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

Solution 1:

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 $A$.

Solution 2:

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 $A$.