Diagonalization to power of matrix

Question

$\left(\array{ 2&3 \\ 5& 1 }\right)^{20}$

$(\lambda-1)(\lambda-2)-15=0$

$\lambda = (3\pm \sqrt{61})/2$

Is this problem should be solved in this method? Lambda is not tidy.

Answer 1

I see two method two calculate the powers of the matrix in your example.

First method : you diagonalize your matrix. Then it is easy.

Second method : If I note $M$ your matrix, you want to calculate $M^{20}$. You know that $M^2 - M + 15 = 0$ (Cayley Hamilton theorem). So by putting $Q = X^2 - X + 15$, and $X^{20} = PQ + R $ where $deg(R) \leq 1$ (so $R = a_{20} + b_{20} X), M^{20} = R(M).$ And it is possible to calculate $R$ by looking its value at your two lamdas. An another way is to make a recursion (but it is not necessary to calculate iterated calculus 20 times...).

If your question is : how I calculate $\lambda^{20} = a_{20} \times 3 \pm b_{20} \times \sqrt{61}$, the answer is you do not have a better way to do that iterating calculus, so don't do it.

EDIT :

I've seen an another answer ; yes, it is a bit faster if you apply exponentiating by squaring, to try to calculate more explicitely $\lambda^{20}$, but it is still dull. I advice you to keep $\lambda^{20}$ in your result.