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.