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Math Help - linear algebra - rank-nullity

  1. #1
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    linear algebra - rank-nullity

    Let V be an n-dimensional vector space where n\geq 1. If T: V\rightarrow Vis a linear transformation, prove that the following statements are equivalent:
    a) im(T)=ker(T)
    b) T^2 = 0, n is even and r(T)=\frac{n}{2}.

    So far I have managed show show that (a) implies (b), but when trying to find (b) implies (a) I get stuck. This is my working so far:

    dim(V) = r(T) + n(T)
    r(T) = \frac{n}{2}, dim(V) = n, so r(T)=n(T)=\frac{n}{2}

    T^2(V)=0, T(T(V))=0, so T(V) \epsilon ker(T)
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by worc3247 View Post
    Let V be an n-dimensional vector space where n\geq 1. If T: V\rightarrow Vis a linear transformation, prove that the following statements are equivalent:
    a) im(T)=ker(T)
    b) T^2 = 0, n is even and r(T)=\frac{n}{2}.

    So far I have managed show show that (a) implies (b), but when trying to find (b) implies (a) I get stuck. This is my working so far:

    dim(V) = r(T) + n(T)
    r(T) = \frac{n}{2}, dim(V) = n, so r(T)=n(T)=\frac{n}{2}

    T^2(V)=0, T(T(V))=0, so T(V) \epsilon ker(T)
    You're almost there. YOu know that since \dim \ker T=\frac{n}{2} and n is even that \dim T(V)=\frac{n}{2}. But, as you noticed since T^2=\bold{0} we have that T(V)\subseteq \ker T. But, since \dim T(V)=\dim \ker T this implies that T(V)=\ker T (a subspace of a finite dimensional space which has the same dimension of the ambient space must be the full space)
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