Does $〈Ax,y⟩=〈x,A^*y⟩$ hold for any inner product? Consider $\Bbb C^n$. Let $A^*$ be the conjugate transpose of matrix $A$, I see in many materials $A^*$ is also called the adjoint of $A$. I am now confused by that does $〈Ax,y⟩=〈x,A^*y⟩$ hold just for the standard inner product (whose proof is simple), or any inner product? Thanks!
 A: It does not hold for every inner product. The analogous situation in the real plane illustrates this. Consider
$$
H = \pmatrix{2 & 0 \\ 0 & 1}
$$
and the inner product $x \cdot y = x^t H y$. 
Let 
$$
A = \pmatrix{0 & 1 \\ 1 & 0}
$$
$x =  \pmatrix{1\\0}$ and $y =  \pmatrix{0\\1}$. 
Then $Ax \cdot y = 1$, but $x \cdot A^t y = 2$. 
Since the real plane is a subspace of the complex plane, this suffices as a counterexample in $\Bbb C^2$ as well (at least if you extend the definition of the dot product by writing $x \cdot y = x^t H \bar{y}$). 
A: Usually, the answer is no, and depends on the structure of the inner product. More specifically, it requires that the orthogonal basis under the inner product is a unitary matrix.
Let's clarify some notations first. Denote by $T$ the operator transforming $x$ to $Ax$. Then $T$ is a linear operator from $\mathbb C^n$ to itself, i.e., $T\in \mathcal L (\mathbb C^n)$, we prefer to use $T^*$ as the "adjoint" operator defined by the equation:
$$
<Tx, y>=<x, T^* y>
$$
for any $x, y \in \mathbb C^n$.
Consider an orthogonal basis for $\mathbb C^n$, $\tilde e=(\tilde e_1, \tilde e_2,...,\tilde e_n)$ under inner product $<.,.>$. Then the matrix of $T$ under this basis is $[m_{ij}]_{n\times n}$, where $m_{ij}=<T\tilde e_j,\tilde e_i>$, which we denoted as $M(T,\tilde e)$. 
On the other hand, since
$$
T(x)=Ax
$$
Then the matrix of $T$ under the trivial basis $e=(e_1, e_2,...,e_n)$ where $e_1=[1,0,...,0]^T$, $e_2=[0,1,...,0]^T$,..., i.e., $M(T, e)=A$.
We use $A^H$ to denote the conjugate transpose of $A$. Then we can show the matrix of the linear operator $T^*$ under basis $\tilde e$ (well, we need to prove it's linear first, which is not very hard) is the conjugate transpose of $M(T,\tilde e)$, i.e.,
$$
 M(T^*,\tilde e)=M(T,\tilde e)^H\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
$$
Now change the basis from $\tilde e$ to $e$, we get
$$
\tilde e^{-1} M(T^*,e) \tilde e= (\tilde e^{-1} A \tilde e)^H=(\tilde e)^H A^H (\tilde e^{-1})^H
$$
If we want $M(T^*,e)=A^H$ holds for all $A$, then
$$
\tilde e^{-1} A^H \tilde e=(\tilde e)^H A^H (\tilde e^{-1})^H
$$
which requires $\tilde e$ being a unitary matrix:
$$
(\tilde e)^H \tilde e = I_n
$$

To see why (1) holds, consider the $j^{th}$ column of $M \equiv M(T,\tilde e)$, call it $M_j$. It's obvious that
$$
M_j=M\tilde e_j
$$
on the other hand,
$$
M\tilde e_j=T\tilde e_j=<T\tilde e_j,\tilde e_1>\tilde e_1+<T\tilde e_j,\tilde e_2>\tilde e_2+...+<Te_j,\tilde e_n>\tilde e_n
$$
so $m_{ij}=<T\tilde e_j,\tilde e_i>$
Similarly, let $M^* \equiv M(T^*,\tilde e)$, then 
$$
m^*_{ji}=<T^* \tilde e_i,\tilde e_j>=< \tilde e_i,(T^*)^*\tilde e_j>=< \tilde e_i,T\tilde e_j>= \overline {< T\tilde e_i,\tilde e_j>}
$$
this holds for each $(i,j)$, which means $M^*=M^H$ 
A: $〈Ax,y⟩=〈x,A^*y⟩ $ this holds for any inner product ! when you assume $A^*$ is adjoint of $A$ in space X equipped with inner product of $〈.,.⟩$
This is how adjoint operators  defined in finite dimension. 
