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?
Best Answer
If you only need an existential proof, just fill in the elements of an infinite matrix as follows. Set the first element to any nonzero real number. Then for every unfilled element, fill it by a number that is transcendental to the field of fractions generated by the previously filled elements (hence this new element is nonzero). Then any finite submatrix of this infinite matrix would be totally invertible.
There are also some meaningful classes of totally invertible matrices. E.g. any submatrix of a totally positive matrix is totally invertible. Some classes of totally positive matrices can be parametrised. See this set of slides entitled "The Grassmannian: total positivity" by Alexander Yong, for instance.