Proving Simultaneous Diagonalizability of nxn Matrices A and B with AB = BA

[Bachelier]

A, B nxn matrices are called simultaneously diagonalizable if there exists P such that both P^-1AP and P^-1BP are diagonal.

Prove if A and B are diagonalizable and AB = BA, then A, B are simultaneousely diagonalizable?

 

[hgfalling]

Suppose A and B commute. Then let v be an eigenvector of A with eigenvalue $\lambda$. Then we have $A(Bv)=(AB)v=BAv=\lambda Bv$.

So $Bv$ is in the eigenspace of A.

Choose a candidate basis ${b_1,b_2,...,b_n}$ consisting of eigenvectors of A such that the eigenvectors are ordered to correspond with the eigenvalues (ie, if $\lambda_1$ has multiplicity 2, then $b_1$ and $b_2$ are eigenvectors corresponding to $\lambda_1$).

Now this isn't necessarily a basis of eigenvectors of B. But because Bv is in the eigenspace of A, we can write B in this basis as a block diagonal matrix (where each block is mxm, where m is the multiplicity of an eigenvalue of A). But B is diagonalizable, so each block can be diagonalized, and if we do that, then we have n independent vectors that are eigenvectors of A and of B, so we win.

 

[trambolin]

Or from the commutativity

$$AB =BA = P \Lambda_A P^{-1}B = BP \Lambda_A P^{-1}$$

Since P is invertible, by a similarity transformation on both sides, (pre multiply with $P^{-1}$ and post multiply with $P$)

$$\Lambda_A P^{-1}BP = P^{-1}BP \Lambda_A$$

Since, $P^{-1}BP$ commutes with arbitrary diagonal matrix, itself is a diagonal matrix. Thus, P diagonalizes simultaneously A and B.

 

[Bachelier]

Thanks.

 

[Landau]

Since, $P^{-1}BP$ commutes with arbitrary diagonal matrix, itself is a diagonal matrix. Thus, P diagonalizes simultaneously A and B.

Why does it commute with an arbitrary diagonal matrix? It commutes with a specific diagonal matrix, namely $\Lambda_A$, the diagonal matrix whose diagonal values are the eigenvalues of A.

I think from this you can only conclude that $P^{-1}BP$ is diagonal if all eigenvalues of A are different... A simple counteraxmple to your proof would be A=P equal to the identity!

 

[trambolin]

If A is already diagonal matrix with arbitrary (which means "any" which then means "choose any diagonalizable A and diagonalize it" by the way also note that the claim is only sufficient not necessary) real numbers as entries, can you give me a nondiagonal matrix B that commutes with A other than identity? Because if you have it, I really need it.

If P = A and P is not diagonal then $P^{-1}AP$ is not diagonal and does not satisfy the assumption in the original claim. If P is diagonal, then A is diagonal (in your case leads to trivial B = B) so back to my question.

 

[Landau]

If I read your proof (post #3) with A equal to the identity (so necessarily P=A), then it says

Or from the commutativity

$$AB =BA = B = B$$

Since I is invertible, by a similarity transformation on both sides, (pre multiply with $I^{-1}$ and post multiply with $I$)

$$B = B$$

Since, $B$ commutes with arbitrary diagonal matrix, itself is a diagonal matrix. Thus, I diagonalizes simultaneously A and B.

which is of course not correct. In your last post you seem to be fixing this by considering different cases (A,B both diagonal, one of them not, or both not), but I am not quite following. Could you elaborate?

 

[trambolin]

Sure. Let's limit the discussion to the commutativity part for now and use the notation $\mathbb{D}$ for the set of all diagonal matrices and $\mathbb{D}_{=} \subset \mathbb{D}$ for the set of all diagonal matrices with identical entries such as identity. What I am trying to say is the following.

Claim: If a matrix B is commuting with any diagonal matrix $A\in\mathbb{D}$. Then B is also diagonal.

My pseudo-proof goes like this. Suppose A is a diagonal 2x2 matrix with distinct elements. Then,
$$AB = \begin{pmatrix} \lambda_1B_{11} &\lambda_1B_{12}\\ \lambda_2B_{21} &\lambda_2B_{22}\end{pmatrix} \neq \begin{pmatrix} \lambda_1B_{11} &\lambda_2B_{12}\\ \lambda_1B_{21} &\lambda_2B_{22}\end{pmatrix} = BA$$
if $\lambda_1 \neq \lambda_2$, or $B_{21}, B_{12} \neq 0$.


Now, your examples are using the elements of $A\in\mathbb{D}_{=}$. But my claim is about the $A\in\mathbb{D}$, hence a bigger set to test with because we can start with any diagonalizable matrix A which might have completely distinct eigenvalues. So if you plug in any element from the bigger set, it puts additional constraints on the off-diagonal entries of B forcing it to be diagonal as I provided a small example above.

 

[trambolin]

Now is the second part about the cases where A is restricted to be $A\in\mathbb{D}_{=}$. Then you can start arguing as follows. I diagonalize B with a matrix Q. And to show that this also diagonalizes A is trivial since $Q^{-1}AQ= Q^{-1}QA = A$ since $A\in\mathbb{D}_{=}$ and commutes with any matrix.

A mixture of these can be done for the matrices that has eigenvalues of multiplicity more than one but also have distinct eigenvaues. But a slightly more tedious proof will lead to similar confirming answer with block diagonal matrix arguments.