# Exterior Algebra: Find a linear operator

I have the following problem:

Let $J: \wedge^{2} \mathbb{R}^{3} \to \wedge^{2} \mathbb{R}^{3}$ isomorphism linear. Find $A: \mathbb{R}^{3} \to \mathbb{R}^{3}$ linear operator such that $\wedge^{2} A = J$.

Recall: If $T: \mathbb{R}^{3} \to \mathbb{R}^{3}$ is linear operator then the linear operator $\wedge^{2} T : \wedge^{2} \mathbb{R}^{3} \to \wedge^{2} \mathbb{R}^{3}$ is given by $\wedge^{2} T(u_{1} \wedge u_{2}) = T(u_{1}) \wedge T(u_{2})$.

One way to do it: is to solve the system equations given by

Let $\alpha = (\alpha_{1},\alpha_{2},\alpha_{3}), \beta = (\beta_{1},\beta_{2},\beta_{3}), \gamma = (\gamma_{1},\gamma_{2},\gamma_{3})$ are linearly independent (LI).

Find $u = (u_{1},u_{2},u_{3}), v = (v_{1},v_{2},v_{3}), u = (w_{1},w_{2},w_{3})$ such that

$\alpha_{1} = u_{1}v_{2} - v_{1}u_{2}$

$\alpha_{2} = u_{1}v_{3} - v_{1}u_{3}$

$\alpha_{3} = u_{2}v_{3} - v_{2}u_{3}$

$\beta_{1} = u_{1}w_{2} - w_{1}u_{2}$

$\beta_{2} = u_{1}w_{3} - w_{1}u_{3}$

$\beta_{3} = u_{2}w_{3} - w_{2}u_{3}$

$\gamma_{1} = v_{1}w_{2} - w_{1}v_{2}$

$\gamma_{2} = v_{1}w_{3} - w_{1}v_{3}$

$\gamma_{3} = v_{2}w_{3} - w_{2}v_{3}$

(We must have $u,v,w$ LI.)

But I can´t solve this. Which kind of theory can I use?

Let $e_1,e_2,e_3$ denote the standard basis of $\Bbb R^3$. It suffices to note that $$(A e_1)\wedge (A e_2) = J(e_1 \wedge e_2)\\ (A e_1)\wedge (A e_3) = J(e_1 \wedge e_3)\\ (A e_2)\wedge (A e_3) = J(e_2 \wedge e_3)$$ $Ae_1$ is perpendicular to both $(A e_1)\wedge (A e_2)$ and $(A e_1)\wedge (A e_3)$. Thus, $J(e_1 \wedge e_2) \wedge J(e_1 \wedge e_3)$ gives us a vector in the direction of $Ae_1$. Similarly, obtain vectors in the directions of $Ae_2$ and $A e_3$.
• If $J: \wedge \mathbb{R}^{4} \to \wedge \mathbb{R}^{4}$ isomorphism linear. How can I find $A: \mathbb{R}^{4} \to \mathbb{R}^{4}$ linear operator such that $\wedge^{2} A = J$? Thank you – Alladin Dec 15 '16 at 23:06
• In general, 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 $\wedge^{k} A = J$ ? Thank you – Alladin Dec 15 '16 at 23:14
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