Determinant-like expression for non-square matrices

Question

I'm interested in whether for any real matrix of size $m \times n$ there is a real number with the following properties:

1. It is a polynomial expression with real coefficients in the entries of the matrix. The expression depends on $m,n$ only.
2. 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?

Answer 1

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.

Answer 2

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

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Answer 3

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.

Answer 4

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