Homework Help: Eigenvalues: Proof that A^3=A

1. Feb 25, 2010

noon0788

Hey, I'm wondering if I have a known set of eigenvalues (-1, +1, 0) for A, if I can prove that the matrix A = A3?

I can prove that if A3 = A, that the eigenvalues would be −1, +1, and 0. The following is the proof:

A*k=lambda*k
A3*k=lambda3*k
Since A=A3, A*k=A3*k
lambda*k=lambda3*k
lambda*k - lambda3*k = 0
lambda - lambda3 = 0
lambda*(1-lambda2) = 0
lambda = 0, -1, +1

Is there any way to prove it the other way around? If I know that the eigenvalues are 0, -1, and +1, can I prove that A3 = A?

Thanks!

2. Feb 25, 2010

tiny-tim

Welcome to PF!

Hi noon0788! Welcome to PF!

Hint: A is linear, so if three eignevectors are p q and r, what is A3(ap + bq + cr) ?

(alternatively, write A = QPQ-1)

3. Feb 25, 2010

noon0788

Thanks tiny-tim, but I'm not quite sure what you mean by A3(ap + bq + cr).

When you say A = QPQ-1, do you mean I should diagonalize the matrix? Here's what I'm thinking:

C-1AC = D (where d is the diagonalized matrix)

D = [1, 0, 0; 0, −1, 0; 0, 0, 0]

det(C-1)*det(A)*det(C) = det(D)
det(C-1)*det(C)*det(A) = det(D)
det(A) = det(D)
det(D) = 0
det(A) = 0

Maybe I'm way off...

4. Feb 25, 2010

tiny-tim

I mean where a b and c are any numbers.

(And why are you using determinants?? )

5. Feb 25, 2010

noon0788

:/ I don't know why I'm using determinants. I guess what I'm confused by is your notation of R3(ap+bq+cr). Is that I just don't recognize that notation. I'm familiar with matrices, but maybe not as familiar as I thought I was.

6. Feb 25, 2010

tiny-tim

A3(ap+bq+cr) is matrix A followed by matrix A followed by matrix A followed by the sum of a times vector p plus b times vector q plus c times vector r, with Ap = p, Aq = -q, Ar = 0.

7. Feb 25, 2010

noon0788

Thanks tiny-tim, but I think my question might just be a bit over my head. I'm confused as to why p q are r are vectors now. I thought they were eigenvalues. Also, the Ap=p stuff is confusing me too. Here's what I've got so far...

A3(a(1) + b(-1) + c(0))
A3(a-b)
A3a - A3b

8. Feb 26, 2010

elibj123

are a &b scalars or vectors?

I'll try to explain the idea better: the three eigenvectors are independent, therefore they create a basis to a three-tuple column vectors. Let's call them $$e_{1},e_{0},e_{-1}$$ (the index corresponds the eigenvalue).
Now any other vector will be:

$$\vec{v}=v_{1}\vec{e}_{1}+v_{0}\vec{e}_{0}+v_{-1}\vec{e}_{-1}$$

Now, what is Av and what is $$A^{3}v$$?

9. Feb 26, 2010

tiny-tim

Hi noon0788!

(just got up :zzz: …)
No, they're eigenvectors.

(That's why Ap = p, with p's eigenvalue = 1 )

ok, now, for any numbers a b and c, what is A(ap + bq + cr) ?

10. Feb 27, 2010

noon0788

A(ap + bq + cr)
= A(ap) + A(bq) + A(cr)
= a(Ap) + b(Aq) + c(Ar)
= a + b
right?

11. Feb 27, 2010

HallsofIvy

Well, no! For one thing, on the left side you have a vector, A(ap+ bq+cr), and on the right you have a number, a+ b. p, b, and q are eigenvectors of A with eigenvalues of 1, -1, and 0, respectively.

A(ap+ bq+ cr)= (1)ap+ (-1)bq+ (0)cr= ap- bq. A2(ap+bq+cr)= A(ap- bq)= ap-(-bq)= ap+ bq. And, finally, A3(ap+ bq+ cr)= A(ap+bq)= ap- bq.

My first thought was to use the fact that every matrix satisfies its own characteristic equation. Since A has eigenvalues 1, -1, and 0, its characteristic equation is x(x+1)(x- 1)= x^3- x.

12. Feb 27, 2010

noon0788

I think I understand it. Just one more question. Am I allowed to distribute A in A(ap + bq + cr)? So it would become aAp + bAq + cAr?

13. Feb 27, 2010

tiny-tim

Yes, because A is linear (and a b and c are scalars).