# Linear independence of vectors $w_1,w_2,w_3$

Suppose that $S = \{v_1,v_2,v_3\}$ is a linearly independent set of vectors in a vector space $V$. Is $T = \{w_1,w_2,w_3\}$, where $w_1 = v_1 + v_2$, $w_2 = v_1 + v_3$, $w_3 = v_2 + v_3$, linearly dependent or linearly independent? Why?

Am I right in claiming that vector space $V$ is a subset of $T$?

Do I set the linear combinations of $w_1$, $w_2$, $w_3$ equal to zero?

$$c_1(v_1+v_2) + c_2(v_1+v_3) + c_3(v_2+v_3) = 0$$

and then solve for $c_1,c_2,c_3$ and see if they have unique solutions?

1. No, $V$ is not a subset of $T$: how can it be?

2. Yes, the strategy for showing linear independence is correct. What can you do next, in order to use the hypothesis on $\{v_1,v_2,v_3\}$? Write $$(c_1+c_2)v_1+(c_1+c_3)v_2+(c_2+c_3)v_3=0$$ and then you know that …

• c1 + c2 = 0, c1 + c3 = 0, c2 + c3 = 0? – antotony Dec 1 '13 at 21:38
• @antotony Yes; isn't this a linear system? – egreg Dec 1 '13 at 22:01
• Oh, now I see it. Because I have three unknowns and three equations, I can solve it right? Thank you for your help! Much appreciated! :D – antotony Dec 1 '13 at 22:27

$$c_1(v_1+v_2) + c_2(v_1+v_3) + c_3(v_2+v_3) = 0$$ $$(*): (c_1+c_2)v_1+(c_1+c_3)v_2+(c_2+c_3)v_3=0$$

Now recall that $$S = \{v_1,v_2,v_3\}$$ is a linearly independent set, hence the coefficients in $$(*)$$ are all 0, from which it follows that $$c_1=c_2=c_3=0$$

Hence W is a linearly independent set. Trivially, V is not a subset of T, as $$|V|\gt |T|$$.