Show that if $\{\vec{v_1},\vec{v_2},\vec{v_3}\}$ is a linearly independent set, then $\{\vec{v_1},\vec{v_2}\}$ is also linearly independent How do you show that if $\{\vec{v_1},\vec{v_2},\vec{v_3}\}$ is a linearly independent set, then $\{\vec{v_1},\vec{v_2}\}$ is also linearly independent. 
I need to explain this in complete sentences. I understand how to show linear independence for specific equations but I don't have a high enough level of understand of linear independence to be able to explain this more broadly.
 A: Suppose that $\{v_1,v_2\}$ is a linearly dependent set.  That means by definition that...

 ...there is some set of constants $c_1,c_2$ where at least one of $c_1,c_2$ is nonzero such that $c_1v_1 + c_2v_2 = 0$.

Then when considering whether or not $\{v_1,v_2,v_3\}$ is a linearly independent set or not we see that...

 ...by using the same $c_1,c_2$ as before and letting $c_3=0$ we have shown that there is an example of $c_1,c_2,c_3$ such that $c_1v_1+c_2v_2+c_3v_3=0$ while at least one of $c_1,c_2,c_3$ is nonzero.

So we have shown that $\{v_1,v_2\}$ linearly dependent implies that $\{v_1,v_2,v_3\}$ is linearly dependent as well.  By contraposition, this is equivalent to have shown that $\{v_1,v_2,v_3\}$ being linearly independent implies $\{v_1,v_2\}$ is linearly independent as well.
A: IIRC,
if $v_{1,2,3}$
are linearly independent then
$\sum a_iv_i = 0
\implies
a_{1,2,3} = 0
$.
If
$v_1,  v_2$
are not linearly independent,
there exist $a, b$
with at least one of
$a, b$ nonzero such that
$av_1+bv_2 =0$.
Therefore
$0
=av_1+bv_2 +0v_3$
so that
$v_{1, 2, 3}$
are not linearly independent.
Contradiction with assumption.
A: Here's a short direct proof, without unnecessary negation:
(equivalent to the other answers) 
Suppose
$a_1 v_1 + a_2 v_2 =0$.
We need to show $a_1=a_2=0$.
Since $0v_3=0$, we also have
$a_1 v_1 + a_2 v_2 + 0 v_3 =0$.
Since $v_1,v_2,v_3$ are linearly independent, $a_1=a_2=0$.
