Let the vectors $v_1 , v_2 , v_3$ be in $R^n$ ($ n \geq 3$) , and assume that these vectors are not zero.
assuming that the sets {${v_1 , v_2}$} , {${v_2 , v_3}$} and {${v_1 , v_3}$} are linearly independent , then does the set {${v_1 , v_2 , v_3}$} have to be independent as well?
My solution:
According to the information given that {${v_1 , v_2}$} , {${v_2 , v_3}$} and {${v_1 , v_3}$} are linearly independent we get that
- $x_1 \cdot v_1+x_2 \cdot v_2=0$ $x_1 ,x_2 = 0$
- $x_2 \cdot v_2+x_3 \cdot v_3=0$ $x_2 ,x_3 = 0$
- $x_3 \cdot v_3+x_1 \cdot v_1=0$ $x_3 ,x_1 = 0$
now we need to check if the set {${v_1 , v_2 , v_3}$} is independent , I did this by counterexample.
assume n=3 and $v_1=$ {$1,0,0$} , $v_2=$ {$0,1,0$} and $v_3=$ {$1,1,0$} then its obvious that {${v_1 , v_2}$} , {${v_2 , v_3}$} and {${v_1 , v_3}$} are linearly independent but {${v_1 , v_2 , v_3}$} is dependent because $v_1 +v_2 = v_3$.
What I mean by general proof is I rather prove this by a way without a countrexample with numbers as I believe it is harder to always find numbers that satisfy what we are looking for , this is what I also tried for the "general" proof.
if the set {${v_1 , v_2 , v_3}$} is dependent then $x_1 \cdot v_1+x_2 \cdot v_2 +x_3 \cdot v_3=0$ while $x_1 , x_2 , x_3$ are different than zero. according to the given information $x_2 \cdot v_2+x_3 \cdot v_3=0$ then I used that to do this $x_1 \cdot v_1+x_2 \cdot v_2 +x_3 \cdot v_3=x_2 \cdot v_2 +x_3 \cdot v_3$ and we $x_1 \cdot v_1 = x_2 \cdot v_2 +x_3 \cdot v_3 $ then we get that it is linearly dependent.. but this way seemed very wrong and uncertain to me is it possible to do this with a general proof? or counterexample is the only way?
Thank you!
Best Answer
$v_1=(1,0,0),v_2=(0,1,0), v_3=(1,1,0)$ is a simple counter-example in $\mathbb R^{3}$. For $\mathbb R^{n}$ you can make an obvious modification