# Proof in vector spacee

I have problem with proofs in vector space. First is

$\vec x+(-(\vec y+\vec z))=(\vec x+(-\vec y))+(-\vec z)$

and the second

$a\cdot \vec x+b \cdot \vec y=b \cdot \vec x+a\cdot \vec y \Leftrightarrow a=b \vee \vec x=\vec y$

Could anyone help me with this? I'm sorry for my bad english.

In the second task I have:

$a\cdot \vec x+b \cdot \vec y$ I have sentece that $\vec x = \vec y$ or $a=b$

So i'm changing $\vec y$ on $\vec x$

$a\cdot \vec x+b \cdot \vec x = (a+b)\vec x = (a+b) \vec y =...$

And i don't know what can i do next.

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I'm not sure what the variables are, can you specify them? –  xiamx Dec 1 '12 at 22:41
I have specified what variables are vectors and which ones aren't. Could you please check that it is correct? –  Arthur Dec 1 '12 at 22:46
What is $\cdot$ ? –  Belgi Dec 1 '12 at 22:46
@Belgi I assume it's multiplication (by scalar) –  Arthur Dec 1 '12 at 22:47

Hints:

For the first one show that, by definition, $$-(v_{1}+v_{2})=-v_{1}-v_{2}$$

where $v_{i}$ are vectors.

For the second use $$\alpha v-\beta v=(\alpha-\beta)v$$

where $\alpha,\beta$ are scalars, $v$ is a vector.

Also use $a-b=-(b-a)$

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Hmm ok I'm trying do first. I'm starting convert left side but I'm ending with $(x+(z-y)$ ;( –  Emil Dec 1 '12 at 23:15
@Emil - How come ? you should have $x+(-y-z)=x-y-z$ –  Belgi Dec 1 '12 at 23:18
$x+(-y+z)=x-(y-z)=x+(z-y)=x-y+z$ (BTW, should I writte it here or in new post?) –  Emil Dec 1 '12 at 23:23
@Emil - thats not he left side on the post...correct me if I'm wrong –  Belgi Dec 1 '12 at 23:25
Oh damn it, you're right, ok now I can do that and start with second ;) –  Emil Dec 1 '12 at 23:30