Also deduce the Jacobi identity for the cross product: u × (v × w) +w × (u × v) +v × (w x u) = 0 их
Also deduce the Jacobi identity for the cross product: u × (v × w) +w × (u × v) +v × (w x u) = 0 их
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
fron Naive Lie theory (John Stillwell)
Need help with attached highligted 1.4.4
Thankyou

Transcribed Image Text:Exercises
The cross product is an operation on Ri+Rj+Rk because u x v is in Ri+Rj+Rk
for any u, v E Ri+Rj+Rk. However, it is neither a commutative nor associative
operation, as Exercises 1.4.1 and 1.4.3 show.
1.4.1 Prove the antisymmetric property u × v = –v × u.
1.4.2 Prove that u × (v × w) = v(u ·w) – w(u ·v) for pure imaginary u, v, w.
1.4.3 Deduce from Exercise 1.4.2 that x is not associative.
1.4.4 Also deduce the Jacobi identity for the cross product:
их (vx w)+wx (ихv) +vx (wx
х и) — 0.
The antisymmetric and Jacobi properties show that the cross product is not com-
pletely lawless. These properties define what we later call a Lie algebra.
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