Hermit 矩阵相似对角化
Hermit 矩阵的定义是满足 $H^* = H$ 的矩阵, 即 $H$ 的共轭转置等于本身。
性质1 Hermit 矩阵酉相似于对角阵。
▶ Proof
利用归纳假设的思路进行证明, 设对于 $n-1$ 阶 Hermit 矩阵命题成立。写出 $n$ 阶矩阵 $A$ 的特征向量 $A\alpha_1 = \lambda_1 \alpha_1; ||\alpha_1||=1$。 可以构造一组 $\mathbb{F}^n$ 上的标准正交基 $\{\alpha_1, \beta_2, \cdots, \beta_n\}$
令矩阵 $$ P_1 = \begin{pmatrix} \alpha_1 & \beta_2 & \cdots & \beta_n\ \end{pmatrix} $$ 则 $$ \begin{align*} P_1^*AP_1 &= \begin{pmatrix} {\alpha_1}^* \\ {\beta_2}^* \\ \cdots \\ {\beta_n}^* \end{pmatrix} A \begin{pmatrix} \alpha_1 & \beta_2 & \cdots & \beta_n\ \end{pmatrix} \\ ~\\ &= \begin{pmatrix} {\alpha_1}^* A \alpha_1 & {\alpha_1}^* A \beta_2 & \cdots & {\alpha_1}^* \beta_n \\ {\beta_2}^* A \alpha_1 & {\beta_2}^* A \beta_2 & ~ & \vdots \\ \vdots & \vdots & \ddots & \vdots \\ {\beta_n}^* A \alpha_1 & {\beta_n}^* A \beta_2 & \cdots & {\beta_n}^* A \beta_n \\ \end{pmatrix} \\ ~\\ &= \left ( \begin{array}{c|ccc} {\alpha_1}^* A \alpha_1 & {\alpha_1}^* A \beta_2 & \cdots & {\alpha_1}^* \beta_n \\ \hline {\beta_2}^* A \alpha_1 & {\beta_2}^* A \beta_2 & ~ & \vdots \\ \vdots & \vdots & \ddots & \vdots \\ {\beta_n}^* A \alpha_1 & {\beta_n}^* A \beta_2 & \cdots & {\beta_n}^* A \beta_n \\ \end{array} \right ) \\ \end{align*} $$ 由归纳假设可得 $n-1$ 阶矩阵 $B_1$ 酉相似于对角阵, 即 $$ {P_2}^* B_1 P_2 = \left ( \begin{array}{c} \lambda_2 & ~ & ~\\ ~ & \lambda_3 & ~ \\ ~ & ~ & \ddots & ~ \\ ~ & ~ & ~ & \lambda_n \\ \end{array} \right)\\ B_1 = \left ( \begin{array}{c} {\beta_2}^* A \beta_2 & ~ & \vdots \\ \vdots & \ddots & \vdots \\ {\beta_n}^* A \beta_2 & \cdots & {\beta_n}^* A \beta_n \\ \end{array} \right ) \\ $$ 从而将 ${P_1}^* A P_1$ 写为 $$ \begin{align*} {P_1}^* A P_1 &= \left ( \begin{array}{c|ccc} {\alpha_1}^* A \alpha_1 & {\alpha_1}^* A \beta_2 & \cdots & {\alpha_1}^* \beta_n \\ \hline {\beta_2}^* A \alpha_1 & ~ & ~ & ~ \\ \vdots & ~ & B_1 & ~ \\ {\beta_n}^* A \alpha_1 & ~ & ~ & ~ \\ \end{array} \right ) \end{align*} $$ 观察其中 $$\begin{align*} {\alpha_1}^* A \beta_2 &= (A^*{\alpha_1})^* \beta_2 \\ &= (A {\alpha_1})^* \beta_2 \\ &= {\lambda_1}^* {\alpha_1}^* \beta_2 \end{align*} $$ 由于 $\{\alpha_1, \beta_2, \cdots, \beta_n\}$ 为一组标准正交基, 所以 ${\alpha_1}^*A\beta_i = 0$。因此 $$ \begin{align*} {P_1}^* A P_1 &= \left ( \begin{array}{c|ccc} \lambda_1 & 0 & \cdots & 0 \\ \hline 0 & ~ & ~ & ~ \\ \vdots & ~ & B_1 & ~ \\ 0 & ~ & ~ & ~ \\ \end{array} \right ) \end{align*} $$ 注意, 此时 ${P_1}^* A P_1$ 仍然不是对角矩阵的形式, 如何将 $B_1$ 转为对角矩阵形式呢。 令 $$ P_3 = \left ( \begin{array}{c} 1 & 0 \\ 0 & P_2 \end{array} \right ) $$ 则 $$ \begin{align*} {P_3}^* ({P_1}^*AP_1)P_3 &= (P_1P_3)^* A (P_1P_3) \\ &= \left ( \begin{array}{c} \lambda_1 & 0 \\ 0 & {P_2}^*B_1P_2 \end{array} \right )\\ ~\\ &= \left ( \begin{array}{c} \lambda_2 & ~ & ~\\ ~ & \lambda_3 & ~ \\ ~ & ~ & \ddots & ~ \\ ~ & ~ & ~ & \lambda_n \\ \end{array} \right)\\ \end{align*} $$ 证毕。
性质2 Hermit 矩阵的特征值都为实数。
▶ Proof
已知 $$ \left \{ \begin{align} A \alpha &= \lambda \alpha ~~~~~~~(1)\\ A^* &= A ~~~~~~~~~(2) \end{align} \right. $$ 由(1)得 $$\alpha^* A^* = \lambda^* \alpha^*$$ 将(2)代入得 $$\alpha^* A = \lambda^* \alpha^*$$ 两边同乘 $\alpha$ 得 $$ \begin{align*} \alpha^* A \alpha &= \lambda^* \alpha^* \alpha \\ \lambda||\alpha||^2 &= \lambda^* ||\alpha||^2 \end{align*} $$ 又 $||a|| \neq0 \Rightarrow \lambda = \lambda^*$, 因此 $\lambda \in \mathbb{R}$.性质3 Hermit 矩阵不同特征值对应的特征向量正交。
