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Show that if ${v_1, v_2, v_3}$ is linearly independent then ${v_1, v_2}$ is also linearly independent.

$$av_1 + bv_2 + cv_3 = 0 \implies a = b = c = 0$$

Thus we need to show that $xv_1 + yv_2 = 0$ if $x=y=0$.

Edit

Assume $v1, v2$ is linearly dependent. Then $xv_1 + yv_2 = 0$, assume WLOG $x \ne 0$. Adding $cv_3$ we have $xv_1 + yv_2 + cv_3 = cv_3$. By (1) we have $a = b = c = 0$, thus $x = 0$ contradiction?

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    Assume it is false: what would you get then? – 2017-01-08
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    @DonAntonio, so the negation would be that either $x\ne0$ or $y \ne 0$? – 2017-01-08
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    No. You would have that $\;v_1,v_2\;$ is linearly **dependent**, but $\;v_1,v_2,v_3\;$ is linearly **independent**. How this leads you smoothly to a harsh contradiction? Think of it and play with this for a while. – 2017-01-08
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    @DonAntonio, edited, is that the contradiction? – 2017-01-08
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    @Ga Almost... After the WLOG what happens if you write $\;xv_1+yv_2+0\cdot v_3=0\;$ ? You **still** have $\;x\neq0\;$ , right? Can you see now the contradiction? – 2017-01-08
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    @DonAntonio, ah yes! So now we have $xv1 + yv2 + 0v3 = 0$, but all coefficients must be $0$ because this statement was Linearly independent, thus $x=0$, contradiction to $x \ne 0$? – 2017-01-08
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    @G Or that, or contradiction to the fact that $\;v_1,v_2,v_3\;$ lin. ind. – 2017-01-08

2 Answers 2

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if $$v_1$$ and $$v_2$$ are linear dependent, the we have wlog. $$x\ne 0$$ and we get $$v_1=-\frac{y}{x}v_2$$ can you proceed here?

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You don't need to show that $xv_1+yv_2=0$ if $x=y=0$. You have to show that

if $xv_1+yv_2=0$, then $x=y=0$

which is quite different.

So, suppose $xv_1+yv_2=0$; then also $$ xv_1+yv_2+0v_3=0 $$ Thus you have $$ av_1+bv_2+cv_3=0 $$ with $a=x$, $b=y$ and $c=0$.

Since $v_1,v_2,v_3$ are linearly independent, this implies $a=b=c=0$.