner product space $V$. If

$$ \mathbf{z} = \sum_{i=1}^n c_i\mathbf{w}_i = c_1\mathbf{w}_1 + c_2\mathbf{w}_2 + \ldots + c_n\mathbf{w}_n $$
then

$$ c_i = \langle \mathbf{z},\mathbf{w}_i \rangle \qquad\qquad\qquad \bar{c}_i = \langle \mathbf{w}_i,\mathbf{z} \rangle $$

Corollary 5.3.3

The Inner Product of Coordinate Vectors in an Orthonormal Basis is the Sum of the Component-wise Product

Let $\left\{\mathbf{w}_1,\mathbf{w}_2,\ldots,\mathbf{w}_n\right\}$ be an orthonormal basis for a complex inner product space $V$. If

$$ \mathbf{z} = \sum_{i=1}^n c_i\mathbf{w}_i \qquad\qquad \mathbf{w} = \sum_{i=1}^n w_i\mathbf{w}_i $$
then

$$ \langle \mathbf{z},\mathbf{w} \rangle = \sum_{i=1}^n c_i \bar{w}_i \qquad\qquad \langle \mathbf{w},\mathbf{z} \rangle = \sum_{i=1}^n w_i \bar{c}_i $$

Corollary 5.3.4

Parseval's Identity

If $\left\{\mathbf{u}_1,\mathbf{u}_2,\ldots,\mathbf{u}_n\right\}$ is an orthonormal basis for a complex inner product space $V$ and

$$ \mathbf{z} = \sum_{i=1}^n c_i\mathbf{w}_i $$
then

$$ \|\mathbf{z}\|^2 = \langle \mathbf{z},\mathbf{z} \rangle = \sum_{i=1}^n c_i\bar{c}_i $$

These theorems all use the inner product defined by

$$ \langle \mathbf{z},\mathbf{w} \rangle = \mathbf{w}^H \mathbf{z} $$
for all $\mathbf{z},\mathbf{w}\in\mathbb{C}^n$. Most of what we know about inner product spaces, including the modified theorems above, comes with little effort from our knowledge of real inner product spaces if we mind the following adjustments:

$ \mathbb{R}^n $ $\qquad$ $ \mathbb{C}^n $
$\langle \mathbf{x},\mathbf{y} \rangle = \mathbf{y}^T\mathbf{x}$ $\langle \mathbf{z},\mathbf{w} \rangle = \mathbf{w}^H\mathbf{z}$
$\mathbf{x}^T\mathbf{y} = \mathbf{y}^T\mathbf{x}$ $\mathbf{z}^H\mathbf{w} = \overline{\mathbf{w}^H\mathbf{z}}$
$|\mathbf{x}|^2 = \mathbf{x}^T\mathbf{x}$ $|\mathbf{z}|^2 = \mathbf{z}^H\mathbf{z}$

Exercise 1

Given

$$ \mathbf{z} = \begin{bmatrix} 1 + 3i \\ -2 -i \end{bmatrix} \qquad\qquad\qquad \mathbf{w} = \begin{bmatrix} 3 + 2i \\ 5 - i \end{bmatrix} $$
compute

  1. $\mathbf{w}^H\mathbf{z}$
  2. $\mathbf{z}^H\mathbf{z}$
  3. $\mathbf{w}^H\mathbf{w}$
Follow Along

$$ \begin{align*} \mathbf{w}^H\mathbf{z} &= \begin{bmatrix} 3 - 2i & 5 + i \end{bmatrix}\begin{bmatrix} 1 + 3i \\ -2 -i \end{bmatrix} \\ \\ &= (3-2i+9i-6i^2) + (-10-2i-5i-i^2) \\ \\ &= (9+7i) + (-9 - 7i) = 0 \\ \\ \mathbf{z}^H\mathbf{z} &= |1+3i|^2 + |-2-i|^2 = 15 \\ \\ \mathbf{w}^H\mathbf{w} &= |3+2i|^2 + |5-i|^2 = 39 \end{align*} $$

7.3.2 Hermitian Matrices

Let $Z\in\mathbb{C}^{m\times n}$ define a $m\times n$ matrix comprised of complex numbers. We can write each entry of $Z$ according to its component form $[z_{ij}]$ where each $z_{ij} = a_{ij} + ib_{ij}$ for $a_{ij},b_{ij}\in\mathbb{R}$.

