Proving Equality of Hermitian matrix

Question

I need to show that the equality $\ A \mathbf u \cdot \mathbf v = \mathbf u \cdot A\mathbf v \ $ holds true for all $\mathbf u, \mathbf v \in \mathbb{C^n}$ if and only if the matrix A is Hermitian

Proving <= direction

$\ A \mathbf u \cdot \mathbf v = A(\mathbf u \cdot \mathbf v)\ $
= $\mathbf u \cdot \bar A \mathbf v$

I can't seem to get rid of the bar on the matrix (bar on the matrix A represents the complex conjugate). I'm not entirely sure how to use the fact that A is hermitian but I think if I am able to prove <= then I can go backwards to prove =>. Any help is appreciated.

Answer 1

The dot product on $\mathbb{C}^n$ us usually defined by $$\boldsymbol{u} \cdot \boldsymbol{v} = \sum_{i=1}^n u_i \overline{v_i},$$ but it is probably more convenient to write it in matrix notation: $$\boldsymbol{u} \cdot \boldsymbol{v} = \boldsymbol{u}^T \overline{\boldsymbol{v}}.$$ Then: \begin{align} A\boldsymbol{u} \cdot \boldsymbol{v} &= (A\boldsymbol{u})^T \overline{\boldsymbol{v}}\\ &= \boldsymbol{u}^T A^T \overline{\boldsymbol{v}}\\ &= \boldsymbol{u}^T \overline{A} \overline{\boldsymbol{v}}\\ &= \boldsymbol{u}^T \overline{A \boldsymbol{v}}\\ &= \boldsymbol{u} \cdot A\boldsymbol{v} \end{align} The second-third line uses the fact that $A$ is Hermitian, while the third-fourth line uses the fact that conjugation preserves addition and multiplication.

Since the complex dot product involves conjugation, you weren't supposed to get rid of the bar at all. By definition, it's supposed to be there. (A reasonable question, already answered elsewhere, is why that's a good definition.)