Powers of a diagonalisable matrix

Summary

Let $F$ be a field, and let $A$ be an $n \times n$ matrix over $F$ with $A = P D P^{-1}$. Here $P$ is invertible and $D$ is diagonal with entries $\lambda_1, \dots, \lambda_n$. Then, for each integer $k \ge 0$, $$A^k = P D^k P^{-1},$$ where $D^k$ is the diagonal matrix with entries $\lambda_1^k, \dots, \lambda_n^k$. So the powers of a diagonalisable matrix are easy to compute.

Remarks

The factors $P^{-1} P$ in $(P D P^{-1})^k$ cancel in pairs. The same idea gives $p(A) = P\, p(D)\, P^{-1}$ for each polynomial $p$. Over $\R$ or $\C$, it also gives the matrix exponential $e^A = P e^D P^{-1}$. It is the standard method to solve linear recurrences, such as the Fibonacci recurrence.

Explanations

L1 human

To raise a diagonalisable matrix to a power, change to the basis of eigenvectors. There the matrix is diagonal, and its power is the diagonal of the powers of the eigenvalues. Then change back to the old basis.