# Rank of the commutator of linear transformations.

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 $rank([A,B])\leq 1$ (Where $[\cdot,\cdot]:\mathcal{L}(V)\times \mathcal{L}(V)\to \mathcal{L}(V)$ is the commutator, defined as $[A,B]\ := AB - BA$). Then $A$ and $B$ have 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 $n\in \mathbb{N}$, 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 or $rank([A,B])= 1$ 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 $B$ to $ker(A)$ 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. I didn't understand why he can take these restrictions and conclude that they have the same eigenvector.

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Fëanor, if I understood your flag correctly, you were asking to have the edit you made to your question undone. Please let me know if you intended something else. – Zev Chonoles Apr 19 '12 at 16:26
@ZevChonoles It's all. Thanks. – Paulo Henrique Apr 19 '12 at 18:59

Doubt 1: since $\mathbb{C}$ is algebraically closed, $A$ possesses an eigenvalue $\lambda$ and thus the kernel of $A-\lambda I$ is not $0$. Now if $v$ is a common eigenvector of $A-\lambda I$ and $B$, then $\mu v=(A-\lambda I)v=Av-\lambda v$, hence $v$ is a common eigenvector of $A$ and $B$.
Doubt 3: here the induction hypothesis can be applied, because it was shown that $B$ is a linear map of $\mathrm{ker}(A)$ into itself, and $\mathrm{ker}(A)$ has a dimension smaller than $n$ (otherwise $A=0$ in which case there is nothing to prove). Moreover the rank of the commutator of $A|_{\mathrm{ker}(A)}=0$ and $B^\prime$ is $0$.
Doubt 4: the restrictions $A^\prime$ and $B^\prime$ are linear maps of $\mathrm{Im}(A)$ into $\mathrm{Im}(A)$, because for $A$ this is trivial and for $B$ this has been shown. The dimension of $\mathrm{Im}(A)$ is smaller than $n$ because by assumption $\mathrm{ker}(A)\neq 0$. Hence again the induction hypothesis can be invoked to conclude that a common eigenvector exists.
The reason for distinguishing the two cases is that the induction hypothesis comes in in two different ways: once via restriction to the kernel, the other time via restriction to the image. One cannot use the kernel of $A$ in the second case, nor can one use the image of $A$ in the first case. – Hagen Knaf Apr 17 '12 at 19:11