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

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

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