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Prove that asquare matrix A is invertible if nad only if A[sup]T[/sup]A is invertible

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that asquare matrix A is invertible if nad only if ATA is invertible


    2. Relevant equations
    Hmm, can't think of any. A A-1 = I maybe?


    3. The attempt at a solution

    I have trouble with these theory questions, so I'm not sure how to approach this.

    If something is invertible, that means it's determinant is not 0? So does it have something to do with that?

    Thanks :eek:
     
  2. jcsd
  3. Oct 22, 2009 #2

    Hurkyl

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    Re: Prove that asquare matrix A is invertible if nad only if ATA is invert

    If you aren't good at "theory" questions, then try rephrasing them as "calculation" questions!

    And vice versa as well, right? If the determinant is not zero, then it's invertible. So we have an identity:
    {X is invertible} = {det X is not zero}​

    What happens if you substitute this into the original question? (There are two things to substitute! Do both!) Is it something that you can solve by doing two calculations?
     
  4. Oct 22, 2009 #3
    Re: Prove that asquare matrix A is invertible if nad only if ATA is invert

    A square matrix A is invertible if and only if ATA is invertible

    Well, If A is invertible, then det(A) is not 0.

    det(A) = det(AT)

    The poduct of two invertible matrices is invertible

    So therefore ATA is invertible if and only if A is invertible.

    This doesn't seem like proof, nor does it seem coherent :( I suck
     
  5. Oct 22, 2009 #4
    Re: Prove that asquare matrix A is invertible if nad only if ATA is invert

    You have the right idea but try to show both directions separately.
    If A is invertible, then det(A) is non zero. So what can you say about det(AAT)?
    Similarly, if det(AAT) is non zero, what can you say about det(A)?
    (Also remember that det(AB) = det(A)det(B))
     
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