(Linear algebra) if $A$ is normal matrix then, eigenvectors of $ A$ are orthogonal. I know that the eigenvectors of a unitary matrix are orthogonal. Then is that also true for a normal matrix? How do I prove?
 A: The spectral theorem states that a matrix $A$ is normal if and only if $A$ is diagonalizable by a unitary matrix $U$, i.e. $A = U D U^\dagger$ for a diagonal matrix $A$.
It is easy to see that, for every unitarily diagonalizable matrix, we can choose a basis of orthogonal eigenvectors:
The columns of $U$ are eigenvectors of $A$, and $U$ is by definition unitary if and only if its columns are orthogonal (with respect to the inner product of $\mathbb C^n$).
A: There is a theorem that states what you ask. But since you asked for a proof, here is one.
The matrix defines a normal operator$~T$ on a complex inner product space$~V$, and I will prove, by induction on the dimension, that such operators admit an orthogonal basis of eigenvectors. (The operator point of view is more flexible than the matrix point of view, and abstracts away from the standard basis of $\Bbb C^n$, which although orthonormal has no specific relation to the problem at hand.)
If the dimension of the space is zero, the basis is empty and we are done. Otherwise, like any linear operator on a finite non-zero dimensional complex vector space, our $T$ has as least one eigenvalue; choose one and call it$~\lambda$. Since $T$ is normal, the eigenspaces of $T$ are $T^*$-stable, so we can restrict $T^*$ to the eigenspace $E_\lambda$ of$~T$ for$~\lambda$, and find an eigenvector $v\in E_\lambda$ that is also an eigenvector for $T^*$. The later maens that $\langle v\rangle$ is nonzero and $T^*$-stable which by simple algebra implies that $\langle v\rangle^\perp$ is $T$-stable. Since we are in an inner product space, $V=\langle v\rangle\oplus\langle v\rangle^\perp$. By restriction $T$ defines a normal operator on the complex inner product space $\langle v\rangle^\perp$ which has dimension one less than$~V$. We can apply induction to get an orthogonal basis of $\langle v\rangle^\perp$ consisting of eigenvectors of (the restriction and therefore of) $T$. Together with $v$, one has an orthogonal basis of$~V$ consisting  of eigenvectors of$~T$.
