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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.

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    $\begingroup$ Can you state the definitions of "$\{v_1, v_2, v_3\}$ is a linearly independent set" and "$\{v_1, v_2\}$ is a linearly independent set"? If you write out the definitions clearly, you should be able to finish the question in one step. $\endgroup$ – angryavian Jan 21 at 22:28
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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.

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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.

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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, $a_1=a_2=0$.

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