🚀 go-pugleaf

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