每週問題 April 13, 2015

方陣的特徵值積與奇異值積有何關係?

Let A be an n\times n matrix. Show that \vert \lambda_1\cdots\lambda_n\vert=\sigma_1\cdots\sigma_n, where \lambda_i and \sigma_i are the eigenvalues and singular values of A, respectively.

 
參考解答:

寫出 A 的奇異值分解 A=U\Sigma V^\ast,其中 \Sigma=\text{diag}(\sigma_1,\ldots,\sigma_n),每一 \sigma_i\ge 0UV 是么正矩陣 (unitary matrix),滿足 U^\ast=U^{-1}V^\ast=V^{-1}。因此,

\displaystyle  A^\ast A=U\Sigma V^\ast V\Sigma U^\ast=U\Sigma^2 U^\ast

使用行列式可乘公式,

\displaystyle  \det(A^\ast A)=\det(U\Sigma^2 U^\ast)=(\det U)\det(\Sigma^2)(\det U^{-1})=\det (\Sigma^2)=\sigma_1^2\cdots\sigma_n^2

另一方面,

\displaystyle  \det(A^\ast A)=(\det A^\ast)(\det A)=(\overline{\det A})(\det A)=\vert \det A\vert^2=\vert\lambda_1\cdots\lambda_n\vert^2

合併以上結果,可得 \vert\lambda_1\cdots\lambda_n\vert=\sigma_1\cdots\sigma_n

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