$\begingroup$

This question is from a masters entrance for which I am preparing.

Let $n\geq 2$ and A be an $n\times n$ matrix with real entries. Let Adj A denote the adjoint of A, ie the (i,j) - th entry of Adj A is the (j,i) - th cofactor of A. Then show that the rank of Adj A is 0,1,n.

If A= 0 matrix then Adj A =0 matrix.

If A is invertible then rank A = rank adj A = n.

If rank is 1 and A is 2$\times$ 2 matrix and if rank A is 1,2 and A is $3 \times 3$ matrix, then I have verified that the rank of AdjA will be 1 but I am unable to generalize it.

Can you shed some light on how to argue that?

$\endgroup$2

1 Answer

$\begingroup$

Consider the following three cases:

  1. If $A$ has rank less than or equal to $n - 2$ then any $n - 1$ columns of $A$ are linearly dependent and in particular, any $(n-1) \times (n-1)$ submatrix of $A$ has linearly dependent columns and so zero determinant. This implies that $\operatorname{adj}(A) = 0$ and so $\operatorname{rank}(\operatorname{adj}(A)) = 0$.
  2. If $\operatorname{rank}(A) = n - 1$ then the identity $A \cdot \operatorname{adj}(A) = \det(A) \cdot I_n = 0_{n \times n}$ shows you that all the columns of $\operatorname{adj}(A)$ must belong to the kernel of $A$ which is one-dimensional. Hence, $\operatorname{rank}(\operatorname{adj}(A)) \leq 1$. You can in fact show that in this case $\operatorname{rank}(\operatorname{adj}(A)) = 1$ but this is not needed to answer your question.
  3. If $\operatorname{rank}(A) = n$ then $A$ is invertible and the identity $\operatorname{adj}(A) \cdot A = \det(A) I_n$ shows you that $\operatorname{adj}(A) = \frac{1}{\det(A)} A^{-1}$ is also invertible and so has rank $n$.
$\endgroup$