If $AB = BA$ and both $A$ and $B$ are Hermitian matrices, then I can show that if $Av = \lambda v$, then $ABv = BAv = B\lambda v = \lambda Bv$. So $Bv$ is an eigenvector of $A$ as well.

Where I am stuck is how this implies that $v$ must also be an eigenvector of $B$?

  • $\begingroup$ It doesn't${}$. $\endgroup$ Apr 25, 2019 at 17:58
  • $\begingroup$ The crucial thing to prove is that each eigenspace of $A$ is invariant under $B$. $\endgroup$ Apr 25, 2019 at 18:23
  • 1
    $\begingroup$ You should always consider trivial cases to check a statement. Here $A=I$ is commuting with any matrix. But it's quite clear that any non zero vector could not be automatically an eigenvector of $B$. $\endgroup$
    – zwim
    Apr 25, 2019 at 18:43

2 Answers 2


As others have pointed out, the exact result stated in your question isn't true. What is true is that $A$ leaves the eigenspaces of $B$ invariant and vice versa. If $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $$ A(Bv) = (AB)v = (BA)v = B(Av) = B(\lambda v) = \lambda (Bv) $$ Thus, $Bv$ is an eigenvector of $A$ with eigenvalue $\lambda$. Let $V_\lambda(A)$ denote the subspace of eigenvectors of $A$ with eigenvalue $\lambda$. We have shown $B(V_\lambda(A)) \subset V_\lambda(A)$. This is the result that holds in general.

If $V_\lambda(A)$ is one dimensional, we easily obtain a result that looks like the result you were originally after. Take $v \in V_\lambda(A)$. Then $Bv \in V_\lambda(A)$, so, since $V_\lambda(A)$ is one dimensional, $Bv = \xi v$. In words, if $v$ is an eigenvector of $A$ with nondegenerate eigenvalue $\lambda$, then $v$ is an eigenvector of $B$.

There's more to say about this, but I think this answer beat me to the punch Matrices commute if and only if they share a common basis of eigenvectors?

In general, the result you're thinking of is true, you just have to phrase it correctly. It's not the case that every eigenvector of $A$ is an eigenvector of $B$. The correct statement is that there exists a basis of vectors $v_1 , \ldots , v_n$ for the space such that $v_i$ is an eigenvector of $A$ and $B$ for all $i$. The truth of this statement requires that $A$ and $B$ are diagonalizable, which is necessarily true if $A$ and $B$ are Hermitian.


What you are trying to prove is in fact false. Consider the commuting Hermitian matrices $$A=\begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} \qquad\text{ and }\qquad B=\begin{pmatrix} 1&0\\ 0&2 \end{pmatrix}.$$ Clearly $(1,1)$ is an eigenvector of $A$ but not of $B$.


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