Matrix Exponent Problem: Find smallest n such that Sn = I

[DmytriE]

Homework Statement

Consider the matrix

cos(3*pi/17) -sin(3*pi/17)
S = sin(3*pi/17) cos(3*pi/17)

Does there exist a positive integer n such that Sⁿ = I where I is the 2x2 identity? If so, what is the smallest such integer? Explain.

Excuse the poor matrix formatting. I do not know how to use the latex formatting to put it into pretty print.

Homework Equations

Not sure...

The Attempt at a Solution

Where should I start? I really have no idea.

 

[tiny-tim]

Hi DmytriE!

Hint: suppose S is

cosθ -sinθ
sinθ cosθ

What is S² ? S³ ? etc?

 

[DmytriE]

Let's suppose that n = 10. I don't think I have to multiply S by S 10 times to get the answer. Unfortunately the answer is not 10. How would S², S³ help me figure it out?

S²:
UL:cos²(θ) - sin²(θ)
UR: -2sin(θ)cos(θ)
LL: 2sin(θ)cos(θ)
LR: -sin²(θ)+cos²(θ)

Each abbreviation represent the place in the matrix that they would appear. UL - Upper left, etc.

S³:
a lot of sines and cosines.

 

[Mark44]

That's not what tiny-tim is suggesting. Your matrix represents a certain kind of transformation.

Instead of thinking about what S, S², S³, etc. are (in terms of their matrix representations), think about what they do to a vector they multiply.

 

[DmytriE]

Thanks for the help! This forum really is the best!

 

[tiny-tim]

Hi DmytriE!

(just got up :zzz:)

S²:
UL:cos²(θ) - sin²(θ)
UR: -2sin(θ)cos(θ)
LL: 2sin(θ)cos(θ)
LR: -sin²(θ)+cos²(θ)

have you got it now?

if not, use standard trigonometric identities