A problem about matrix $A\in \mathbb{C}^{n\times n}$. Set：
$$ r(A) = \max_{||u||_2=1}|u^* Au| $$
Prove the following statements：
(1) $||A||_2 \leq 2 r(A)$
(2)if $A^*A=AA^*$ then $r(A)=||A||_2$
 A: Write $A=\frac{A+A^*}2+\frac{A-A^*}2$. Let $H_1:=\frac{A+A^*}2$, then $H_1$ is hermitian and is therefore diagonalizable. Thanks to that we can see that $||H_1||=r(H_1)$. Since $H_2:=\frac{A-A^*}2$ is anti-hermitian it's in particular a normal matrix, so $H_2$ is unitary diagonalizable and $||H_2||=r(H_2)$. Since $r(H_1)\leq r(A)$ and $r(H_2)\leq r(A)$ we have 
$$||A||_2\leq ||H_1||_2+||H_2||_2=r(H_1)+r(H_2)\leq 2r(A).$$
A: For part (2) note that $A$ is normal in this case and hence is diagonalizable with respect to an orthonormal basis.  (See here for more about normal operators.)  As noted above, this implies that $||A||_2 = r(A)$: If $||v||_2=1$ and $v = a_1e_1 + \dotsc + a_ne_n$ in this basis then
$$
|v^{\ast}Av| = |\langle Av, v \rangle| = \left|\lambda_1 |a_1|^2 + \dotsc + \lambda_n |a_n|^2 \right| \leq \max_k(|\lambda_k|) \cdot (|a_1|^2 + \dotsc + |a_n|^2) = \max_k(|\lambda_k|)
$$
and this maximum is attained for some $e_k$ and so $r(A) = \max_k(|\lambda_k|)$.  A similar computation shows that this indeed equals $||A||_2$.
