Let $A \in \mathbb{C}^{n \times n}$ be such that $A = A^{*} $ and $<x,Ax> > 0$ whenever $x \ne 0$. We define $$\forall u,v \in \mathbb{C}^n: (u|v) := <u,Av>$$ where $<.,.>$ is the standard where the first argument should be complex conjugated. Prove that $(u|v)$ is an inner product.
So, I have to show positive definiteness: Let $u \in \mathbb{C}^n$ such that $u \ne o$. By definition the following holds: $$(u|u) = <u,Au> > 0$$ and if $u=0$: $$(u|u) = <u,Au> = 0$$ Now I have to show linearity:
Let $a,b \in \mathbb{R}$ and $u,x,v \in \mathbb{C}^n$, then: $$ (au+bx,v)= <au+bx,Av>= <au,Av> + <bx, Av> = a<u,Av> + b<x,Av> = a(u|v)+b(x|v)$$
Now, I have to show complex symmetry but I am not sure how to show complex symmetry here. Could anyone show me how to show complex symmetry in detail? I think I have to use he fact that $A=A^*$ somehow but I am not quite sure how it should look like.
$A=A^*$ means that for all $u,v$, you have $$\langle u,Av \rangle = \langle Au, v \rangle$$
Therefore, $$(u|v)=\langle u, Av \rangle = \langle Au, v \rangle = \langle v, Au \rangle = (v|u)$$