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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?

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  • $\begingroup$ What do you mean by "polynomial expression?" How is the determinant a polynomial expression? It looks like a sum of products to me. $\endgroup$ – DanielV Aug 19 '14 at 14:33
  • $\begingroup$ @DanielV, the determinant of an $n\times n$ (real, say) matrix is indeed a polynomial of order $n$ on $\mathbb R^{n\times n}$. If you fix $n^2-1$ elements and let one vary, the resulting expression is always a polynomial. $\endgroup$ – Joonas Ilmavirta Aug 19 '14 at 14:36
  • $\begingroup$ @DanielV, it is a multivariate polynomial of degree $n$ in $n^2$ variables (the matrix entries). The monomials are the products of "generalized diagonals" in the matrix and the coefficients are $\pm 1$. $\endgroup$ – Yoni Rozenshein Aug 19 '14 at 14:37
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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.

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  • $\begingroup$ This is an interesting approach! Can it be generalized to the complexes, or a general field? $\endgroup$ – Yoni Rozenshein Aug 19 '14 at 14:41
  • $\begingroup$ Answering myself: Consider complex $2 \times 1$ matrices. Only the zero matrix isn't full rank. So we need a polynomial $P(z, w) : \mathbb{C}^2 \to \mathbb{C}$ whose only zero is at $z=w=0$. This is impossible. $\endgroup$ – Yoni Rozenshein Aug 19 '14 at 14:45
  • $\begingroup$ @YoniRozenshein, it can be generalized to the complex case if you allow polynomials on $(A,A^*)$: just replace $\det(A_k)^2$ with $|\det(A_k)|^2$. I don't know how to extend this idea to the complex case without conjugation or to fields of finite characteristic. At least naively this seems to work only if there is reasonable notion of positivity in the field. Of course, if you allow polynomials with values in $\mathbb F^N$, take $D(A)=(\det(A_1),\dots,\det(A_N))$. $\endgroup$ – Joonas Ilmavirta Aug 19 '14 at 14:46
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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.

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  • $\begingroup$ Thank you, this is informative. In my use, I cannot use conjugation. However, I found that I can work with several polynomial expressions being $0$ simultaneously and not just one (I didn't write it in the original post because I didn't think about it when I posted it). So I ended up comparing the determinants of all $n \times n$ submatrices to zero, which works in any field, as detailed by Joonas's answer. $\endgroup$ – Yoni Rozenshein Aug 21 '14 at 11:38
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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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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

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  • $\begingroup$ @HectorBladin: Is there anyway to define a determinant which depends holomorphically on the coefficients ? $\endgroup$ – Curiosity Mar 7 '19 at 15:26
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See 1904.08097 for a review I authored of generalised determinant functions of tall matrices, and their properties -- this should provide a self-contained introduction to three different generalised determinants.

The function mentioned by Joonas Ilmavirta is the square of the "determinant-like function" that I first wrote about in 2013, albeit with an erroneous factor of $\sqrt{|m-n|!}$ at the front, which is corrected in the above review. It is also the norm-squared of the vector determinant, and the product of the singular values of the matrix.

If you want a non-trivial determinant for "wide matrices", i.e. flattenings, you will need to be a bit creative in the definition of the determinant, such as by defining it as the scaling of $m$-volumes where $m$ is the dimension of the flattened space.

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