每週問題 March 17, 2014

Posted on 03/17/2014 by ccjou

這是運用 Hermitian 矩陣的正交對角化證明 Cayley-Hamilton 定理。

Let A be an n\times n Hermitian matrix with eigenvalues \lambda_i, 1\le i\le n. Prove that

(A-\lambda_1I)(A-\lambda_2I)\cdots(A-\lambda_nI)=0.

 
參考解答：

Hermitian 矩陣 A 可么正對角化 (unitarily diagonalizable)，設為 A=UDU^\ast，其中 D=\text{diag}(\lambda_1,\ldots,\lambda_n) 且 U^\ast=U^{-1}。因此，

\displaystyle\begin{aligned} (A-\lambda_1I)\cdots(A-\lambda_nI)&=(UDU^\ast-\lambda_1I)\cdots(UDU^\ast-\lambda_nI)\\ &=U (D-\lambda_1I)U^\ast \cdots U(D-\lambda_nI)U^\ast\\ &=U(D-\lambda_1I)\cdots(D-\lambda_nI)U^\ast\\ &=U\begin{bmatrix} \prod_{i=1}^n(\lambda_1-\lambda_i)&&\\ &\ddots&\\ &&\prod_{i=1}^n(\lambda_n-\lambda_i) \end{bmatrix}U^\ast\\ &=0. \end{aligned}

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2 Responses to 每週問題 March 17, 2014

  1. levinc417 says:
    03/17/2014 at 10:16 am

    似乎可用最小多項式說明？

    Reply

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