Note: The subscript $i$ is the index for the row and the regular script $i$ is the imaginary unit $i=\sqrt{-1}$.

Using the definition of matrix addition, it is possible for us to write $Z$ as the sum

$$ Z = [z_{ij}] = [a_{ij} + ib_{ij}] = [a_{ij}] + i[b_{ij}] = A + iB $$
where $A = [a_{ij}]$ and $B = [b_{ij}]$ are both in $\mathbb{R}^{m\times n}$. This allows us to define the conjugate of the matrix $Z$ as

$$ \overline{Z} = A - iB $$
Simply put, this works in the way we are accustomed to conjugates operating, just negate all of the imaginary terms. For complex-valued matrices, we also need to make sure that we use the conjugate transpose, so $Z^H = \overline{Z}^T$ just as with complex vectors. The conjugate transpose follows rules similar to what we are familiar with for the typical transpose

  • $\left(Z^H\right)^H = Z$
  • $\left( \alpha Z + \beta W \right)^H = \bar{\alpha}Z^H + \bar{\beta}W^H$
  • $\left( ZW \right)^H = W^H Z^H $

Definition of Hermitian matrix

A matrix $Z$ is called Hermitian if $Z = Z^H$.

Example 1

Both

$$ Z = \begin{bmatrix} -1 & 2 - 2i \\ 2 + 2i & 3 \end{bmatrix} \qquad\qquad\qquad W = \begin{bmatrix} 3 & 5 + i & 4 - 2i \\ 5 - i & -7 & -3 + 2i \\ 4 + 2i & -3 - 2i & 2 \end{bmatrix} $$
are Hermitian since

$$ \begin{align*} Z^H = \begin{bmatrix} \overline{-1} & \overline{2 + 2i} \\ \overline{2 - 2i} & \overline{3} \end{bmatrix} &= \begin{bmatrix} -1 & 2 - 2i \\ 2 + 2i & 3 \end{bmatrix} = Z \\ \\ W^H = \begin{bmatrix} \overline{3} & \overline{5 - i} & \overline{4 + 2i} \\ \overline{5 + i} & \overline{-7} & \overline{-3 - 2i} \\ \overline{4 - 2i} & \overline{-3 + 2i} & \overline{2} \end{bmatrix} &= \begin{bmatrix} 3 & 5 + i & 4 - 2i \\ 5 - i & -7 & -3 + 2i \\ 4 + 2i & -3 - 2i & 2 \end{bmatrix} = W \end{align*} $$

Theorem 7.3.1

Hermitian matrices have real eigenvalues and eigenvectors corresponding to distinct eigenvalues are orthogonal.

Proof:

Suppose $A$ is a Hermitian matrix, and let $\lambda$ be an eigenvalue of $A$ with corresponding eigenvector $\mathbf{x}$. Set the scalar $\alpha = \mathbf{x}^H\! A\mathbf{x}$, then

$$ \bar{\alpha} = \alpha^H = \left(\mathbf{x}^H\! A\mathbf{x}\right)^H = \mathbf{x}^H A^H \left(\mathbf{x}^H\right)^H = \mathbf{x}^H\! A\mathbf{x} = \alpha $$
So $\alpha\in\mathbb{R}$. Also,

$$ \alpha = \mathbf{x}^H\! A\mathbf{x} = \mathbf{x}^H\! \lambda\mathbf{x} = \lambda\mathbf{x}^H \mathbf{x} = \lambda\|\mathbf{x}\|^2 $$
which implies that $\lambda\in\mathbb{R}$ since

$$ \lambda = \dfrac{\alpha}{\|\mathbf{x}\|^2} $$
Next, we want to show that eigenvectors for distinct eigenvalues are orthogonal, so let $\mathbf{x}_1$ and $\mathbf{x}_2$ be eigenvectors for $\lambda_1$ and $\lambda_2$, respectively. The expressions

$$ \begin{align*} \left(A\mathbf{x}_1\right)^H \mathbf{x}_2 = \mathbf{x}_1^H A^H \mathbf{x}_2 &= \mathbf{x}_1^H A \mathbf{x}_2 = \lambda_2 \mathbf{x}_1^H \mathbf{x}_2 \\ \\ \left(A\mathbf{x}_1\right)^H \mathbf{x}_2 = \left(\mathbf{x}_2^H A\mathbf{x}_1\right)^H &= \left(\lambda_1\mathbf{x}_2^H \mathbf{x}_1\right)^H = \lambda_1 \mathbf{x}_1^H \mathbf{x}_2 \end{align*} $$
show that


