How to prove that the spectral norm is unitarily invariant?

Question

How to prove that $\| U A \|_2 = \| A U \|_2 = \| A \|_2$ for any unitary matrix $U$?

It can be proved that

$$\| UA\|_2 = \sqrt{\lambda_{\max}({(UA)}^*(UA))} = \sqrt{\lambda_{\max}(A^*U^*UA)} = \sqrt{\lambda_{\max}(A^*A)} = \| A\|_2\\$$

but how to prove $\| A U \|_2 = \| A \|_2$?

Answer 1

Hint: Prove $\|M\|_2 = \|M^*\|_2$ for square matrices. Then $\|AU\| = \|U^* A^*\| = \|A^*\| = \|A\|$.

Answer 2

This comes from the fact that the eigenvalues of $AB$ and $BA$ are the same:

Proof:

If $\lambda\neq0$ is an eigenvalue of $AB$ then there is some nonzero eigenvector $v$ such that $ABv=\lambda v$. Now consider $(BA)Bv=B(ABv)=\lambda Bv$. So $w=Bv$ is a nonzero eigenvector (since $ABv=A(Bv)=\lambda v$) with $\lambda$ as eigenvalue. Of course you repeat the argument for $BA$.

If $\lambda=0$ is an eigenvalue of $AB$ then $\det(AB)=0$, so $\det(BA)=\det(B)\det(A)=\det(AB)=0$ and $\lambda=0$ is an eigenvalue of $BA$. Again, this can be repeated for $BA$.

We conclude they have the same eigenvalues. $\Box$

Knowing this, can you use it to prove what you want?