How to calculate Rotation Matrix?

How do I calculate a given matrix n-th's power formula?

• I wish to calculate a given matrix n-th power. Suppose the matrix is 3x3 and cannot be diagonalized. I've read there is a method using Cayley-Hamilton theorem to expand the matrix as a linear combination of its lower powers. Even when I do that, I can't find a generalization formula for the matrix n-th power. Is there any mathematical tool I can use to generalize it? I wish to know for the general case, of course, but in this particular case I have:A³=4A²-5A+2I, with eigenvalues 1, 1, and 2.I've been for a lot of time in this problem, and can't find a way to solve it... It's from Apostol's Calculus book, volume 2, exercise 7.12-10.

• Answer:

  If you have [math]A^3 = 4A^2 - 5A + 2I[/math], then multiply both sides by [math]A^n[/math] to get [math]A^{n+3} = 4A^{n+2} - 5A^{n+1} + 2A^n[/math] for any non-negative integer [math]n[/math]. This is a linear recurrence relation, whose characteristic polynomial is [math]\lambda ^3 = 4\lambda^2 - 5\lambda + 2[/math]. The roots of this characteristic polynomial are [math]\lambda = 1, 1, 2[/math]. Therefore, the solution to this recurrence relation is [math]A^n = (C_1 + C_2 n) \cdot 1^n + C_3 \cdot 2^n[/math] for some constant matrices [math]C_1, C_2, C_3[/math]. For [math]n = 0, 1, 2[/math], we get: [math]C_1 + C_3 = I[/math][math]C_1 + C_2 + 2C_3 = A[/math][math]C_1 + 2C_2 + 4C_3 = A^2[/math]Solving this system yields: [math]C_1 = 2A-A^2[/math][math]C_2 = -2I+3A-A^2[/math][math]C_3 = I-2A+A^2[/math]Plugging this in yields the formula: [math]A^n =[/math] [math](2^n-2n)I -[/math] [math](2^{n+1}-3n-2)A +[/math] [math](2^n-n-1)A^2[/math].This method can easily be applied to other matrices, assuming we know the eigenvalues.