Normal Matrices and the Spectral Theorem

Normal Matrices

$A$ is normal when $A^*A = AA^*$. Diagonal, Hermitian, skew-Hermitian and unitary matrices are normal.

For unitarily similar $A = UBU^*$, $A$ is normal $\iff$ $B$ is normal, since

$$A^*A = UB^*BU^*, \qquad AA^* = UBB^*U^*$$

An upper triangular $T$ is normal $\iff$ $T$ is diagonal.

Proof $(\Rightarrow)$ Compare the diagonals of $T^*T$ and $TT^*$. $(T^*T)_{ii}$ is the squared norm of the $i$th column of $T$, and $(TT^*)_{ii}$ is the squared norm of the $i$th row. The first column is just $t_{11}$, so $|t_{11}|^2 = \sum_j |t_{1j}|^2$ and $t_{1j} = 0$ for $j > 1$. Continuing row by row, $T$ is diagonal.

Spectral Theorem for Normal Matrices

$A$ is unitarily diagonalizable when $A = UDU^*$ with $U$ unitary and $D$ diagonal. The following are equivalent

1. $A$ is normal,
2. $A$ is unitarily diagonalizable,
3. $\|A\|_F^2 = \sum |\lambda_i|^2$.

Proof

$(2 \Rightarrow 1, 3)$ $A = UDU^*$. $D$ is normal so $A$ is normal, and $\|A\|_F^2 = \|UDU^*\|_F^2 = \|D\|_F^2 = \sum |\lambda_i|^2$.

$(1 \Rightarrow 2)$ $A = UTU^*$ by Schur. $A$ normal $\Rightarrow$ $T$ normal $\Rightarrow$ $T$ diagonal.

$(3 \Rightarrow 2)$ $A = UTU^*$ and

$$\sum_{i < j} |t_{ij}|^2 + \sum_{i} |t_{ii}|^2 = \|T\|_F^2 = \|A\|_F^2 = \sum |\lambda_i|^2 = \sum_i |t_{ii}|^2$$

so $t_{ij} = 0$ for $i < j$ and $T$ is diagonal.

Spectral Theorem for Hermitian Matrices

$A$ is Hermitian $\iff$ $A$ is unitarily diagonalizable and all eigenvalues are real.

Proof

$(\Rightarrow)$ $A$ Hermitian $\Rightarrow$ normal $\Rightarrow$ $A = UDU^*$. Then $A = A^* \Rightarrow UDU^* = UD^*U^* \Rightarrow D = D^*$, so $D$ is real.

$(\Leftarrow)$ $(UDU^*)^* = UD^*U^* = UDU^*$ for real $D$.

A unitary diagonalization is a sum of rank-1 matrices,

$$UDU^* = \sum_i d_{ii} \, U^{(i)} U^{(i)*}$$

Classes of Normal Matrices

For a normal $A$, the location of the eigenvalues decides the class:

• $A$ is Hermitian iff all $\lambda \in \mathbb{R}$,
• $A$ is skew-Hermitian iff all $\lambda$ are purely imaginary,
• $A$ is unitary iff all $|\lambda| = 1$.

So Hermitian and unitary means $\lambda \in \{1, -1\}$, unitary and skew-Hermitian means $\lambda \in \{i, -i\}$, and the only matrix that is Hermitian and skew-Hermitian is 0.

Matrix Exponential

$$e^B = I + B + \frac{1}{2!}B^2 + \frac{1}{3!}B^3 + \cdots$$

$A$ is unitary $\iff$ there is a skew-Hermitian $B$ with $A = e^B$.

Proof

$(\Leftarrow)$ $B$ is skew-Hermitian, so $B = U \mathrm{diag}(i\theta_1, \ldots, i\theta_n) U^*$ with $\theta_j \in \mathbb{R}$. Then

$$e^B = U e^{\mathrm{diag}(i\theta_1, \ldots, i\theta_n)} U^* = U \mathrm{diag}(e^{i\theta_1}, \ldots, e^{i\theta_n}) U^*$$

which is unitary.

$(\Rightarrow)$ $A = UDU^* = U \mathrm{diag}(e^{i\omega_1}, \ldots, e^{i\omega_n}) U^*$ since $|\lambda| = 1$. Let $B := U \mathrm{diag}(i\omega_1, \ldots, i\omega_n) U^*$. Then $B^* = U \mathrm{diag}(-i\omega_1, \ldots, -i\omega_n) U^* = -B$, so $B$ is skew-Hermitian and $e^B = A$.

Real Symmetric Matrices

For $A \in M_n(\mathbb{R})$, $A$ is symmetric $\iff$ there are $Q, D \in M_n(\mathbb{R})$, $Q$ orthogonal and $D$ diagonal, with $A = QDQ^T$.

Proof

$(\Leftarrow)$ $A^T = QD^TQ^T = QDQ^T = A$.

$(\Rightarrow)$ $A$ is Hermitian, so $\sigma(A) \subseteq \mathbb{R}$. $A$ is real with real eigenvalues, so $A = QTQ^T$ with $Q$ and $T$ real. $A$ Hermitian $\Rightarrow$ $A$ normal $\Rightarrow$ $T$ normal $\Rightarrow$ $T$ diagonal.