TITLE: Functoriality of adjugate matrix
QUESTION [5 upvotes]: 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!

REPLY [3 votes]: Copied from math.stackexchange: Another way to approach this avoids the use of duality via inner-products. Functorality should be clear. It may be that one requires finite rank free modules here so that the bilnear pairing below is perfect.
The bilinear pairing is,
$$
V \times \Lambda^{n-1} V \to \Lambda^n V
$$
sending
$$
(v, \eta) \to v \wedge \eta
$$
and written
$$
\langle v, \eta \rangle = v \wedge \eta.
$$
Then given $T : V \to V$, we define,
$$
\Lambda^{n-1} T : \Lambda^{n-1} V \to \Lambda^{n-1} V
$$
on indecomposable elements by
$$
\Lambda^{n-1} T (v_1 \wedge \cdots v_{n-1}) = T(v_1) \wedge \cdots \wedge T(v_{n-1})
$$
and extend to all of $\Lambda^{n-1} V$ by alternating multilinearity as usual.
The adjugate $\operatorname{adj}(T) : V \to V$ is the adjoint of $\Lambda^{n-1} T$ with respect to the pairing:
$$
\langle \operatorname{adj}(T) (v), \eta \rangle = \langle v, \Lambda^{n-1} T (\eta) \rangle,
$$
or using the definition of the pairing,
$$
\operatorname{adj} (T) (v) \wedge \eta = V \wedge \Lambda^{n-1} T \eta
$$
Now one observes that,
$$
\begin{split}
\langle \operatorname{adj} (T) \circ T (v), \eta\rangle &= \langle T(v), \Lambda^{n-1} T(\eta)\rangle \\
&= T(v) \wedge \Lambda^{n-1} T (\eta) \\
&= \Lambda^{n} T (v \wedge \eta) \\
&= \det T v \wedge \eta \\
&= \langle \det T v, \eta \rangle
\end{split}
$$
If the pairing is perfect, this implies that,
$$
\operatorname{adj} (T) \circ T = \det T \operatorname{Id}.
$$
In particular, if $\det T$ is an invertible element of the underlying ring, then $\det T \operatorname{Id}$ is invertible and $\operatorname{adj} (T)$ commutes with $T$ (which both are invertible) as well as satisfying,
$$
\operatorname{adj} (T) = \det T T^{-1}
$$
which is the usual formula.
This all explained (sections 5 to 8) here:
http://people.reed.edu/~jerry/332/27exterior.pdf