Rank of Matrix Determined by Square Submatrices

Question

In an answer to this post, it is argued that an $n\times m$ matrix $A$ is of full rank $r=\min\{n,m\}$ if and only if every $r\times r$ submatrix $A'$ is invertible. This seams intuitive but does anyone have a reference for this result?

Answer 1

As you have stated it, this is wrong. Consider the matrix $A=(1 \ 0)$ which has full rank $1$ but the $0$ submatrix is not invertible.

I assume you are referring to jlewk's answer where he uses that an $n\times m$ matrix has full rank $r:=\min(m,n)$ if and only if there exists an invertible $r\times r$ submatrix $A'.$ The rank of a matrix is the same as the rank of its transpose so we may assume that $n\geq m$ by transposing the matrix, if necessary. Then we get this chain of equivalencies: $A$ has rank $r=m$ $\Leftrightarrow$ There exist $m$ linearly independent columns of $A$ $\Leftrightarrow$ $A$ has an invertible $m\times m$ submatrix.