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If $m,n,r\in \Bbb N$ and $A$ is an $m\times n$ matrix satisfying $(AA^T)^r=I$ is it true that $m=n\implies A$ is invertible.
I think it's true since $\det (AA^T)^r=1\implies \det A^{2r}=1\implies (\det A)^{2r}=1\implies \det A\neq 0\implies A $ is invertible.
So the result is true. Is this correct?
Your proof is fine, but you can be more explicit by actually finding the inverse of $A$: as long as $r\ge1$ we have $$AA^T(AA^T)^{r-1}=I$$ and $A$ is square, so $$A^{-1}=A^T(AA^T)^{r-1}\ .$$
You can use contraposition method as well. First suppose $A$ is not invertible. Then either $\text{det}(A)=0$ or $A$ is not a square matrix (i.e. $m\ne n$). Since $(AA^T)^r=I$, it is not the case $\text{det}(A)=0$. Therefore $m\ne n$. Hence $m=n$ implies $A$ is invertible.
