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I have matrix $A$ of $2 \times 2$ and want to raise it to a power of $n$.

I was thinking about getting $J$, the Jordan form of $A$ and then separating $J$ to two matrices: $S$ (diagonal matrix) and $N$ (nilpotent matrix) and then using the binomial to get $J^n = (S + N)^n$.

Is that the right way of doing it ?

Thanks :)

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  • $\begingroup$ Yes, that makes things tractable. $\endgroup$ – Hagen von Eitzen May 12 '14 at 10:28
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If your matrix is diagonalizable we're done (as $\;N=0\;$ in your notation), otherwise we have that

$$\exists\,a\in\Bbb F:=\text{ the definition field}\;\;s.t.\;\;A=P^{-1}JP\;\;\text{and}\;\; J=\begin{pmatrix}a&1\\0&a\end{pmatrix}\;\;\text{(why?)}\;\implies$$

but then it's easy to prove (simple induction) that

$$J^n=\begin{pmatrix}a^n&na^{n-1}\\0&a^n\end{pmatrix}\;,\;\;n\in\Bbb N$$

and certainly

$$A^n=P^{-1}J^nP\;\;\ldots$$

I think the above is simpler than using the decomposition $\;A=J+N\;$ , at least when the matrix order is $\;2\times 2\;$ .

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  • $\begingroup$ thank you I forgot about that $\endgroup$ – Matilda May 12 '14 at 10:48
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Hint: Consider diagonalizing the matrix. If this is not possible, then your approach works.

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