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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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    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?2012-11-30

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.2012-12-01
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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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    @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:http://meta.math.stackexchange.com/questions/32$8$6/how-do-i-accept-an-answer2012-12-01