Editing Adjugate matrix (section)

== Relation to exterior algebras ==
The adjugate can be viewed in abstract terms using [[exterior algebra]]s. Let {{math|''V''}} be an {{math|''n''}}-dimensional [[vector space]].  The [[exterior product]] defines a bilinear pairing
<math display=block>V \times \wedge^{n-1} V \to \wedge^n V.</math>
Abstractly, <math>\wedge^n V</math> is [[isomorphic]] to {{math|'''R'''}}, and under any such isomorphism the exterior product is a [[perfect pairing]]. That is, it yields an isomorphism
<math display=block>\phi \colon V\ \xrightarrow{\cong}\ \operatorname{Hom}(\wedge^{n-1} V, \wedge^n V).</math>
This isomorphism sends each {{math|'''v''' ∈ ''V''}} to the map <math>\phi_{\mathbf{v}}</math> defined by
<math display=block>\phi_\mathbf{v}(\alpha) = \mathbf{v} \wedge \alpha.</math>
Suppose that {{math|''T'' : ''V'' &rarr; ''V''}} is a [[linear transformation]]. [[Pullback]] by the {{math|(''n''&nbsp;− 1)}}th exterior power of {{math|''T''}} induces a morphism of {{math|Hom}} spaces. The '''adjugate''' of {{math|''T''}} is the composite
<math display=block>V\ \xrightarrow{\phi}\ \operatorname{Hom}(\wedge^{n-1} V, \wedge^n V)\ \xrightarrow{(\wedge^{n-1} T)^*}\ \operatorname{Hom}(\wedge^{n-1} V, \wedge^n V)\ \xrightarrow{\phi^{-1}}\ V.</math>

If {{math|1=''V'' = '''R'''<sup>''n''</sup>}} is endowed with its [[canonical basis]] {{math|'''e'''<sub>1</sub>, ..., '''e'''<sub>''n''</sub>}}, and if the matrix of {{math|''T''}} in this [[basis (linear algebra)|basis]] is {{math|'''A'''}}, then the adjugate of {{math|''T''}} is the adjugate of {{math|'''A'''}}. To see why, give <math>\wedge^{n-1} \mathbf{R}^n</math> the basis
<math display=block>\{\mathbf{e}_1 \wedge \dots \wedge \hat\mathbf{e}_k \wedge \dots \wedge \mathbf{e}_n\}_{k=1}^n.</math>
Fix a basis vector {{math|'''e'''<sub>''i''</sub>}} of {{math|'''R'''<sup>''n''</sup>}}. The image of {{math|'''e'''<sub>''i''</sub>}} under <math>\phi</math> is determined by where it sends basis vectors:
<math display=block>\phi_{\mathbf{e}_i}(\mathbf{e}_1 \wedge \dots \wedge \hat\mathbf{e}_k \wedge \dots \wedge \mathbf{e}_n)
= \begin{cases} (-1)^{i-1} \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n, &\text{if}\ k = i, \\ 0 &\text{otherwise.} \end{cases}</math>
On basis vectors, the {{math|(''n''&nbsp;− 1)}}st exterior power of {{math|''T''}} is
<math display=block>\mathbf{e}_1 \wedge \dots \wedge \hat\mathbf{e}_j \wedge \dots \wedge \mathbf{e}_n \mapsto \sum_{k=1}^n (\det A_{jk}) \mathbf{e}_1 \wedge \dots \wedge \hat\mathbf{e}_k \wedge \dots \wedge \mathbf{e}_n.</math>
Each of these terms maps to zero under <math>\phi_{\mathbf{e}_i}</math> except the {{math|1=''k'' = ''i''}} term. Therefore, the pullback of <math>\phi_{\mathbf{e}_i}</math> is the linear transformation for which
<math display=block>\mathbf{e}_1 \wedge \dots \wedge \hat\mathbf{e}_j \wedge \dots \wedge \mathbf{e}_n \mapsto (-1)^{i-1} (\det A_{ji}) \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n.</math>
That is, it equals
<math display=block>\sum_{j=1}^n (-1)^{i+j} (\det A_{ji})\phi_{\mathbf{e}_j}.</math>
Applying the inverse of <math>\phi</math> shows that the adjugate of {{math|''T''}} is the linear transformation for which
<math display=block>\mathbf{e}_i \mapsto \sum_{j=1}^n (-1)^{i+j}(\det A_{ji})\mathbf{e}_j.</math>
Consequently, its matrix representation is the adjugate of {{math|'''A'''}}.

If {{math|''V''}} is endowed with an [[inner product]] and a volume form, then the map {{math|''φ''}} can be decomposed further. In this case, {{math|''φ''}} can be understood as the composite of the [[Hodge star operator]] and dualization. Specifically, if {{math|ω}} is the volume form, then it, together with the inner product, determines an isomorphism
<math display=block>\omega^\vee \colon \wedge^n V \to \mathbf{R}.</math>
This induces an isomorphism
<math display=block>\operatorname{Hom}(\wedge^{n-1} \mathbf{R}^n, \wedge^n \mathbf{R}^n) \cong \wedge^{n-1} (\mathbf{R}^n)^\vee.</math>
A vector {{math|'''v'''}} in {{math|'''R'''<sup>''n''</sup>}} corresponds to the linear functional
<math display=block>(\alpha \mapsto \omega^\vee(\mathbf{v} \wedge \alpha)) \in \wedge^{n-1} (\mathbf{R}^n)^\vee.</math>
By the definition of the Hodge star operator, this linear functional is dual to {{math|*'''v'''}}. That is, {{math|ω<sup>∨</sup>∘ φ}} equals {{math|'''v''' ↦ *'''v'''<sup>∨</sup>}}.