# Prove A is Diagonalizable (Actual Question)

1. Mar 21, 2008

### pezola

[SOLVED] Prove A is Diagonalizable (Actual Question)

1. The problem statement, all variables and given/known data

Suppose that A $$\in$$ M$$^{nxn}$$(F) and has two distinct eigenvalues, $$\lambda$$$$_{1}$$ and $$\lambda$$$$_{2}$$, and that dim(E(subscript $$\lambda$$$$_{1}$$ ))= n-1. Prove that A is diagonalizable.

3. The attempt at a solution

So far, I know that dim(E subscript $$\lambda$$2) $$\geq1$$
and that
dim(E subscript $$\lambda$$1) + dim(E subscript $$\lambda$$2) $$\leq$$ n.
So dim(E subscript $$\lambda$$2) = 1.

I am not exactly sure how this helps me to show A is digonalizable. Maybe I am thinking of something else and don't need this to prove A is diagonalizable. Please help.

(Also, sorry about my prevous blank post; I am new)

2. Mar 21, 2008

### tiny-tim

Welcome to PF!

Hi pezola! Welcome to PF!

Your reasoning seems fine.

Can't you now use proof by induction - that is, assume the theorem is true for all numbers up to n - 1, and then prove it for n?
… no problemo! …

3. Mar 21, 2008

### StatusX

A matrix is diagonalizable iff its eigenvectors form a basis for the space (do you understand why?). You've pretty much shown this, maybe just another word or too on why the eigenvectors are independent.