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Heyo, I'm just wondering if I'm correct in assuming that the following three vectors in four dimensional space are linearly dependent.. I simplified them using elemental row operations and ended up with a matrix that has two rows full of zeros, and based on that I concluded that there are infinitely many solutions to the equation system, therefore, the vectors must be dependent. BUT, if you write out the original three vectors as a system of equations, and then calculate the three variables, you come to the conclusion that the only solution is one where all three variables are zero, therefore making the three vectors independent :S a tad confuseddd

$v_1=(1,1,-1,0)^T$

$v_2=(0,-1,1,-2)^T$

$v_3=(3,1,-1,-4)^T$

maany thanks

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    $\begingroup$ It holds that $v_3=3v_1+2v_2.$ Therefore, the vectors are NOT independent. $\endgroup$ – user376343 May 31 at 11:23
  • $\begingroup$ @user376343 how are you able to deduce that purely from the fact that $v_3=3v_1+2v_2$ ? $\endgroup$ – Biaaach May 31 at 11:25
  • $\begingroup$ What does “linearly independent” mean to you? $\endgroup$ – José Carlos Santos May 31 at 11:26
  • $\begingroup$ Did you learn a method for this? Perhaps called "Gaussian reduction" or "echelon form"? $\endgroup$ – GEdgar May 31 at 11:27
  • $\begingroup$ @JoséCarlosSantos to me it means the following: a set of vectors is linearly independent if none of its vectors can be represented as a scalar multiple of the other/remaining vectors.. $\endgroup$ – Biaaach May 31 at 11:28
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They are dependent: $$ 3v_1 + 2v_2 - v_3 = 0$$

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  • $\begingroup$ oh damn alright, thanks. do you think that would be considered sufficient evidence/proof of their dependence? $\endgroup$ – Biaaach May 31 at 11:26
  • $\begingroup$ Yes, because this is how "linearly dependent" is defined / direct application of the definition. $\endgroup$ – farud May 31 at 11:30
  • $\begingroup$ great, thanks for taking the time to answer it $\endgroup$ – Biaaach May 31 at 11:32
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    $\begingroup$ I'll add to @farud's comment that a system of vectors is linearly independent , by definition, if there no non-trivial linear relation between them. $\endgroup$ – Bernard May 31 at 11:33

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