每週問題 March 17, 2014

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)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}

This entry was posted in pow 二次型, 每週問題 and tagged , . Bookmark the permalink.

2 則回應給 每週問題 March 17, 2014

  1. levinc417 說:

    似乎可用最小多項式說明?

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