Prove if $\operatorname{adj}(A) =0$, then $\operatorname{rank}A \leq n-2$
My initial idea is to setup an induction. Base case $A \in M_{2×2}$ which obviously implies that $A = [O]_{2×2} \rightarrow \operatorname{rank}A \leq 0 = n-2$. However, I can't seem to reason how assuming it worked in the $M_{n\times n}$ case allows for the $M_{(n+1)×(n+1)}$ case to work….
Best Answer
One approach is to use the notion of "determinantal rank", as explained in this post. In particular: if the adjoint of $A$ is zero, then all size $n-1$ submatrices of $A$ have determinant zero which means that $A$ has rank at most $n-2$.
For a direct proof: we are given that $\operatorname{adj}(A) = 0$. Suppose for the purpose of contradiction that $\operatorname{rank}(A) \geq n-1$. It follows that we may select $n-1$ linearly independent rows of $A$. This submatrix has rank $n-1$, which means that we may select $n-1$ linearly independent columns. These columns form a square, size $(n-1)$ submatrix of $A$.
Because this submatrix is invertible, its determinant is non-zero. However, this means that one of the entries of $\operatorname{adj}(A)$ is non-zero, contradicting our premise.