Cartan's Criterion for Solvability -- Lie Algebras 13

[@lexinwonderland5741]

i'm always thrilled to see an update to this series!!! i have no doubt you're busy as hell with the channel AND teaching AND god knows what else, but for what its worth the MathMajor series have fans who can't get to academia like me and your courses here are SO valuable.

[@2tehnik]

28:05 what earlier trace relationship? I'm especially confused because I don't see how, in the end, we used the assumption about the Killing form being 0.

[@arnouth5260]

The earlier trace relationship is the assumption that k([x,y],z) = 0 for all x,y,z in L. This is where we use the assumption of the Killing form being zero.

[@Calcprof]

There's a nice geometric proof of the Jordan Decomposition Theorem. Consider a matrix A. Look at S = Intersection over n of range(A^n) and T = union over n of ker(A^n). Easy to see R^n = S +T, and that's and T are invariant subspaces of A. A is nilpotent on T, and invertible on S. With a little more work (look at det(A^k - lambda I) for large k, must have a zero..), A is diagonalizable on S.