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Thread: kernel: proof

  1. September 28th 2009, 03:46 PM #1
    altave86
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    kernel: proof

    prove Null(A)\subset Null(A^2)

    I just don't know how to tackle this problem without making a computational mess. Can anyone tell me what's the trick? please
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  2. September 28th 2009, 04:20 PM #2
    Bruno J.
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    Well if x \in \mbox{Nullspace }(A) , then Ax=0, so also A^2x = A(Ax)=A(0)=0, so x \in \mbox{Nullspace }(A^2) . Therefore \mbox{Nullspace }(A) \subset \mbox{Nullspace }(A^2) .
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