I've got a non Diagonalizable Matrix, for example: $A =\begin{bmatrix} 3&4&3\\-1&0&-1\\ 1&2&3\end{bmatrix}$
The corresponding characteristic polynomial should be $(x-2)^3$ so that means we only have one eigenvalue $\lambda_1 = 2 $.
I tried to get the Jordan–Chevalley decomposition.
So I computed: $Eig(A,2)$ which has a dimension of 1, that, as expected tells me that the matrix is not diagonalizable, then I computed $dim(null(2 I – A)^2)$ which has a dimension of 1 and then $dim(null(2 I – A)^3)$ which has a dimension of 2, therefore I'm not even able to compute the Jordan–Chevalley decomposition at this time because I would have a matrix in 4 dimensions?!
So the question is, how can I from there compute $A^n$ ? Jordan Canonical Form?
Best Answer
The power of the Jordan blocks can be written as: $$ J_k(\lambda)^n = \begin{bmatrix} \lambda^n & \binom{n}{1}\lambda^{n-1} & \binom{n}{2}\lambda^{n-2} & \cdots & \cdots & \binom{n}{k-1}\lambda^{n-k+1} \\ & \lambda^n & \binom{n}{1}\lambda^{n-1} & \cdots & \cdots & \binom{n}{k-2}\lambda^{n-k+2} \\ & & \ddots & \ddots & \vdots & \vdots\\ & & & \ddots & \ddots & \vdots\\ & & & & \lambda^n & \binom{n}{1}\lambda^{n-1}\\ & & & & & \lambda^n \end{bmatrix}$$
Hence if $A = PJP^{-1}$ is the Jordan canonical form,
then we have $A^n = PJ^n P^{-1}$