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Math Help - rank of f^2 = rank of f

  1. #1
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    rank of f^2 = rank of f

    {Let X be a finite dimensional vector space over F, and f:X\to X a linear transformation. If the rank of f^2=f\circ f is equal to the rank of f, prove that \text{ker} f \cap \text{Im}f=\{0\}

    Since \text{rank}f\circ f=\text{rank} f\Rightarrow \text{Im}f\circ f=\text{Im} f.

    ???????
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    MHF Contributor Drexel28's Avatar
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    Re: rank of f^2 = rank of f

    Quote Originally Posted by dwsmith View Post
    {Let X be a finite dimensional vector space over F, and f:X\to X a linear transformation. If the rank of f^2=f\circ f is equal to the rank of f, prove that \text{ker} f \cap \text{Im}f=\{0\}

    Since \text{rank}f\circ f=\text{rank} f\Rightarrow \text{Im}f\circ f=\text{Im} f.

    ???????
    Well, I think a direct method may perhaps be a better choice. Suppose that \text{span}(f(x))\in\ker f\cap\text{im }f for some non-zero [/tex]f(x)[/tex]. You can see then that f(f(x))=0 now try using this and the ideas behind why \text{rank}(f^2)\leqslant \text{rank}(f) to conclude that \text{rank}(f^2)\leqslant\text{rank}(f)-1.
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    Re: rank of f^2 = rank of f

    Quote Originally Posted by Drexel28 View Post
    Well, I think a direct method may perhaps be a better choice. Suppose that \text{span}(f(x))\in\ker f\cap\text{im }f for some non-zero [/tex]f(x)[/tex]. You can see then that f(f(x))=0 now try using this and the ideas behind why \text{rank}(f^2)\leqslant \text{rank}(f) to conclude that \text{rank}(f^2)\leqslant\text{rank}(f)-1.
    Why the span(f(x))??
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    MHF Contributor Drexel28's Avatar
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    Re: rank of f^2 = rank of f

    Quote Originally Posted by dwsmith View Post
    Why the span(f(x))??
    You want to show that \text{im }f^2 misses an entire dimension one subspace of [tex]\text{im }f
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    Re: rank of f^2 = rank of f

    Quote Originally Posted by Drexel28 View Post
    You want to show that \text{im }f^2 misses an entire dimension one subspace of [tex]\text{im }f
    I am still lost. Did you look at the pdf of the notes yesterday? Nothing is in there to help us (the students understand).
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    Re: rank of f^2 = rank of f

    apply the rank-nullity theorem to the 2nd f in fof, to get:

    dim(f(X)) = dim(f(f(X)) + dim(ker(f)∩f(X)).

    since dim(f(X)) = dim(f(f(X)), dim(ker(f)∩f(X)) = 0.
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