Finding $B_{2\times 2}$ such that $A^{51}=B$ Assuming  $$ A=\left(\begin{matrix}0 & \sqrt{3}\\\sqrt{3}\ &4\end{matrix}\right)$$ how to find B such that $$A^{51}=B$$ 
My attempts: 
if find $P,D$ that $D$ be diagonal matrix $A=P^{-1} DP$ then $$A^{51}=P^{-1} D^{51}P$$  therefore $B=P^{-1} D^{51}P$ but  how to find $D$ and $P$?
Thanks for any hints.

You would find the eigenvalues and eigenvectors for the matrix $A$.
You should get:
$$\lambda_1 = 2 + \sqrt{7}, v_1 = \left(\left(\frac{-2 + \sqrt{7}}{\sqrt{3}}\right), ~1\right)$$
$$\lambda_2 = 2 - \sqrt{7}, v_2 = \left(-\left(\frac{2 + \sqrt{7}}{\sqrt{3}}\right), ~ 1\right)$$
This forms the Jordan Normal Form, with 
$$J = P^{-1}DP$$
You have the columns of $D$ as a linear combination of the eigenvalues.
$$D = [ \lambda_1 | \lambda_2]$$
You have the columns of $P$ as a linear combination of the eigenvectors.
$$P = [ v_1 | v_2]$$
And of course, you can easily find $P^{-1}$.
That should help you move forward.
Regards
