## 聯繫特徵值與奇異值的引線矩陣

$A$ 為一個 $m\times n$ 階矩陣，$A$ 的奇異值定義為 Gramian 矩陣 $A^{\ast}A$$AA^{\ast}$ 的特徵值的非負平方根 (見“奇異值分解(SVD)”)，並以遞減排序表示：$\sigma_1\ge\cdots\ge\sigma_q\ge 0$$q=\min\{m,n\}$。我們知道任一方陣的特徵值可能是複數，但 Hermitian 矩陣的所有特徵值都是實數 (見“特殊矩陣(9)：Hermitian 矩陣”)，因此 Hermitian 矩陣的特徵值也可按遞減方式排列。本文介紹一個建立於任意 $m\times n$ 階矩陣 $A$$(m+n)\times(m+n)$ 階 Hermitian 矩陣

$\tilde{A}=\begin{bmatrix} 0&A\\ A^{\ast}&0 \end{bmatrix}$

$\sigma_1,\ldots,\sigma_n,0,\ldots,0,-\sigma_n,\ldots,-\sigma_1$

$A=\begin{bmatrix} 1&0\\ 0&3\\ 2&0 \end{bmatrix},~\tilde{A}=\begin{bmatrix} 0&0&0&1&0\\ 0&0&0&0&3\\ 0&0&0&2&0\\ 1&0&2&0&0\\ 0&3&0&0&0 \end{bmatrix}$

$\begin{bmatrix} 0&A\\ A^{\ast}&0 \end{bmatrix}\begin{bmatrix} \mathbf{x}\\ \mathbf{y} \end{bmatrix}=\lambda\begin{bmatrix} \mathbf{x}\\ \mathbf{y} \end{bmatrix}$

\begin{aligned} A\mathbf{y}&=\lambda\mathbf{x}\\ A^{\ast}\mathbf{x}&=\lambda\mathbf{y}\end{aligned}

$A^{\ast}A\mathbf{y}=\lambda A^{\ast}\mathbf{x}=\lambda^2\mathbf{y}$

$p_{\tilde{A}}=p_{A^{\ast}A}(t^2)t^{m-n}$

$\Sigma=\begin{bmatrix} D\\ 0 \end{bmatrix}$

$A=\begin{bmatrix} U_1&U_2 \end{bmatrix}\begin{bmatrix} D\\ 0 \end{bmatrix}V^{\ast}=U_1DV^{\ast}$

$W=\begin{bmatrix} \frac{1}{\sqrt{2}}U_1&-\frac{1}{\sqrt{2}}U_1&U_2\\ \frac{1}{\sqrt{2}}V_1&\frac{1}{\sqrt{2}}V_1&0 \end{bmatrix}$

\begin{aligned} W^{\ast}\tilde{A}W&=\begin{bmatrix} \frac{1}{\sqrt{2}}U_1^{\ast}&\frac{1}{\sqrt{2}}V_1^{\ast}\\ -\frac{1}{\sqrt{2}}U_1^{\ast}&\frac{1}{\sqrt{2}}V_1^{\ast}\\ U_2^{\ast}&0 \end{bmatrix}\begin{bmatrix} 0&U_1DV^{\ast}\\ VDU_1^{\ast}&0 \end{bmatrix}\begin{bmatrix} \frac{1}{\sqrt{2}}U_1&-\frac{1}{\sqrt{2}}U_1&U_2\\ \frac{1}{\sqrt{2}}V_1&\frac{1}{\sqrt{2}}V_1&0 \end{bmatrix}\\ &=\begin{bmatrix} D&~&~\\ ~&-D&~\\ ~&~&0 \end{bmatrix}\end{aligned}

$\lambda_k(A)+\lambda_1(B)\ge\lambda_k(A+B)\ge\lambda_k(A)+\lambda_p(B)$

$\lambda_k(\tilde{A})+\lambda_1(\tilde{B})\ge\lambda_k(\tilde{A}+\tilde{B})\ge\lambda_k(\tilde{A})+\lambda_{m+n}(\tilde{B})$

\begin{aligned} \lambda_k(\tilde{A})&=\sigma_k(A)\\ \lambda_k(\tilde{B})&=\sigma_k(B)\\ \lambda_k(\tilde{C})&=\sigma_k(C)\end{aligned}

$\sigma_k(A)+\sigma_1(B)\ge\sigma_k(A+B)\ge\sigma_k(A)-\sigma_1(B)$

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