Hermitian Matrix Inequality

Question

If we have {$A_{ij}\}_{n*n}$ a Hermitian matrix. v=($v_1,v_2..v_n$), w=($w_1,w_2...w_n$) are two complex vectors. Then how can I show the inequality

|$\sum_{i,j=1}^nA_{ij}v_i\overline{w_j}$|$\leq \sqrt{\sum_{i,j=1}^nA_{ij}v_i\overline{v_j}} \sqrt{\sum_{i,j=1}^nA_{ij}w_i\overline{w_j}}$

Thanks for any hint!

Answer 1

This is the Cauchy-Schwarz Inequality (with a square root taken on both sides).

Answer 2

I shall expand on @TravisWillse answer. It is well known that given a positive-definite, Hermitian matrix $A$, one can define an inner product $$\langle x|y\rangle_A=x^*Ay=\sum_{i,j=1}^n x_i^* A_{ij}y_j.$$ The Cauchy-Schwarz inequality, for any valid inner product, states $$|\langle x|y\rangle|^2\leq \langle x|x\rangle\cdot\langle y|y\rangle.$$ Using our inner product and taking square-root gives the desired result.