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Thread: This is tricky..Linear algebra again

  1. #1
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    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
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    Hi

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

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