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Guys I am given a set of vectors $(v_1,\dots,v_{10}), \mathbb{R}^{10}, n=10$, which are linearly independent. It doesn't give me the values of the vectors but tells me to find if these are independent or not.

a. $v_1+v_2, v_2+v_3, v_3+v_4, v_4+v_1$

b. $v_1+v_2, 2(v_2+v_3), 3(v_3+v_4),..., 10(v_{10}+v_1)$

Not sure how to solve this. I know the definition of independency but I am having trouble applying it.

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    $\begingroup$ can I put a1, a2, a3, a4 before the vectors, solve in respect to the vectors and if it can equal 0 its independent? I dont want to confuse myself too much here. $\endgroup$ – D-Man Mar 19 '13 at 18:07
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Since $v_1+v_2 = (v_2+v_3)-(v_3+v_4)+(v_4+v_1)$ we see that the vector $v_1+v_2$ is a linear combination of the other $3$ vectors thus $v_1+v_2$ is in span$\{v_2+v_3,v_3+v_4,v_4+v_1\}$ and so the set in part a cannot be linear independent. You can adapt this argument to show that, in part b, $v_1+v_2$ is a linear combination of $v_2+v_3, v_3+v_4, .., v_{10}+v_1$ which shows that this set is not linearly independent either.

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  • $\begingroup$ Just pick $v_1+v_2 = \frac{1}{2}2(v_2+v_3)-\frac{1}{3}3(v_3+v_4)+\cdots+\frac{1}{10}10{v_{10}+v_1)$ and you have a linear combination that shows the set of vectors is dependent. $\endgroup$ – Wintermute Mar 19 '13 at 18:31
  • $\begingroup$ You need the dollar signs around the LaTeX. Also great answer. $\endgroup$ – Thomas Mar 19 '13 at 18:33
  • $\begingroup$ wouldn't b be independent? I worked it out and found that it was unique so that signals independency? correct me if I am wrong. $\endgroup$ – D-Man Mar 20 '13 at 2:15
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For a: So consider: $$ a_1(v_1 + v_2) + a_2(v_2 + v_3) + a_3(v_3+v_4) + a_4(v_4 + v_1) = 0 . $$ Can you prove from this that all the $a_i$'s are zero? You get that this is $$ (a_1 + a_4)v_1 + (a_1 + a_2)v_2 + (a_2 + a_3)v_3 + (a_3 + a_4)v_4 = 0. $$

So since the $v_i$'s are linearly independent, you get $$ \begin{align} a_1 + a_4 &= 0 \\ a_1 + a_2 &= 0 \\ a_2 + a_3 &= 0 \\ a_3 + a_4 &= 0. \end{align} $$ Does this imply that all the $a_i$'s are zero? This set of equations will have a solution if and only if the matrix $$ \pmatrix{1 & 0 & 0 &1 \\ 1 & 1 & 0 & 0 \\0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1\\} $$ is invertible. Is it?

For b: Try the same thing. Start with $$ a_1(v_1 + v_2) + a_22(v_2+ v_3) + \dots + a_{10}10(v_{10} + v_1) = 0. $$ Again just expand the left hand side and try to do as in part a.

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  • $\begingroup$ alright cool it makes a little bit more sense now! is it the same type of idea for say the vectors: V1, V1+c1v2, V2+c2v3, v3+c3v4,... v10 +c10v1, c1,....c10? just want to make sure there is no exceptions to the ways shown above. $\endgroup$ – D-Man Mar 20 '13 at 2:09
  • $\begingroup$ so based off this it is invertible.... which means there is a solution which means its independent? $\endgroup$ – D-Man Mar 20 '13 at 3:28
  • $\begingroup$ @D-Man: The determinant of the $4\times 4$ matrix is $0$, so ... $\endgroup$ – Thomas Mar 20 '13 at 14:07
  • $\begingroup$ @D-Man: Also: the other answer is better since here it is shown directly that the vectors are linearly dependent: We find a linear combination that is zero. $\endgroup$ – Thomas Mar 20 '13 at 14:11

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