Determinant-like expression for non-square matrices I'm interested in whether for any real matrix of size $m \times n$ there is a real number with the following properties:


*

*It is a polynomial expression with real coefficients in the entries of the matrix. The expression depends on $m,n$ only.

*It is zero precisely when the matrix is not of full rank ($\min\left\{m,n\right\}$).


For square matrices, the determinant has these properties.
If this is a known thing, what is it called and where can I read about it?
 A: There is such a thing, at least over the reals.
Suppose $m>n$.
Then an $m\times n$ matrix has full rank if and only if it contains an $n\times n$ submatrix of full rank.
Let $A$ be an $m\times n$ matrix and let $A_1,\dots,A_N$ be its $n\times n$ submatrices.
(The exact value of the number $N$ is irrelevant here; it only depends on $m$ and $n$.)
Now let $D(A)=\sum_{k=1}^N\det(A_k)^2$.
Clearly $D(A)$ is polynomial in each element since the determinant is, and $D(A)=0$ if and only if none of the $n\times n$ submatrices of $A$ has full rank.
I don't know if such things have been studied or given a name.
A: For an $n\times m$ real or complex matrix $A$ with $n\le m$, the matrix has full rank if and only if $\det(A\cdot A^*)\ne 0$. It is a simple corollary of the Cauchy-Binet formula that 
$$
\det(A\cdot A^*) = \sum_{1\le i_1<\cdots<i_n\le m} 
\left| \det\begin{pmatrix}
a_{1,i_1} & \dots & a_{1,i_n} \\
\vdots & \ddots & \vdots \\
a_{n,i_1} & \dots & a_{n,i_n} \\ \end{pmatrix}
\right|^2.
$$
This also provides an efficient method to compute the sum mentioned by Joonas Ilmavirta above.
A: You can look at Radić M.:
A Definition of Determinant of Rectangular Matrix
. Glas. Mat.
1
(21) (1966),
17–22.
Zbl 0168.02703, MR 0209303 , Google Scholar

A: For a real matrix $A$ of size $m\times n$ you have two cases:
1) If $m\leq n$, then you consider the matrix $A^{t}A$ which is square. In this case you have the following
$$ A \ \text{is of full row rank} \ \Longleftrightarrow \ \det(A^{t}A)\neq 0.$$
2) If $m\geq n$, then you consider the matrix $AA^{t}$ which is square. In this case you have the following
$$ A \ \text{is of full column rank} \ \Longleftrightarrow \ \det(AA^{t})\neq 0.$$
So the number is $A$ is a $m\times n$ rectangular matrix with $m\leq n$, then the number $\det(A^{t}A)$ plays the role of the usual determinant of a square matrix. Conversely, if $m\geq n$, then the number $\det(AA^{t})$ acts like the usual determinant of a square matrix. In the complex case as it was explained in a previous answer you replace $A^{t}$ by $A^{*}$. More details in 
http://www.seas.ucla.edu/~vandenbe/133A/lectures/inverses.pdf
