Given two $n \times n$ Hermitian matrices $T$ and $S$ each with $n$ distinct eigenvalues, show that:
The matrix $N = T + iS$ is diagonalisable.
I know a matrix is diagonalisable if it has $n$ distinct linearly independent eigenvectors. Alternatively, it is diagonalisable if it has n distinct eigenvalues (this is sufficient but not necessary).
I realise a key point must be that the eigenvalues of the matrices are distinct, as every matrix can be decomposed as the sum of a Hermitian and an Antihermitian matrix but not every matrix is diagonalisable, but I can't find how to use this to show the statement. I have tried showing $N$ has distinct eigenvalues as well but I do not know how to relate the eigenvalues of $N$ to those of $T$ and $S$.
Any hints on how to show this?
$\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}0&1/2\\1/2&0\end{pmatrix}+i\begin{pmatrix}0&-i/2\\i/2&0\end{pmatrix}$