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Thread: This is tricky..Linear algebra again

  1. #1
    Apr 2008

    This is tricky..Linear algebra again

    Let A be an n x n matrix with the property that the intersection of its column space and its null space is the 0 vector. That is: nullA(intersection)colA={0}. Show that A and A^2 have the same rank........
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  2. #2
    Super Member flyingsquirrel's Avatar
    Apr 2008

    First, you can show that \ker A \oplus \mathrm{im}\, A = E, which might be useful. (rank-nullity theorem)
    Then, showing that A and A^2 share the same rank is equivalent to proving that \dim \ker A = \dim \ker A^2 (rank-nullity theorem again) Here, we are in a special case : as \ker A\subset \ker A^2, (find a proof ) you can try to show that \ker A\supset \ker A^2 which will give you \ker A = \ker A^2 and the equality of the dimensions.

    For \ker A\supset \ker A^2, you might take X\in E which can be decomposed as X=X_k+X_i with X_k\in\ker A and X_I\in \mathrm{im}\,A (because \ker A \oplus \mathrm{im}\, A = E) and develop A^2X=\ldots
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