$$ \begin{align*} \lambda_1 \mathbf{x}_1^H \mathbf{x}_2 &= \lambda_2 \mathbf{x}_1^H \mathbf{x}_2 \\ \\ \lambda_1 \mathbf{x}_1^H \mathbf{x}_2 - \lambda_2 \mathbf{x}_1^H \mathbf{x}_2 &= 0 \\ \\ \left(\lambda_1 - \lambda_2\right)\, \langle \mathbf{x}_2, \mathbf{x}_1 \rangle &= 0 \\ \end{align*} $$
which can only mean that $\langle \mathbf{x}_2, \mathbf{x}_1 \rangle = 0$, since $\lambda_1 \neq \lambda_2$. Therefore $\mathbf{x}_1$ and $\mathbf{x}_2$ are orthogonal.
$\tombstone$

7.3.3 Unitary Matrices

Orthogonal matrices have a complex-valued counterpart called unitary matrices .

Definition of Unitary Matrix

A matrix $U\in\mathbb{C}^{n\times n}$ is unitary if its columns form an orthonormal set in $\mathbb{C}^n$.

An equivalent statement is that a matrix is unitary if and only if $U^H U = I$. Since $U$ is square with orthonormal columns it is rank $n$ and hence invertible. These facts also imply that $U^H = U^{-1}$.

Corollary 7.3.2

If a Hermitian matrix $A$ has distinct eigenvalues, then there exists a unitary matrix $U$ which diagonalizes $A$.

Proof:

Let $\mathbf{x}_i$ be an eigenvector corresponding to eigenvalue $\lambda_i$ of $A$. For each $i=1,2,\ldots,n$, set

$$ \mathbf{u}_i = \dfrac{\mathbf{x}_i}{\|\mathbf{x}_i\|} $$
so $\mathbf{u}_i$ is a unit eigenvector for $\lambda_i$. By Theorem 5.3.1 , $\left\{\mathbf{u}_1,\mathbf{u}_2,\ldots,\mathbf{u}_n\right\}$ is an orthonormal set in $\mathbf{C}^n$. Let

$$ U = \begin{bmatrix}\ \mathbf{u}_1 & \mathbf{u}_2 & \ldots & \mathbf{u}_n\ \end{bmatrix} $$
so $U$ is unitary and diagonalizes $A$.
$\tombstone$

Exercise 2

Let

$$ A = \begin{bmatrix} -1 & 2 + i \\ 2 - i & 3 \end{bmatrix} $$
Find a unitary matrix $U$ that diagonalizes $A$.

Follow Along We begin by finding the eigenvalues of $A$

$$ \begin{align*} p(\lambda) = |A - \lambda I| &= \begin{bmatrix} -1 - \lambda & 2 + i \\ 2 - i & 3 - \lambda \end{bmatrix} \\ \\ &= (-1-\lambda)(3-\lambda) - (2-i)(2+i) \\ \\ &= \lambda^2 - 2\lambda - 8 \\ \\ &= (\lambda + 2)(\lambda - 4) \end{align*} $$
So the eigenvalues of $A$ are $\lambda_1 = -2$ and $\lambda_2 = 4$. We want to find the associated unit eigenvectors $\mathbf{u}_1$ and $\mathbf{u}_2$, so we determine bases for $N(A + 2I)$ and $N(A - 4I)$ by determine the reduced row echelon form of each of these matrices.

