mcanally@kurims.kyoto-u.ac.jp (David Scott McAnally) (05/15/91)
In article <1991May14.102145.5659@bnlux1.bnl.gov> kyee@bnlux1.bnl.gov (kenton yee) writes: >hi, >we know any hermitian matrix H and any antihermitian matrix >A is diagonalizable. i want to prove that any linear combination >M = aH + bA are diagonalizable (a,b real). one way is to >note that M' = aH + ibA is hermitian and diagonalizable, >so one can continue M' to M by taking b->-ib. this should >be ok unless one of the eigenvectors has a singularity blocking >the analytic continuation. does anyone have a better proof? >also, is there a relation between the eigenvectors of H, A >and M? > >thanks, --ken Surely, this requires specific knowledge of the matrices involved, eg, the (hermitian) Pauli matrices sigma , sigma and sigma are 1 2 3 diagonalizable, but sigma +/- i sigma are not. 1 2 If H and A commute, then there are no problems with diagonalizability. David McAnally kurims.kurims.kyoto-u.ac.jp ``Other kings said I was daft to build a castle on a swamp." King of Swamp Castle: Monty Python and the Holy Grail
jwunsch@phoenix.Princeton.EDU (Jared Wunsch) (05/15/91)
In article <1991May14.102145.5659@bnlux1.bnl.gov> kyee@bnlux1.bnl.gov (kenton yee) writes: >hi, >we know any hermitian matrix H and any antihermitian matrix >A is diagonalizable. i want to prove that any linear combination >M = aH + bA are diagonalizable (a,b real). Can this be true? It seems to me that given an arbitrary matrix A, if we let A* be its Hermitian transpose, then we have A = (A+A*)/2 + (A-A*)/2 ie. A is a linear combination of a Hermitian and an anti-Hermitian matrix. So if the conjecture were true, we'd have every matrix diagonalizable.