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Suppose we know that $S = \{v_1, v_2, v_3\}$ is a linearly independent set of vectors in some vector space $V$ . Let $T = \{v_1 + v_2, v_2 + v_3, v_3 + v_1\}$. Prove that $T$ is also linearly independent.

This proof makes sense to me that it would be true, but I cannot figure out the steps of the proof that make this a true statement?

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  • $\begingroup$ Why does it make sense that it would be true? See if you can translate your intuition to a proof. Remember the definition of linear independence. Or maybe the definition of linear dependence. $\endgroup$ – Matthew Leingang Sep 27 '16 at 18:10
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If $v_1,\, v_2,\, v_3$ are linearly independent, then $av_1 + bv_2 + cv_3 = 0 \implies a = b = c = 0 $. Now suppose $$ a(v_1+v_2) + b(v_2+v_3) + c(v_3+v_1) = 0, $$ then this would imply $$ (a+c)v_1 + (a+b)v_2 + (b+c)v_3 = 0. $$ Because this is a linear combination of our original linear independent vectors, this implies $ a+c = a+b = b+c = 0$, from which you easily get $a = b = c = 0$.

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$\det \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix} = 2 \neq 0$.

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Suppose $\{v_1,v_2,v_3\}$ are Linearly Independent. Then whenever $a_1v_1+a_2v_2+a_3v_3=0$ for scalars $a_1,a_2,a_3$ we have $a_1=a_2=a_3=0$.

Now let $b_1(v_1+v_2)+b_2(v_2+v_3)+b_3(v_3+v_1)=0$ for scalars $b_1,b_2,b_3$.

Then since a vector space is afterall a commutative group (additive), you can write the above as $(b_1+b_3)v_1+(b_1+b_2)v_2+(b_2+b_3)v_3=0$.

But $v_1,v_2,v_3$ are LI. From what I wrote in first line, $b_1+b_3=b_1+b_2=b_2+b_3=0$.

Thus we have from above that $b_1=b_2=b_3=0$.

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