Find large power of a non-diagonalisable matrix 
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?

Notice the characteristic polynomial of $A$ is 
$$\chi_A(\lambda) \stackrel{def}{=}\det(\lambda I_3 - A) = \lambda^3-\lambda^2-\lambda+1
= (\lambda^2-1)(\lambda-1)$$
By Cayley-Hamilton theorem, we have
$$\chi_A(A) = (A^2 - I)(A-I) = 0 \quad\implies (A^2-I)^2 = (A^2-I)(A-I)(A+I) = 0$$
This means $A^2-I$ is nilpotent. In following binary expansion of $A^{30}$
$$A^{30} = (I + (A^2 - I))^{15} = \sum_{k=0}^{15} \binom{15}{k}(A^2-I)^k$$
only the term $k = 0$ and $1$ contributes. i.e.
$$A^{30} = I + 15 (A^2-I) = 
\begin{bmatrix}
1 & 0 & 0\\
15 & 1 & 0\\
15 & 0 & 1
\end{bmatrix}
$$
