In Strang's Linear algebra, he proves the following (which has been asked and answered on SE, but my question is about a particular part of his proof):

Let $A$ and $B$ be complex $n\times n$ matricies. Prove that if $AB=BA$, then $A$ and $B$ share a common eigenvector.

His proof is as follows: Let $\lambda$ be a eigenvalue of $A$. Starting from $Ax=\lambda x$, we have $ABx=BAx=B\lambda x=\lambda Bx$. So, $x$ and $Bx$ are both eigenvectors of $A$ sharing the same $\lambda$ (or else $Bx=0$). If we assume the eigenvalues of $A$ are distinct, so the eigenspaces are one dimensional, then $Bx$ must be a multiple of $x$. In other words, $x$ is an eigenvector of $B$ as well as $A$.

My question is, and I am probably overthinking something quite elementary, but when he says "if we assume the eigenvalues of $A$ are distinct...", I agree then with the rest of the argument.... but why can he do that?

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    $\begingroup$ I'm assuming the proof continues in some way? If the eigenspace is not one-dimensional, then this argument shows that, letting $E_{\lambda}$ denote the eigenspace of $A$ corresponding to $\lambda$, then $B$ sends $E_{\lambda}$ to itself. Thus, the restriction of $B$ ot $E_{\lambda}$ is a complex operator on a vector space of dimension greater than $1$, and $B|_{E}$ has an eigenvector. That eigenvector must also be an eigenvector of $A$ since it lies in $E_{\lambda}$. You cannot just assume all eigenspaces are one-dimensional (take $A$ to be $\lambda I_n$ for some $\lambda$). $\endgroup$ Dec 17, 2021 at 22:40
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    $\begingroup$ 4th edition of the text has the line "The proof with repeated eigenvalues is a little longer", after the argument that OP gives. $\endgroup$
    – podiki
    Dec 17, 2021 at 22:53
  • $\begingroup$ Have a look at this answer comments and other answers. $\endgroup$
    – user376343
    Dec 17, 2021 at 22:55

3 Answers 3


I think it's sloppy and I'm not sure what Strang meant. So let me try to fix that proof.

Theorem. For any vector $v$, if $v$ is a $\lambda$-eigenvector of $A$, then $Bv$ is also a $\lambda$-eigenvector of $A$. Strang has proved this.

Corollary. If $x$ is a $\lambda$-eigenvector of $A$, then $Bx, B^2 x, B^3 x, \dots$ are also $\lambda$-eigenvectors of $A$.

Now, let $m$ be the first index s.t. $B^m x \in \operatorname{span}(x, Bx, \dots, B^{m-1} x)$. Denote $U = \operatorname{span}(x, Bx, \dots, B^{m-1} x)$. Then $U$ is an invariant subspace of $B$, i.e. $B[U] \subseteq U$. Consider $B$ as a linear operator rather than a matrix, now consider its restriction to $U$ - it must have an eigenvector in $U$ - this eigenvector will be an eigenvector of both $B$ and $A$.


I find it strange that Strang writes it like that. It doesn't seem like you may assume that. However, the proof shows that $B$ acts on the eigenspace of $A$, i.e. it sends a vector with eigenvalue $\lambda$ to another such vector. Viewing $B$ as a linear map on this subspace, it must have an eigenvector (as any linear map has at least one eigenvector). This eigenvector of $B$ is by construction also an eigenvalue of $A$.


Since you are working over the field of complex numbers, which is algebraically closed, you can argue as follows.

Pick any eigenvalue $\lambda$ of $A$ and consider the subspace $V = \{x \in \Bbb C^n: Ax = \lambda x\}$. The proof you quoted shows that, for any vector $x \in V$, we have $Bx \in V$.

Therefore we may view $B$ as a linear map from $V$ to $V$, which again has at least one eigenvector $v \in V$. The vector $v$ is then a common eigenvector of both $A$ and $B$.


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