For any n>2 and any n by n matrix A over an arbitrary field, the adjugate of the adjugate of A equals the determinant of A to power n-2 times A. Is there a unified way, without dividing into two cases - A invertible and A non-invertible, to prove this result ?
the adjugate of the adjugate
0
$\begingroup$
linear-algebra
-
0are you referring to the second proof (because as far as i can tell, the first proof **is** by dividing into the two cases) ? – 2012-06-25
1 Answers
6
Exactly the same universal approach in the related problem below works on your similar problem.
Hint $\ $ Denote the adjoint of $\rm\:A\:$ by $\rm\:A^*.\:$ Then
$\rm\: A A^* = |A|\: I_n \ \Rightarrow\ |A|\, |A^*| = |A|^n\ \Rightarrow\ |A^*| = |A|^{n-1}$
where the cancellation of $\rm\:|A|\:$ is done universally, i.e. considering the matrix extries as indeterminates, so the determinant is a nonzero polynomial in a domain $\rm\:\mathbb Z[a_{ij}],$ hence is cancellable. For further discussion of such universal cancellation of "apparent singularities" see here and here and here.