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Let $A$ be a $n\times n$ matrix, we define its operator norm (or just spectral norm) as $$||A||=\max_{||x||=1}||Ax||.$$ If $||A||\leq 1$, we say $A$ is a contraction.

Show that $$||A||\leq 1\Rightarrow A^*(I-AA^*)^{1/2}=(I-A^*A)^{1/2}A^*.$$

This is Bhatia, Matrix Analysis, Exercise I.3.6. I have no idea on it...

xldd
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  • hi why can you say that the norm is less than 1? Doesnt it depend on the definition of your matrix norm? – Ka Wa Yip Feb 09 '17 at 00:37

1 Answers1

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First of all, note that $\|A\| = \sqrt{\rho(A^*A)} = \sqrt{\rho(AA^*)} \leq 1$, so that $I - A^*A$ and $I - AA^*$ are indeed positive (semi)definite. So, we may take PSD square roots $(I - A^*A)^{1/2}$ and $(I - AA^*)^{1/2}$.

Now, verify that $$ A^*(I - AA^*) = (I - A^*A)A^* $$ At this point, the key observation is that there exists a polynomial $p$ such that $p[(I - AA^*)] = \sqrt{(I - AA^*)}$ and $p[(I-A^*A)] = \sqrt{(I - A^*A)}$. From there, the conclusion follows.

Ben Grossmann
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