Exercise 2.1.2

Answers

It’s a linear transformation since we have

T ((a_{1}, a_{2}, a_{3}) + (b_{1}, b_{2}, b_{3})) = T (a_{1} + b_{1}, a_{2} + b_{2}, a_{3} + b_{3}) = (a_{1} + b_{1} - a_{2} - b_{2}, 2 a_{3} + 2 b_{3})

= (a_{1} - a_{2}, 2 a_{3}) + (b_{1} - b_{2}, 2 b_{3}) = T (a_{1}, a_{2}, a_{3}) + T (b_{1}, b_{2}, b_{3})

and

T (c a_{1}, c a_{2}, c a_{3}) = (c (a_{1} - a_{2}), 2 c a_{3}) = cT (a_{1}, a_{2}, a_{3}) .

$N (T) = {(a_{1}, a_{1}, 0)}$ with basis ${(1, 1, 0)}$ ; $R (T) = ℝ^{2}$ with basis ${(1, 0), (0, 1)}$ . Hence $T$ is not one-to-one but onto.

jephianlin

2011-06-27 00:00