Exercise 2.1.31

Answers

1.: If $x \in W$ , we have $T (x) \in R (T)$ and $T (x) \in W$ since $W$ is $T$ -invariant. But by the definition of direct sum, we have $T (x) \in R (T) \cap W = {0}$ and hence $T (x) = 0$ .
2.: By Dimension Theorem we have dim $(N (T)) =$ dim $(V) -$ dim $(R (T))$ . And since $V = R (T) \oplus W$ , we have dim $W =$ dim $(V) -$ dim $(R (T))$ . In addition with $W \subset N (T)$ we can say that $W = N (T)$ .
3.: Take $T$ be the mapping in Exercise 2.1.21 and $W = {0}$ . Thus $W \neq N (T) = {(a_{1}, 0, 0, \dots)}$ .

jephianlin

2011-06-27 00:00