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Let $A: V \to V$ linear operator where $V$ is $n$-dimensional vector space. Consider $\wedge^{k}A: \wedge^{k}V \to \wedge^{k}V$ given by $u_{1}\wedge ... \wedge u_{k} \mapsto A(u_{1})\wedge ... \wedge A(u_{k})$.

When $k=n$, we know that $A(u_{1})\wedge ... \wedge A(u_{n}) = det(A) (u_{1}\wedge ... \wedge u_{n})$.

But what can we say when $k<n$?

(In wikipedia, they say that " Minors of a matrix can also be cast in this setting, by considering lower alternating forms $\wedge^{k}V$ with $k < n$.", but they don't give any reference.)

I'd like some reference to study it. Thanks

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    See my answer here. A good reference is Birkoff and MacLane's Algebra. – symplectomorphic Dec 27 '16 at 20:34
  • I looked it, but it doesn´t have this problem. Thanks – Alladin Dec 27 '16 at 21:09
  • @Alladin what exactly does "this problem" refer to? At the very least, the answer linked talks about where the minors of a matrix come in. – Ben Grossmann Dec 27 '16 at 21:10
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    Another useful reference is Bhatia's Matrix Analysis. One notable point is that the eigenvalues of $\wedge^k A$ are all (non-repeating) products of $k$ of the eigenvalues of $A$. – Ben Grossmann Dec 27 '16 at 21:11
  • When $k<n$, what we can say about $A(u_{1})\wedge...\wedge A(u_{k})$? – Alladin Dec 27 '16 at 21:13
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    @Alladin what do you mean by "What can I say about..."? That's really vague. The answer link has certainly told you something about all that, but you're implying that there's something missing. What's missing? – Ben Grossmann Dec 27 '16 at 21:29
  • Sorry. It's satisfactory. Thanks – Alladin Dec 27 '16 at 21:31
  • I like the answer below but would like to point out that this whole thing can be cast into the analysis of linear independence. This is not relevant to the question but is relevant to the "meaning" of minors and forms. – rrogers Jan 13 '18 at 15:37

1 Answers1

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See my answer here.

Two good references are:

Bourbaki, Algebra I: Chapters 1-3, Proposition 10, page 529.

Birkhoff and MacLane, Algebra, pages 563-564.