Prove that $A^{10}$ is equal to linear combination of $A^k, k = 1,...,9$ and identity matrix. Let $A=\begin{bmatrix}2&0&1\\0&1&0\\1&0&1\end{bmatrix}$. 
I did this with brute force and it was messy, is there a more theoretical way?  

The characteristic polynomial of $A$ is 
$$P_A(t)= \det (t I_3 - A) = t^3- 4 t^2 + 4 t  -1$$
The matrix $A$ satisfies the polynomial equality
$$P_A(A) = A^3 - 4A^2 + 4A - 1 =0$$
and therefore $A^3 = 4A^2 - 4A + 1$ and so $A^{10} = A^7(2A-1)^2$, a polynomial of degree $9$. 
We can find a unique polynomial $R(t)$ of degree $2$ that satisfies $A^{10}= R(A)$; $R(t)$ is the remainder of the division of $t^{10}$ by $P_A(t)$. We have 
$$t^{10} = ( t^3- 4 t^2 + 4 t  -1) ( t^7+4 t^6+12 t^5+33 t^4+88 t^3+232 t^2+609 t+1596)+\\ +\,4180 t^2-5775 t+1596$$ 
and so 
$$A^{10} = 4180\, A^2-5775\, A+1596$$
Obs: $4180 t^2-5775 t+1596$ is the unique polynomial of degree $\le 2$ that coincides with $t^{10}$ on the set of zeroes $\{1, \frac{3 + \sqrt{5}}{2}, \frac{3 - \sqrt{5}}{2}\}$ of the polynomial $P_A(t)$ -- the eigevalues of $A$.
