Exercise 2.3.12

Answers

1.: If $UT$ is injective, we have that $UT (x) = 0$ implies $x = 0$ . Thus we have that if $T (x) = 0$ we also have $UT (x) = 0$ and hence $x = 0$ . So $T$ is injective. But $U$ may not be injective. For example, pick $U (x, y, z) = (x, y)$ , a mapping from $ℝ^{3}$ to $ℝ^{2}$ , and $T (x, y) = (x, y, 0)$ , a mapping from $ℝ^{2}$ to $ℝ^{3}$ .
2.: If $UT$ is surjective, we have that for all $z \in Z$ there is a vector $x \in V$ such that $UT (x) = z$ . Thus we have that if for all $z \in Z$ we have $z = U (T (x))$ and hence $U$ is surjective. But $T$ may not surjective. The example in the previous question could also be the example here.
3.: For all $z \in Z$ , we can find $z = U (y)$ for some $y \in W$ since $U$ is surjective and then find $y = T (x)$ for some $x \in V$ since $T$ is surjective. Thus we have $z = UT (x)$ for some $x$ and hence $UT$ is surjective. On the other hand, if $UT (x) = 0$ , this means $T (x) = 0$ since $U$ is injective and $x = 0$ since $T$ is injective.

jephianlin

2011-06-27 00:00