Homepage › Solution manuals › Stephen Friedberg › Linear Algebra › Exercise 1.6.12

Exercise 1.6.12

Answers

If ${u, v, w}$ is a basis then the dimension of $V$ would be three. So it’s enough to check ${u + v + w, v + w, w}$ is lineaerly independent. Assuming $a (u + v + w) + b (v + w) + cw = au + (a + b) v + (a + b + c) w = 0$ we have $a = a + b = a + b + c = 0$ and hence $a = b = c = 0$ .

jephianlin

2011-06-27 00:00

Exercise 1.6.12

Answers

Comments

Add answer