For given $n\times n$ matrix $A$ singular matrix, prove that $\operatorname{rank}(\operatorname{adj}A) \leq 1$ For given $n\times n$ matrix $A$ singular matrix, prove that $\operatorname{rank}(\operatorname{adj}A) \leq 1$
So from the properties of the adjugate matrix we know that
$$ A \cdot \operatorname{adj}(A) = \operatorname{det}(A)\cdot I$$
Since $A$ is singular we know that $\operatorname{det}(A) = 0$, thus
$$ A \cdot \operatorname{adj}(A) = 0$$
This is where I'm getting lost, I think I should say that for the above to happen one of the two, $A$ or $\operatorname{adj}(A)$ would have to be the $0$ matrix, but if $A = 0$ then $\operatorname{adj}(A) = 0$ for sure, which means I said nothing.
A leading hint is needed.
 A: This is an old question, but I would like to add a solution that only uses properties of matrices i.e. no abstract algebra.
Sylvester's rank inequality states that for two matrices $X,Y \in \mathcal{M_n(\mathbb{C})}$ we have that $\operatorname{rank}(XY)\ge \operatorname{rank}X+\operatorname{rank} Y-n$.
Since $A$ is singular, we have $\operatorname{rank}A\le n-1$.
We have the equality $A\cdot \operatorname{adj}A=O_n$.
Hence, from Sylvester's rank inequality $0=\operatorname{rank}(A\cdot \operatorname{adj}A)\ge \operatorname{rank}A + \operatorname{rank}(\operatorname{adj}A)-n$.(1)
Case 1. $\operatorname{rank}A\le n-2=>\operatorname{adj}(A)=O_n$ as in Julien's solution.
Case 2. $\operatorname{rank}A=n-1$.
From (1) we have that $\operatorname{rank}(\operatorname{adj}A)\le 1$, so $\operatorname{rank}(\operatorname{adj}A)\in \{0,1\}$.
If $\operatorname{rank}(\operatorname{adj}A)=0$, then $\operatorname{adj}A=O_n$ and this contradicts $\operatorname{rank} A=n-1$.
Hence, $\operatorname{rank}(\operatorname{adj}A)=1$ in this case.
In both cases $\operatorname{rank}(\operatorname{adj}A)\le 1$.
A: Since $A$ is singular, $\mbox{rank}A\leq n-1$.
Case 1: $\mbox{rank}A\leq n-2$. Then $A$ contains no invertible submatrix of order $n-1$. So every minor of order $n-1$ is zero. What can you conclude about $\mbox{adj}(A)$?
Case 2: $\mbox{rank}A= n-1$. By rank-nullity, we get $\dim\ker A=1$. Now $A\cdot \mbox{adj}(A)=0$ means that the range of $\mbox{adj}(A)$ is contained in $\ker A$. So...
A: Since the adjugate is anti-multiplicative, and every matrix $A$ is a product $A = T D S$, where $T$, $S$ are of determinant $1$, and $D$ is a diagonal matrix, it's enough to check for $A$ diagonal. Now it's easy.
Another approach, pure algebra. It is enough to check that every minor of order $2$ of the matrix $\operatorname{adj}(A)$ is $0$ if $\det A = 0$. Now we have the complement formula for the minors of the adjoint matrix
$$M_{ij, kl}( \operatorname{adj}(A)) = \pm \cdot M_{\overline{ij}, \overline{kl}}(A) \cdot \det A$$
(works for any commutative ring)
( Jacobi formula for minors- see Aitken-- Determinants and Matrices)
