Prove A is Diagonalizable (Actual Question)

[pezola]

[SOLVED] Prove A is Diagonalizable (Actual Question)

Homework Statement

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.

The Attempt at a Solution

So far, I know that dim(E subscript \lambda2) \geq1
and that
dim(E subscript \lambda1) + dim(E subscript \lambda2) \leq n.
So dim(E subscript \lambda2) = 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)

 

[tiny-tim]

Welcome to PF!

So far, I know that dim(E subscript \lambda2) \geq1
and that
dim(E subscript \lambda1) + dim(E subscript \lambda2) \leq n.
So dim(E subscript \lambda2) = 1.

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?

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

… no problemo! … ​

 

[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.