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Inner products

  1. Jul 11, 2005 #1
    Suppose that A is a square. Show that A is invertible if and only if A^T*A is invertible.

    I know that if A is an m X n matrix, m>=n, and rkA=n, then the n X n matrix A^T*A is invertible, and that rk(A^TA)= rkA, but I'm still not sure how to start the proof...

    TIA
     
  2. jcsd
  3. Jul 11, 2005 #2

    robphy

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    Can you use the determinant?
     
  4. Jul 11, 2005 #3
    Yes, but how does that help me?
     
  5. Jul 11, 2005 #4

    matt grime

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    M is invertible if and only if det(M) is not zero.

    det(M) is the same as det(M^t)

    det(MN)=det(M)det(N)

    if x and y are real (or complex) numbers and xy=0 then one of x or y is zero.
     
  6. Jul 11, 2005 #5
    det(A) = det(A^t) and det(AB) = det(A)det(B) (for all matrices A, B of the proper size).

    If A is invertible, then det(A) != 0, so that det(A^t) != 0, and therefore det(AA^t) = det(A)det(A^t) != 0, i.e. AA^t is invertible.

    The converse is similar.
     
  7. Jul 11, 2005 #6

    matt grime

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    snap. why was the title inner products though?
     
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