Can the Eigenvalues of A Determine if A^3=A?

[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!

 

[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)

 

[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...

 

[tiny-tim]

I mean where a b and c are any numbers.

(And why are you using determinants?? )

 

[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.

 

[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.

 

[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

Thanks for your help.

 

[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?

 

[tiny-tim]

Hi noon0788!

(just got up :zzz: …)

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.

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) ?

 

[noon0788]

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

 

[HallsofIvy]

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

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.

 

[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?

 

[tiny-tim]

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