Unitary Matrices and Schur Decomposition

Unitary Matrices

$U$ is unitary when $U^*U = I$.

$U$ is unitary $\iff$ the columns of $U$ are orthonormal, since

$$U^*U = \begin{bmatrix} U_1^* \\ U_2^* \\ \vdots \end{bmatrix} \begin{bmatrix} U_1 & U_2 & \cdots \end{bmatrix} = \big[\, U_i^* U_j \,\big]_{ij}$$

• If $U, V$ are unitary, $UV$ is unitary.
• For all $x \in \mathbb{C}^n$, $\|Ux\|_2 = \|x\|_2$.

$A, B$ are unitarily similar when $A = UBU^*$ with $U$ unitary. Then $\|A\|_F = \|B\|_F$, since for unitary $U, V$

$$\|UAV^*\|_F^2 = \mathrm{tr}\big((UAV^*)^* UAV^*\big) = \mathrm{tr}(VA^*U^*UAV^*) = \mathrm{tr}(VA^*AV^*) = \mathrm{tr}(A^*A) = \|A\|_F^2$$

Householder Matrices

For $w \in \mathbb{C}^n$ with $\|w\|_2 = 1$,

$$H_w := I - 2ww^*$$

$H_w$ is unitary and Hermitian.

For $x \neq y \in \mathbb{R}^n$ with $\|x\|_2 = \|y\|_2$, if

$$w = \frac{1}{\|x - y\|_2}(x - y)$$

then $H_w x = y$ and $H_w y = x$.

For $x \in \mathbb{C}^n$ with $\|x\|_2 = 1$, there is a unitary $U$ whose first column is $x$, by extending $x$ to an orthonormal basis with Gram-Schmidt.

Schur Decomposition

Let $A \in M_n$ have eigenvalues $\lambda_1, \ldots, \lambda_n$ in any order. Then there are a unitary $U$ and an upper triangular $T$ with diagonal $\lambda_1, \ldots, \lambda_n$ such that

$$A = UTU^*$$

$U$ and $T$ can be taken real if $A$ and its eigenvalues are real.

Proof Let $x$ be a normalized eigenvector with eigenvalue $\lambda_1$, and $U$ unitary with first column $x$. Then

$$U^*AU = \begin{bmatrix} x^* \\ U_2^* \\ \vdots \end{bmatrix} [\, Ax \mid AU_2 \mid \cdots \,] = \begin{bmatrix} \lambda_1 & * \\ 0 & B \end{bmatrix}$$

and $\sigma(B) = \{\lambda_2, \ldots, \lambda_n\}$. Let $y$ be a normalized eigenvector of $B$ for $\lambda_2$, and $V \in M_{n-1}$ unitary with first column $y$. Then

$$\left(U \begin{bmatrix} 1 & 0 \\ 0 & V \end{bmatrix}\right)^* A \left(U \begin{bmatrix} 1 & 0 \\ 0 & V \end{bmatrix}\right) = \begin{bmatrix} \lambda_1 & * \\ 0 & V^*BV \end{bmatrix} = \begin{bmatrix} \lambda_1 & * & * \\ 0 & \lambda_2 & * \\ 0 & 0 & C \end{bmatrix}$$

Continuing, in the end the middle matrix is upper triangular with $\lambda_1, \ldots, \lambda_n$ on the diagonal, and the product of the unitaries $\Omega$ is unitary, so $A = \Omega T \Omega^*$.

Cayley-Hamilton

$$p_A(A) = 0$$

Proof Let $A = UTU^*$ and $p_A(t) = (t - \lambda_1) \cdots (t - \lambda_n)$. Then

$$p_A(A) = U p_A(T) U^* = U \big[(T - \lambda_1 I)(T - \lambda_2 I) \cdots (T - \lambda_n I)\big] U^*$$

$T - \lambda_1 I$ has its first column zero. Since $T - \lambda_k I$ is upper triangular with a zero in position $(k,k)$, multiplying on the right by it keeps the first $k - 1$ columns zero and makes the $k$th column zero too. So the product of the first $k$ factors has its first $k$ columns zero, and the full product is 0.

Diagonalizable Matrices are Dense

For all $A \in M_n$ and $\varepsilon > 0$ there is $E \in M_n$ with $\|E\|_F < \varepsilon$ such that $A + E$ is diagonalizable.

Proof

1. $A = UTU^*$.
2. Let $D$ be diagonal such that $T + D$ has distinct diagonal entries and $\|D\|_F < \varepsilon$.
3. $E := UDU^*$, so $\|E\|_F = \|D\|_F < \varepsilon$.
4. $A + E = UTU^* + UDU^* = U(T + D)U^*$.
5. The diagonal of $T + D$ is distinct, so $A + E$ has distinct eigenvalues and is diagonalizable.

Almost Diagonal Similarity

For all $\varepsilon > 0$ there are an invertible $S$, a diagonal $D$, and $E$ with $\|E\|_F < \varepsilon$ such that

$$A = S(D + E)S^{-1}$$

Proof Let $A = UTU^*$ and for $\delta > 0$ let $D_\delta = \mathrm{diag}(\delta, \delta^2, \ldots, \delta^n)$. Then

$$(D_\delta^{-1} T D_\delta)_{ij} = t_{ij} \delta^{j-i}$$

so the diagonal is unchanged and the entries above it are $O(\delta)$. Hence

$$A = (U D_\delta)(D_\delta^{-1} T D_\delta)(U D_\delta)^{-1}$$

with $D_\delta^{-1} T D_\delta = D + E$, $D = \mathrm{diag}(\lambda_1, \ldots, \lambda_n)$ and $\|E\|_F \to 0$ as $\delta \to 0$.

Schur's Inequality

$$\|A\|_F^2 \ge \sum_{i=1}^{n} |\lambda_i|^2$$

Proof With $A = UTU^*$,

$$\|A\|_F^2 = \|T\|_F^2 = \sum_{i \neq j} |t_{ij}|^2 + \sum_{i=1}^{n} |t_{ii}|^2 \ge \sum_{i=1}^{n} |t_{ii}|^2 = \sum_{i=1}^{n} |\lambda_i|^2$$

Equality holds iff $T$ is diagonal, i.e. iff $A$ is normal.