Exercise 1.5.13

Answers

1.: Sufficiency: If ${u + v, u - v}$ is linearly independent we have $a (u + v) + b (u - v) = 0$ implies $a = b = 0$ . Assuming that $cu + dv = 0$ , we can deduce that $\frac{c + d}{2} (u + v) + \frac{c - d}{2} (u - v) = 0$ and hence $\frac{c + d}{2} = \frac{c - d}{2} = 0$ . This means $c = d = 0$ if the characteristc is not two. Necessity: If ${u, v}$ is linearly independent we have $au + bv = 0$ implies $a = b = 0$ . Assuming that $c (u + v) + d (u - v) = 0$ , we can deduce that $(c + d) u + (c - d) v = 0$ and hence $c + d = c - d = 0$ and $2 c = 2 d = 0$ . This means $c = d = 0$ if the characteristc is not two.
2.: Sufficiency: If $au + bv + cw = 0$ we have $\frac{a + b - c}{2} (u + v) + \frac{a - b + c}{2} (u + w) + \frac{- a + b + c}{2} (v + w) = 0$ and hence $a = b = c = 0$ . Necessity: If $a (u + v) + b (u + w) + c (v + w) = 0$ we have $(a + b) u + (a + c) v + (b + c) w = 0$ and hence $a = b = c = 0$ .

jephianlin

2011-06-27 00:00