Let A = [cos θ -sin θ]
[sin θ cos θ]
Show that A^n = [cos(nθ) -sin(nθ)]
[sin(nθ) cos(nθ)]. Is this plausible? Explain.

(Note: Those are matrices mentioned above.)

Question

Let A = [cos θ -sin θ]
            [sin θ cos θ] 
Show that A^n = [cos(nθ)  -sin(nθ)]
                           [sin(nθ)   cos(nθ)]. Is this plausible? Explain.

(Note: Those are matrices mentioned above.)

kinggeorge · Answer

It certainly seems plausible. Try the case where n=2 first, and then you should be able to prove it for all n.

anonymous · Answer

I tried n = 2 as well and it works as well. How can I explain this if it is plausible?

kinggeorge · Answer

When you tried n=2, simply erase every 2, and rewrite n in it's place. This should be sufficient to show that it for A^n.

kinggeorge · Answer

A better way to explain in that I just noticed, would be to say that this is a rotation matrix. It rotates a vector by $\theta$ in the x-y plane. Thus, if you squared it, you would apply the rotation twice, and would rotate by $2\theta$. Then, you can say that if you applied the transformation 
n-times (A^n), you would be rotating by $n\theta$. So you should have $n\theta$ in place of $\theta$.

anonymous · Answer

I get it now. Thanks for the help. :)

kinggeorge · Answer

Sorry that took so long.

anonymous · Answer

No prob it's alright. It was a bit challenging lol.