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First off, I have this matrix A:

1 0 3
1 0 2
0 5 0

I have calculated the eigenvalues, which are (11-sqrt(141))/2 and (11+sqrt(141))/2. From what I understand, if I don't have 3 distinct eigenvalues then the matrix is not diagonalizable in R. Is this matrix diagonalizable in R?

The second part of my question is: if it's not diagonalizable then what other option do I have for calculating A^n? n is really big, something like the order of 10^12. I need to calculate this for a programming problem.

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    $\begingroup$ I think you need a different approach to your problem. $A^n$ is going to have impossibly big entries for $n=10^{12}$. Already for $n=100$ we have the row 1, col 1 entry as 21569359182880372085408476588692190023999240141617216817121. $\endgroup$ – almagest Sep 6 '14 at 11:06
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    $\begingroup$ The eigenvalues are false, check on wolframalpha $\endgroup$ – Bman72 Sep 6 '14 at 11:11
  • $\begingroup$ you may be able to prove something by induction. Do a Jordan normal decomposition, as suggested, then see what you can do with it. $\endgroup$ – ShakesBeer Sep 6 '14 at 11:11
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    $\begingroup$ The sum of the eigenvalues equals the sum of the diagonal entries. If your two eigenvalues are correct, then $-10$ must also be an eigenvalue. $\endgroup$ – Gerry Myerson Sep 6 '14 at 11:12
  • $\begingroup$ The solution of the problem will be mod 1000000006. Thanks for pointing out that my calculations were wrong. :) First things first. I need to find out where I went wrong on calculating the eigenvalues and then proceed with the solution. $\endgroup$ – Ariel Sep 6 '14 at 11:38
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Check the Jordan decomposition (diagonal matrix + nilpotent). But as pointed out in the comments, if you raise something to the power $10^{12}$, don't expect your computer to be able to handle it.

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  • $\begingroup$ The solution of the problem will be mod 1000000006. $\endgroup$ – Ariel Sep 6 '14 at 11:39
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To calculate the eigenvalues:

$$\det(xI-A)=\begin{vmatrix}x-1&0&-3\\ -1&x&-2\\ 0&-5&x\end{vmatrix}=x^2(x-1)-15-10(x-1)=x^3-x^2-10x-5$$

The above cubic has three rather ugly roots, all of them different...and thus the matrix is diagonalizable.

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