Is there a Way to Think of the Adjugate Matrix Invariantly.

Solution 1:

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

Solution 2:

Suppose $T:V \to V$ where $V$ is $n$-dimensional. This induces a map $T^\sharp:\Lambda^{(n-1)}(V^*) \to \Lambda^{(n-1)}(V^*)$. where $V^*$ denotes the dual space. If $e_1,\dots,e_n$ is a basis of $V$, then $(e^*_2\wedge\cdots\wedge e^*_n)$, $-(e^*_1\wedge e^*_3\wedge\cdots \wedge e^*_n),\dots$, $(-1)^{n-1}(e^*_1\wedge \cdots \wedge e^*_{n-1})$ forms a basis of $\Lambda^{(n-1)}(V^*)$, where $e^*_1,\dots,e^*_n$ is the usual dual basis of $V^*$. (This is the Hodge star operator of the basis on $V^*$.) Then the matrix representation of $T^\sharp$ is the adjugate matrix of the matrix representation of $T$.