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Suppose that the vectors in the set $[(x_1,x_2,x_3),(y_1,y_2,y_3),(z_1,z_2,z_3)]$ where each vector belongs to $R^3$ are linearly independent. Show that the vectors in the set $[(x_1,y_1,z_1),(x_2,y_2,z_2),(x_3,y_3,z_3)]$ are linearly independent. Can this problem be generalized for $R^n$ where $n$ is any positive integer? Please do not use any concept of matrices here, only use properties of vector spaces in general.

I may want to add that I tried but the correct mathematical procedure is not hitting me.

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  • $\begingroup$ Can you show why you are interested in this problem? Seems a little bit artificial to exchange components between vectors. $\endgroup$ – enzotib Aug 17 '14 at 8:57
  • $\begingroup$ This problem came in an examination conducted by our college. I am not particularly interested in this problem because I don't find any beauty in it. I don't think I get the result. $\endgroup$ – AnonymousMaths Aug 17 '14 at 13:07
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We prove if $[(x_1,y_1,z_1),(x_2,y_2,z_2),(x_3,y_3,z_3)]$ are linearly dependent, then $[(x_1,x_2,x_3),(y_1,y_2,y_3),(z_1,z_2,z_3)]$ is linearly dependent too.

WLOG, suppose $(x_1,y_1,z_1) = a(x_2,y_2,z_2) + b(x_3,y_3,z_3)$, then in order to find $c, d, e$ such that$c(x_1,x_2,x_3)+d(y_1,y_2,y_3)+e(z_1,z_2,z_3) =0$, we only need to find $c,d,e$ such that $c(x_2,x_3)+d(y_2,y_3)+e(z_2,z_3) =0$ since $$cx_1 + dy_1 + ez_1 = a(cx_2 +dy_2 + ez_2) + b(cx_3+dy_3 + ez_3)$$

Then with two equation and three variables, you can solve it.

This can be generalized for $R^n$, without using matrix concept, if you admit linear system with more variables than equations have non-zero solution

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  • $\begingroup$ I am sorry I could not follow from "we only need to find c,d,e such that..." Could you take the pain of explaining a bit more? Thank you so much! I had so much trouble juggling with these 9 equations and finally going nowhere. $\endgroup$ – AnonymousMaths Aug 17 '14 at 13:02
  • $\begingroup$ @AnonymousMaths Because the equation following "since", we automatically get $cx_1 + dy_1 + ez_1 = 0$ when $c(x_2,x_3)+d(y_2,y_3)+e(z_2,z_3) =0$ $\endgroup$ – Petite Etincelle Aug 17 '14 at 13:11
  • $\begingroup$ Okay, let me go over it again then. Thanks! Now, due to the symmetry, can I conclude that the reverse also holds? $\endgroup$ – AnonymousMaths Aug 17 '14 at 13:16
  • $\begingroup$ @AnonymousMaths Yes of course you can $\endgroup$ – Petite Etincelle Aug 17 '14 at 13:17
  • $\begingroup$ A curiosity just. How would you rate this question? I could not do it so obviously it was hard for me. $\endgroup$ – AnonymousMaths Aug 17 '14 at 13:20

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