Show that $\frac12\|A\|_2\leq \sup_{\|u\|_2=1}|u^HAu|$.

Question

Let $A\in\mathbb{C}^{n\times n}$, show that $$\frac12\|A\|_2\leq \sup_{\|u\|_2=1}|u^HAu|,$$ where $\|A\|_2$ is the 2-norm of $A$.

Answer 1

1.) all vectors are constrained to have a length (2 norm) of one
2.) for notational simplicity, I assume WLOG that the Hermitian part of $A$ has a maximal singular value at least as large as the maximal singular value of the skew-hermitian part of $A$. (If this wasn't true, the only alteration would be on the 3rd to last line, we'd change the plus sign to a minus sign)

$\Big \Vert A \Big \Vert_2$
$= \Big \Vert \frac{1}{2}\big(A+A^*\big) + \frac{1}{2}\big(A-A^*\big) \Big \Vert_2 $
$\leq \frac{1}{2}\Big \Vert \big(A+A^*\big)\Big\Vert_2 + \frac{1}{2}\Big\Vert \big(A-A^*\big) \Big \Vert_2 $
$= \max_{x} \frac{1}{2}\Big\{\Big \vert \mathbf x^*\big(A+A^*\big)\mathbf x\Big \vert\Big\} + \max_{y} \frac{1}{2}\Big\{\Big \vert \mathbf y^*\big(A-A^*\big)\mathbf y\Big \vert\Big\} $
$\leq 2 \cdot \max_{x} \frac{1}{2}\Big\{\Big \vert \mathbf x^*\big(A+A^*\big)\mathbf x\Big \vert\Big\} $
$\leq 2 \cdot \max_{x} \frac{1}{2}\Big\{\Big \vert \mathbf x^*\big(A+A^*\big)\mathbf x +\mathbf x^*\big(A-A^*\big)\mathbf x\Big \vert\Big\} $
$= 2 \cdot \max_{x} \Big \vert \mathbf x^*A\mathbf x \Big \vert $

justifications:
sub-additivity of operator 2 norm, normal matrices have maximal singular values given by modulus of a quadratic form. Finally recall that a quadratic Hermitian form is purely real and a quadratic skew-hermitian form is purely imaginary