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Does [itex]A^T A[/itex] have an inverse?

  1. Nov 19, 2012 #1
    For any [itex]A \in \mathcal{R}^{n \times m}[/itex], does [itex]A^T A[/itex] have an inverse?

    From the wikipedia article for transpose ( http://en.wikipedia.org/wiki/Transpose ), I found that [itex]A^T A[/itex] is positive semi-definite (which means for any [itex]x [/itex] which is a column vector, [itex]x^T A^T A x \ge 0[/itex] ). And the Wikipedia article for positive-definite matrix ( http://en.wikipedia.org/wiki/Positive_definite_matrix ) , (which means for all [itex]x[/itex] which is a non-zero column vector, [itex]x^T A^T A x \gt 0[/itex] ) says that for any positive definite [itex]A^T A[/itex], [itex]A^T A[/itex] is invertible.

    So for any [itex]A \in \mathcal{R}^{n \times m}[/itex], [itex]A^T A[/itex] has an inverse for the case when [itex]x^T A^T A x \gt 0[/itex] for any non-zero column vector [itex]x[/itex], but what about the case when [itex]x^T A^T A x = 0[/itex] ?

    Or is there any way that I can get a proof that [itex]A^T A[/itex] has an inverse?

    Thanks.
     
    Last edited: Nov 19, 2012
  2. jcsd
  3. Nov 19, 2012 #2

    D H

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    Think of an n×m matrix that consists entire of zeros.
     
  4. Nov 19, 2012 #3
    Oh, OK. So [itex]A^T A[/itex] cannot have an inverse when [itex]A[/itex] is a matrix of all zeros.

    What about the case when [itex]A[/itex] is a non-zero matrix?
     
  5. Nov 19, 2012 #4
    In the case of square matrices, if [itex]det A = 0[/itex] then [itex]A^TA[/itex] is singular.
     
  6. Nov 19, 2012 #5

    AlephZero

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    "Positive semi-definite" is not the same as "positive definite". Changing ##x^TA^TAx \ge 0## to ##x^TA^TAx \gt 0## makes a BIG difference here.
     
  7. Nov 19, 2012 #6

    D H

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    Think of an n×m matrix that consists entire of ones and m is greater than 1.

    As AlphaZero already noted, there's a huge difference between positive definite and positive semi-definite.
     
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