I have this problem :

If $\{v_1,v_2,v_3\}$ is independent is $\{v_1+v_2,v_2+v_3,v_1+v_3\}$ also independent?

My solution

For $\lambda_1,\lambda_2,\lambda_3 \in R$

We need to find if the only solution for this equation :


Is the trivial solution.

$$\lambda_1v_1+\lambda_1v_2+\lambda_2v_2+\lambda_2v_3+\lambda_3v_1+\lambda_3v_3= \\ v_1(\lambda_1+\lambda_3)+v_2(\lambda_1+\lambda_2)+v_3(\lambda_2+\lambda_3)=0$$

So for :




Since we know that for any $\delta_1,\delta_2,\delta_3 \in F \implies \delta_1v_1+\delta_2v_2+\delta_3v_3=0$ only if $\delta_1=\delta_2=\delta_3=0$ and for the $\delta_1,\delta_2,\delta_3$ we chosen.

Therefore we can conclude that $\{v_1+v_2,v_2+v_3,v_1+v_3\}$.

I'm not sure if my proof is correct, I'll glad to hear feedbacks.

Thank you!

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    $\begingroup$ What information do you have about the characteristic of the field $F$? $\endgroup$ – Mark Dickinson Jan 11 '15 at 15:58
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    $\begingroup$ but your aim is to show $\lambda_1 = 0, \lambda_2= 0, \lambda_3 = 0.$ have you done that? i don't think so. $\endgroup$ – abel Jan 11 '15 at 16:02
  • $\begingroup$ @MarkDickinson Should be $R$ instand. Edited, Thanks $\endgroup$ – JaVaPG Jan 11 '15 at 16:11

So far you've shown that your coefficients must satisfy \begin{align*} 0 &= \lambda_1+\lambda_3 \\ 0 &= \lambda_1 + \lambda_2 \\ 0 &= \lambda_2 + \lambda_3. \end{align*} At this point you're almost done, but you still need to show that this implies that each of the $\lambda_i$'s is equal to zero. This is not so hard to see but it needs to be verified for the proof to be complete.

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    $\begingroup$ We'd also need to know that we're not working in characteristic $2$... $\endgroup$ – Mark Dickinson Jan 11 '15 at 16:01
  • $\begingroup$ That's a good point. $\endgroup$ – Tom Jan 11 '15 at 16:08
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    $\begingroup$ @Tom Oh I thought its pretty clear since $\lambda_1+\lambda_3=\lambda_2+\lambda_2 \implies \lambda_2=\lambda_3=0$. so $\lambda_1=0$. $\endgroup$ – JaVaPG Jan 11 '15 at 16:19

what is left to do is to show that $A = \pmatrix{1&0&1\\1&1&0\\0&1&1}$ in invertible. there are several ways to do this:

(a) row reduce and see that you have a pivot on each row,

(b) take the determinant of $A$ and show that it is not zero,

(c) write $A$ as the sum of identity matrix $I$ and the permutation matrix $P = \pmatrix{0&0&1\\1&0&0\\0&1&0}$ know that the eigenvalues of $P$ are $1, \omega, \omega^2,$ where $\omega = -1/2 + i\sqrt 3/2$ cube root of unity. what that means is only $P - \lambda I$ are invertible. therefore $P + I$ is not invertible.

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    $\begingroup$ In fact, $\det(A)=2$, so that $A$ is not invertible for fields of characteristic $2$ (e.g. appearing in coding theory). $\endgroup$ – Dietrich Burde Jan 11 '15 at 16:18

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