Re: al matrices
- From: Robert Israel <israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: 25 Jun 2007 21:52:13 -0400
loup blanc <bourgeois_gerald@xxxxxxxx> writes:
Let u be a real number,v be a non real complex number and w be its
conjugate.
Let A,B,C be real (n,n) matrices which are simultaneously diagonalizable
over the complex field:
A=Pdiag(ai)P^-1, B=Pdiag(bi)P^-1, C=Pdiag(ci)P^-1 where, for all i,
(ai,bi,ci) is a permutation of (u,v,w).
I think that necessarily n is even.
Is it true?
Yes.
I'll use CC for the complex number field to distinguish it from the matrix C.
Let U = Ker(A-uI) = {x in CC^n: A x = u x}. Then U is the direct sum
U_1 + U_2 where U_1 = {x in U: B x = v x} and U_2 = {x in U: B x = w x}.
The complex conjugation operator J, which is linear over the reals R,
leaves U invariant and interchanges U_1 and U_2. Thus U_1 and U_2 must
have the same dimension over R, and therefore also over CC (since they are
vector spaces over CC). So the dimension of U over CC is even. Similarly,
Ker(B-uI) and Ker(C-uI) have even dimensions. Since CC^n is the direct
sum of Ker(A-uI), Ker(B-uI) and Ker(C-uI), its dimension n is also even.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada V6T 1Z2
.
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