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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?2012-12-01
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    I have specified what variables are vectors and which ones aren't. Could you please check that it is correct?2012-12-01
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    What is $\cdot$ ?2012-12-01
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    @Belgi I assume it's multiplication (by scalar)2012-12-01
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    Yeah that's right, I forgot about that, sorry.2012-12-01

1 Answers 1

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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 $(x+(z-y)$ ;(2012-12-01
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    @Emil - How come ? you should have $x+(-y-z)=x-y-z$2012-12-01
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    $x+(-y+z)=x-(y-z)=x+(z-y)=x-y+z$ (BTW, should I writte it here or in new post?)2012-12-01
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    @Emil - thats not he left side on the post...correct me if I'm wrong2012-12-01
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    Oh damn it, you're right, ok now I can do that and start with second ;)2012-12-01
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    In the second I have condition that a=b and x=y. So it's only change the names of scalars and vectors, but how I can writte it in math language?2012-12-01
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    @Emil - I don't understand your question, can you try to explain ? where are you stuck ?2012-12-01
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    At the beginning becouse I have $a\cdot x+ b \cdot y = a \cdot x + b \cdot x = a \cdot y + b \cdot x$ That's ok?2012-12-01
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    @Emil but you wrote $ax+by=bx+ay$ and not $ax+by=ax+bx$. where do you get that last $x$ that should be $y$ ?2012-12-01
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    I change it becouse I'm using from if and only if $\vec x=\vec y$ so can i change $x $ with $y$ or not?2012-12-01
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    @Emil - I think you should edit the post and add the details on what you did so it will be possible to comment n it. its hard to say what you did there2012-12-02