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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- Sep 3rd 2011, 10:31 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) - Sep 3rd 2011, 06:51 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 $\displaystyle V$ is a finite dimensional vector space, $\displaystyle T$ an operator on $\displaystyle V$, $\displaystyle \lambda$ an eigenvalue of $\displaystyle T$ and $\displaystyle W$ the generalized eigenspace corresponding to $\displaystyle \lambda.$

suppose that $\displaystyle W$ is an eigenspace. then there exists a scalar $\displaystyle \mu$ such that $\displaystyle W = \{x \in V: \ Tx = \mu x \}$. now if $\displaystyle (T-\lambda I)^k x = 0,$ for some $\displaystyle k \geq 1$ and

$\displaystyle x \neq 0$, then $\displaystyle \sum_{i+j=k} (-1)^i \binom{k}{i}\lambda^i T^j(x)=0$ and, since $\displaystyle T^j(x)=\mu^j x,$ we'll get $\displaystyle \sum_{i+j=k}(-1)^i \binom{k}{i}\lambda^i \mu^jx = 0.$ thus $\displaystyle (\mu - \lambda)^kx = 0$ which implies $\displaystyle \mu = \lambda.$

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

for example, consider the linear transformation $\displaystyle T: \mathbb{C}^2 \longrightarrow \mathbb{C}^2$ defined by $\displaystyle T(x,y)=(x,0).$ then the generalized eigenspace corresponding to $\displaystyle \lambda = 0$ is the eigenspace corresponding to $\displaystyle \lambda=0.$ - Sep 4th 2011, 07:55 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? - Sep 4th 2011, 04:49 PMHallsofIvyRe: Generalized eigenspace
Every matrix satisfies its own characteristic equation. That is, if the characteristic equation for matrix A is $\displaystyle (\lambda- a)^n= 0$ 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.