# Proving Equality of Hermitian matrix

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.

• Do you mean the complex conjugate when you write a bar? Remember that you can write the inner product of vectors as $u \cdot v = u^*v$ where $*$ denotes the complex inner product. If you use this identity, then you should be able to prove the $\Leftarrow$ direction. Nov 26, 2014 at 1:28
• @Svinepels Yes, the bar is supposed to represent the complex conjugate. Sorry I'm not sure how I would use that identity, could you explain how I would go about doing that? Nov 26, 2014 at 1:43

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.