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 ?
As requested, I'll provide my example and a little bit of exposition:
The claim your professor makes, namely
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:
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.