Hermitian matrix the only diagonizable During the last lecture one of my professors claimed that the hermitian matrix is the ONLY complex matrix which was diagonizable.
This seems strange to mee (not to say a very very strong claim to make), I know that Hermitian matrices are the only ones diagonizable by unitary transformations, but that they're the only diagonizable ones seems strange.
Could someone elaborate on this ? Or provide some kind of proof for this ?
 A: As requested, I'll provide my example and a little bit of exposition:
The claim your professor makes, namely 

All diagonalizable (in $\mathbb{C}$) matrices are Hermitian.

is false. Let us quickly recall what it means for a square complex matrix $A$ to be Hermitian: it means that $A=A^\ast$, where $A^\ast$ is the conjugate transpose of $A$,  A simple counter-example then is the following: $\left[ \begin{matrix} 1+i & 0 \\ 0 & 1+i \end{matrix}\right]$ which is clearly diagonalizable (it is diagonal already) but is not Hermitian because $A\neq A^\ast$.
Perhaps your professor meant to make a weaker claim:

All matrices that are unitary diagonalizable are Hermitian.

Again, this is false, as the above example points out: $$\left[ \begin{matrix} 1+i & 0 \\ 0 & 1+i \end{matrix}\right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right] \left[ \begin{matrix} 1+i & 0 \\ 0 & 1+i \end{matrix}\right] \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right]$$  The identity matrix $\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right]$ is indeed a unitary matrix, so that $\left[ \begin{matrix} 1+i & 0 \\ 0 & 1+i \end{matrix}\right]$ has a unitary diagonalization.
In general, the Spectral Theorem tells us that all normal complex matrices are unitary diagonalizable, and there are normal matrices that are not Hermitian.
