A be a matrix satisfying ${(AA^t)}^r=I$ then when it is invertible This question was asked in my linear algebra mid term and I was unable to solve it.

Let m ,n ,r be natural numbers . Let A be an m*n matrix with real entries such that ${(AA^t)}^r =I $ where I is m * m identity matrix. Then which  of following is / are true:

A m =n
B $ AA^{t}$ is invertible
C $A^{t} A $ is invertible
D if m=n , then A is invertible
I am really confused in this question due to $A A^{t}$a nd was unable to think which option are true or any contradictions in options . It's my humble request to guide you through this question.
I followed Hoffman and Kunze Linear algebra.
Thanks!!
 A: Correction: as it is written, $I$ must be an $m \times m$ matrix.

$M = A\,A^T$ is an $m\times m$ symmetric matrix.
Since it is symmetric, it is diagonalizable, with orthonormal eigenvectors: $M = V\,D\,V^T$
$$M^r = V\,D^r\,V^T = I \Rightarrow D^r = V^T\,V = I$$
Therefore, the eigenvalues of $M$ will all be $\pm 1$, which means $M$ is invertible.

A very simple example that shows that $m$ may be different than $n$; and that $A^T\,A$ may be singular is:
$$A = \begin{bmatrix}1&0\end{bmatrix}$$

If $m=n$ and $B = (A\,A^T)^{-1}$
$$(A\,A^T)\,B = I \Rightarrow A\,(A^T\,B) = I$$
If a square matrix has a right inverse then it has a left inverse (and  they are the same). So $A$ must be invertible and $A^{-1} = A^T\,B$.

$\boxed{\text{Answer: A=False; B=True; C=False; D=True}}$
A: The question has already been answered but you can also prove (B) and (D) in a different way as follows: 

Let $(AA^T)^r=I$. If $r=1$ then $(AA^T)=I$ , since $I$ is invertible so $(AA^T)$ is invertible. 
If $r>1$ then $(AA^T)(AA^T)^{r-1}=I$. So $(AA^T)^{r-1}$  is the inverse of $(AA^T)$ and hence $(AA^T)$ is invertible. This proves (B). 

Now suppose that $n=m$ then both $A$ and $A^T$ are $n \times n$ matrices.
Again if $r=1$ then $AA^T=I$ , so $A^T$ is the inverse of $A$ and hence $A$ is invertible.
Now suppose $r>1$. Since $(AA^T)(AA^T)^{r-1}=I$ $\Rightarrow$ $AA^T(AA^T)^{r-1}=I$ . 
Take $B=A^T(AA^T)^{r-1}$ then $B$ is also an $n \times n$ matrix and it satisfies $AB=I$ so $B$ is the inverse of $A$ and hence $A$ is invertible. This proves (D).
