# 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.

• 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. – angryavian Jan 21 at 22:28

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.

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.

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, $$a_1=a_2=0$$.