Confusion with normal linear transformations and their matrices I'm aware that given a normal map $T:V \to V$ ($V$ an inner product space) it is possible to find an orthonormal basis of eigenvectors of $T$ for $V$ (provided the $K$ is algebraically closed), however if I am given a particular matrix representation for $T$ after choosing some basis for $V$, say $A: \mathbb{C}^n \to \mathbb{C}^n$ is it then true that the eigenvectors of this matrix have to form an orthonormal basis for $\mathbb{C}^n$?
I'm sure this has a simple answer, but it has been confusing me! I'm asking the question because I am trying to decide whether certain matrices represent self-adjoint (or more generally normal transformations) in some basis.
Thanks for any help
 A: Adding to Peter Franek's answer:
Let $T$ be normal, and $B$ be an orthonormal basis of $V$. Then the matrix representation $A\in \mathbb C^{n,n}$ of $T$ with respect to $B$ is normal, too,
$$
AA^H=A^HA.
$$
And there exists an orthonormal basis of eigenvectors of $A$.
The reason for this is that the matrix representation of the adjoint $T^*$ is equal to $A^H$, which is not necessarily true if the basis $B$ is not orthonormal.
A: I'm not sure if I understand the question correctly but I think it can't hold. Take, for example,
$A=\begin{pmatrix} 0 & 1\\ -1 & 0\end{pmatrix}$ which is normal, and a similar matrix
$$
\begin{pmatrix}
1 & -i \\
0 & 1
\end{pmatrix}
\begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix}
\begin{pmatrix}
1 & i \\
0 & 1
\end{pmatrix}=
\begin{pmatrix}
i & 0 \\
-1 & -i
\end{pmatrix}
$$
which has eigenvectors $(-2i, 1)$ and $(0,1)$; those are not "orthogonal" to each other, if you still use the standard hermitian product on $\mathbb{C}^2$. Of course, if you have new coordinates, you should transform your "formula" for the hermitian product as well; then they would still be orthonormal. 
There is no reason to expect that if you compute eigenvectors in the new basis and do the hermitian product using the "old formula", you get compatible results.
