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!


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 $A e_3$.

Now, it's sufficient to compute the length of each vector.

  • $\begingroup$ 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 $\endgroup$ – Alladin Dec 15 '16 at 23:06
  • $\begingroup$ 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 $\endgroup$ – Alladin Dec 15 '16 at 23:14
  • $\begingroup$ If you'll answer, please do here ( math.stackexchange.com/questions/2060585/… ). $\endgroup$ – Alladin Dec 15 '16 at 23:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.