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?
 A: 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 $A$.
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 $A$.
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 $B$.
