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My task is to show that the set of vectors: $\bf x_1, x_2, x_3, x_4$ where

$\bf x_1=[1,0,0]$

$\bf x_2=[t,1,1]$

$\bf x_3=[1,t,t^2]$ and

$\bf x_4=[t+2,t+1,t^2+1]$ are linearly dependent. (Note: $x_i$ can also be written in matrix format.)

To show that they are linearly dependent I form the equation:

$\bf c_1x_1+c_2x_2+c_3x_3+c_4x_4=0$ and will show that there is a nonzero solution to it. That is I will show that aside from $\bf c_1,c_2,c_3,c_4=0$ there is some other solution to it.

However solving puts me in a system of 3 equations in 4 unknowns which seems new to me. They are:

$\bf c_1+c_2t+c_3+c_4(t+2)=0$

$\bf 0+c_2+c_3t+c_4(t+1)=0$

$\bf 0+c_2+c_3t^2+c_4(t^2+1)=0$

Can someone help me to find a non trivial solution to the given system of equation? or Will you help me showing that the 4 vectors above are linearly dependent?

Thank you so much for your help.

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    $\begingroup$ Here $x_1+x_2+x_3=x_4$. Isn't it? So they are linearly dependent. $\endgroup$ – Extremal Feb 13 '15 at 22:35
  • $\begingroup$ Yes I see it, but I don't know if I can express a vector in terms of the other will mean linear dependence, is that true? Thanks $\endgroup$ – Jr Antalan Feb 13 '15 at 22:41
  • $\begingroup$ Of course! Because that implies you can find a combination of $c_i$ where $c_i\neq 0$ for all $i$. See other's answers too. $\endgroup$ – Extremal Feb 13 '15 at 22:43
  • $\begingroup$ we have shown you that substituting C1=1,C2=1,C3=1,C4=-1, the equality will be satisfied with all the C's different from zero! thus the vectors are linearly dependent(Hint: Only 1 constant C shall be different from zero so we can say the 4 vectors are L.D) $\endgroup$ – Mistos Feb 13 '15 at 22:47
  • $\begingroup$ It is so clear to me now, thanks @Mathi. $\endgroup$ – Jr Antalan Feb 13 '15 at 22:48
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Hint:

You can make an easy solution if you use the fact that if some vector in a list of vectors is a linear combination of other vectors in that same list, then the list is linearly dependent.

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  • $\begingroup$ if I can express a vector in terms of the other will mean linear dependence, is that true? Thanks $\endgroup$ – Jr Antalan Feb 13 '15 at 22:41
  • $\begingroup$ @JrAntalan Yes. If you have a list of vectors $v_0,v_1,\ldots,v_n$, and if you can can write for instance $$v_0=c_1v_1+\cdots+c_nv_n,\tag{for some scalars $c_j$}$$ then by adding to both sides the additive inverse of $v_0$ we get : $$0=-v_0+c_1v_1+\cdots+c_nv_n,$$ which means that the list of those vectors is linearly dependent. $\endgroup$ – Workaholic Feb 13 '15 at 22:43
  • $\begingroup$ Now I know, thanks Workaholic $\endgroup$ – Jr Antalan Feb 13 '15 at 22:46
  • $\begingroup$ @JrAntalan You're welcome. $\endgroup$ – Workaholic Feb 13 '15 at 22:47
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Note that $$\bf x_4=x_1+x_2+x_3 \Rightarrow -x_1-x_2-x_3+x_4=0$$ so $\bf c_1=c_2=c_3=-1$ and $\bf c_4=+1$

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Notice that $X$2 is a linear combination of $X$4, $X$3 and $X$1! where $X$2= $X$4-$X$3-$X$1! that would prove that the four vectors are linearly dependent!

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  • $\begingroup$ You'r Welcome!. $\endgroup$ – Mistos Feb 13 '15 at 22:49
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four vectors in $R^3$ are always linearly dependent. you don't need anything more.

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    $\begingroup$ These vectors dont live in $\mathbb R ^3$ though, as they have a mixture of variables and real numbers in them. This looks more like it comes from an infinite dimensional vector space of the polynomial ring over the reals times itself times itself $\endgroup$ – Alan Feb 14 '15 at 0:43

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