I found this theorem on Prasolov's Problems and Theorems in Linear Algebra:(adsbygoogle = window.adsbygoogle || []).push({});

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

He gives this proof:

The proof will be carried out by induction on [itex]n=dim(V)[/itex]. He states that we can assume that [itex]ker(A)\neq \{0\}[/itex], otherwise we can replace [itex] A[/itex] by[itex] A - \lambda I[/itex]; doubt one: why can we assume that? For [itex]n=1[/itex] it's clear that the property holds, because [itex] V = span(v) [/itex] for some [itex]v[/itex]. Supposing that holds for some [itex]n[/itex]. Now he divides in to cases:

1.[itex]ker(A)\subseteq ker(C)[/itex]; and

2.[itex]ker(A)\not\subset ker(C)[/itex].

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

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

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

-Thanks

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# Rank of the commutator.

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