# T-Invariant Subspaces

1. Jun 4, 2010

### hitmeoff

1. The problem statement, all variables and given/known data
Show that W is a T-invariant subspace of T for:
W = E$$_{\lambda}$$

2. Relevant equations

3. The attempt at a solution

Ok, so I know that I need to show that T maps every element in E$$_{\lambda}$$ to .

E$$_{\lambda}$$ = N(T-$$\lambda$$I)

so T must map every eigenvector related to $$\lambda$$ to another eigenvector in E$$_{\lambda}$$

T(x) maps to zero vector, when x is an eigenvector associated with $$\lambda$$ which is in the eigenspace of $$\lambda$$, correct?

Last edited: Jun 4, 2010
2. Jun 4, 2010

### Office_Shredder

Staff Emeritus
Your post is just a huge blank space where you messed up the tex for making a lambda. Can you edit it?

3. Jun 4, 2010

fixed

4. Jun 4, 2010

### Office_Shredder

Staff Emeritus
Ok I get it now. $$E^{\lambda}$$ is the set of all eigenvectors with eigenvalue $$\lambda$$. Let's say v is an eigenvector, and $$Tv=\lambda v$$. What is $$T(\lambda v)$$ and how does this help you answer the question?

5. Jun 4, 2010

### hitmeoff

T($$\lambda$$v) = $$\lambda$$2v

and this is just a multiple of T, so T(v) = $$\lambda$$v maps to the eigenspace?

6. Jun 4, 2010

### hitmeoff

and another question...

Show that W is a T-invariant subspace when

W = N(T)

So N(T) : { x E W: T(x) = 0} , since W is a subspace it contains the zero vector, thus any vector in N(T) will map to zero which is in W?