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Exercise 2.2.14

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It can be checked that differentiation is a linear operator. That is, $T_{i}$ is an element of $L (V)$ for all $i$ . Now fix some $n$ , and assume $\sum_{i = 1}^{n} a_{i} T_{i} = 0$ . We have $T_{i} (x^{n}) = \frac{n!}{(n - i)!} x^{n - i}$ and thus ${T_{i} (x^{n})}_{i = 1, 2, \dots, n}$ would be an independent set. Since $\sum_{i = 1}^{n} a_{i} T_{i} (x^{n}) = 0$ , we have $a_{i} = 0$ for all $i$ .

jephianlin

2011-06-27 00:00

Exercise 2.2.14

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