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Rank/Image/Kernel proof

  1. Jun 17, 2011 #1
    1. The problem statement, all variables and given/known data

    Let V be a vector space. Show that for every three linear operators
    A,B,C: V -> V, we have
    rk(ABC) =< rk(B)

    2. Relevant equations

    V = rk(A) + dimKer(A)

    rk(A) = dimIm(A)

    3. The attempt at a solution

    Im(ABC) = {ABC(v) | vEV}
    = {AB(C(v)) | vEV}

    So Im(ABC) is a subset of Im(AB)

    So dimIm(ABC) =< dimIm(AB)

    So rk(ABC) =< rk(AB)

    Im(AB) = {A(B(w)) | wEV}

    Ker(B) subset of ker(AB) because if Bx=0, then ABx = A0 = 0

    By the dimension formula, this leads to rk(B) >= rk(AB)

    So putting the two results together we get rk(ABC) =< rk(B)

    Is this correct? Thanks!
  2. jcsd
  3. Jun 17, 2011 #2
    Hi Maybe_Memorie! :smile:

    Your proof seems correct! But it seems you'll need V to be finite dimensional for it to work.
  4. Jun 17, 2011 #3
    We've only dealt with finite dimensional vector spaces.
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