We define $\langle\cdot, \cdot\rangle $: $\mathbb{C}^{n\times n} \times \mathbb{C}^{n\times n}\longrightarrow \mathbb{C} $ by $ \langle A, B\rangle := \operatorname{Tr}(AB^{∗})$. Show that this mapping gives an inner product on the vector space $\mathbb{C}^{n\times n}$.
Does someone have an idea how I can prove that?
An inner product space is a vector space $\mathbb{C}^{n×n} \times \mathbb{C}^{n×n}$ over the field $\mathbb{C}$ together with an inner product, i.e., with a map $$\mathbb{C}^{n×n} \times \mathbb{C}^{n×n}\longrightarrow \mathbb{C}$$ such that satisties
Conjugate symmetry:
$\langle A,B\rangle ={\overline {\langle B,A\rangle }}$
Linearity in the first argument:
$\langle aA,B\rangle =a\langle A,B\rangle \\\langle A+B,C\rangle =\langle A,C\rangle +\langle B,C\rangle $
Positive-definiteness:
$\langle x,x\rangle \geq 0\\\langle x,x\rangle =0\Leftrightarrow x=\mathbf {0} \,.$
where $A,B,C \in \mathbb{C}^{n×n} \times \mathbb{C}^{n×n}$ and $a\in \mathbb{C}$.
HINT:
Use the following facts.
$$Tr(aA)=aTr(A)$$ $$Tr(A)=Tr(A)^T$$ $$Tr(AB)=Tr(BA)$$ $$Tr(AB^T)=Tr(A^TB)=Tr(B^TA)=Tr(BA^T)$$
