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, $$\lambda$$₁ and $$\lambda$$₂, and that dim(E_$$\lambda$$1) = 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_$$\lambda$$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 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.