Generalized eigenspace

Printable View

September 3rd 2011, 10:31 AM
erich22

Generalized 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, 06:51 PM
NonCommAlg

Re: Generalized eigenspace

Quote:

Originally Posted by erich22

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)

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 $V$ is a finite dimensional vector space, $T$ an operator on $V$ , $\lambda$ an eigenvalue of $T$ and $W$ the generalized eigenspace corresponding to $\lambda.$

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

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

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

for example, consider the linear transformation $T: \mathbb{C}^2 \longrightarrow \mathbb{C}^2$ defined by $T(x,y)=(x,0).$ then the generalized eigenspace corresponding to $\lambda = 0$ is the eigenspace corresponding to $\lambda=0.$
September 4th 2011, 07:55 AM
erich22

Re: 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, 04:49 PM
HallsofIvy

Re: Generalized eigenspace

Every matrix satisfies its own characteristic equation. That is, if the characteristic equation for matrix A is $(\lambda- a)^n= 0$ 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.