If all minors are $0$, then rank is at most $n-2$

Question

Can anyone prove/disprove the following statement?

Given a square matrix of size $n\ge2$, if all of its $(n-1)$-rowed minors are zero, then $\operatorname{rank}(A)\leq n-2$.

I'm having trouble connecting the information on the minors to what I have to prove.

Thank you in advance.

Answer 1

HINT:

If all $n-1$-sized minors are $0$, so are the $n$-sized one (think row expansion of determinant), so all minors of size $\ge n-1$.

Now, let $r$ the largest possible size of a non-zero minor. It is $\le n-2$. (Note, that if all elements of the matrix are $0$, we take $r=0$. Otherwise, $r\ge 1$). Now, you have to show that the columns which contain this minor are a basis for the space of columns. It is a bit involved, but the linear dependence is explicit, using the row expansion of minors of size $r+1$ containing this given minor.