Determinant of a non-square matrix

Question

I wrote an answer to this question based on determinants, but subsequently deleted it because the OP is interested in non-square matrices, which effectively blocks the use of determinants and thereby undermined the entire answer. However, it can be salvaged if there exists a function $\det$ defined on all real-valued matrices (not just the square ones) having the following properties.

1. $\det$ is real-valued
2. $\det$ has its usual value for square matrices
3. $\det(AB)$ always equals $\det(A)\det(B)$ whenever the product $AB$ is defined.
4. $\det(A) \neq 0$ iff $\det(A^\top) \neq 0$

Does such a function exist?

Answer 1

Such a function cannot exist. Let $A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0\end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$. Then, since both $AB$ and $BA$ are square, if there existed a function $D$ with the properties 1-3 stated there would hold \begin{align} \begin{split} 1 &= \det \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \det(BA) = D(BA) = D(B)D(A) \\ &= D(A)D(B) = D(AB) = \det(AB) = \det \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} = 0. \end{split} \end{align}

Answer 2

This extension of determinants has all 4 properties if A is a square matrix (except it loses the sign), and retains some attributes of determinants otherwise.

If A has more rows than columns, then $$|A|^2=|A^{T}A|$$ If A has more columns than rows, then $$|A|^2=|AA^{T}|$$

This has a valid and useful geometric interpretation. If you have a transformation $A$, this can still return how the transformation will scale an input transformation, limited to the smaller of the dimensions of the input and output space.

You may take this to be the absolute value of the determinant. It's always positive because, when looking at a space embedded in a higher dimensional space, a positive area can become a negative area when looked at from behind.