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.
 A: 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
A: Here's another way.  The generating function of $A^n$ is $$G(t) = \sum_{n=0}^\infty t^n A^n = (I-tA)^{-1} = 
\dfrac{1}{-1+4t+3t^2}  \left( \begin {array}{cc} -1+4\,t&-t\sqrt {3}\\ -t
\sqrt {3}&-1\end {array} \right) 
$$
Now using a partial fraction decomposition
$$
\eqalign{\frac{1}{-1+4t+3t^2} &={\frac {\sqrt {7}}{14(t+2/3-\sqrt {7}/3)}}-{\frac {\sqrt {7
}}{14(t+2/3+\sqrt {7}/3)}}\cr
&= -\frac{\sqrt{7}}{14} \sum_{n=0}^\infty \frac{t^n}{(-2/3 + \sqrt{7}/3)^{n+1}}
+\frac{\sqrt{7}}{14} \sum_{n=0}^\infty \frac{t^n}{(-2/3 - \sqrt{7}/3)^{n+1}}
}$$
Thus with $\alpha = -2/3 + \sqrt{7}/3$ and $\beta = -2/3 - \sqrt{7}/3$,
the coefficient of $t^{51}$ in $G(t)$ is
$$ \dfrac{\sqrt{7}}{14} \left(\frac{-1}{\alpha^{52}} + \frac{1}{\beta^{52}}\right) \pmatrix{-1 & 0\cr 0 & -1\cr} + \dfrac{\sqrt{7}}{14} \left(\frac{-1}{\alpha^{51}} + \frac{1}{\beta^{51}}\right) \pmatrix{4 & -\sqrt{3}\cr -\sqrt{3} & 0\cr}$$
A: What you need is called an Eigendecomposition of the matrix $A$. See for instance http://en.wikipedia.org/wiki/Eigendecomposition_%28matrix%29 or almost any textbook on linear algebra.
A: You may proceed as follows: Consider the characteristic polynomial $p_A$ of $A$:
$$
p_A(\lambda)=\det(A-\lambda I)=\lambda(\lambda-4)-3=\lambda^2-4\lambda-3=(\lambda-2-\sqrt{7})(\lambda-2+\sqrt{7})
$$
Next, for $n \ge 3$, we find $a_n, b_n\in \mathbb{R}$ such that 
$$
\lambda^n=q(\lambda)p_A(\lambda)+a_n\lambda+b_n.
$$
We have
$$
(2+\sqrt{7})a_n+b_n=(2+\sqrt{7})^n,\ (2-\sqrt{7})a_n+b_n=(2-\sqrt{7})^n,
$$
i.e.
$$
a_n=\frac{(2+\sqrt{7})^n-(2-\sqrt{7})^n}{2\sqrt{7}},\ b_n=3\frac{(2+\sqrt{7})^{n-1}-(2-\sqrt{7})^{n-1}}{2\sqrt{7}}.
$$
Thanks to the Cayley-Hamilton theorem we have
$$
A^n=a_nA+b_nI.
$$
Now just take $n=51$.
