Rank of an Adjoint of A

Question

Show that if rank$(A)=n-1$, then adj$(A)$ has a nonzero entry and rank(adj$(A)$)=1.

What I know is that if rank$(A)=n-1$, then it must be true that $A$ has a non-singular $(n-1)\times(n-1)$ submatrix so $\text{adj}\, A\ne0$. I just don't know how to go on from here. Please help.

Answer 1

let "adj" denote the adjugate or 'classical adjoint'.
$B:= \text{adj}\big(A\big)$

$BA = \det\big(A\big)\cdot I =0$

applying Sylvester's Rank Inequality we have
$\text{rank}\big(B\big) + \big(n -1\big) = \text{rank}\big(B\big) + \text{rank}\big(A\big) \leq \text{rank}\big(BA\big) + n = 0 +n = n$
$\implies \text{rank}\big(B\big) \leq 1$

and you already have $\text{rank}\big(B\big) \geq 1$