## 每週問題 July 6, 2015

Let $A$ be an $n\times n$ matrix with all eigenvalues equal to 1 in absolute value. Show that $A$ is a unitary matrix if, for all $\mathbf{x}\in\mathbb{C}^n$,

$\displaystyle \Vert A\mathbf{x}\Vert\le\Vert\mathbf{x}\Vert$.

$\displaystyle \Vert A\Vert_2=\max_{\Vert\mathbf{x}\Vert\neq\mathbf{0}}\frac{\Vert A\mathbf{x}\Vert}{\Vert\mathbf{x}\Vert}=\sigma_{\max}\le 1$

$\displaystyle \Vert A\mathbf{x}\Vert=\Vert UTU^\ast U\mathbf{e}_n\Vert=\Vert UT\mathbf{e}_n\Vert=\Vert T\mathbf{e}_n\Vert=\left(\vert t_{1n}\vert^2+\cdots+\vert t_{n-1,n}\vert^2+\vert\lambda_n\vert^2\right)^{1/2}$

$A^\ast A=UT^\ast U^\ast UTU^\ast=UT^\ast TU^\ast=UU^\ast =I$