Let $A \in F^{nxn}$ be a matrix and the determinant of each minor is $0$. Prove that $rank(A) \leq (n-2)$.
Could anyone explain why this is true? I know that if the determinant is 0 then the matrix is not invertible. So I can say that $rank(A) < n$ but why is it less than $n-2$?
Let $A$ be a matrix $m\times n$ and $k\in \mathbb{Z}$ such that $0<k\leq \min\{m,n\}$. We define a minor of order $k$ of $A$, denoted as $\Delta_k$, as the determinant of a matrix $k\times k$ that results by the elimination of $m-k$ rows and $n-k$.
Also, we know that a number $r$ is said to be the rank of a matrix if it satisfies these two conditions:
$\square$ There is at least one minor of order $r$ that not vanishes.
$\square$ Every minor of order $r+1$ or higher vanishes.
Now, we get back to the problem. If all the $(n-1)-$order minors vanish, then the determinant vanishes. Indeed, we can express the determinant for some constants $c_i$ as $$\det(A)=\sum_{i=1}^n c_i \Delta_k^i=0$$ since $\Delta_k^i=0$. Now, the rank just can be $n-2$ or lower since
$\square$ Every minor of order $(n-2)+1=n-1$ or higher (the determinant of the matrix) vanishes.
$\square$ If there are minors that not vanishes, they are from order $n-2$ or lower because we are told that all $n-1$ and $n$ order vanishes.