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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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    $\begingroup$ See my answer here. A good reference is Birkoff and MacLane's Algebra. $\endgroup$ – symplectomorphic Dec 27 '16 at 20:34
  • $\begingroup$ I looked it, but it doesn´t have this problem. Thanks $\endgroup$ – Alladin Dec 27 '16 at 21:09
  • $\begingroup$ @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. $\endgroup$ – Omnomnomnom Dec 27 '16 at 21:10
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    $\begingroup$ 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 $A$. $\endgroup$ – Omnomnomnom Dec 27 '16 at 21:11
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    $\begingroup$ @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? $\endgroup$ – Omnomnomnom Dec 27 '16 at 21:29
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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.

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