# Homework Help: Diagonalizability of matrix

1. Jun 5, 2010

### hitmeoff

1. The problem statement, all variables and given/known data
Suppose the A $$\in$$ Mn X n(F) has two distinct eigenvalues, $$\lambda$$1 and $$\lambda$$2, and that dim(E$$\lambda$$1) = n -1. Prove A is diagonalizable.

2. Relevant equations

3. The attempt at a solution

1. The charac poly clearly splits because we have eigenvalues.
2. need to show m = dim (E).

Ok, we are given that dim(E$$\lambda$$1) = n - 1

we know multiplicity has to be 1 $$\leq$$ dim(E$$\lambda$$1) $$\leq$$ m.

so: 1 $$\leq$$ n - 1 $$\leq$$ m.

But im stuck now, not sure how to show that m = dim(E$$\lambda$$)

Last edited: Jun 5, 2010
2. Jun 5, 2010

### Office_Shredder

Staff Emeritus
Re: Diagonalizability

We're told that $$dimE^{\lambda_1}$$ is $$n-1$$. What is $$dimE^{\lambda_2}$$ then, and what can you say about $$E^{\lambda_1} \oplus E^{\lambda_2}$$?

3. Jun 5, 2010

### hitmeoff

Re: Diagonalizability

$$dimE^{\lambda_2}$$ = 1 then right?

$$E^{\lambda_1} \oplus E^{\lambda_2}$$ = V?

4. Jun 5, 2010

### hitmeoff

Re: Diagonalizability

A linear operator T on a finite-dimensional vector space V is diagonalizable iff V is the direct sum of eigenspaces of T.