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Find $A^8$ using Cayley Hamilton Therorem, when $$A = \begin{pmatrix} 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 0 \end{pmatrix}$$

I found that $P(t) =t^4-2x^2+1 = (t-1)^2(t+1)^2$.

But how can I use Cayley-Hamilton to find $A^8$?

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Given that $P(t) = t^4-2t^2+1$, the Cayley-Hamilton Theorem yields that $$P(A) = O,$$ where $O$ is $4$ by $4$ zero matrix. Then $$O = A^4 - 2A^2 + I \iff A^4 = 2A^2 - I \implies A^8 = (2A^2 - I)^2.$$

EDIT: As suggested, this can be further simplified such that

$$ A^8 = 4A^4 - 4A^2+ I = 4(2A^2 - I) - 4A^2 + I = 4A^2 -3I .$$

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  • $\begingroup$ $=4(2A^2-I)-4A^2+I=4A^2-3I$ $\endgroup$ – Empy2 Feb 6 '16 at 13:59

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