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Exercise 2.1.32

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We have $N (T_{W}) \subset W$ since $T_{W}$ is a mapping from $W$ to $W$ . For $x \in W$ and $x \in N (T_{W})$ , we have $T_{W} (x) = T (x) = 0$ and hence $x \in N (T)$ . For the converse, if $x \in N (T) \cap W$ , we have $x \in W$ and hence $T_{W} (x) = T (x) = 0$ . So we’ve proven the first statement. For the second statement, we have

R (T_{W}) = {y \in W : T_{W} (x) = y, x \in W} = {T_{W} (x) : x \in W} = {T (x) : x \in W} .

jephianlin

2011-06-27 00:00

Exercise 2.1.32

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