Let a matrix $A \in \mathbb{R}^{m \times n}$ be given, with $m \geq n$ and $\mathrm{rank}(A) = n$. We know that there exists at least one non-vanishing minor of order $n$.
Now, let $A$ be appended with a matrix $B \in \mathbb{R}^{p \times n}$, getting $\left[ \begin{array}{c} A\\ B \end{array}\right] \in \mathbb{R}^{(m+p) \times n}$, with $\mathrm{rank}(B) < n \leq p$. Does the number of non-vanishing minors of order $n$ possibly increase?
Best Answer
Yes, it's possible. To see this, let's pick an extreme example: $A$ is the identity matrix appended vertically to a zero matrix. Hence there is exatly one nonzero minor of order $n$.
You then append vertically a matrix $B$ that is made of one column vector of ones and all the other column vectors are null. $B$ has rank $1<n$ as required, as long as $n\ge2$.
Then, by taking the rows $2\dots n$ together with any of the $p$ last rows, you get a nonzero minor (the submatrix is the identity, apart from a permutation of the rows).
More generally, with any $A$ and $B$ satisfying the conditions in the question, any nonzero row $u$ of $B$ will add at least one more nonzero minor: you can always remove one row of the known nonzero minor and replace it with $u$, to get a nonzero minor. But not always an arbitrary one: $u$ is a linear combination of the rows of the minor (since these row vectors form a base of $\Bbb R^n$), i.e. $u=\sum \lambda_i r_i$ with at least one nonzero $\lambda_i$. Then for any $\lambda_i\ne0$, if you exchange $u$ and $r_i$ you will get a nonzero minor.