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.

  • $\begingroup$ 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. $\endgroup$
    – Ulrik
    Nov 26, 2014 at 1:28
  • $\begingroup$ @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? $\endgroup$
    – lllll
    Nov 26, 2014 at 1:43

1 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.)

  • $\begingroup$ Thank you! I forgot you had to take the complex conjugate of v when removing the dot product $\endgroup$
    – lllll
    Nov 26, 2014 at 2:12

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