TITLE: Find large power of a non-diagonalisable matrix
QUESTION [14 upvotes]: If $A = \begin{bmatrix}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$, then find $A^{30}$.

The problem here is that it has only two eigenvectors, $\begin{bmatrix}0\\1\\1\end{bmatrix}$ corresponding to eigenvalue $1$ and $\begin{bmatrix}0\\1\\-1\end{bmatrix}$ corresponding to eigenvalue $-1$. So, it is not diagonalizable.
Is there any other way to compute the power?

REPLY [11 votes]: Given $\mathrm A$, we compute one Jordan decomposition, $\mathrm A = \mathrm P \mathrm J \mathrm P^{-1}$, where
$$\mathrm J = \begin{bmatrix} -1 & 0 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\end{bmatrix}$$
Note that $\mathrm A^{30} = \mathrm P \mathrm J^{30} \mathrm P^{-1}$. Since $(-1)^{30} = 1$ and $\begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix}^{30} = \begin{bmatrix} 1 & 30\\ 0 & 1\end{bmatrix}$, we have
$$\mathrm J^{30} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 30\\ 0 & 0 & 1\end{bmatrix}$$
Thus,
$$\mathrm A^{30} = \mathrm P \mathrm J^{30} \mathrm P^{-1} = \cdots = \begin{bmatrix} 1 & 0 & 0\\ 15 & 1 & 0\\ 15 & 0 & 1\end{bmatrix}$$