Why is the set of all full rank $\mathbb{R}^{m \times n}$ (rectangular) matrices a Zariski open set?

Note: this question is almost certainly a duplicate. I have spent more than an hour on Google and searching for the original question, but can't find it. Please feel free to close this question when you find the original question as long as you link to the original question.

Question: Why is the set of all full rank $$m \times n$$ matrices, with $$m\not=n$$, a Zariski open set?

Equivalent Question: Why does the set of all rank-deficient $$m\times n$$ matrices (with $$m \not=n$$) the zero set/locus of a polynomial equation (i.e. Zariski closed)?

Note: these two other questions (1)(2) are about the Euclidean topology, so don't answer the question.

Similarly, this question is about square matrices, so doesn't answer the question. (Clearly singular matrices are the zero set of the determinant polynomial.)

This question suggests using the sum of the absolute values of the determinants of the $$p \times p$$ submatrices (where $$p = \min \{m,n\}$$), however this is not a polynomial, so can't be used to show that the set of rank deficient matrices is Zariski closed.

What I had been thinking was using the product of the determinants of the $$p \times p$$ submatrices as the polynomial of which rank-deficient matrices would be the zero set. However, that does not make sense, since it is full-rank if and only if at least one of those determinants is non-zero, not if and only if at least of those determinants is zero.

Slide 8 here says that this set is Zariski open, although it does not seem to explain why. I believed for some reason that it is true, but when I tried to show it, I realized I don't know why it's true (if it is).

• This question also seems relevant: math.stackexchange.com/questions/1074452/… -- but it's only about Lebesgue measure zero, not Zariski closed set. Commented Mar 6, 2020 at 4:59
• $\det(A^\top A)$ (if $m\ge n$) or $\det(A A^\top)$ (if $m< n$) is a polynomial thus its vanishing set is closed thus the set where it doesn't vanish is open. Commented Mar 6, 2020 at 7:53
• @reuns This also a really helpful way of thinking about it. To confirm my understanding of your comment, you are using the fact mentioned e.g. in this question: math.stackexchange.com/questions/349738/… (that the Gramian of a matrix always has the same rank as the original matrix)? So the idea is to just choose the Gramian which would therefore invertible (if the original matrix is full rank), and use its determinant. This is similar to the distinctions between "thin" and "fat" SVDs, correct? Commented Mar 6, 2020 at 23:37

A matrix is of rank at least $$k$$ if and only if it has a $$k\times k$$ minor with nonvanishing determinant. This corresponds to those $$k$$ columns included in the minor being linearly independent in the image. The condition that a specific $$k\times k$$ minor has nonvanishing determinant is a Zariski-open condition by the linked question about square matrices (it's equivalent to the determinant, a polynomial in the entries, of the minor being nonzero), and asking about whether there exists any such $$k\times k$$ minor is a union of sets of this form and is thus again an open set. So the condition that a matrix is of rank at least $$k$$ is Zariski-open.
As a rectangular matrix has rank at most $$\min(m,n)$$, we see that the set of rank $$\min(m,n)+1$$ matrices is empty, so the set of matrices of full rank is just the set of matrices with rank at least $$\max(m,n)$$, which is Zariski-open by the above paragraph, so we're done.