If $\mathbf{AA}^T=\mathbf{I}$, is $\mathbf A$ necessarily square?

Question

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 $\textbf{A}^T\textbf{A}=\textbf{I}$?

Answer 1

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.

Answer 2

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 $m$).

Answer 3

$$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$