Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rank of the commutator.

  1. Apr 16, 2012 #1
    I found this theorem on Prasolov's Problems and Theorems in Linear Algebra:

    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
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Rank of the commutator.
  1. Rank and Trace (Replies: 3)

  2. Rank of a matrix (Replies: 3)

  3. Rank of a matrix (Replies: 1)

Loading...