How to show that $\| QA\|_2=\| A \|_2$ where $Q$ is unitary (for a matrix A)

Question

I want to show that for a unitary matrix $Q$ and a matrix $A$ that

$$ \|QA\|_2=\|A\|_2$$

I start with the definition of matrix induced norms:

$$\| QA \|_2 = \sup_{x \neq 0}\frac{\|QAx\|_2}{\|x\|_2}$$

Using the Cauchy-Schwarz inequality,

$$\| QA \|_2 = \sup_{x \neq 0}\frac{\|QAx\|_2}{\|x\|_2} \leq \sup_{x \neq 0}\frac{\|Q\|\|Ax\|_2}{\|x\|_2}=\|Q\|_2\sup_{x \neq 0}\frac{\|Ax\|_2}{\|x\|_2}=\|A\|_2$$

since $\|Q\|_2$ and $\sup_{x \neq 0}\frac{\|Ax\|_2}{\|x\|_2}=\|A\|_2$ by definition.

This results in $\|QA\|_2 \leq \|A\|_2$. I am not sure how I get the strict equality.

Answer 1

This is Joel's comment, expanded into an answer. I'm not looking to steal his points, so if he posts an answer, I'll delete mine. $$\|QA\|_2=\sup_{x\neq 0}\frac{\|QAx\|_2}{\|x\|_2}=\sup_{x\neq 0}\frac{\|Ax\|_2}{\|x\|_2}=\|Ax\|_2.$$ Now, this second step obviously requires $\|QAx\|_2=\|Ax\|_2$. But this is easier to prove because we're now working with a vector norm. That is, for any compatible vector $y$, $$\|Qy\|_2=\left( (Qy)^*(Qy)\right)^{1/2} = \left(y^*Q^*Qy \right)^{1/2}=(y^*y)^{1/2}=\|y\|_2.$$

Answer 2

In order to go the other direction, you can simply note that $$ A = Q^*(QA) $$ where $Q^*$ is unitary. From there, apply your inequality to find $$ \|Q^*(QA)\|_2 \leq \|Q^*\|_2\|QA\|_2 = \|QA\|_2 $$