Re: Diagonalization of an orthogonal matrix with determinant 1

"Lucillon" <nospam@xxxxxx> writes:

I have just seen that every matrix in the special orthogonal group
SO(n,Complexnumbers)={A in M(n,C) where
Transpose(A)A=ATranspose(A)=identity-matrix and det(A)= 1}, can be
diagonalized using Lie-group theory. Does any body know how to show this
in
a simpler way for example using linear algebra?

Orthogonal matrices are unitary, and thus normal. Use the Spectral Theorem.

If that doesn't count as "simpler", use the following fact:
if U is any unitary matrix and w a complex number with |w| = 1,
T = i (U + w I)(U - w I)^(-1) is hermitian. A matrix that diagonalizes
T will also diagonalize U = w (T + i I) (T - i I)^(-1).
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

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