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Let $u$ and $v$ be non-zero vectors in $V$. Prove or disprove the following claim. $u$ and $v$ are linearly dependent $\implies$ $(u+v)$ and $(u-v)$ are linearly dependent.

Is the following proof correct? Otherwise can someone answer the given question? Thanks in advance. :)

Here is my answer.

Since $u , v$ are two linearly dependent vectors, $au + bv = 0$ with $a,b \neq 0$

we can write, $[(a+b)/2]\cdot(u+v) + [(a-b)/2] \cdot(u-v) = 0$

case 1 ($|a|=|b|$) : if $a=b$ then $(a+b)/2 \neq 0 \implies (u+v)$ and $(u-v)$ are linearly dependent.

if $a= -b$ then $(a-b)/2 \neq 0\implies (u+v)$ and $(u-v)$ are linearly dependent.

case 2 ($|a| \neq|b|$) : then $(a+b)/2, (a-b)/2 \neq 0\implies(u+v)$ and $(u-v)$ are linearly dependent.

So, u and v are linearly dependent $\implies(u+v)$ and $(u-v)$ are linearly dependent.

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Is $V$ a vectorspace over some arbitrary field? Then you cannot divide by $2$ (or at least you have to handle this case separately). – born Jan 7 '13 at 9:30
In the question it just says V is a vector space. Sorry I didn't get what you said. – Snitch Jan 7 '13 at 9:37
When you speak about a vetorspace you always have to say what your "base field" is. That is the set where the scalars come from. In high school we are normally used to the vectorspace $\mathbb{R}^n$ which we normally consider "over the field $\mathbb{R}$", but in general the base field can be everything (for example $\mathbb{F}_2$ or $\mathbb{Q}$) and $V$ does not need to be a power of the field $\mathbb{F}^n$. – born Jan 7 '13 at 9:42

Your proof is correct. However, I don't see why you need to break it up into those 2 cases.

Like you said, $au + bv = 0 \Leftrightarrow (a, b)=(0,0)$.

Now, assume that $a(u+v) + b(u-v) = 0$, so $(a+b) u + (a-b) v = 0$. Hence, we know that $a+b = 0 $ and $a-b=0$, which gives that $2a=0, 2b=0$. (Assuming that your field does not have characteristic 2) hence $a=0, b=0$.

In the event that your field has characteristic 2, then $u+v = u-v$ so these vectors are not linearly independent.

Edit: OP wanted a direct proof along the lines he started, and I pointed out that he doesn't need to split it out.

Assume that $u, v$ are not linearly independent. Hence, there exists $(a, b) \neq (0,0)$ such that $au + bv = 0$.

This implies that $ \frac {a+b}{2} (u+v) + \frac {a-b}{2} (u-v) = au +bv = 0$ (assuming that the field does not have characteristic 2). For this to actually show that $(u+v), (u-v)$ are not linearly independent, we have to show that $(\frac {a+b}{2}, \frac {a-b}{2}) \neq (0,0)$. This is true because $ (\frac {a+b}{2}, \frac {a-b}{2}) = (0,0) \Leftrightarrow (a, b) = (0,0)$.

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But this question was to prove or disprove if u and v are linearly dependent ⟹ (u+v) and (u−v) are linearly dependent. Not independence. There I need to to consider the cases when u = v and u = -v. isn't it? – Snitch Jan 7 '13 at 9:51
@TavionPotter Think contrapositive. You could do it directly like you did, in which case since $au+bv = 0$, hence $\frac {a+b}{2}(u+v) - \frac {a-b}{2}(u-v) = 0$, which is just reversing the steps above. You then have to verify that at least one of them is non-zero, which as you realize, splits it into several cases. – Calvin Lin Jan 7 '13 at 16:07
yeah. Thank you very much. I mean if I am gonna prove this in the way I started I can't do it with out splitting it right? That was my problem. :) – Snitch Jan 8 '13 at 6:19
@Travion Alright, I've added a solution along your lines, with doesn't split into cases. – Calvin Lin Jan 8 '13 at 14:26

since {u,v} are linearly independent, the set {u+v,u-v} is LI if and only if the determinant formed by the coefficients that is 1 1 1 -1

is non zero

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refer LINEAR ALGEBRA by L MIRSKY – venkatesh sripad Jan 7 '13 at 9:36
{u,v} is linearly dependent. – Snitch Jan 7 '13 at 9:52

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