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I understand that vectors $ x_1, x_2, $ and $x_3$ are always linearly dependent if each vector only has an x and y component. What I don't understand is why this is true if vectors $x_1$ and $x_2$ are linearly dependent. If the 2 aforementioned vectors are linearly dependent then they don't span an area and therefore it would be $x_1, x_2$, and $x_3$ to span a single area, where span($x_1, x_3$) = span($x_2,x_3$). Then it would reduce down to $x_3$ having to be a scalar multiple of either $x_1$ or $x_2$ which is not guaranteed.

I am sure that I am just missing something simple (or multiple things), but I would appreciate it if you could point it out.
Also let me know if I need to clarify my question or reasoning.

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  • $\begingroup$ With $x_1,x_2,x_3$ are linearly dependent means that one vector can be written as a linear combination of the remaining vectors. $\endgroup$ Commented Oct 8, 2019 at 22:58

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When in doubt, go back to definitions: a set of vectors $\{v_1,\dots,v_n\}$ is linearly dependent iff there is a nontrivial linear combination of them that vanishes, that is, if there is some set $\{c_1,\dots,c_n\}$ of scalars, not all zero such that $c_1v_1+\cdots+c_nv_n=0$. The bold phrase is the key here. If you have $c_1x_1+c_2x_2=0$ and not $c_1=c_2=0$, then you can add whatever and as many vectors as you like to this set, and it will still be linearly dependent, because you can always form the nontrivial linear combination $c_1x_1+c_2x_2+0x_3+\cdots+0x_n= c_1x_1+c_2x_2 =0$.

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As @kingW3 and @Joe pointed out, "$x_1$ and $x_2$ are linearly dependent" means (1) $x_1 = ax_2+bx_3$ and (2) $x_2 = cx_1+dx_3$ for $a,b,c,d \in \mathbb{K}$, it does NOT NECESSARILY have to mean $x_1=ex_2$ for $e \in \mathbb{K}$. If you now plug (1) into (2) (or the other way around) you get your answer that $x_2$ is a scaled vector of $x_3$, if that was what you meant(?).

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