Why the matrix $\displaystyle AA^T$ is invertible, if the generic real matrix $\displaystyle A$ has full rank?

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- Jan 3rd 2011, 10:34 AMRieszInvertibility of the product matrix-transpose
Why the matrix $\displaystyle AA^T$ is invertible, if the generic real matrix $\displaystyle A$ has full rank?

- Jan 3rd 2011, 10:41 AMsnowtea
If a matrix has full rank, then its transpose also has full rank.

The product of two invertible matrices is invertible. - Jan 3rd 2011, 11:41 AMHallsofIvy
I don't think that quite answers the question! The OP didn't say they are invertible, just of "full rank". A square matrix is invertible if and and only if it has full rank but I imagine this problem is for non-square matrices.

If the matrix A is, say, m by n, then we can think of it as representing a linear transformation from a vector space of dimension n to a vector space of dimension m. If A has full rank, then m> n and A maps the vector space of dimension n one-to-one onto a subspace of dimension n. - Jan 3rd 2011, 12:33 PMDrexel28
Assuming that these are real matrices.

In general $\displaystyle \text{rk}\left(AA^{\top}\right)=\text{rk}\left(A\r ight)$. To prove this we first note that in general $\displaystyle \text{rk}(M)=\text{rk}\left(M^{\top}\right)$ so it suffices to prove that $\displaystyle \text{rk}\left(A^{\top}A\right)=\text{rk}\left(A\r ight)$. But, for this (by first isomorphism theorem [the rank-nullity theorem if that previous theorem is unfamiliar to you]) it is sufficent to prove that $\displaystyle \ker A^{\top}A=\ker A$. Evidently $\displaystyle \ker A\subseteq\ker A^{\top}A$ and the reverse inclusion follows by noticing that if $\displaystyle v\in\ker A^{\top}A$ then $\displaystyle 0=\left\langle A^{\top}Av,v\right\rangle=\left\langle Av,Av\right\rangle$ and thus $\displaystyle Av=\bold{0}$ and thus $\displaystyle v\in\ker A$. - Jan 3rd 2011, 01:18 PMRiesz
Sorry, I have to specify matrix dimensions:

$\displaystyle A \in \mathbb{R}^{nxm}$, $\displaystyle n \le m$.