i'm interested in geometric interpretations of many linear algebra notions (check also related geometric interpretation of matrix minors). it came to me recently that geometric description of adjugate matrix (for example in case 3×3-matrix) might be quite hard—feel welcome to fill the gap!—but what catched my attention is functoriality of adjugate matrix ($\scriptstyle \mathbf I^\mathrm D = \mathbf I$ and $\scriptstyle (\mathbf{AB})^\mathrm D = \mathbf B^\mathrm D \mathbf A^\mathrm D$); my question is:

what kind of functor is the adjugating (for linear endomorphisms)?

it seems to have strong relationship with (hermitian) adjoint but has slightly different properties (it commutes with transpose). thanks in advance!

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 What does "what kind of functor" mean? – Igor Rivin Sep 27 2011 at 11:18 hmm… i know for example that matrix transpose corresponds to dualization functor or rather it is it's special case. for transpose i know that it's functorial: $I^\mathrm T = I$, $(\mathbf{AB})^\mathrm T = \mathbf B^\mathrm T \mathbf A^\mathrm T$ and it commutes with inverse. this functor is $(\cdot)^\star\colon \mathrm{Hom}(M, N) \to \mathrm{Hom}\left(N^\star, M^\star\right)$ for $R$-modules $M$ and $N$, where $M^\star, N^\star$ are their dual modules respectively, and $\mathrm{Hom}(M, N)$ denotes set of all homomorphisms $M \to N$. i'm looking for similar explanation! :p – joel Sep 27 2011 at 13:09

Well, if your matrix happens to be invertible, the adjugate is the inverse. Otherwise, the adjugate gives the action on the $n-1$st exterior power (you can use the $*$ operator to map it back to the space itself. @Qiaochu has alluded to this in his answer to the question you cite. You can also check out Section 9 of my paper http://arxiv.org/pdf/math/0403375v1 to see some other geometric results on the subjects (the published version has fewer typos, so you might want to look at that too, if you have access).
@Manny-Reyes: i know that $\mathbf{AA}^\mathrm D = \mathbf A^\mathrm D \mathbf A = (\det \mathbf A) \mathbf I$ which gives our result if $\mathbf A$ is invertible, because then $\mathbf{AA}^{-1} = \mathbf A^{-1} \mathbf A = \mathbf I$. – joel Sep 27 2011 at 13:57