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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...

Note 1: see the comments turned into cw answer

Note 2: I know rank 0 iff the zero matrix. I don't really see how this helps though. My concern is (mainly?) about how you say determinant of nothing is 1 or at least nonzero instead of 0. In this case, refer to Note 1.

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  • $\begingroup$ 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 $\det(A)=\operatorname{sgn}(\operatorname{id})\times1=1$. $\endgroup$
    – user1551
    Oct 23 '21 at 22:17
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If a matrix has rank 0, then it is the zero matrix.

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    $\begingroup$ i know. but anyway my question then is about the 0x0 minors of, well, i guess any matrix and not just the zero matrix. see my edit to OP please. $\endgroup$
    – BCLC
    Oct 23 '21 at 15:17

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