$$ \begin{align*} A + 2I &= \begin{bmatrix} 1 & 2+i \\ 2-i & 5 \end{bmatrix}\xrightarrow[-(2-i)R_1 + R_2]{~} \begin{bmatrix} 1 & 2+i \\ 0 & 0 \end{bmatrix} \\ \\ A - 4I &= \begin{bmatrix} -5 & 2+i \\ 2-i & -1 \end{bmatrix}\xrightarrow[\frac{1}{2+i}R_1 + R_2]{~} \begin{bmatrix} -5 & 2+i \\ 0 & 0 \end{bmatrix} \end{align*} $$
From these relationships, we see that eigenvectors associated with $\lambda_1$ and $\lambda_2$ are

$$ \mathbf{x}_1 = \begin{bmatrix} 2 + i \\ -1 \end{bmatrix} \qquad\qquad\qquad \mathbf{x}_2 = \begin{bmatrix} 2 + i \\ 5 \end{bmatrix} $$
respectively. We can normalize each of these vectors to determine the unit eigenvectors

$$ \mathbf{u}_1 = \begin{bmatrix} \frac{2}{\sqrt{6}} + \frac{i}{\sqrt{6}} \\ -\frac{1}{\sqrt{6}} \end{bmatrix} \qquad\qquad\qquad \mathbf{u}_2 = \begin{bmatrix} \frac{2}{\sqrt{30}} + \frac{i}{\sqrt{30}} \\ \frac{5}{\sqrt{30}} \end{bmatrix} $$
Hence the unitary matrix $U$ that diagonalizes $A$ is given by

$$ U = \begin{bmatrix} \frac{2}{\sqrt{6}} + \frac{i}{\sqrt{6}} & \frac{2}{\sqrt{30}} + \frac{i}{\sqrt{30}} \\ -\frac{1}{\sqrt{6}} & \frac{5}{\sqrt{30}} \end{bmatrix} $$
and we have that

$$ A = UDU^T = \begin{bmatrix} \frac{2}{\sqrt{6}} + \frac{i}{\sqrt{6}} & \frac{2}{\sqrt{30}} + \frac{i}{\sqrt{30}} \\ -\frac{1}{\sqrt{6}} & \frac{5}{\sqrt{30}} \end{bmatrix} \begin{bmatrix} -2 & 0 \\ 0 & 4 \end{bmatrix} \begin{bmatrix} \frac{2}{\sqrt{6}} + \frac{i}{\sqrt{6}} & -\frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{30}} + \frac{i}{\sqrt{30}} & \frac{5}{\sqrt{30}} \end{bmatrix} $$

7.3.4 Schur Decomposition and the Spectral Theorem

Theorem 7.3.3

Schur Decomposition
For each $n\times n$ matrix $A$, there is a unitary matrix $\,U$ such that $\,U^H\! AU$ is an upper triangular matrix.

Proof:

We prove the theorem by mathematical induction . For the case of $n=1$, no work is required and the result clearly holds. Now, suppose that the result holds for the $(n-1)^\text{st}$ case of $(n-1)\times (n-1)$ matrices. We need to show that this implies the result holds for $n\times n$ matrices.

Let $A$ be an $n\times n$ matrix with eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_n$ including multiplicities. Let $\mathbf{v}_1$ be the eigenvector corresponding to $\lambda_1$ with $\|\mathbf{v}_1\| = 1$. By the Gram-Schmidt process, we use this $\mathbf{v}_1$ to construct an orthonormal basis $\left\{\mathbf{v}_1,\mathbf{v}_2,\ldots,\mathbf{v}_n\right\}$ for $\mathbb{C}^n$. Define $V$ such that

$$ V = \begin{bmatrix} \ \mathbf{v}_1 & \mathbf{v}_2 & \ldots & \mathbf{v}_n \end{bmatrix} $$
so $V$ is a unitary matrix.

The first column of the matrix $V^H\! AV$ is given by $V^H\! A\mathbf{v}_1$, so we know that

$$ V^H\! A\mathbf{v}_1 = \lambda_1 V^H \mathbf{v}_1 = \lambda_1\mathbf{e}_1 $$
and therefore

$$ V^H\! AV = \left[\begin{array}{c|ccc} \lambda_1 & x & \ldots & x \\ \hline 0 \\ \vdots & & \overset{\sim}{A} \\ 0 \end{array}\right] $$
where $\overset{\sim}{A}$ is an $(n-1)\times (n-1)$ matrix. Since $V$ is a unitary matrix, it must be the case that $A$ and the interior matrix have the same eigenvalues. Hence, $p_A(\lambda) = (\lambda_1 - \lambda)p_{\overset{\sim}{A}}(\lambda)$ and the eigenvalues of $\overset{\sim}{A}$ (including multiplicities) are $\lambda_2,\ldots,\lambda_n$.

