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Is true that for all $J:\wedge^{k} \mathbb{R}^{n} \to\wedge^{k} \mathbb{R}^{n}$ isomorphism linear there exists $A:\mathbb{R}^{n} \to \mathbb{R}^{n}$ linear operator such that $\wedge^{k} A =J$?

When $k=(n-1)$, we have a positive answer, just do like this.

Related: Exterior Algebra: Find a linear operator

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    $\begingroup$ I don't think that my approach will work for $n \neq 3$, since I'm exploiting the isomorphism between $\Bbb R^3$ and $\wedge^2 \Bbb R^3$. $\endgroup$ – Omnomnomnom Dec 15 '16 at 23:53
  • $\begingroup$ I agree, but your approach will work to $\wedge^{n-1} \mathbb{R}^{n}$ because it has isomorphism with $\mathbb{R}^{n}$. It's the same argument. $\endgroup$ – Alladin Dec 15 '16 at 23:56
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Not in general. The construction $A \mapsto \Lambda^k(A)$ gives us a nice (smooth) homomorphism $\varphi \colon \operatorname{GL}_n(\mathbb{R}) \rightarrow \operatorname{GL}(\Lambda^k(\mathbb{R}^n))$ and merely by considering dimensions, we can see that it can't be onto.

The first non-trivial case in which it fails is when $n = 4$ and $k = 2$ where we have

$$ \dim \operatorname{GL}_4(\mathbb{R}) = 4^2 = 16, \\ \dim \operatorname{GL}(\Lambda^2(\mathbb{R}^4)) = { 4 \choose 2 }^2 = 6^2 = 36 $$

so there are (infinitely) many isomorphisms $J \colon \Lambda^2(\mathbb{R}^4) \rightarrow \Lambda^2(\mathbb{R}^4)$ which are not of the form $J = \Lambda^2(A) = \varphi(A)$.

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