I have to find the matrix $U$ that simultaneously diagonalize two hermitian matrices $A$ and $B$ which commute, i.e. $AB=BA$. I know of the theorem that assures the existence of such a matrix and that there are many questions on Mathematics SE about its proof, I'm not interested in it but in how to find said matrix explicitly, for which I did not, strangely, find any existing question.
The solution to this exercise proceeds as follows: find the unitary matrix $V$ which diagonalises $A$, i.e. $V^tAV=D_A$ where $D_A$ is diagonal. The apply the same change of basis to $B$ and define $B'=V^tBV$. Then find $W$ unitary which diagonalises $B'$, i.e. $W^tB'W= D_B$ diagonal. Then $U=VW$.
Obviously $VW$ diagonalises $B$ because $(VW)^tB(VW)=W^tV^tBVW=W^tB'W=D_B$. But it is not clear to me why $VW$ diagonalises $A$. We have $$(VW)^tAVW=W^tV^tAVW=W^tD_AW$$
Explicit calculations reveal $W^tD_AW=D_A$. I don't know if this is true in general or if in general we obtain a different, albeit diagonal matrix, and I don't know how to justify this fact. I still have not used the fact that $A$ and $B$ commute, so they have a common orthonormal eigenbasis, thus the matrices $V$ and $W$ must be somehow related, but I don't see how, any help?
Thank you.
There is a unitary matrix $V$ s.t. $D=V^*AV,H=V^*BV$ where $D=diag(\lambda_1 I_{n_1},\cdots,\lambda_k I_{n_k})$, the $\lambda_i\in\mathbb{R}$ are distinct and $H$ is a hermitian matrix; since $DH=H D$, one can write $H=diag(H_1, \cdots,H_k)$ where $H_i$ is $n_i\times n_i$ hermitian. For every $i$, there is a unitary $n_i\times n_i$ matrix $W_i$ s.t. $\Delta_i=W_i^*H_iW_i$ is real diagonal.
Finally, the matrix $U=Vdiag(W_1,\cdots,W_k)$ satisfies $U^*AU=D,U^*BU=diag(\Delta_1,\cdots,\Delta_k)$ and we are done.