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An equation related to the dimension of linear operator

  1. Dec 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Let V be a finite vector space, and A, B be any two linear operator. Prove that,
    rank A = rank B + dim(Im A [tex]\cap[/tex] Ker B)

    3. The attempt at a solution
    Since rank A = dim Im A
    dim(Im B)+ dim(Ker B)=dim V
    dim(Im A + Ker B)=dim(Im A)+dim(Ker B)-dim(Im A [tex]\cap[/tex] Ker B)
    It seems equivalent to prove that dim(Im A + Ker B)=dim V

    it is obvious that Im A + Ker B [tex]\subseteq[/tex] V. So it should be true that V [tex]\subseteq[/tex] Im A + Ker B.

    If I let Ker B=<e_1,...,e_k> and extend it to a basis of V=<e_1,..e_k, e_k+1,...e_n>. Choose any [tex]x=\sum^{n}_{i=1} x_{i}e_{i}=\sum^{k}_{i=1} x_{i}e_{i} + \sum^{n}_{i=k+1} x_{i}e_{i}[/tex]. The first term is in Ker B, the second is not in Ker B and therefore should be in Im A. This is what confuses me a lot since it is not necessary to be true IMO...

    What's your idea? Where I made the mistakes? Can you give me any hint to prove it? Thanks very much
  2. jcsd
  3. Dec 2, 2008 #2


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    Are A and B arbitrary operators on V? If so this isn't true unless V={0}. For a counterexample, take A=0 and let B be any nonzero operator.
  4. Dec 2, 2008 #3
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