How to show that if $V \in C^{m \times m}$ unitary and $A \in C^{m \times m}$ then $||AV||_2=||A||_2$

Question

How to show that if $V \in C^{m \times m}$ is unitary and $A \in C^{m \times m}$ then $$||AV||_2=||A||_2$$

Where $$||A||_2 = \sup_{ ||x||_2=1 } ||Ax||_2$$ and $$||AV||_2 = \sup_{ ||x||_2=1 } ||AVx||_2$$

Answer 1

You have $$\Vert V x\Vert^2 = \langle V x,V x \rangle = \langle x , V^* V x \rangle = \langle x,x\rangle = \Vert x\Vert ^2 $$ for any $x$. So for any $x$ you have $\Vert Vx\Vert = \Vert x\Vert$.

Hence $V$ has norm $1$. $V^{-1}$ as well is unitary and so it has norm $1$.

Clearly $\Vert A V\Vert \leq \Vert A \Vert \Vert V\Vert = \Vert A\Vert$.

For the other inequality you have

$$\Vert A\Vert = \Vert AV V^{-1}\Vert \leq \Vert A V\Vert \Vert V^{-1}\Vert = \Vert AV\Vert\,.$$ Then $\Vert AV \Vert = \Vert A\Vert$.