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Calculate $$\left(\begin{matrix} 6&2&1\\ 0&4&-12\\0&-1&5 \end{matrix}\right)^{50}$$.

It got all messed up when I tried finding the characteristic polynomial, but anyway I think the direction is trying to diagonalize the matrix...

Thanks in advance for any assistance!

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    $\begingroup$ The matrix is block upper triangular. One eigenvalue is obvious and the other two are the eigenvalues of the 2x2 matrix at bottom right. $\endgroup$ – user1551 Dec 5 '13 at 16:22
  • $\begingroup$ The characteristic polynomial here is not messy - it factors as $(6-\lambda)(8-\lambda)(1-\lambda)$. With an eigenvalue of $8$, raising to the $50$th power is going to get out of hand unless you leave it written as $8^{50}$. $\endgroup$ – alex.jordan Dec 5 '13 at 16:23
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Hint:

Find the Jordan Normal Form and use $A^{50} = PJ^{50}P^{-1}$.

You have three distinct eigenvalues, so the exponential is just each eigenvalue to the 50-th power.

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  • $\begingroup$ Spot on! $+\uparrow$ $\endgroup$ – Namaste Dec 6 '13 at 2:17
  • $\begingroup$ Indeed it is! ;-) $\endgroup$ – Namaste Dec 6 '13 at 2:21
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Let $A$ be the matrix whose 50th power is given to compute. The Eigen values of $A$ are 6, 8, 1. Hence we have $(x-1)(x-6)(x-8)=: \psi(x)$ as charactoristic polynomial. Now if we divide $x^{50}$ by $\psi(x)$ then the remainder will be of the form $ax^2+bx+c$ where we need to determine $a, b, c$.

So we have $x^{50}=\psi(x)Q(x)+ax^2+bx+c$ where $Q(x)$ is quotient in the divison. Then we must have the syatem of three equations in three unknowns $a, b, c$ as $$\begin{align} 64a+8b+c&=8^{50}\\ 36a+6b+c&=6^{50}\\ a+b+c&=1 \end{align}$$

Solve them and obtain the values of $a, b, c$ separately, say $a_0, b_0, c_0$. Then we have $x^{50}=a_0 x^2+b_0 x+c$. Hence $A^{50}=a_0 A^2+b_0 A+c_0 I_3$. Since now we know the values of $a_0, b_0, c_0$ so computations you can finish yourself.

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