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Operator powers

  1. May 10, 2012 #1
    Consider a linear operator τ:V→V (where V is finite-dimensional) such that rk τ2=rk τ. Show that (im τ) [itex]\cap[/itex] (ker τ) is the zero space.

    Here's where I am:

    Its easy to see that im τ=im τ2, since it is a subspace with the same dimension. I also know that if (im τ) [itex]\cap[/itex] (ker τ) contains a nonzero vector, then it has positive dimension.

    My intuition is that I can use that positive dimension and the rank plus nullity theorem to show a contradiction, but I just can't seem to figure out how. Rank-plus-nullity gives me null τ = null τ2, and I know that dim ((im τ) [itex]\cap[/itex] (ker τ)) ≤ null τ. Any idea where to go next?
  2. jcsd
  3. May 11, 2012 #2
    I think I got it:

    Let σ denote the restriction of τ to its own image, and consider it as a function im τ→im τ2.

    It's easy to see that this function is surjective. Since it is a surjective map between finite-dimensional spaces of equal dimension, it is also injective, so ker σ = {0}.

    However, ker σ = ker τ [itex]\cap[/itex] im τ, so this completes the proof.
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