An extension of the determinant to non square matrices I'm an undergraduate student in mathematics and today I've been asked the following question by a friend of mine: let $v, w$ be vectors in $\mathbb{R}^3 $and $ A=\left( \begin{array}{cc} v_1 \ v_2 \ v_3 \\ w_1 \ w_2 \ w_3 \end{array} \right)$, she asked for a geometrical interpretation of the fact that $\det (AA^T)=$area of the parallelogram enclosed between $v$ and $w$ squared. The algebra turns out to be correct (we proved it in this particular case), but I started wondering: is this true in general? Or, to put it in a better way: does it make sense to extend $\det$ to non square matrices as $\sqrt{\det(AA^T)}$?
My reasoning is the following: we know that $\det(AA^T)=\det(A)^2 \ \forall A \in \mathbb{R}^{n\times n}$. For now, let's just pretend that there exists a function $\operatorname{Area}: M(m, n, \mathbb{R}) \mapsto \mathbb{R}$ such that $Area(A) = \det(A)$ when $A$ is square and that behaves similarly to $\det$ (that is, is invariant under transposition, obeys Binet theorem ecc.). Thus, $\operatorname{Area}(M)^2=\operatorname{Area}(MM^T)=\det(MM^T) \  \forall M \in M(n, m, \mathbb{R})$. So such a function must be identically equal to $\pm\sqrt{\det(M M^T)}$ (note that $\det(MM^T)$ is always positive, giving a hint that this extension is very likely to be sensible). Now comes the question: does any of this actually make sense? I.e.:

*

*Unfortunately the definition is not so straight forward, in the sense that $\operatorname{Area}(A^T)=\sqrt{\det(A^TA)}$ which is generally different from $\sqrt{\det(AA^T)}$, so it looks like the definition might be flawed BUT I noticed (see edit) that one of the two is always zero, so $\operatorname{Area}$ could just be defined as the one that is not zero.

*Is it true that $\operatorname{Area}(A) \neq 0 \iff A$ has full rank?

*More specifically, does this actually still represent the area of the parallelogram (or hyperparallelogram in general) enclosed between the vectors $Ae_j$ where $\{e_j: 1\leq j \leq m\}$ is the standard basis of $\mathbb{R}^m$?

*If the previous point was true, how would one decide the sign of the result such that it still reflects the "flipping" somehow? Does "flipping" even make sense for a function $:\mathbb{R}^m \mapsto \mathbb{R}^n$?

Any comment is appreciated!
Edit: as was pointed out in the comments, if $m<n, A \in M(m, n, \mathbb{R})$ then $\det{A^TA}=0$, because $A^TA \in M(n, n, \mathbb{R})$ and $rank(A^TA) \leq rank(A) \leq m < n$ which means that $rank(A) < n$ and therefore $A^TA$ is not invertible, so $\det(A^TA)=0$.
 A: There are many wonderful comments already in this discussion and I wanted to add my own, but it turned out to be too long for a comment so I post it here.
The determinant may be thought of as fundamentally a measurement of volumes${}^*$. If I have a volume in $n$-dimensional space and a $n\times n$ square matrix $A$, and if I apply the linear transformation $A$ to the entire space (i. e., replace every element $x$ in the shape by $Ax$), then the volume of the new shape is $\mid\det A\mid$ times larger than the original volume, with the sign of the determinant indicating whether the transformation is orientation preserving or inverting.
If we are mapping from $\mathbb{R}^n$ to $\mathbb{R}^k$ for $k < n$, then since the two spaces under question are no longer the same, it doesn't make sense to ask about the orientation-preserving properties of the transformation. However, it still makes sense to ask about how the volume is changed by the transformation. For example, if I have a ball of radius $1$ centered at the origin of my initial space, then what happens when I map this ball into $\mathbb{R}^k$ by the transformation $A$? What is its shape and what is its volume?
Enter the singular value decomposition, which gives a very concrete answer to this question. The singular value decomposition states that the mapped ball is an ellipsoid in $k$ dimensions. Like an ellipse in two dimensions with a major and minor axis, an ellipsoid in $k$ dimensions has $k$ axes with lengths $\sigma_1\geqslant\sigma_2\geqslant\cdots\geqslant\sigma_k$. Remarkably, the singular value decomposition says that there are perpendicular unit vectors $u_1,\ldots,u_k$ in $\mathbb{R}^n$ such that $Au_1,\ldots,Au_k$ are the perpendicular axes of our ellipsoid. We can write these as $Au_j = \sigma_jv_j$, where the $v_j$'s are perpendicular unit vectors.
Pick additional orthonormal vectors $u_{k+1},\ldots,u_n$ to complete a perpendicular set of basis vectors for our original space. Now write down matrices $U$ and $V$ whose columns are the vectors $u_1,\ldots,u_n$ are $v_1,\ldots,v_k$ and define a diagonal rectangular matrix $\Sigma$ with diagonal entries $\sigma_1,\ldots,\sigma_k$. For example, if $n = 4$ and $k = 2$,
$$
\Sigma = \begin{bmatrix} \sigma_1 & 0 \\ 0 & \sigma_2 \\ 0 & 0 \\ 0 & 0 \end{bmatrix}.
$$
Then, this geometric idea that $A$ sends the unit ball to a certain ellipsoid can now be encoded in a matrix factorization $A = U\Sigma V^T$, where $U$ and $V$ are orthogonal matrices $U^TU = UU^T = I_n$ and $V^TV = VV^T = I_k$ and $\Sigma$ is an $n\times k$ diagonal matrix.
This is a powerful tool and you may encounter it later in your mathematical career or perhaps not at all. My explanation was a rather brief one, but hopefully it gets the idea across. The reason it is helpful in answering your question is the following

Theorem. Let $A$ be a $n\times k$ matrix for $k\le n$. The matrices $A^TA$ and $AA^T$ have the same nonzero eigenvalues, which are precisely the squares of the nonzero singular values of $A$.${}^\%$

In particular, if $k\le n$ and $A$ has $k$ nonzero singular values${}^\dagger$, then $\sqrt{\det(A^TA)}$ is the product of the singular values of $A$ and represents the amount the volume of the ellipsoid in $k$-dimensional space divided by the volume of the unit ball!
As suggested by a commenter, the most general concept of the volume or area of a linear transformation between two spaces is the product of all the nonzero singular values, which also is defined and nonzero for nonzero rectangular matrices of all sizes, even with $k > n$.
I hope this answer gives you a sense for "why" this seemingly-arbitrary formula you found does indeed naturally encode information about the "volume" or "area" of a linear transformation and introduces (or reminds) you of an interesting tool for further study in linear algebra.

${}^*$ By volume, I really mean hypervolume or (extra-technically) Lebesgue measure.
${}^\%$ This answers why one of $\det(A^TA)$ and $\det(AA^T)$ is zero for $k < n$: $AA^T$ must have $n - k$ zero eigenvalues and thus $\det(AA^T) = 0$.
${}^\dagger$ One can fairly straightforwardly show that an $n\times k$ matrix $A$ has $k$ nonzero singular values if, and only if, $A$ has full (column) rank ($k\le n$)
