rank $k$ implies nonzero minor of size $k \times k$ but for for the vacuous case of $k=0$

Note 0: I use 'minor' to refer to both sub-matrix and its determinant. So when you delete row k and column $$l$$, you get a minor matrix. Its determinant is a minor determinant.

Is this true for $$k=0$$ instead of $$k=1, ..., \min\{m,n\}$$?

Proposition: Let $$F$$ be a field. For any $$A \in F^{m \times n}$$ with $$1 \le rank(A)=k \le \min\{m,n\}$$, we have that some minor determinant $$\det(M_{(i,j)})$$ of a minor matrix of $$M_{(i,j)}$$, of size $$k \times k$$, is nonzero.

I was thinking vacuously yes, but I think it's still no because the empty sum is zero or something (unlike the empty product which is 1). But I think it should be yes because there's no such thing as a $$0 \times 0$$ matrix...

• The determinant of an empty matrix is usually defined to be $1$. In view of Leibniz's formula $\det(A)=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i\in\{1,2,\ldots,n\}}a_{i\sigma(i)}$, recall that a permutation is a bijective map on a set. Since there is exactly one bijective map (namely, the identity map) on the empty set, the sum in Leibniz's formula isn't empty and it contains exactly one summand. However, the product $\prod_{i\in\emptyset}a_{i\sigma(i)}$ is empty. Hence $\det(A)=\operatorname{sgn}(\operatorname{id})\times1=1$. Oct 23 '21 at 22:17