Is a Matrix with Two Distinct Eigenvalues Always Diagonalizable?

[hitmeoff]

Homework Statement

Suppose the A \in M_{n X n}(F) has two distinct eigenvalues, \lambda₁ and \lambda₂, and that dim(E_\lambda1) = n -1. Prove A is diagonalizable.

Homework Equations

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_\lambda1) = n - 1

we know multiplicity has to be 1 \leq dim(E_\lambda1) \leq m.

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

But I am stuck now, not sure how to show that m = dim(E_\lambda)

 

[Office_Shredder]

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}?

 

[hitmeoff]

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}?

dimE^{\lambda_2} = 1 then right?

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

 

[hitmeoff]

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