Q:

If $ker(A)=\{0v\}$ then $AA^{T}$ is invertible?

Let $A\in M_m,_n$ where $m>n$ and if $ker(A)=\{0v\}$ then $AA^{T}$ is invertible?
I think it is not. Because $dimker(A)+dimIm(A)=n$, $dimker(A)=0$, then $dimIm(A)=n$, but then $dimIm(AA^{T})=n$. Now we have $dimker(AA^{T})+dimIm(AA^{T})=m$, then $dimker(AA^{T})=m-n$. Because $dimker(AA^{T})\not =0$ then $AA^{T}$ is not invertible. Is this ok?

A:

Counterexample. Let
$$
A 
= 
\begin{pmatrix}
1 \\
1
\end{pmatrix} \ .
$$
Then $\mathrm{ker} A = {0}$ (btw, I am not sure why you need that "v" in $0v$), but 
$$
AA^t 
= 
\begin{pmatrix}
1 \\
1
\end{pmatrix} \ 
\begin{pmatrix}
1  &  1
\end{pmatrix}
=
\begin{pmatrix}
1  &  1  \\
1  &  1
\end{pmatrix}  \ .
$$
Which, as you can easily check, has $\mathrm{ker} (AA^t) \neq 0 $, so it's not invertible.

