Does the Rank of a Commutator Determine Common Eigenvectors?

[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 byA - \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 into 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 statement 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

 

[fresh_42]

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 byA - \lambda I; doubt one: why can we assume that?

We need a common eigenvector, not necessarily to the same eigenvalues. If $A$ is injective and $\lambda$ an eigenvalue of $A$, which always exists over $\mathbb{C}$, then
$$
\operatorname{rank}([A-\lambda I, B])=\operatorname{rank}([A,B]) \leq 1
$$
and the condition still holds, and $A-\lambda I$ is not injective. Now let's assume we found an eigenvector $v \in \mathbb{C}^n$ for both, i.e. $(A-\lambda I)(v)=\nu v\, , \, B(v)=\mu v$. Then $A(v)=(\nu+\lambda)v$ and $v$ is an eigenvector for $A$, too. Hence we may assume that $A$ is not injective, since a result in this case delivers a result for the injective case, too.

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 into 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?

What is $C$? I assume it stands for $C=[A,B]$?

Anyway. The answer is no, since we can always distinguish these two cases. There is simply no third possibility, regardless of how $C$ is defined.

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 statement that v is also a eigenvector of A. This proves the case 1.

Yes. But where is the problem?

Say $v\in\ker A \subseteq \ker [A,B]$ by assumption of case 1. Thus $AB(v)=[A,B](v)+B(A(v)) = 0+0=0$ and $B$ can be restricted to $B':=\left.B\right|_{\ker A}\in \mathcal{L}(\ker A) \neq \{\,0\,\}$, and $\dim (\ker A)<n$, because $\dim (\ker A)=n$ means $A=0$ and all vectors are eigenvectors to the eigenvalue $0$, i.e. any eigenvector of $B$ will do. Hence we have a vector $v\in \ker A$ by induction, which is an eigenvector of $B'$ and and an eigenvector of $A$: $B'(v)=B(v)=\mu v$ and $A(v)=0\cdot v=0$. Remains to check the rank condition for the induction step, but
$$
\left. \operatorname{rank} [A,B']\right|_{\ker A}=\left.AB'\right|_{\ker A}=\left.AB\right|_{\ker A}= \left. \operatorname{rank} [A,B]\right|_{\ker A} \leq \operatorname{rank} [A,B] \leq 1
$$

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).

$y=v :=[A,B](x)\, , \,x \in \ker A - \{\,0\,\}$

Now, comes doubt four, that is similar to three: he takes the restrictions A&#039;,B&#039; of A,B to Im(A) and the states that rank[A&#039;,B&#039;]\leq 1 and therefor by the inductive hypothesis the operators A&#039; and B&#039; have a common eigenvector. And this proves the case 2, concluding the entire proof.

-Thanks

$[A,B](V)$ is at most one dimensional, and $[A,B](x)\neq 0$. Hence
$$
[A,B](V)=[A,B](x)=A(B(x)) = v = [\left.A\right|_{\mathbb{C}x},\left.B\right|_{\mathbb{C}x}] \neq 0
$$
and $\operatorname{rank}[A',B'] = 1$ so the induction hypothesis applies. Since we only restricted the transformations to invariant subspaces, all vectors, including the solution eigenvector are still vectors in $V$. Note that we remained in the non injective case for $A$ the entire proof!