Let's call a real matrix of size $m \times n$ totally invertible
if for every $k$ rows and $k$ columns that we choose, we get an invertible matrix. I am curious about the following:
Is there a totally invertible matrix for all sizes $m \times n$?
By taking the transpose, we can assume $m \leq n$.
For $m=1$ we can take any vector in $\Bbb R^m$ without a zero entry.
For $m=2$ we can take $\begin{pmatrix} 1 &2 & 3 & ... &n \\ 2 & 3 & 4 & ... & n+1 \end{pmatrix}$. Indeed, $\det\begin{pmatrix} a &b \\ a+1 & b+1 \end{pmatrix}=a-b$ is nonzero when $a \neq b$. This does not generalize, since $a_{ij}=i+j-1$ fabulously fails for all submatrices of size $\geq 3$.
I also tried taking the rows to be $m$ of the cyclic shifts of the vector $(1,2, ... ,n)$. This does not work either because I found a $k=3, m=n=5$ counterexample.
I do feel that the answer should be positive, however. Is it?