Eigenvalue Proof: Proving A^2=A has 0 or 1 as an eigenvalue

[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

 

[lanedance]

start with eigenvalue [itex] lambda [/itex] with a correpsonding eigenvector u.

Is u also an eigenvector of A^2?

if so what is the corresponding eigenvalue?

 

[nickw00tz]

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

 

[lanedance]

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

 

[lurflurf]

^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

 

[HallsofIvy]

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