Diagonalizable matrices

A matrix $A$ is diagonalizable if there exists a basis $x_1$, …, $x_n$ of eigenvectors. In that case, we can write:

$$A=X\Lambda X^{-1}$$

where column $i$ of $X$ is $x_i$ and $\Lambda$ is a diagonal matrix with ${\lambda }_i$ on the diagonal:

$$A{x}_i={\lambda }_i{x}_i$$

A key result is that $A^k$ has a very simple expression:

$$A^k=X{\Lambda }^k{X}^{-1}$$

where ${\Lambda }^k$ is diagonal with ${\lambda }_i^k$ on the diagonal.

Matrix-vector and matrix-matrix product, Invertible matrix, Eigenvalues