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Matrix Proof

  1. Sep 24, 2012 #1
    Show that if A is invertible, then At is invertible
    and (At )-1
    = (A-1)t .


    If A is invertible, then det(A)≠0
    det(A)=det(A^t)
    thus det(A^t)≠0



    I'm not really sure how to prove the second part. It's an identity that I remembered, but don't know how to prove.

    I'll take a crack at it though:

    Multiply A^t to both sides
    gives I=A^t (A^-1)^t
    For the RHS, can i multiply A^t to A^-1 or does the ^t prevent me from doing that?
     
  2. jcsd
  3. Sep 24, 2012 #2

    jbunniii

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    If A is invertible, then there exists a matrix B such that AB = I and BA = I. What is the natural thing to do here, to both sides of each equation?
     
  4. Sep 24, 2012 #3
    I'm not sure what you are asking.

    (At )-1 = (A-1)t .

    If AB=I, then obviously, A is the inverse of B or vise versa.

    So In this case I let A=(A^t)^-1
    Thus A^=((A^t)^-1)^-1 ? <<<<multiply this to both sides? If so, the left said reduces to the I matrix, but the right side still looks complicated.
     
    Last edited: Sep 24, 2012
  5. Sep 24, 2012 #4

    jbunniii

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    If A is invertible, then there exists a matrix B such that AB = I and BA = I. Try transposing both sides of each of these equations.
     
  6. Sep 24, 2012 #5

    Dick

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    This can be a little hard to see. You already have I=A^t (A^-1)^t. This tells you that A^t has an inverse. What is it? Now take a deep breath. The inverse of A^t is called (A^t)^(-1).

    Now that I look back I'm not sure you even got that far. Take junniii's advice. Find the transpose of A^(-1)A=I.
     
    Last edited: Sep 24, 2012
  7. Sep 24, 2012 #6
    OH!

    So in this form

    I=A^t [(A^-1)^t]

    To get to the identity matrix A^T has to be multiplied by its inverse (A^T)^-1. The quantity [(A^-1)^t] must equal (A^T)^-1.


    So simple, yet hard.
     
  8. Sep 24, 2012 #7

    Dick

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    Yes, really simple. But it can be completely opaque if you don't remember what inverse means and that inverses are unique.
     
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