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Math Help - Nonsingular Matrix Proof.

  1. #1
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    Nonsingular Matrix Proof.

    The Proof:
    a. Show that if A(nxn) is a nonsingular matrix, so is A^2.
    b. Generalize to n: Show that if A is a nonsingular matrix, so is A^n.

    I understand that a nonsingular matrix is one that has an inverse (in our course we have not yet talked about determinants so please do not use those in responses), but I cannot think how I would generally any matrix A to always be nonsingular. If I could figure this out, I think I could prove how A^2 or A^n would follow as being nonsingular as well. Any help on this proof you could offer me is greatly appreciated! Thanks!
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Perhaps (perhaps) the definition you have covered is: A\in \mathbb{K}^{n\times n} is non singular iff Ax=0 implies x=0\;\;(x\in\mathbb{K}^{n\times 1}) .

    If so, suppose A is non singular then,

    A^2x=0\Rightarrow A(Ax)=0\Rightarrow Ax=0\Rightarrow x=0

    which implies A^2 is non singular.

    Try the second part.


    Fernando Revilla
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    I'm not sure I'm understanding your definition of nonsingularity. My professor taught us that if you can find an inverse of a matrix, then it is nonsingular. This is not necessarily when talking about systems of equations which is what I'm getting from your definition, talking about x=0. Why would A or x have to equal 0 to be nonsingular?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by tommiegirl13 View Post
    My professor taught us that if you can find an inverse of a matrix, then it is nonsingular.

    For that reason I said perhaps . So, you have covered: A non singular iff there exists B such that AB=BA=I . Suppose A non singular then,

    A^2B^2=A(AB)B=AIB=AB=I

    B^2A^2=B(BA)A=BIA=BA=I

    which implies A^2 is non singular.


    Fernando Revilla
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  5. #5
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    Thank you so much!! I get it now. You really made it click for me!
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