[Math] Do commuting matrices share the same eigenvectors

eigenvalues-eigenvectorslinear algebramatrices

In one of my exams I'm asked to prove the following

Suppose $A,B\in \mathbb R^{n\times n}$, and $AB=BA$, then $A,B$ share the same eigenvectors.

My attempt is let $\xi$ be an eigenvector corresponding to $\lambda$ of $A$, then $A\xi=\lambda\xi$, then I want to show $\xi$ is also some eigenvector of $B$ but I get stuck.

Best Answer

The answer is in the book Linear Algebra and its Application by Gilbert Strang. I'll just write down what he said in the book.

Starting from $Ax=\lambda x$, we have

$$ABx = BAx = B \lambda x = \lambda Bx$$

Thus $x$ and $Bx$ are both eigenvectors of $A$, sharing the same $\lambda$ (or else $Bx = 0$). If we assume for convenience that the eigenvalues of $A$ are distinct – 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 $A$.

There's another proof using diagonalization in the book.

Related Question