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Forum index » Science and Technology » Math » Undergraduate
generalized eigenvector
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Jeremy Watts
science forum Guru Wannabe


Joined: 24 Mar 2005
Posts: 239

PostPosted: Fri Jun 16, 2006 5:18 pm    Post subject: generalized eigenvector Reply with quote

the definition of a generalized eigenvector (not the same generalized
eigenvector of the generalized eigenvalue problem), is according to
'Schaum's Outlines' :-

A vector Xm is a generalized vector if :-

(A - lambdaI)^m Xm = 0

but,

(A - lambdaI)^(m-1) Xm =/= 0

where Xm is an eigenvector of rank 'm', 'A' is a square matrix and 'lambda'
an associated eigenvalue.

But according to wikipedia, the definition is simply :-

http://en.wikipedia.org/wiki/Generalized_eigenvector

why do these definitions differ??
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G.E. Ivey
science forum Guru


Joined: 29 Apr 2005
Posts: 308

PostPosted: Fri Jun 16, 2006 6:31 pm    Post subject: Re: generalized eigenvector Reply with quote

Quote:
the definition of a generalized eigenvector (not the
same generalized
eigenvector of the generalized eigenvalue problem),
is according to
'Schaum's Outlines' :-

A vector Xm is a generalized vector if :-

(A - lambdaI)^m Xm = 0

but,

(A - lambdaI)^(m-1) Xm =/= 0

where Xm is an eigenvector of rank 'm', 'A' is a
square matrix and 'lambda'
an associated eigenvalue.

But according to wikipedia, the definition is simply
:-

http://en.wikipedia.org/wiki/Generalized_eigenvector

why do these definitions differ??


I don't see any difference! The definition you give essentially says that v is a generalized eigenvector if and only if (A- lambdaI)^m v= v for some number m. The wikpedia reference talks about there being (A- lambdaI) x_k= x_(k-1) and, finally, (A- lambdaI)^m x_(m-1)= 0. Of course, if (A- lambdaI)^m x= 0 then you can define the x_k as (A- lambdaI)^k x.
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