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

Question

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?

Answer 1

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.