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 $M$.

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    $\begingroup$ en.wikipedia.org/wiki/Exterior_algebra is what you need. You will want to consider the $n-1$ forms if your space is $n$ dimensional. It's a bit of a steep learning curve, so I don't think I can give an answer that fits here. $\endgroup$ Commented Jul 19, 2015 at 22:01
  • $\begingroup$ @StephenMontgomery-Smith I am fairly comfortable with exterior algebra. So it'll be great if you can write at least a short answer to this question. If I do not understand something, I will open follow up threads. $\endgroup$ Commented Jul 19, 2015 at 22:49
  • $\begingroup$ This MO post has some related links in comments, specifically this and this. $\endgroup$ Commented May 30, 2016 at 12:59
  • $\begingroup$ @MartinSleziak Thank you. I will sure read these posts. $\endgroup$ Commented May 30, 2016 at 19:38

3 Answers 3


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:


  • $\begingroup$ Great answer. Thanks. $\endgroup$ Commented Dec 4, 2016 at 5:17

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 $T$.

  • $\begingroup$ Thank you. I understand it. Do you also know if this works for modules over commutative rings with identity? $\endgroup$ Commented Jul 19, 2015 at 23:37
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    $\begingroup$ I don't know. I guess the issue is to see whether $\Lambda^{(n-1)}(V^*)$ is isomorphic to $V$. It all makes sense to me for vector spaces. I don't know enough module theory to answer your question. Is there even a notion of determinant for such maps? (Determinant is the induced map $\Lambda^{(n)}(V) \to \Lambda^{(n)}(V)$.) $\endgroup$ Commented Jul 19, 2015 at 23:46
  • $\begingroup$ Given an $R$-linear map $f:R^n\to R^n$, we can form the matrix of $f$ with respect to some basis of $R^n$ and find out the determinant of this matrix. I doubt the determinant will be basis dependent. So probably all this can be extended to free modules. But anyway. For now I should concentrate on vector spaces. $\endgroup$ Commented Jul 21, 2015 at 4:53
  • $\begingroup$ Can you have a look at this? math.stackexchange.com/questions/1368399/… $\endgroup$ Commented Jul 21, 2015 at 5:57

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 $T$.


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