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I have the following Linear Algebra question:

Prove/disprove: if $(x_1,y_1),(x_2,y_2)$ are 2 linearly independent vectors then $(x_1,y_1,z_1),(x_2,y_2,z_2)$ are also linearly independent vectors

thank you, Dor

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Write out a generic linear combination of these 3d vectors. What do you get in the first two coordinates? How does it relate to your 2d vectors? –  WimC Nov 30 '12 at 11:42
    
I don't think i completely understand. I know that a*(x1,y1)+b*(x2,y2) = 0 only if both a,b = 0. But does that necessarily means that a*(x1,y1,z1)+b*(x2,y2,z2) =0 also only if both a,b = 0? –  Dor Shalom Nov 30 '12 at 11:50
    
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. –  Julian Kuelshammer Nov 30 '12 at 12:04

2 Answers 2

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Two vectors $(x_1,y_1)$ and $(x_2,y_2)$ are independent if there is no scalar $c$ such that both $cx_1 = x_2$ and $cy_1=y_2$. If these two equations cannot be simultaneously satisfied by a single $c$, then they are independent.

The criterion is basically the same in three dimensions: $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are independent unless there is some scalar $c$ such that $cx_1=x_2$, $cy_1=y_2$, and $cz_1 = z_2$ simultaneously. In your problem, the vectors cannot satisfy the first two equations simultaneously, the certainly cannot satisfy all three. Following this logic, adding components cannot make any number of independent vectors dependent.

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In layman's terms: a pair of vectors is independent if they are not parallel. Adding $z$ components to two non-parallel plane vectors cannot make them parallel in three dimensions. –  orlandpm Dec 1 '12 at 1:33

Write down the definition of linear independence: $$(x_1,y_1,z_1),(x_2,y_2,z_2) {\mbox{ are linearly independent}}\Leftrightarrow\left(\forall\alpha,\beta\hspace{3pt} \alpha(x_1,y_1,z_1)+\beta(x_2,y_2,z_2)=0\rightarrow\alpha=\beta=0\right)$$ Now expand it to the coordinates. If $\alpha(x_1,y_1,z_1)+\beta(x_2,y_2,z_2)=0$ then, in particular $\alpha(x_1,y_1)+\beta(x_2,y_2)=0$. What does this imply?

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first of all, thank you all for your help, I am new here... so first let my correct my question, i'm not sure if it matters: prove/disprove: \begin{pmatrix} x1 \\ y1 \end{pmatrix} and \begin{pmatrix} x2 \\ y2 \end{pmatrix} are linearly independent --> \begin{pmatrix} x1 \\ y1 \\ z1 \end{pmatrix},\begin{pmatrix} x2 \\ y2 \\ z2 \end{pmatrix} are independent. i'm not sure if i can say that z1+z2 = 0 only if $\alpha = \beta = 0$ –  Dor Shalom Nov 30 '12 at 12:15
    
That's exactly the same question I gave the answer to. Look again at my answer - do you understand it? –  Dennis Gulko Nov 30 '12 at 14:34
    
Yes. Thanks alot! –  Dor Shalom Dec 1 '12 at 11:20
    
@Dor Shalom: Since you are new to the site, I wanted to let you know that you should accept the answer that answers your question (don't get me wrong - I'm not telling you to accept my answer, but in general, once you got a satisfying answer to a question asked on this site you should accept it, so that other users will know that you got your answer). More on accepting answers can be found here:meta.math.stackexchange.com/questions/3286/… –  Dennis Gulko Dec 1 '12 at 11:30

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