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

[omoplata]

For any A \in \mathcal{R}^{n \times m}, does A^T A have an inverse?

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

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

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

Thanks.

 

[D H]

Think of an n×m matrix that consists entire of zeros.

 

[omoplata]

Oh, OK. So A^T A cannot have an inverse when A is a matrix of all zeros.

What about the case when A is a non-zero matrix?

 

[Dead Boss]

In the case of square matrices, if det A = 0 then A^TA is singular.

 

[AlephZero]

"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.

 

[D H]

What about the case when A is a non-zero matrix?

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.