If all minors are $0$, the 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

Original Post: Show that $\text{rank}(A) \leq 2$.

This doesn't appear to be true. Consider the following matrix: $$A = \begin{bmatrix} 0 & 0 & 0 & 0&0 \\ 0 & 1 & 0 &0 &0 \\ 0 & 0 & 1 &0 &0 \\ 0 & 0 & 0 &1 &0 \\ 0 & 0 & 0 & 0 &0 \end{bmatrix}$$ $A$ clearly has rank $3$. To see why all of $A$'s minors have determinant zero, just note that to create a minor, you eliminate a column and a row. But it's not possible to remove both row $1$ and row $5$ (which are both zero rows), and thus any minor must have a row of zeroes (which implies its determinant is nonzero).

Updated Version: Show that $\text{rank}(A) \leq n - 2$.

Proof: Proceed by contradiction and suppose the rank is $n - 1$ (it clearly can't be $n$, because Laplace expanding along any row or column would yield a zero determinant). If the rank is $n-1$, then it must mean that there exists some column we can remove that doesn't change the rank (because there must exist $n-1$ linearly independent columns). Assume this is the $k$th column. Upon removing this column, we have a $n \times (n-1)$ matrix, supposedly with rank $n - 1$. As this matrix has $n$ rows, it follows that we must be able to remove exactly one row without changing the rank. Let this be the $j$th row. Then we are left with a rank $n-1$ matrix that has dimensions $(n-1) \times (n-1)$, and thus it has nonzero determinant. But this matrix is exactly the $(k,j)$th minor of $A$, a contradiction.

Answer 2

It's well-known that row-rank = column-rank = determinantal-rank where the determinantal rank is the size of the largest square submatrix with non-zero determinant. Your question is an easy corollary

But let's give an ad hoc proof for your example. I'll prove the contrapositive, that if $\text{rank}\, A\ge n-1$ then $A$ has a nonzero minor of size $n-1$.

As the rank of $A$ is $n$ or $n-1$, then $n-1$ rows of the matrix are linearly independent. We can delete one row to make an $n-1$-by-$n$ matrix $B$ with linearly independent rows. It's rank is $n-1$, so $n-1$ of its columns are linearly independent. So we can delete one column to get an $n-1$-by-$n-1$ matrix $C$ with linearly independent columns. Then $C$ is non-singular, so $\det C\ne0$. Then $\det C$ is a non-zero minor of $A$.