Let a 2 x 2 matrix A = [ cos x sin x , -sin x cos x].?

Question

Let a 2 x 2 matrix A = [ cos x sin x , -sin x cos x]. Prove or disprove the conjecture that (A)^k = [cos kx sin kx ,
-sin kx cos kx].

Answer 1

this is true
prove using
Mathematical induction
[[first]]
[ cos x sin x , -sin x cos x] = [ cos1x sin1x , -sin1x cos1x]
so then n = 1 is true
[[second]]
assume n = k is true
show
A*A^k that is
[ cos x sin x , -sin x cos x] * [ coskx sinkx , -sinkx coskx]
= [ cos(k+1)x sin(k+1)x , -sin(k+1)x cos(k+1)x]
(that is = A^(k+1))
by direct calculation

this proves is is true for all k Є IN

the end
.

Answer 2

The easiest proof is inductive, the interesting point is that the statement is valid for every integer k (even k<0). Let's begin with k = 0 (a little bit unusual induction base, but it works!), You have:
A^0 = [cos 0, sin 0, -sin 0, cos 0] = [1,0,0,1] the unary matrix.
Let's assume A^k = [cos kx sin kx , -sin kx cos kx] for k ≥ 0, then
A^(k+1) = A^k*A = [cos kx; sin kx; -sin kx; cos kx]*[ cos x; sin x; -sin x; cos x] =
=[ (cos kx)*(cos x) - (sin kx)*(sin x) ,
(cos kx)*(sin x) + (sin kx)*(cos x) ,
(-sin kx)*(cos x) + (cos kx)*(-sin x) ,
(cos kx)*(cos x) - (sin kx)*(sin x) ] =
=[cos (k+1)x; sin (k+1)x; -sin (k+1)x; (cos (k+1)x] according additional formulas for the sine and cosine, A^k and A being commutative, so A^(k+1) is expressed the same way what completes the proof for k ≥ 0.
Now note that A^(-1) = [cos x; -sin x; sin x; cos x] /e.g. by multiplying A*A^(-1) You obtain the identity matrix/ - this is the so called ORTHOGONAL matrix /its inverse is equal to its transpose!/. Assuming for k ≥ 0 that
A^(-k) = [cos kx; -sin kx; sin kx; cos kx], multiplying
A^(-k)*A^(-1) same way as above You complete the induction downwards /from A^(-k) to A^(-(k+1))/, so the statement is true for every integer k.

Answer 3

Matrix A =
cosα sinα
-sinα cosα
rotates vector x by angle α counter-clockwise.

Consequently (A^k)x = AAAA..AAx rotates vector x by angle kα.

Answer 4

How come your 2X2 matrix only has 1 row?