By the induction hypothesis, it is possible to write $\overset{\sim}{A}$ as

$$ \overset{\sim}{A} = \overset{\sim}{W}\begin{bmatrix} \lambda_2 & x & \ldots & x \\ 0 & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & x \\ 0 & \ldots & 0 & \lambda_n \end{bmatrix} \overset{\sim}{W}^H $$
so

$$ \tilde{W}^H\! \tilde{A}\tilde{W} = \begin{bmatrix} \lambda_2 & x & \ldots & x \\ 0 & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & x \\ 0 & \ldots & 0 & \lambda_n \end{bmatrix} $$
We may now write

$$ \begin{align*} \left[\begin{array}{c|ccc} 1 & 0 & \ldots & 0 \\ \hline 0 \\ \vdots & & \overset{\sim}{W} \\ 0 \end{array}\right]^H & \left[\begin{array}{c|ccc} \lambda_1 & x & \ldots & x \\ \hline 0 \\ \vdots & & \overset{\sim}{A} \\ 0 \end{array}\right]\left[\begin{array}{c|ccc} 1 & 0 & \ldots & 0 \\ \hline 0 \\ \vdots & & \overset{\sim}{W} \\ 0 \end{array}\right] \\ \\ &= \left[\begin{array}{c|ccc} 1 & 0 & \ldots & 0 \\ \hline 0 \\ \vdots & & \overset{\sim}{W}^H \\ 0 \end{array}\right]\left[\begin{array}{c|ccc} \lambda_1 & x & \ldots & x \\ \hline 0 \\ \vdots & & \overset{\sim}{A}\overset{\sim}{W} \\ 0 \end{array}\right] \\ \\ &= \left[\begin{array}{c|ccc} \lambda_1 & x & \ldots & x \\ \hline 0 \\ \vdots & & \overset{\sim}{W}^H\!\overset{\sim}{A}\overset{\sim}{W} \\ 0 \end{array}\right] \\ \\ &= \left[\begin{array}{c|ccc} \lambda_1 & x & \ldots & x \\ \hline 0 & \lambda_2 & \ldots & x \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & \lambda_n \end{array}\right] \end{align*} $$
Defining the matrix $W = \left[\begin{array}{c|c} 1 & 0 \\ \hline 0 & \overset{\sim}{W} \end{array}\right]$, which is unitary, we have

$$ \left[\begin{array}{c|ccc} \lambda_1 & x & \ldots & x \\ \hline 0 \\ \vdots & & \overset{\sim}{A} \\ 0 \end{array}\right] = W \left[\begin{array}{cccc} \lambda_1 & x & \ldots & x \\ 0 & \lambda_2 & \ldots & x \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & \lambda_n \end{array}\right] W^H $$
which proves the theorem since

$$ A = V\left[\begin{array}{c|ccc} \lambda_1 & x & \ldots & x \\ \hline 0 \\ \vdots & & \overset{\sim}{A} \\ 0 \end{array}\right] V^H = VW \left[\begin{array}{cccc} \lambda_1 & x & \ldots & x \\ 0 & \lambda_2 & \ldots & x \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & \lambda_n \end{array}\right] (VW)^H $$
and the product of unitary matrices is unitary.
$\tombstone$

In the special case that $A$ is a Hermitian matrix, the triangular matrix from a Schur decomposition will be diagonal.

Theorem 7.3.4

Spectral Theorem
If $A$ is a Hermitian matrix, then there exists a unitary matrix $U$ that diagonalizes $A$.

By Theorem 6.4.3, we know that there is a matrix $U$ such that $T = U^H\! A U$ is upper triangular. In addition,

$$ T^H = (U^H\! A U)^H = U^H\! A^H U = U^H\! A U = T $$
so $T$ is Hermitian and therefore must be diagonal.
$\tombstone$

Creative Commons Logo - White


Your use of this self-initiated mediated course material is subject to our Creative Commons License .


Creative Commons Attribution-NonCommercial-ShareAlike 4.0

Creative Commons Logo - Black
Attribution
You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use.

Creative Commons Logo - Black
Noncommercial
You may not use the material for commercial purposes.

Creative Commons Logo - Black
Share Alike
You are free to share, copy and redistribute the material in any medium or format. If you adapt, remix, transform, or build upon the material, you must distribute your contributions under the same license as the original.