If $\mathbf{AA}^T=\mathbf{I}$, is $\mathbf A$ necessarily square? If $\mathbf{AA}^T=\mathbf{I}$, is $\mathbf A$ necessarily square?
I am starting to learn about matrices, and had the above question. When I have tried to think about this, I have not been able to progress using matrix multiplication, since $\textbf{A}$ and its transpose do not have inverses unless they are square. The only conclusion I could come to using matrix multiplication is that the product of a matrix and its transpose, whatever the dimensions, is square and symmetric.
I also tried to consider this component-wise; for a 1x3 case, it was easy to see that there are no solutions. But the algebra for a 2x3 case was quite messy because it involved 6 variables.
I am not sure how else to think about this. I have seen the proofs that a matrix must be square to have and inverse (here), but the answers all rely on the additional defining property of an inverse being that $\textbf{AA}^{-1}=\textbf{A}^{-1}\textbf{A}$, and if $\textbf{A}$ was not square, $\textbf{AA}^{-1}$ could theoretically be equal to $\textbf{I}$ but then it would not have the same dimensions as $\textbf{A}^{-1}\textbf{A}$, violating the above property. As a similar constraint is applied to orthogonal matrices, these qould also have to be square.
However is it possible for a non-square matrix to be such that  $\textbf{AA}^T=\textbf{I}$, whether or not $\textbf{A^TA}=\textbf{I}$, where the identity matrix here could be of a different dimension? If so, does $\textbf{AA}^T=\textbf{I}$ mandate that $\textbf{A}^T\textbf{A}=\textbf{I}$?
 A: This is definitely possible. For instance, take $A = (1,0,0)$. Then $A A^T = (1) = I$. In fact, $vv^T = I$ is true for any row vector $v$ such that $||v|| = 1$. 
Edit: As pointed out in the comments, this only works if $A$ is an $m \times n$ matrix where $m < n$. To see why, assume $m > n$. Then $A$ is an $m \times n$ matrix and $A^T$ is an $n \times m$ matrix, so the product $A A^T = I$ is an $m \times m$ matrix. Obviously, the rank of $I$ is $m$. But $r(A) \leq n < m$. Similarly, $r(A^T) < m$. So $r(A A^T) < m = r(I)$, which is a contradiction.
This also explains why if $A A^T = I$ and $A$ is not square, then $A^T A$ will NEVER be equal to the identity.
A: Take any square matrix $B$ such that $BB^T=I$. Consider the matrix $A$ obtained from $B$ by inserting as many zero columns to $B$ as you want. Then $AA^T=I$.
Trivial example: $B$ is the $2\times 2$ identity; consider
$$
A=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0
\end{bmatrix}
$$
Less trivial example:
$$
B=\begin{bmatrix}
1/\sqrt{2} & 1/\sqrt{2} \\ -1/\sqrt{2} & 1/\sqrt{2}
\end{bmatrix}
$$
and
$$
A=\begin{bmatrix}
0 & 1/\sqrt{2} & 0 & 1/\sqrt{2} & 0 & 0 \\
0 & -1/\sqrt{2} & 0 & 1/\sqrt{2} & 0 & 0
\end{bmatrix}
$$
However, if $A$ is not square, say $m\times n$ and $AA^T=I$, then the rank of $A$ is $m$ and $n>m$). Then $A^TA\ne I$, because the rank of $A^TA$ is at most $m$ (actually, exactly equal to $m$).
A: $$
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0
\end{pmatrix}
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
0 & 0
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
$$
