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 $A$ 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 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 $A$ by$A - \lambda I$; 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 $A',B'$ 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

$[A,B](V)$ 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!