Let $A$ be a $d \times d$ real matrix, and let $1<k<d$. Suppose that all the $k$-minors of $A$ are non-zero.
Is it necessarily true that $\text{rank}(A)>k$?
I am looking for a counter-example. When $k=1$, this is obviously false; take $A=\begin{pmatrix} 1 & 2 \\\ 1 & 2 \end{pmatrix}$. All the entries of $A$ (the $1$-minors) are non-zero, but $\text{rank}(A)=1$.
Is there an easy way to generalize this example to $k>1$?
Comment: In the converse direction, we do know that $\text{rank}(A)>k$ implies there are many non-zero $k$-minors; There exist a non-zero $k+1$-minor; omitting any row and column participating in this minor, we obtain a non-zero $k$-minor. Thus we have at least $(k+1)^2$ non-zero $k$-minors. This lower bound is sharp; there exist matrices of rank $k+1$ with exactly $(k+1)^2$ non-zero $k$-minors- e.g. take $\text{Id}_{k+1 \times k+1} \otimes O_{d-k-1 \times d-k-1}$.
Best Answer
A simpler version of your example (if I may) is to take $A=\begin{pmatrix} 1 & 1 \\\ 1 & 1 \end{pmatrix}$, or rather $A=\begin{pmatrix} 1\\\ 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ \end{pmatrix}$.
Now for any $k$ let $u$ be the all-1's $k$-vector, $u=(1\ 1\ \dots 1)$; and let $I$ be the $k\times k$ identity matrix.
Let $V$ be the $(k+1)\times k$ matrix $\begin{pmatrix} I \\ u \end{pmatrix}$, for $A$ take the $(k+1)\times(k+1)$ matrix $VV^{T}$. (Note that your example as I have modified it is just the case $k=1$.)
The claim is now that (a) $A$ is singular but (b) every $k\times k$ minor of $A$ is non-zero.
The statement (a) is clear, as $A$ has at most $k$ linearly independent rows.
For (b) we use Binet-Cauchy. The matrix of $k\times k$ minors of $A$, which we write as $A^{(k)}$, is given by $$A^{(k)}=V^{(k)}(V^T)^{(k)}=V^{(k)}(V^{(k)})^{T}. $$
It is easy to compute $V^{(k)}$, it is just a $(k+1)$-column-vector whose entries are either $\pm 1$. Hence the $k\times k$ minors of $A$ are all $\pm 1$.