Re: Eigenvalue Problem

On 31 Jul., 19:11, jonathan.eth...@xxxxxxxxx wrote:
Greetings Everyone,

If I have the following eigenvalue problem: Ax = ax
where A is a square matrix, a is an eigenvalue and x is an
eigenvector.

and I can write A = B + jC (B and C real symmetric matrices, j is
sqrt(-1))

Can I state the following:

Ax = ax
(B + jC)x = (b+jc)x
Bx + jCx = bx + jcx
And solve the two eigenvalue problems separately as follows:
Bx = bx
Cx = cx


This would require x to be a simultaneous eigenvector of B and C.
Unless BC=CB, I wouldn't bet on that.

E.g. let C= [[1 0] [0 2]], B=[[1 2] [2 1]].
Then (1 0) and (0 1) are eigenvectors of C with eigenvalues 1, 2,
(1 1), (1 -1) are eigenvectors of B with eigenvalues 3, -1.
However, for A = [[1+j 2] [2 1+2j]], none of these is an eigenvector.



If this is a valid operation, is the set of eigenvectors 'x'
determined in both problems the same?

thanks
-Jon


.

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