Columns and rows are independent $\implies$ minor is invertible

Question

Let $A$ be an $n\times m$ matrix and let $r$ be its rank. Let $I$ be a subset of $\{1,\cdots,n\}$ of size $r$ such that the corresponding rows are independent and $J$ be a subset of $\{1,\cdots,m\}$ of size $r$ such that the corresponding columns are independent. It is well-known that one such pair $(I,J)$ exists such that $A_{I,J}$ is invertible; prove that the minor $A_{I,J}$ is invertible no matter which pair $(I,J)$ we choose.

Answer 1

$\DeclareMathOperator{\rank}{rank}$

Given $I = \{i_1, i_2, \ldots, i_r\}, J = \{j_1, j_2, \ldots, j_r\}$ such that the conditions in the question are satisfied, where $1 \leq i_1 < i_2 < \cdots < i_r \leq n$, $1 \leq j_1 < j_2 < \cdots < j_r \leq m$, let $L$ be the $r \times n$ matrix such that

1. For $j \in \{1, \ldots, n\}\backslash I$, the $j$-th column is an $r$-long zero column vector.
2. For $1 \leq k \leq r$, the $i_k$-th column is an $r$-long column vector whose $k$-th element $1$ and the remaining elements $0$.

Similarly, let $R$ be the $m \times r$ matrix such that

1. For $i \in \{1, \ldots, m\}\backslash J$, the $i$-th row is an $r$-long zero row vector.
2. For $1 \leq l \leq r$, the $j_l$-th row is an $r$-long row vector whose $l$-th element $1$ and the remaining elements $0$.

It is easy to verify that $A_{I, J} = LAR$. Hence by Frobenius rank inequality (the $11$-th inequality in the link): \begin{align*} \rank(A_{I, J}) = \rank(LAR) \geq \rank(LA) + \rank(AR) - \rank(A) = r + r - r = r. \end{align*}

That $\rank(LA) = r$ is because $LA$ is the $r \times m$ matrix consisting of the $I$ rows of $A$, which by condition are linearly independent. By the same token, $\rank(AR) = r$.

On the other hand, clearly $\rank(A_{I, J}) \leq r$. This shows that $\rank(A_{I, J}) = r$, i.e., the order $r$ matrix $A_{I, J}$ is invertible.