Show linear independence of sum of vectors from a base

Question

let $V$ be a vectorspace and $\{a_1,a_2,a_3,a_4\}$ a base of $V$; Moreover $b_1:=2a_1-a_2,$ $b_2:=a_2+a_3+a_4,$ $b_3:=a_3-a_4$; Show that $b_1,b_2,b_3$ are linear independent;

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By definition of linear independence:

(1) $x\cdot b_1+y\cdot b_2+z\cdot b_3 = 0 \Rightarrow x=y=z=0$

so

(2) $x\cdot_(2a_1-a_2)+y\cdot (a_2+a_3+a_4)+z\cdot(a_3-a_4) = 0 \Rightarrow x=y=z=0$.

This is where I'm stuck already. I've tried resorting (2) so that I have $a_1\cdot(2x)+a_2\cdot(\dots$ but that doesn't get me anywhere.

Any help or pointers are much welcome!

Thanks Anton

Answer 1

Suppose that $x(2a_1-a_2)+y(a_2+a_3+a_4)+z(a_3-a_4)=0$. This is the same thing as asserting that $2xa_1+(-x+y)a_2+(y+z)a_3+(y-z)a_4=0$. Since $\{a_1,a_2,a_3,a_4\}$ is a basis, then$$\left\{\begin{array}{l}2x=0\\-x+y=0\\y+z=0\\y-z=0.\end{array}\right.$$It is easy to see that the only solution of this system is $x=y=z=0$.

Answer 2

Consider your equation $x(2a_1-a_2)+y*(a_2+a_3+a_4)+z*(a_3-a_4)=0$.

It gives

$2xa_1 + (-x+y) a_2+ (y+z)a_3 + (y-z)a_4=0$.

Since $\{a_1,\ldots,a_4\}$ is a basis, it follows that $2x=0$, $y-x=0$, $y+z=0$ and $y-z=0$. Thus $x=0$ and then in order $y=0$ and $z=0$, as required.