# Eigenvalue proof

nickw00tz

## Homework Statement

Proof: Prove that if A is an nxn (square mtx) such that A^2=A, then A has 0 or 1 as an eigenvalue.

## The Attempt at a Solution

A=A^2
A^2-A=0
A(A-I)=0
A=0 or A=1
and then plugging the A solutions into the characteristic equation and solving for λ

Homework Helper
start with eigenvalue $lambda$ with a correpsonding eigenvector u.

Is u also an eigenvector of A^2?

if so what is the corresponding eigenvalue?

Last edited:
nickw00tz
can we assume, for the proof, the eigenvalues are both equal to λ?

Homework Helper
can we assume, for the proof, the eigenvalues are both equal to λ?

no, don't assume, but can you show it?

Also though I get your meaning please be explicit in you question (eg. what do you mean by "both")

nickw00tz
Sorry about that, what I meant was could we associate λ as an eigenvector for A and A^2. For example:

If Au=λu
then (A^2)u=λu, where u=/=0

Homework Helper
^What do you mean both there are n eigenvalues.
Are you working over a splitting field?
Sketch of proof
1)since A=A2
A and A2 have the same eigenvalues
2)find out when A and A2 have the same eigenvalues

If v is an eigenvector of A with eigenvalue $\lambda$, then $A^2v= A(Av)= A(\lambda v)= \lambda Av= \text{what?}$