0
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – WimC
    Nov 30, 2012 at 11:42
  • $\begingroup$ 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? $\endgroup$
    – Dor Shalom
    Nov 30, 2012 at 11:50

2 Answers 2

0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Dec 1, 2012 at 1:33
0
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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 $\alpha = \beta = 0$ $\endgroup$
    – Dor Shalom
    Nov 30, 2012 at 12:15
  • $\begingroup$ That's exactly the same question I gave the answer to. Look again at my answer - do you understand it? $\endgroup$ Nov 30, 2012 at 14:34
  • $\begingroup$ Yes. Thanks alot! $\endgroup$
    – Dor Shalom
    Dec 1, 2012 at 11:20
  • $\begingroup$ @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/… $\endgroup$ Dec 1, 2012 at 11:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .