Can someone help?

1. Is a generalized eigenspace an eigenspace?

2. How can we be sure that a generalized eigenspace has a full rank?

(Bow)

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- September 3rd 2011, 11:31 AMerich22Generalized eigenspace
Can someone help?

1. Is a generalized eigenspace an eigenspace?

2. How can we be sure that a generalized eigenspace has a full rank?

(Bow) - September 3rd 2011, 07:51 PMNonCommAlgRe: Generalized eigenspace
regarding your second question, i'm not sure what you mean by a "full rank eigenspace". so i'll wait until you define that term.

your first question is easy to answer though:

suppose that is a finite dimensional vector space, an operator on , an eigenvalue of and the generalized eigenspace corresponding to

suppose that is an eigenspace. then there exists a scalar such that . now if for some and

, then and, since we'll get thus which implies

so the only way that a generalized eigenspace becomes an eigenspace is just the trivial one.

for example, consider the linear transformation defined by then the generalized eigenspace corresponding to is the eigenspace corresponding to - September 4th 2011, 08:55 AMerich22Re: Generalized eigenspace
Thank u so much for the 1st answer. It really2 helps:).

For the second question, I mean how can we be sure that a generalized eigenspace W has the same dimension as the algebra multiplicity of the eigenvalue,lambda? - September 4th 2011, 05:49 PMHallsofIvyRe: Generalized eigenspace
Every matrix satisfies its own characteristic equation. That is, if the characteristic equation for matrix A is then (A- aI)^nv= 0 for

**every**v. It might happen that there are n independent eigenvectors in which case the "eigenspace" has dimension n. If not, if the eigenspace has dimension m< n, for v NOT in that eigenspace, we must have (A- aI)^(n-m)v= 0 so that v is a "generalized" eigen value.