<s>[INST] Hermitian matrices in an arbitrary inner product I was wondering if anyone can give me any hints for how to prove the following proposition: 
Proposition: Let $V = F^n$ and let $A\in M_{n\times n}(F)$. Prove that $\langle A^* x,y \rangle = \langle x, A y \rangle$. 
The problem that I am having is how to prove this for an arbitry inner product on $F^n$, an arbitrary field. Some textbooks take this as an axiom or the definition of the adjoint of $A$, but the book that I got this from does not (Linear Algebra by Friedberg). 
 [/INST] This is true only for $V$ with the standard inner product. 
Counter-example:
Assume $V=\mathbb{R}^2$ and we define the inner product in the following way:
$$\langle x | y\rangle = x^T \begin{pmatrix}1 & \frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2} & 1 \end{pmatrix} y$$
Lets take for simplicity $x = \begin{pmatrix} 1\\0 \end{pmatrix} , y =  \begin{pmatrix} 0\\1 \end{pmatrix} $
Lets take $A$ to be $\begin{pmatrix}1 & 2\\3 & 4 \end{pmatrix} $
Now some calculations:
$$\langle x | Ay \rangle = x^T\begin{pmatrix}1 & \frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2} & 1 \end{pmatrix} \begin{pmatrix}1 & 2\\3 & 4 \end{pmatrix} y = x^T\begin{pmatrix}1+\frac{3\sqrt{3}}{2} & 2(1+\sqrt{3})\\\frac{1}{2}(6+\sqrt{3}) & 4+\sqrt{3} \end{pmatrix}  y = 2(1+\sqrt{3}) $$
$$\langle A^*x | y \rangle = \left (\begin{pmatrix}1 & 3\\2 & 4 \end{pmatrix} 
 x\right )^T\begin{pmatrix}1 & \frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2} & 1 \end{pmatrix} y = \begin{pmatrix} 1& 2 \end{pmatrix}\begin{pmatrix}1 & \frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2} & 1 \end{pmatrix} \begin{pmatrix} 0\\ 1 \end{pmatrix} = 2 + \frac{\sqrt{3}}{2} $$
$$ \langle x | Ay \rangle \neq \langle A^*x | y \rangle  $$
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