# If diagonalizable matrices commute does it neccesarily mean that they can be simultaneously diagonalized?

If matrices $M_1$ and $M_2$ can be simultaneously diagonalized, than they commute, which can be easily shown: \begin{align} M_1M_2&=P^{-1}D_1PP^{-1}D_2P \\ &=P^{-1}D_1D_2P \\ &=P^{-1}D_2D_1P \\ &=P^{-1}D_2PP^{-1}D_1P \\ &=M_2M_1 \end{align}

But is converse also true? If diagonalizable matrices $M_1$,...,$M_n$ all mutually commute, does this mean that they can be simultaneously diagonalized?

If so, how to show this? If not, what is the simplest counterexample?

• possible duplicate of Simultaneously Diagonalizable Proof – Jonas Dahlbæk Aug 9 '14 at 10:35
• Not a duplicate. The linked question is only a partial duplicate of this one. – Git Gud Aug 9 '14 at 10:37
• The answer generalizes to finitely many matrices. – Jonas Dahlbæk Aug 9 '14 at 10:50

$M$ can be diagonalized iff the minimal polynomial $m$ for $M$ splits completely into linear, non-repeated factors $m(\lambda)=(\lambda-\lambda_1)(\lambda-\lambda_2)(\cdots)(\lambda-\lambda_N)$. The usual proof of this involves the unique ($N-1$)-st order polynomials $p_{k}$ such that that $p_{k}(\lambda_{j})=\delta_{j,k}$. Then $\sum_{k=1}^{N}p_{k}\equiv 1$ because the sum is an $(N-1)$-st order polynomial which is $1$ in $N$ places. Therefore, $$I = p_1(M)+p_2(M)+\cdots+p_N(M).$$ Furthermore $p_j(M)p_k(M)=0$ for $j \ne k$ because $m$ divides $p_j p_k$ for $j \ne k$. Therefore each $p_j(M)$ is a projection matrix; to see this, apply $p_j(M)$ to the above identity: $$p_j(M)=p_j(M)^{2}.$$ Furthermore $(M-\lambda_k I)p_k(M)=0$ which implies $$M = \lambda_1 p_1(M)+\lambda_2 p_2(M)+\cdots+ \lambda_N p_N(M).$$ If $M_1,M_2,M_3,\cdots,M_J$ are commuting diagonalizable matrices, you can perform the above construction for each $M_j$ in order to obtain eigenvalues $\lambda_{j,1},\lambda_{j,2},\cdots,\lambda_{j,K_{j}}$ and polynomials $p_{j,1},p_{j,2},\cdots,p_{j,K_j}$ for each $1 \le j \le J$. Because the $M_j$ commute, then the same is true of all of the $p_{j,k}(M_j)$. Now form all of the distinct products $$P_{k_1,k_2,\cdots,k_J}=p_{1,k_1}(M_1)p_{2,k_2}(M_2)\cdots p_{L,k_J}(M_J).$$ The sum of all such products is $I$, and every such $P$ is a projection. Discard the products that turn out to $0$. Because the order of the factors may be rearranged without changing $P$, it follows that $$(M_{j}-\lambda_{j,k_{j}}I)P_{k_1,k_2,\cdots,k_J}=0,\;\;\; 1 \le j \le J.$$ So there are non-zero projections $Q_{1},Q_{2},\cdots,Q_{m}$ whose sum is $I$, whose products are $0$ for distinct factors, and such that every $M_{j}$ is a scalar multiple of the identity on the range of a given $Q_{k}$. Choose a basis for each of the ranges of $Q_{j}$. Combining these bases produces a basis with respect to which each $M_j$ has a diagonal representation.