Proving linear independence of vectors which are functions of other independent vectors

Question

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...

Answer 1

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.

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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).

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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.

Answer 2

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?

Answer 3

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 $(u-v,v+w,u+w)$.