Eigenvalue proof

  • Thread starter nickw00tz
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  • #1
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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 in to the characteristic equation and solving for λ
 

Answers and Replies

  • #2
lanedance
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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?
 
Last edited:
  • #3
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can we assume, for the proof, the eigenvalues are both equal to λ?
 
  • #4
lanedance
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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")
 
  • #5
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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
 
  • #6
lurflurf
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^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
 
  • #7
HallsofIvy
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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]
 

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