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Math Help - Linear Algebra Proof, Am I on the right track?

  1. #1
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    Linear Algebra Proof, Am I on the right track?

    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(T), 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(T), then N(T) = N(T).

    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(T) + nullity(T)

    This implies that nullity(T) = nullity(T). Furthermore, N(T) and N(T) 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(T) and that N(T) is a subspace of N(T). 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(T). □

    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 T(y) = T(T(y)) = T(x) = 0 which means that y ∈ N(T) 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!!
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  2. #2
    MHF Contributor

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    I suspect you are saying more than you need to but what you say is correct.
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  3. #3
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    Usually when I ask such questions on the internet, I make sure the reader knows exactly what I'm referring to so I do tend to make things longer. But I'd be a lot shorter if this were being handed in for marking. I'm just studying linear algebra on my own. Thanks for your help
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