# Simultaneous Jordanization

Hello everyone

I would like to have a detailed reference to the statement bellow:

Let $A,B\in \mathbb{R}^{n\times n}$ such that $AB=BA$. Suppose $A$ has real eigenvalues only and $B$ is diagonalizable. Then, there exists a nonsingular matrix $P$ such that $P^{-1}AP$ is of Jordan canonical form and $P^{-1}BP$ is diagonale. Precisely \begin{equation*} P^{-1}AP=\mathrm{diag}(J_{1},\cdots,J_{r}) \quad\text{and}\quad P^{-1}BP=\mathrm{diag}(\underset{p_{1}}{\underbrace{\mu_{1},\cdots,\mu_{1}}},\cdots,\underset{p_{r}}{\underbrace{\mu_{r},\cdots,\mu_{r}}}) \end{equation*} where $J_{k}$ is a Jordan block of size $p_{k}$.

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@Chris Godsil: the OP made the explicit assumption that $B$ is diagonalizable. –  Noah Stein Nov 2 '12 at 12:20
@Noah: I've deleted my silly comment. –  Chris Godsil Nov 2 '12 at 15:24

Let $\lambda$ be an eigenvalue of $B$, and $V_\lambda$ the corresponding eigenspace (the set of all eigenvectors of $B$ corresponding to $\lambda$ and the zero vector). Then the whole space is the direct sum of those $V_\lambda$. (This is your condition that $B$ is diagonalizable).
Claim. $A$ maps each $V_\lambda$ into itself. Proof. Let $x\in V_\lambda$ this means that $Bx=\lambda x$. Now let $Ax=y$. We have $By=BAx=ABx=\lambda Ax=\lambda y$. So $y\in V_\lambda$. This proves the claim.
Now by the Jordan theorem, each $V_\lambda$ contains a Jordan basis for $A$. This basis (as the whole $V_\lambda$) consists of eigenvectors of $B$. Taking all these bases together we obtain a basis of the whole space, which is a Jordan basis for $A$ and each vector is an eigenvector of $B$. Take this basis as columns of your $P$.