Let Prove that where is the cross product of vectors.

Note: The proof should be asindirectas possible.

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- Sep 7th 2009, 05:56 PMNonCommAlgJacobi identity
Let Prove that where is the cross product of vectors.

__Note__: The proof should be as__indirect__as possible. - Sep 7th 2009, 06:25 PMalexmahone
- Sep 7th 2009, 09:36 PMsimplependulum
My solution was wrong , Sorry!

- Sep 8th 2009, 08:41 AMpankaj
- Sep 8th 2009, 03:49 PMNonCommAlg
- Sep 8th 2009, 06:30 PMluobo
- Sep 8th 2009, 07:19 PMluobo
- Sep 9th 2009, 05:17 AMNonCommAlg
- Sep 12th 2009, 08:49 AMhalbard
How about this.

Warning: I'm relying on the identity . This is proved by considering the identity .

First assume that , and are linearly dependent. WLOG assume there are scalars , with .

Then and and so

, proving the result in this case.

Now assume that , and are linearly independent, thus forming a basis for .

We have and .

Hence .

Similarly and .

Since , and form a basis and is orthogonal to all of them, the result follows. - Sep 13th 2009, 02:16 AMNonCommAlg
- Oct 7th 2009, 10:38 PMBruno J.
Here is a proof. Consider the special orthogonal group as a Lie group. Its Lie algebra is the set of skew-symmetric matrices, and the Lie bracket is the commutator . A skew-symmetric matrix can be associated in a natural way with a vector in , and it is easy to check that the vector associated to the bracket is . By the Jacobi identity for the Lie bracket, the result follows.

- Oct 8th 2009, 12:20 AMNonCommAlg
this is a clever solution with lots of details! (Nod) in simple words: first we note that if where is an associative ring, and if we define for all then the

Jacobi identity holds, i.e. now let be the set of skew-symmetric matrices with real entries, i.e. consists of all matrices

in the form define the map by clearly is an isomorphism of vector spaces and it's easy to see that call this result since is an associative ring, we have

taking of both sides and using the linearity of and gives us: which completes the proof.