Re: Quadratic Matrix Equation

From: Jon Slaughter (Jon_Slaughter_at_Hotmail.com)
Date: 10/22/04 

Date: Thu, 21 Oct 2004 20:54:14 -0500

"Joshua Marshall" <joshua.marshall@gmail.com> wrote in message
news:7383a587.0410211700.43183c37@posting.google.com...
> Consider the matrix equation:
>
> \lambda^2*I_n - \lambda*D - H = 0,
>
> where \lambda is a scalar, I_n the n x n identity, D a diagonal n x n
> matrix and H an n x n matrix whose eigenvalues are known.
>
> Is it possible to solve for the 2*n possible values for \lambda? Note
> that the problem simplifies if the diagonal entries of D are all the
> same, but I'm interested in the general solution. Any ideas?!
>
> Josh

If the eigenvalues are distinct, then you can write H = P D'/P

so, use multiply both sides of your equation by P and /P, and you get
something like this:

L^2*I - L M - D' = 0

Which implies that M and H are similar(well, they both are diagonalizable,
so ofcourse)

subtracting the two equations give

L(M - D) + (H - D') = 0

or LB - C = 0

then if B is similar to a diagonal so that B = SD''/S

you get

LD'' - E = 0

now, the "crucial" step where L^2 was removed didn't actually depend on H
being diagonalizable, you can actually use any two matricies P and D' such
that H = PD'/P and you can get rid of L^2.... though, you will have to check
that L works cause maybe its possible to pick up extraneous solutions(
haven't really checked this method out, since I just came up with it... and
so I could be missing the point...).

Relevant Pages

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