# If {$v_1,v_2,v_3$} are linear independent is {$v_1+3v_3,v_2+v_1,v_1+v_3$} also linear independent?

If {$v_1,v_2,v_3$} are linear independent is {$v_1+3v_3,v_2+v_1,v_1+v_3$} also linear independent ?

I tried $a(v_1+3v_3)+b(v_2+v_1) +c(v_1+v_3)=0$

thus by grouping i got $(a+b+c)v_1+(3a+c)v_3+bv_2=0$

thus because $v_1,v_3,v_2$ are linear independent this has only a solution if $a,b, c$ are 0

• Yes. WLOG you can assume $v_i = e_i$ where $e_i$ is the canonical basis, after that you just have a determinant to compute. – user171326 Jan 13 '17 at 0:51
• $V \supseteq S$ is lin ind if $\forall x \in S: x \notin <S-\{x\}>$.. – mle Feb 15 '17 at 22:29

## 1 Answer

Your solution is correct; however, I'd suggest inserting one piece of detail that you skipped over.

The fact that $$(a+b+c)v_1+bv_2+(3a+c)v_3=0$$ does not, immediately, imply that you only have a solution for $a=b=c=0$; it has a solution only for $a+b+c=0$, $b=0$, and $3a+c=0$.

Now, clearly, $b=0$; then the system reduces to $a+c=0$ and $3a+c=0$, and you can finish the algebra from here to show that $a=b=c=0$.