Given a square matrix $M$ with entries from a field $F$, the adjugate of $M$ is defined as the transpose of the cofactor matrix.
Is there an interpretation of this concept in terms of linear operators on vector spaces?
As an example of what I am trying to ask, consider the operation of taking the transpose of a matrix (with entries from a field).
This can be thought of in terms of linear operators in the following way:
Let $T:V\to V$ be a linear operator on a finite dimensional vector space $V$. We define the transpose of $T$ as the linear map $T^t:V^*\to V^*$ which sends a member $\omega\in V^*$ to the member $(v\mapsto \omega(Tv))$ of $V^*$.
Now if $\mathcal B$ is a basis of $V$ and $M$ is the matrix representation of $T$ with respect to the basis $\mathcal B$, then the matrix representation of $T^t$ with respect to the dual basis of $\mathcal B$ is same as the matrix transpose of $M$.
Best Answer
Another way to approach this (that works for modules) and avoiding the use of the dual space. It may be that one requires finite rank free modules here so that the bilnear pairing below is perfect. I haven't checked.
The bilinear pairing is, $$ V \times \Lambda^{n-1} V \to \Lambda^n V $$ sending $$ (v, \eta) \to v \wedge \eta $$ 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, $$ \operatorname{adj} (T) \circ T = \det T \operatorname{Id}. $$
This all explained (sections 5 to 8) here:
http://people.reed.edu/~jerry/332/27exterior.pdf