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Linear Algebra Proof, Is it correct?

  1. Jul 5, 2010 #1
    I would like to have someone go over my proof and see if its correct or at least on the right track. Here's the problem:

    Let V be a finite-dimensional vector space, and let T: V → V be linear. If rank(T) = rank(T²), prove that R(T) ∩ N(T) = {0}.

    PROOF:

    Lemma: Let V be a finite-dimensional vector space, and let T: V → V be linear. If rank(T) = rank(T²), then N(T) = N(T²).

    Proof: By the Nullity-Rank Theorem (i.e. Let V and W be vector spaces, and let T: V → W be linear. If V is finite-dimensional, then nullity(T) + ran(T) = dim(V)) I have,

    dim(V) = rank(T) + nullity(T)
    dim(V) = rank(T²) + nullity(T²)

    This implies that nullity(T) = nullity(T²). Furthermore, N(T) and N(T²) are both subspaces of V and as a matter of fact, they are both vector spaces. Now it is easily shown that N(T) ⊂ N(T²) and that N(T) is a subspace of N(T²). Therefore, by the Dimension Theorem (i.e. Let W be subspace of a finite-dimensional vector space V. Then W is finite-dimensional and dim(W) ≤ dim(V). Moreover, if dim(W) = dim(V), then V = W) we have N(T) = N(T²). □

    Seeking a contradiction, suppose that R(T) ∩ N(T) ≠ {0}. Therefore, there is an x ≠ 0 ∈ R(T) ∩ N(T). This implies that x ∈ R(T) and x ∈ N(T). Since x ∈ R(T) then T(y) = x for some y ∈ V. Note that T²(y) = T(T(y)) = T(x) = 0 which means that y ∈ N(T²) and hence by the lemma y ∈ N(T) also. However, this implies that T(y) = 0 and hence a contradiction since x = T(y) ≠ 0. ♦

    I'd appreciate if someone could look at it. Thanks a lot!!
     
  2. jcsd
  3. Jul 5, 2010 #2

    HallsofIvy

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    You've probably said more than is necessary but I find nothing wrong with it.
     
  4. Jul 5, 2010 #3
    lol you're the same person as on mathhelpforum...thanks again :)
     
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