I wanted to know how to prove that, given a matrix $A \in \mathbb{M}_n(\mathbb{R})$ that is block diagonal, if the matrix $A$ is diagonalizable, then each block is diagonalizable too.
I am using this fact to prove that if two diagonalizable matrices commute, then they are simultaneously diagonalizable. That follows because: if $D$ is the diagonal matrix of a matrix $A$ $(D=V^{-1}AV)$, and $AB=BA$, that can be transformed into $DC=CD$ where $C=V^{-1}BV$ and then C must be block diagonal. (Now here is where I use the proof I asked) Each block $C_i$ is diagonalizable, so let $S_i$ be an invertible matrix such that $S_{i}^{-1}C_{i}S_{i}$ is diagonal. If $S$=diag{$S_1,S_2, \dots S_p$}, then $S^{-1}CS=S^{-1}(V^{-1}BV)S=$diag{$S_{1}^{-1}C_{1}S_{1},\space S_{2}^{-1}C_{2}S_{2}, \space \dots,\space S_{p}^{-1}C_{p}S_{p}$} is therefore a diagonal matrix. Computing $S^{-1}(V^{-1}AV)S=S^{-1}DS=D$ is also diagonal. So both A and B are simultaneously diagonalizable by the matrix $SV$. (Also if you see any mistakes in this proof sketch, please tell me!)
Thanks in advanced :)
