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Exercise 8.B.6

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Proof. The formula for $N$ doesn’t really matter here. We only care about the dimension of $𝔽^{5}$ , which is $5$ . Using the same reasoning from the proof of 8.31, and because $N^{j} = 0$ for $j \geq 5$ , we have

\begin{aligned} I + N & = (I + a_{1} N + a_{2} N^{2} + a_{3} N^{3} + a_{4} N^{4}) \\ = I + 2 a_{1} N + (2 a_{2} + a_{1}^{2}) N^{2} + (2 a_{3} + 2 a_{1} a_{2}) N^{3} + (2 a_{4} + 2 a_{1} a_{3} + a_{2}^{2}) N^{4} \end{aligned}

for some $a_{1}, a_{2}, a_{3}, a_{4} 𝔽$ . Choose

\begin{aligned} a_{1} & = \frac{1}{2} \\ a_{2} & = \frac{- 1}{8} \\ a_{3} & = \frac{1}{16} \\ a_{4} & = \frac{- 5}{128} \end{aligned}

and the terms on the second line will collapse to $N + I$ . Hence

I + \frac{1}{2} N - \frac{1}{8} N^{2} + \frac{1}{16} N^{3} - \frac{5}{128} N^{4}

is a square root of $N + I$ . □

celiopassos

2017-10-06 00:00

Exercise 8.B.6

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