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?
Thanks in advance for your help!