I am looking for a simple proof that two hermitic matrix that commutes are diagonalisable in a same base.
I know that we only need $[A,B]=0$ to ensure that it exists a basis in which both are diagonalisable, and I have seen the proof associated to it.
But I want a simpler proof of it in the restrictive case where $A$ and $B$ are hermitic.
Like not a proof by recurrence but using some properties about orthogonality of vectors. I know such a proof is possible because I have seen it in a quantum mechanics course but I don't find it on my own.
You can do something like this:
Let $$V = \ker(A- \lambda_1I) \dot+ \ker(A- \lambda_2I) \dot+ \cdots \dot+ \ker(A- \lambda_kI)$$
be the eigenspace decomposition for $A$.
Since $AB = BA$, every $\ker (A-\lambda_jI)$ is $B$-invariant. Let $B_j = B|_{\ker (A-\lambda_jI)}$. Trivially $B_j$ is also hermitian so there exists the decomposition of the space $\ker (A-\lambda_jI)$ onto eigenspaces for $B_j$:
$$\ker (A-\lambda_jI) = \ker(B_j - \mu^{(j)}_1I) \dot+ \cdots \dot+\ker(B_j - \mu^{(j)}_{m(j)}I)$$
Since these eigenspaces of $B_j$ are also $A$-invariant, we can pick an orthonormal basis for each of them and thus construct the basis for $V$ as
$$(b) = \Big(\text{basis for}\ker(A - \lambda_1 I),\ldots, \text{basis for}\ker(A - \lambda_k I)\Big)$$
where each basis for $\ker (A - \lambda_j I)$ is a union of bases for $\ker(B_j - \mu^{(j)}_iI)$.
Both $A$ and $B$ should diagonalize in the basis $(b)$.