## The rank of the adjugate matrix

In dependence on the rank of a square matrix $$\boldsymbol{A}$$, determine the rank of its adjugate matrix $$\operatorname{adj}\boldsymbol{A}$$.
• #### Hint

Consider separately the case $$\operatorname{rank}(\boldsymbol{A})= n-1$$ .
• #### Solution

If $$\boldsymbol{A}$$ is regular, then $$\operatorname{adj}\boldsymbol{A}$$ is also regular, because $$\boldsymbol{A}\operatorname{adj}\boldsymbol{A}=\det\boldsymbol{A}\cdot\mathbf{I}$$ and the rank cannot rise during the matrix product.

If $$\boldsymbol{A}$$ is singular, two cases can occur: If $$\operatorname{rank}(\boldsymbol{A})\le n-2$$, then each of its submatrices also has rank at most $$n-2$$ and thus each submatrix created by removing one row and one column is singular and $$\operatorname{adj}\boldsymbol{A}=\boldsymbol{0}$$ of rank $$0$$.

If $$\operatorname{rank}(\boldsymbol{A})=n-1$$, then at least one submatrix is regular and thus $$\operatorname{adj}\boldsymbol{A}\ne \boldsymbol{0}$$ and it has rank at least 1. On the other hand, from the equality $$\boldsymbol{A}\operatorname{adj}\boldsymbol{A}=\det\boldsymbol{A}\cdot\mathbf{I}=0\mathbf{I}=\boldsymbol{0}$$ it follows that all columns of $$\operatorname{adj}\boldsymbol{A}$$ belong to the solution subspace of the system $$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{0}$$ of dimension 1, so the rank of $$\operatorname{adj}\boldsymbol{A}$$ is exactly 1.

Separately, the zero matrix of rank 1 has the adjugate matrix $$\mathbf{I}_1$$ of rank 1.