If $(AA^t)^r=I$ then $A^tA$ is invertible? Let $m,n,r$ be natural numbers. Let $A$ be an $m\times n$ matrix with real entries such that $(AA^t)^r=I$, where $I$ is the $m\times m$ identity matrix and $A^t$ is the transpose of the matrix $A$. We can conclude that 


*

*$m=n$

*$AA^t$ is invertible

*$A^tA$ is invertible

*If $m=n$, then $A$ is invertible


If $r=1$, $AA^t=I$ hence $\vert{AA^t}\vert=\vert{I}\vert=1$ hence option 2 correct
If $m=n$, $\vert{AA^t}\vert=\vert{A}\vert\vert{A^t}\vert=1$ hence $A$ is invertible hence option 4 correct
Let $$A=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
\end{bmatrix}
$$
then $AA^t=I$ but $A$ is not square matrix hence option 1 wrong
how to conclude for option 3
 A: Let 
$$
A=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
\end{bmatrix}
$$
then $A^{t}A = I$ but $m \neq n$, option 1 is wrong.
and 
$$
AA^{t}=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}
\neq I
$$
thus, option 3 is wrong.
$(AA^{t})^r=I$, then $\det (AA^{t})^r = 1$, it implies that $\det(AA^t) \neq 0$, thus option 2 is correct.
if $n=m$, then $0 \neq \det(AA^{t}) =\det(A)^2$, thus $\det(A) \neq 0$, option 4 is correct.
However, if $n=m$, then we can conclude that $A$, $A^{t}A$ and $AA^{t}$ are invertible.
A: Your proof that conclusion 1. holds is wrong; you can't suppose that $r=1$. What happens is that$$(AA^t)^r=\operatorname{Id}\Longrightarrow\det\bigl((AA^t)^r\bigr)=1\Longleftrightarrow\det(AA^t)^r=1$$and therefore $\det(AA^t)\neq0$. Therefore $AA^t$ is invertible.
If you don't want to use determinants, you can argue like this: if $AA^t$ wasn't invertible, there would be a vector $v$ such that $(AA^t).v=0$ But then $(AA^t)^r$ would be $0$ too.
A: I assume that $m,n,r\ge1$.
Set $B=AA^t$. Since $B^r=I$, we have $B^{r-1}B=BB^{r-1}=I$, so $B$ is invertible and $B^{-1}=B^{r-1}$.
Now note that $AA^t$ and $A^tA$ have the same rank as $A$ (it's a very interesting exercise, use the rank-nullity theorem). Therefore


*

*if $m=n$, then $A$ is invertible, so also $A^tA$ is invertible;

*if $m\ne n$, then $m<n$ and $A^tA$ is not invertible.


A simple example with $m\ne n$ is
$$
A=\begin{bmatrix} 1 \\ 0 \end{bmatrix}
$$
