# Rank of the commutator.

1. Apr 16, 2012

### Hurin

I found this theorem on Prasolov's Problems and Theorems in Linear Algebra:

Let V be a $\mathbb{C}$-vector space and $A,B \in \mathcal{L}(V)$such that $rank([A,B])\leq 1$. Then $A$ and $B$ has a common eigenvector.

He gives this proof:
The proof will be carried out by induction on $n=dim(V)$. He states that we can assume that $ker(A)\neq \{0\}$, otherwise we can replace $A$ by$A - \lambda I$; doubt one: why can we assume that? For $n=1$ it's clear that the property holds, because $V = span(v)$ for some $v$. Supposing that holds for some $n$. Now he divides in to cases:
1. $ker(A)\subseteq ker(C)$; and
2. $ker(A)\not\subset ker(C)$.

Doubt two: the cases 1 and 2 come from (or is equivalent to) the division $rank([A,B])= 1$ or $rank([A,B])=0$?

After this division he continues for case one: $B(ker(A))\subseteq ker(A)$, since if $A(x) = 0$, then $[A,B](x) = 0$ and $AB(x) = BA(x) + [A,B](x) = 0$. Now, the doubt three is concerning the following step in witch is considered the restriction $B'$ of $B$ in $ker(A)$ and a selection of an eigenvector $v\in ker(A)$ of $B$ and the statment that $v$ is also a eigenvector of $A$. This proves the case 1.

Now, if $ker(A)\not\subset ker(C)$ then $A(x) = 0$ and $[A,B](x)\neq 0$ for some $x\in V$. Since $rank([A,B]) = 1$ then $Im([A,B]) = span(v)$, for some $v\in V$, where $v=[A,B](x)$, so that $y = AB(x) - BA(x) = AB(x) \in Im(A)$. It follows that $B(Im(A))\subseteq Im(A)$. Now, comes doubt four, that is similar to three: he takes the restrictions $A',B'$ of $A,B$ to $Im(A)$ and the states that $rank[A',B']\leq 1$ and therefor by the inductive hypothesis the operators $A'$ and $B'$ have a common eigenvector. And this proves the case 2, concluding the entire proof.

-Thanks