4
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ 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
3
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.