Let A be an $n \times n$ matrix. Show that $A$ is not invertible and $\text{rank } A + \text{rank}(\text{adj}(A)) = n$ if and only if $\text{col}(\text{adj}(A)) = \text{null } A$.
The rank-nullity states that $\text{rank }A + \text{nullity} = n$, so we could assume $\text{nullity} = \text{rank}(\text{adj}(A))$ right? Then you can derive $\text{rank}(\text{adj}(A) = \text{dim}(\text{col}(\text{adj}(A)))$ given that $\text{rank}(A) = \text{dim}(\text{col}(A)$. Where can I go from there?
To finish the 'only if' part: since $A\cdot {\rm adj}(A)=\det(A)\cdot I=0$, each column of ${\rm adj}(A)$ is in the nullspace $\ker(A)$, so ${\rm col}({\rm adj}(A)) \subseteq \ker(A)$.
For the 'if' part, we get $A\cdot{\rm adj}(A)=0$, so $A$ is not invertible, and the dimension calculation you made yields $$n={\rm rank}(A)+{\rm null}(A)={\rm rank}(A) +{\rm rank}({\rm adj}(A))$$