# 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.

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Compute the eigenvectors of $A$. – David Wheeler Jan 27 '13 at 19:32
...together with the eigenvalues. The eigenvalues constitutes $D$, and eigenvectors...$P$. – Tapu Jan 27 '13 at 19:33
A has real eigen value but let A have complex eigen value then eigen vector correspond to this eigen vector has two part (imaginary and real ) how find p now ? – Maisam Hedyelloo Jan 27 '13 at 19:45

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 $$\dfrac{\sqrt{7}}{14} \left(\frac{-1}{\alpha^{52}} + \frac{1}{\beta^{52}}\right) \pmatrix{-1 & 0\cr 0 & -1\cr} + \dfrac{\sqrt{7}}{14} \left(\frac{-1}{\alpha^{51}} + \frac{1}{\beta^{51}}\right) \pmatrix{4 & -\sqrt{3}\cr -\sqrt{3} & 0\cr}$$

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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 $P^{-1}$.

Regards

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Clear proof, Amzoti. – S. Snape Jan 27 '13 at 20:03
@BabakSorouh: Thank you, I really like your responses to problems! Regards – Amzoti Jan 27 '13 at 20:03
Great work here! +1 – amWhy May 6 '13 at 0:12

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.

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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 $n=51$.

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