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Let $P$ be a $2 \times 2$ complex matrix such that Trace(P) = 1, Determinant(P) = -6, Find Trace$(P^4 - P^3).$

What I have done, if $\lambda_1, \lambda_2$ is the eigen value, then I find $\lambda_1 = 3$ and $\lambda_2 = -2.$ So the eigen value of $P^4$ will be 81 and 16. From this point we can find Trace$(P^4).$ Similarly for Trace$(P^3).$ So we can find the desired result.

My question is am I right? Also What is the role of complex matrix here?

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    $\begingroup$ This looks good to me. The complex field plays almost no role, but the question setter has to specify the ground field anyway. $\endgroup$ – user1551 Dec 14 '16 at 10:09
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By application of Cayley–Hamilton theorem, we have $$P^2-P-6I=0$$ Thus $$\text{Trace}(P^2)=\text{Trace}(P)+6\,\text{Trace}(I_{2\times 2})=13$$ On the othe hand $$P^3=P^2+6P$$ therefore $$\text{Trace}(P^3)=13+6=19$$ and $$P^4=P^2+12P+36I^2$$ so $$\text{Trace}(P^4)=13+12+72=97$$

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    $\begingroup$ Or write $x^4 - x^3 = (x^2+6)(x^2-x-6)+6x+36$ and so $P^4 - P^3 =6P+36I$ $\endgroup$ – lhf Dec 14 '16 at 10:23
  • $\begingroup$ Yes exactly.... $\endgroup$ – Behrouz Maleki Dec 14 '16 at 10:25

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