每週問題 October 3, 2016

Let $A$ and $B$ be $n\times n$ Hermitian matrices. Suppose $A$ is invertible. Show that there exists a nonsingular matrix $P$ so that $P^\ast AP$ and $P^\ast BP$ are diagonal if and only if $A^{-1}B$ is diagonalizable and all its eigenvalues are real.

$P^\ast AP=D_1$$P^\ast BP=D_2$，其中 $D_1$$D_2$ 為對角矩陣。因為 $P^\ast AP$$P^\ast BP$ 都是 Hermitian，$D_1$$D_2$ 是實對角矩陣。寫出

$A^{-1}B=PD_1^{-1}P^\ast (P^\ast)^{-1}D_2P^{-1}=PD_1^{-1}D_2P^{-1}$

$D=\begin{bmatrix} \lambda_1I_{n_1}&&\\ &\ddots&\\ &&\lambda_kI_{n_k}\end{bmatrix}$

$\lambda_1,\ldots,\lambda_k$ 為兩兩互異的實數。因此，$BS=ASD$，即有 $S^\ast BS=(S^\ast AS)D$，其中 $S^\ast B S$$S^\ast AS$ 是 Hermitian。使用分塊矩陣表達，$S^\ast AS=[\tilde{A}_{ij}]$$S^\ast BS=[\tilde{B}_{ij}]$，這裡 $\tilde{A}_{ij}$$\tilde{B}_{ij}$$n_i\times n_j$ 階分塊，$i,j=1,\ldots,k$。使用 $\tilde{A}_{ij}=\tilde{A}_{ji}^\ast$$\tilde{B}_{ij}=\tilde{B}_{ji}^\ast$，且 $\tilde{B}_{ij}=\lambda_j\tilde{A}_{ij}$$i\neq j$，推得

$\lambda_j\tilde{A}_{ij}=\tilde{B}_{ji}^\ast=\overline{\lambda_i}\tilde{A}_{ji}^\ast=\lambda_i\tilde{A}_{ij}$

$S^\ast AS=\begin{bmatrix} \tilde{A}_{11}&&\\ &\ddots&\\ &&\tilde{A}_{kk}\end{bmatrix},~~S^\ast BS=\begin{bmatrix} \lambda_1\tilde{A}_{11}&&\\ &\ddots&\\ &&\lambda_k\tilde{A}_{kk}\end{bmatrix}$

$U=\begin{bmatrix} U_1&&\\ &\ddots&\\ &&U_k\end{bmatrix}$

\begin{aligned} P^\ast AP&=U^\ast S^\ast ASU=\begin{bmatrix} D_1&&\\ &\ddots&\\ &&D_k\end{bmatrix}\\ P^\ast BP&=U^\ast S^\ast BSU=\begin{bmatrix} \lambda_1D_1&&\\ &\ddots&\\ &&\lambda_kD_k\end{bmatrix}.\end{aligned}

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