A rotation matrix turns every vector in the plane through a fixed angle. So what happens when you apply it again and again? Applying a rotation of θ twice is just a rotation of 2θ; three times, 3θ; and so on. Matrix power is exactly this idea written algebraically: An means "apply the transformation A a total of n times."
The intuition
Multiplying a vector by a matrix A transforms it once. Multiplying by A again transforms the result once more. Hence
An=n timesA⋅A⋯A,
with the conventions A1=A and A0=I (the identity), just as x0=1 for numbers.
Watch out
An is not raising each entry to the power n. You must carry out full matrix multiplication. For example, with B=(1011), B2=(1021) — the top-right entry becomes 2, not 12.
The rotation case
The cleanest example is the rotation matrix through angle θ (counterclockwise):
Rθ=(cosθsinθ−sinθcosθ).
Because stacking two rotations adds their angles,
Rθn=Rnθ=(cosnθsinnθ−sinnθcosnθ).
Proving it by induction
This is a classic exam result, proved by mathematical induction on n.
Base case (n=1): Rθ1=Rθ=R1⋅θ, true.
Inductive step: assume Rθk=Rkθ. Then
Rθk+1=RθkRθ=RkθRθ.
Multiplying the two matrices and using the addition formulas
By induction: true for n=1; assuming it for n=k and multiplying by A, the angle-addition formulas give the (k+1) form. Hence An=[cosnθ−sinnθsinnθcosnθ] for all n∈N. …
Use mathematical induction: the base n=1 is the matrix itself, and multiplying the n=k form by A and applying cos(kθ+θ), sin(kθ+θ) gives the n=k+1 form.
Concept. Prove a statement P(n) for all natural n by (a) base case and (b) P(k)⇒P(k+1).
Base case n=1.A1=[cosθ−sinθsinθcosθ] — the formula holds.
Inductive step. Assume Ak=[coskθ−sinkθsinkθcoskθ]. Then