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i'm interested in geometric interpretations of many linear algebra notions (check also related geometric interpretation of matrix minors). it came to me recently that geometric description of adjugate matrix (for example in case 3×3-matrix) might be quite hard—feel welcome to fill the gap!—but what catched my attention is functoriality of adjugate matrix ($\scriptstyle \mathbf I^\mathrm D = \mathbf I$ and $\scriptstyle (\mathbf{AB})^\mathrm D = \mathbf B^\mathrm D \mathbf A^\mathrm D$); my question is:

what kind of functor is the adjugating (for linear endomorphisms)?

it seems to have strong relationship with (hermitian) adjoint but has slightly different properties (it commutes with transpose). thanks in advance!

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What does "what kind of functor" mean? –  Igor Rivin Sep 27 '11 at 11:18
    
hmm… i know for example that matrix transpose corresponds to dualization functor or rather it is it's special case. for transpose i know that it's functorial: $I^\mathrm T = I$, $(\mathbf{AB})^\mathrm T = \mathbf B^\mathrm T \mathbf A^\mathrm T$ and it commutes with inverse. this functor is $(\cdot)^\star\colon \mathrm{Hom}(M, N) \to \mathrm{Hom}\left(N^\star, M^\star\right)$ for $R$-modules $M$ and $N$, where $M^\star, N^\star$ are their dual modules respectively, and $\mathrm{Hom}(M, N)$ denotes set of all homomorphisms $M \to N$. i'm looking for similar explanation! :p –  joel Sep 27 '11 at 13:09

2 Answers 2

i found quite a satisfying answer to this question which gives enough trails for further enquiry. the answer involves exterior algebra and gradation (check also hodge duality).

if $\mathrm A\colon V \to V$ is a linear map, then it naturally induces a graded operator $\mathrm A^*$ in the exterior algebra of $V$. when $V$ has finite dimension, there are two pieces of the exterior algebra which are isomorphic to $V$, namely $\Lambda^1(V)$, the dual space of $V$, and $\Lambda^{n-1}(V)$, being the space of $(n-1)$-multivectors of $V$.

the restriction of $\mathrm A^*$ to $\Lambda^1(V)$ is the dual map (also called adjoint map). generally the dual space is not canonically isomorphic to $V$ if it has only linear structure; this changes when some sort of duality is established thanks to an extra structure, e.g. inner product. this is the adjoint linear map acting in $V$ with matrix $\mathbf A^\mathrm T$ depending on the choice of basis. on the other hand $\Lambda^{n-1}(V)$ is naturally isomorphic to $V$ (the isomorphism is given by the determinant), and so the restriction of $\mathrm A^*$ to $\Lambda^{n-1}(V)$ naturally induces an linear map on $V$ denoted $\operatorname{adj} \mathrm A$ with matrix $\mathbf A^\mathrm D$. since this is natural, an matrix adjugate can be defined in terms of this.

both notions of adjointness (classical and conjugate transpose) are quite related. the simple one is just duality and makes more sense for operators in general vector spaces (or better in inner product space). the seemingly not-so-simple one is natural, makes sense for operators or matrices, is usually expressed in terms of matrices, and appears often in relation to quadratic forms.

source: google groups on sage developement

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Well, if your matrix happens to be invertible, the adjugate is the inverse. Otherwise, the adjugate gives the action on the $n-1$st exterior power (you can use the $*$ operator to map it back to the space itself. @Qiaochu has alluded to this in his answer to the question you cite. You can also check out Section 9 of my paper http://arxiv.org/pdf/math/0403375v1 to see some other geometric results on the subjects (the published version has fewer typos, so you might want to look at that too, if you have access).

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If the matrix is invertible, I think the adjugate would actually be the inverse times the determinant. Right? –  Manny Reyes Sep 27 '11 at 13:39
    
i'll check it out right away—if it's possible would you be kind enough to explain in plain language (basic linear algebra if it's possible) how does action of exterior power works? :) thanks in advance for your effort! –  joel Sep 27 '11 at 13:55
    
@Manny: yes, that's correct. –  Igor Rivin Sep 27 '11 at 13:56
    
@Manny-Reyes: i know that $\mathbf{AA}^\mathrm D = \mathbf A^\mathrm D \mathbf A = (\det \mathbf A) \mathbf I$ which gives our result if $\mathbf A$ is invertible, because then $\mathbf{AA}^{-1} = \mathbf A^{-1} \mathbf A = \mathbf I$. –  joel Sep 27 '11 at 13:57

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