# [SOLVED] I don't know

• October 22nd 2009, 04:15 PM
Noxide
[SOLVED] I don't know
what to do for this one:

Given: 3 vectors, a, b, c, are linearly independent in a vector space V
Asked: Show that 3a, 2a-b, a + c are also linearly independent
• October 22nd 2009, 04:29 PM
Defunkt
Quote:

Originally Posted by Noxide
what to do for this one:

Given: 3 vectors, a, b, c, are linearly independent in a vector space V
Asked: Show that 3a, 2a-b, a + c are also linearly independent

Assume 3a, 2a-b, a+c are dependent. Then, there exist scalras $x_1,x_2,x_3$, not all zero, such that $x_1(3a) + x_2(2a-b) + x_3(a+c) = 0 \Rightarrow 3x_1a + 2x_2a -x_2b + x_3a + x_3c = 0$ $\Rightarrow (3x_1+2x_2+x_3)a + (-x_2)b + (x_3)c = 0$

If $x_2=x_3=0$ then $x_1 \neq 0$. Therefore not all the coefficients are zero.

Let $\lambda_1 = 3x_1+2x_2+x_3$
$\lambda_2 = -x_2$
$\lambda_3 = x_3$

Then $\lambda_1a + \lambda_2b + \lambda_3c = 0$ but not all of the coefficients are zero, in contradiction to the fact that a,b,c are linearly independent... ==> the assumption can not hold, therefore 3a, 2a-b and a+c are linearly independent.
• October 22nd 2009, 04:51 PM
Noxide
wow, that's a really awesome proof by contradiction!

thanks for the help (Clapping)