Proving that $\overline{\underline v}^T\overline{A}^TA\underline v\geq0$

Question

I want to prove that, for any invertible matrix $A\in\mathbb{C}_{n\times n}$, we have $$ \overline{\underline v}^T\overline{A}^TA\underline v\geq0 $$ and it is $0$ iff $\underline v=\underline 0$. Here $\overline{A}^T$ denotes the conjugate transpose of $A$.

My attempt. In the real case, I know that $\underline v^TA^TA\underline v\geq0$ and it is $0$ iff $\underline v=\underline 0$. But how can I prove this for the complex case? In particular, I want to prove that $(\overline{A\underline v})^T(A\underline v)\geq0$...

Thank You

Answer 1

You get $\bar v^T\bar A^TAV=\|Av\|^2$ where the norm is the standard Hermitean norm $\|x\|=\sqrt{\sum_k|x_k|^2}$. Now conclude from $Av=0$ to $v=0$.

Answer 2

Let $Av = [z_1, z_2, \dots, z_n]^T$ Then

\begin{align} v^*A^*Av &= [z_1^*, z_2^*, \dots, z_n^*] \ [z_1, z_2, \dots, z_n]^T \\ &= \sum_{k=1}^n z_k^* z_k^{\phantom{*}} \\ &= \sum_{k=1}^n \|z_k^{\phantom{*}}\|^2 \\ &\ge 0 \end{align}