Is there a way to think of the adjugate matrix invariantly?

Question

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 $M$.

Answer 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

Answer 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 $T$.

Answer 3

Let $R$ be a commutative unitary ring, $M$ be a finite free module of rank $n\geq1$ and $A:M\to M$ be a linear map. Then $\det A\in R$ is defined as the unique element s.t. $\bigwedge^n A:\bigwedge^nM\to \bigwedge^nM$ maps $x$ to $(\det A)\cdot x$.

The map $A$ induces an $R[t]$-linear map $A:M\otimes_R R[t]\to M\otimes_R R[t]$ and the characteristic polynomial of $A\in \DeclareMathOperator{End}{\mathrm{End}}\End_R(M)$ is defined as $\chi_A(t):=\det(tI_n-A)\in R[t]$ where $tI_n-A\in \End_{R[t]}(M\otimes_R R[t])$.

Expanding $\chi_A(t)=t^n+c_{n-1}t^{n-1}+\cdots+c_1t+(-1)^n \det A$ and substituting $t=A$, by Cayley-Hamilton theorem (the version for commutative ring), we have $$A^n+c_{n-1}A^{n-1}+\cdots+c_1A+(-1)^n \det A\cdot I_n=0$$ In particular, $$A\cdot(-1)^n(A^{n-1}+c_{n-1}A^{n-2}+\cdots+c_0)=\det A \cdot I_n=(-1)^n(A^{n-1}+c_{n-1}A^{n-2}+\cdots+c_0)\cdot A$$ So it suggests we can define $\mathop{\mathrm{Adj}}A:=(-1)^n(A^{n-1}+c_{n-1}A^{n-2}+\cdots+c_0)$ or $g(A)$ where $g(t):=(-1)^n \frac{\chi_A(t)-\chi_A(0)}{t}\in R[t]$. Then the property of $\mathop{\mathrm{Adj}}A$ follows from Cayley-Hamilton.

This definition relies heavily on the Cayley-Hamilton theorem and can easily cause circular arguments, so be care to use it.

Answer 4

On a real or complex finite dimensional vector space $V$, for an invertible linear operator $T:V\to V$, one could define \begin{equation} \operatorname{Adj}T:=(\det T) T^{-\ast}:V^\ast\to V^\ast, \end{equation} where $\det T$ is the determinant defined as the product of the eigenvalues of $T$, $T^\ast:V^\ast\to V^\ast$ the transpose of $T$, which must be also invertible and $T^{-\ast}$ its inverse. Unfortunately, when the operator $T$ is not invertible this relation cannot be used to define the adjugate of $T$.