Extremal values of Hermitian bilinear and quadratic forms

59 Views Asked by At

Let $A$ be an $n \times n$ complex Hermitian matrix. I recall having read before that $\sup\left\{\left<Ax,y \right> \ : \ \| x\| = \| y\| = 1\right\} = \sup\left\{\left<Ax,x \right> \ : \ \| x\| = 1\right\}$. Is it also true that $\inf\left\{\left<Ax,y \right> \ : \ \| x\| = \| y\| = 1\right\} = \inf\left\{\left<Ax,x \right> \ : \ \| x\| = 1\right\}$? Thanks.

1

There are 1 best solutions below

3
On BEST ANSWER

Let $\lambda_1\leq ... \leq \lambda_n$ be the real eigenvalues.

The sup statement (possible taking real part) is correct precisely when $\lambda_n+\lambda_1\geq 0$ (sup over unit vectors): $$\sup {\rm Re} \;\{\langle Ax,y\rangle \} = \sup \;\{\langle Ax,x\rangle \} $$

and the inf statement is correct precisely when $\lambda_1+\lambda_n\leq 0$. $$\inf {\rm Re} \;\{\langle Ax,y\rangle \} = \inf \;\{\langle Ax,x\rangle \} $$