## If [A,B] = rA + sB, then A and B have a common eigenvector

Posted: December 1, 2009 in Elementary Algebra; Problems & Solutions, Linear Algebra
Tags: ,

It is a well-known fact that commuting matrices over $\mathbb{C}$ share a common eigenvector. This fact is a special case of the following result.

Problem. Let $A,B \in M_n(\mathbb{C}).$ Prove that if $AB-BA=rA + sB,$ for some $r,s \in \mathbb{C},$ then $A$ and $B$ share a common eigenvector.

Solution. The case $r=s=0$ is well-known. Suppose that $s \neq 0.$ Let $\lambda$ be any eigenvalue of $A.$ Let

$k =\max \{\ell \in \mathbb{Z}: \ \ \lambda + \ell s \ \ \text{is an eigenvalue of} \ A \}.$

Let $u$ be a non-zero element of $\mathbb{C}^n$ such that $Au = (\lambda + ks)u$ and put

$\displaystyle v=Bu + \frac{r(\lambda + ks)}{s}u.$

Then $Av = (\lambda + (k+1)s)v.$  Therefore, by the maximality of $k,$ we  have $v=0$ and so $u$ is an eigenvector of both $A$ and $B. \ \Box$