压缩映射$A$, $σ_1(A)≤1$, $H(A)=I$, 则$A=I$

hbghlyj

Topics in matrix analysis by Horn, Roger A
3.1 The singular value decomposition
page 157
原图

25. Suppose $A \in M_n$ and $\sigma_1(A) \leq 1$, that is, $A$ is a contraction. If $\mathrm{H}(A) \equiv\frac{1}{2}\left(A+A^*\right)=I$, show that $A=I$.

其中H(A)为Hermitian Part

Czhang271828

按此贴记号,  $A+A^\ast =2I$ 即
$$
I=\dfrac12R(\Theta^\ast +\Theta)=R\cdot \mathrm{Re}(\Theta).
$$
由于 $\mathrm{Re}(\Theta)$ 各项小于等于 $1$, 从而 $R$ 与 $\Theta$ 只能为 $\{\pm 1\}^n$-矩阵. 简单讨论知 $R\Theta=I$.