Proof $\langle Ax,y\rangle = \langle x,A^*y\rangle$ when $A$ Hermitian I was trying to understand a proof of why a Hermitian $A$ matrix has its eigenvectors orthogonal. As part of the proof they state
$$\langle Ax,y\rangle = \langle x,A^*y\rangle$$
From which property this is derived
 A: Hint: For any vectors $u,v$, we have $\langle u,v \rangle = v^*u$.  Thus,
$$
\langle Ax,y \rangle = y^*(Ax)
$$
We'd like to show that the above is equal to
$$
\langle x,A^*y \rangle = (A^*y)^*x
$$
How do we get from one to the other?
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{A = A\+\quad\imp\quad\mbox{real eigenvalues}\ \braces{\lambda_{i}}}$. With a pair of eigenvectors $v_{i}$ and eigenvalues $\lambda_{i}$ of $A$ $\pars{~Av_{i} = \lambda_{i}v_{i}\,,\quad i = 1,2~}$:
$$\color{#0000ff}{\large%
v_{1}\+v_{2}}
= {1 \over \lambda_{2}}\,v_{1}\+\lambda_{2}v_{2}
= {1 \over \lambda_{2}}\,v_{1}\+A v_{2}
= {1 \over \lambda_{2}}\,\pars{v_{2}\+ A v_{1}}\+
= {1 \over \lambda_{2}}\,\pars{v_{2}\+\lambda_{1}v_{1}}\+
= \color{#0000ff}{\large{\lambda_{1} \over \lambda_{2}}\,v_{1}\+ v_{2}}
$$
$$\color{#0000ff}{\large%
\lambda_{1} \not= \lambda_{2}\quad\imp\quad v_{1}\+v_{2} = 0}$$
A: What exactly is your question?
The adjoint matrix $A^*$ of $A$ is by definition a matrix that satisfies
\begin{align}
\langle Ax,y \rangle = \langle x,A^*y \rangle,
\end{align}
for all $x$ and $y$. Here $\langle\cdot,\cdot \rangle$ is called the inner product, and in a complex vector space it is defined as
\begin{align}
\langle x,y\rangle = \sum\limits_{i = 1}^{n}x_i \overline{y}_i,
\end{align}
where $\overline{y}$ is the complex conjugate of $y$ and $n$ is the length of the vectors $x$ and $y$.
A: This result is based on Riesz representation theorem. The theorem states that any linear functional $ f(a) \in  \mathbb F , \forall a \in \mathbb F^n,$ can be represented as $f(a)=(a|b)$ where $b \in \mathbb F^n$. 
For a given $y, (Ax|y)=f(x)$ therefore $\exists b:(Ax|y)=(x|b),$ let $A^*y=b$ we get $(Ax|y)=(x,A^*y)$, $A^*$ can be proved to be unique. 
Also this is true for all $A$. If $A$ is Hermitian, then $(Ax|y)=(x|Ay)$ because $A=A^*$
