## 每週問題 April 13, 2015

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.

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

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

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