If the set $\{v_1,v_2,v_3\}$ as a subset of $\mathbf R^n$ (doesn't have to be in $\mathbf R^n$) is linearly independent, then so is the set $\{2v_1 + v_2 +v_3 , v_1 + 2v_2 + v_3 , v_1 + v_2 + 2v_3\}$. Give a proof if true, or a counterexample if false.

  • $\begingroup$ What do you think? What work have you done on this problem? $\endgroup$ – Larry B. Feb 17 '17 at 21:05
  • $\begingroup$ I am not sure, but I was thrown off by the fact that it said the set {v1,v2,v3} are a subset of R^n, and not in R^n. What difference does this make? @LarryB. $\endgroup$ – user417408 Feb 17 '17 at 21:06
  • $\begingroup$ Doesn't make a difference; it's a set of vectors, and that set is a subset of $\mathbb{R}^n$, which means each vector is in $\mathbb{R}^n$. $\endgroup$ – Larry B. Feb 17 '17 at 21:14
  • $\begingroup$ Do you see how $\{2v_1+v_2+v_3,v_1+2v_2+v_3,v_1+v_2+2v_3\}$ relates to the matrix $\left[\begin{smallmatrix}2&1&1\\1&2&1\\1&1&2 \end{smallmatrix}\right]$? Do you know how the reduced row echelon form relates to the linear independence or dependence of a set of vectors? $\endgroup$ – JMoravitz Feb 17 '17 at 21:29


By the definition of linear independence, you have to prove that $$ 1)\qquad a(2v_1+v_2+v_3)+b(v_1+2v_2+v_3)+c(v_1+v_2+2v_3)=0 $$ iff $a=b=c=0$

reordering $1)$ we have: $$ v_1(2a+b+c)+v_2(a+2b+c)+v_3(a+b+2c)=0 $$ and, if $\{v_1,v_2,v_3\}$ are linearly independent this is equivalent to: $$ \begin{cases} 2a+b+c=0\\ a+2b+c=0\\ a+b+2c=0 \end{cases} $$

Now solve the system.


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