Exercise 2.1.1

Answers

1.: Yes. That’s the definition.
2.: No. Consider a map $f$ from $ℂ$ over $ℂ$ to $ℂ$ over $ℂ$ by letting $f (x + iy) = x$ . Then we have $f (x_{1} + i y_{1} + x_{2} + i y_{2}) = x_{1} + x_{2}$ but $f (iy) = 0 \neq = if (y) = iy$ .
3.: No. This is right when $T$ is a linear trasformation but not right in general. For example, $\begin{array}{l} T : & ℝ \to ℝ \\ x \mapsto x + 1 \end{array}$
It’s one-to-one but that $T (x) = 0$ means $x = - 1$ . For the counterexample of converse statement, consider $f (x) = | x |$ .
4.: Yes. We have $T (0_{V}) = T (0 x) = 0 T {(0_{V})}_{W} = 0_{W}$ , for arbitrary $x \in V$ .
5.: No. It is dim $(V)$ . For example, the transformation mapping the real line to ${0}$ will be.
6.: No. We can map a vector to zero.
7.: Yes. This is the Corollory after Theorem 2.6.
8.: No. If $x_{2} = 2 x_{1}$ , then $T (x_{2})$ must be $2 T (x_{1}) = 2 y_{1}$ .

jephianlin

2011-06-27 00:00