# Question about a proof: If $AB=BA$ then $A$ and $B$ share a common eigenvector, from Strang's Linear Algebra

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?

• 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 $\lambda$). Dec 17, 2021 at 22:40
• 4th edition of the text has the line "The proof with repeated eigenvalues is a little longer", after the argument that OP gives. Dec 17, 2021 at 22:53

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 $$B$$.