Proving linear independence of vectors which are functions of other independent vectors If the $n$-component vectors $a,b,c$ are linearly independent, show that $a+b, b+c, a+c$ are also linearly independent, Is this true of $a-b,b+c,a+c$? 
What I did was write the new vectors as sums and set it equal to zero
$$\begin{align} \sum(\lambda_i(a_i+b_i))+\sum(\lambda_i(b_i+c_i))+\sum(\lambda_i(a_i+c_i))=0 \end{align}$$
Then I assumed you could just add these terms up, but I don't actually know if that's allowed in this case. Anyways you obviously get 
$$\sum\lambda_i(2a_i+2b_i+2c_i)=0$$
Seems to me that since the problem tells you $a,b$,and $c$ are linearly independent that these respectiev $\lambda$'s on these new terms must also be $0$.
Again, I doubt that works so I'm asking here. Wish this book had answers...
 A: A set of vectors $\{v_i\}$ is linearly independent if and only if
$$\alpha_i = 0$$
whenever
$$\sum \alpha_i v_i = 0.$$
Suppose, for contradiction, that $a+b$, $b+c$, $a+c$ are not linearly independent. Then there exist scalars $x,y,z$, not all zero, with
$$x(a+b) + y(b+c) + z(a+c) = 0$$
$$(x+z)a + (x+y)b+(y+z)c = 0.$$
Since $a,b,c$ are linearly independent, all of these coefficients must be zero. Therefore $x=-z$, $x=-y$, $y=-z$. What can you conclude?
A: Your setup needs to be modified some.
The key idea here is that any equation of linear dependence for $a+b$, $b+c$, and $a+c$ results in an equation of linear dependence for $a$, $b$, and $c$.
When you wrote
$$
\sum\lambda_i(a_i+b_i) + \sum\lambda_i(b_i+c_i) + \sum\lambda_i(a_i+c_i) = 0,
$$
you're not multiplying your vectors $a+b$, $b+c$, and $a+c$ by scalars; you're actually computing the dot product $(\lambda_1,\ldots,\lambda_n)\cdot ((a+b)+(b+c)+(a+c))$. What you really want is to show that if
$$
\alpha(a+b)+\beta(b+c)+\gamma(a+c) = 0
$$
for scalars $\alpha,\beta,\gamma$, then $\alpha=\beta=\gamma=0$. You rewrite this as a sum of (different) scalars time the vectors $a,b,c$, and then use the linear independence of the latter to complete the argument.

An alternative "slick" way is as follows. Notice that
\begin{align*}
a &= \frac{1}{2}\lbrack (a+b) - (b+c) + (a+c)\rbrack,\\
b &= \frac{1}{2}\lbrack (b+c) - (a+c) + (a+b)\rbrack,\\
c &= \frac{1}{2}\lbrack (c+a) - (a+b) + (b+c)\rbrack.
\end{align*}
So the "span" (set of all linear combinations of) $a+b$, $b+c$, and $c+a$ contains the lienarly independent vectors $a$, $b$, and $c$. You can show, in general, that
three vectors $u,v,w$ are linearly independent if you can find three linear combinations of $u$, $v$, and $w$ that are linearly independent. So this completes the proof (assuming you accept this last fact...you should prove it if you haven't already).

For the second problem, notice that the third vector $c+a = (a-b)+(b+c)$ is a linear combination of the first two vectors. So no, they're not independent.
A: Suppose that vectors $u,v$, and $w$ are linearly independent in $V$, this implies that for the equality, $$a_1u+a_2v +a_3w = 0,$$ we have $a_1 = a_2 = a_3 = 0$ for all $a_1,a_2,a_3
 \in \mathbb{F}$.
To check if the set of vectors $(u+v,v+w,u+w)$ is linearly independent: 
$\begin{align} a_1u+a_2v+a_3w & =(a_1u+a_2v -a_2v)+(a_2v+a_3w-a_3w)+(a_1u+a_3w-a_1u) \\ &=(a_1u+a_2v)-a_2v+(a_2v+a_3w)-a_3w+(a_1u+a_3w)-a_1u \\& =(a_1u+a_2v)+(a_2v+a_3w)+(a_1u+a_3w)-(a_1u+a_2v+a_3w) \\ 2(a_1u+a_2v+a_3w) &= (a_1u+a_2v)+(a_2v+a_3w)+(a_1u+a_3w)  \\ &=a_1(u+u)+a_2(v+v)+a_3(w+w)\end{align}$
Since $(u,v,w)$ is linearly independent, we obtain $a_1 = a_2 = a_3 = 0$, which implies $a_1+a_2 = a_2+a_3 = a_1+a_3 = 0,$ which proves that the new vector combination is linearly independent as well.
The same can be said for $(u-v,v+w,u+w)$.
