For any [itex]A \in \mathcal{R}^{n \times m}[/itex], does [itex]A^T A[/itex] have an inverse?(adsbygoogle = window.adsbygoogle || []).push({});

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.

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

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