The rank of the adjugate matrix
Task number: 4468
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.
