The rank of the adjugate matrix

Idea. The entries of $\operatorname{adj}A$ are the $(n-1)\times(n-1)$ minors of $A$ (up to sign), so the pattern of vanishing depends on the rank. The identity $A\operatorname{adj}A=(\det A)I$ controls the kernel.

Case 1: $\operatorname{rank}A=n$. Then $\det A\ne0$, $\operatorname{adj}A=\det A\cdot A^{-1}$ is invertible: rank $n$.

Case 2: $\operatorname{rank}A\le n-2$. Every $(n-1)\times(n-1)$ minor has size larger than the rank, hence vanishes (Problem 83, Step 1). All entries of $\operatorname{adj}A$ vanish: $\operatorname{adj}A=0$, rank $0$.

Case 3: $\operatorname{rank}A=n-1$.
- *Nonzero.* Some $(n-1)\times(n-1)$ minor is nonzero (Problem 83), so $\operatorname{adj}A\ne0$: rank $\ge1$.
- *Columns.* $\det A=0$, so $A\operatorname{adj}A=0$: every column of $\operatorname{adj}A$ lies in $\ker A$, which has dimension $n-\operatorname{rank}A=1$. So the column space of $\operatorname{adj}A$ has dimension $\le1$, i.e. $\operatorname{rank}\operatorname{adj}A\le1$.

Hence $\operatorname{rank}\operatorname{adj}A=1$. (By symmetry, $\operatorname{adj}A\,A=0$ puts the rows into $\ker A^T$, also of dimension $1$: $\operatorname{adj}A=c\,uv^T$ with $u\in\ker A$, $v\in\ker A^T$.)

Example. $n=2$, $A=\begin{pmatrix}1&2\\2&4\end{pmatrix}$ (rank $1=n-1$): $\operatorname{adj}A=\begin{pmatrix}4&-2\\-2&1\end{pmatrix}$, rank $1$. For $n=4$, random matrices of rank $4,3,2$ gave adjugate ranks $4,1,0$ numerically.

Remark. The case $n=1$ is excluded since $\operatorname{adj}A=(1)$ always. The result implies that $\operatorname{adj}$ is continuous but its rank pattern jumps. Also $\det(\operatorname{adj}A)=(\det A)^{n-1}$.