Let $V$ be a vector space over a subfield $F$ of the complex numbers. Suppose $\alpha$, $\beta$ and $\gamma$ are linearly independent vectors in $V$. Prove that $(\alpha +\beta)$, $(\beta +\gamma)$ and $(\gamma +\alpha)$ are linearly independent.
My attempt: Since $\alpha \neq \gamma$, $\beta \neq \alpha$, $\beta \neq \gamma$, we have $(\alpha +\beta) \neq (\beta +\gamma)$, $(\beta +\gamma) \neq (\gamma +\alpha)$, $(\alpha +\beta)\neq (\gamma +\alpha)$. This is necessary condition for linear independence and “unnecessary detail” in proof. If $x_1 \cdot (\alpha +\beta)+x_2 \cdot (\beta +\gamma) +x_3 \cdot (\gamma +\alpha)=0_V$. So $(x_1 \cdot \alpha +x_1 \cdot \beta) +(x_2\cdot \beta +x_2 \cdot \gamma )+(x_3 \cdot \gamma +x_3 \cdot \alpha) =0_V$. By distributive axiom of vector space, $(x_1+x_3) \cdot \alpha+(x_1+x_2) \cdot \beta+(x_2+x_3)\cdot \gamma=0_V$. Since $\{\alpha, \beta, \gamma\}$ is independent, $(x_1+x_3)=(x_1+x_2)=(x_2+x_3)=0_F$. Which implies $x_1=x_2=x_3=0_F$. Hence $\{(\alpha +\beta) ,(\beta +\gamma) ,(\gamma +\alpha)\}$ is linearly independent. Is my proof correct?
Best Answer
Your proof is correct. You used the linearity of $\alpha, \beta$ and $\gamma$, to show that the pairwise summation of elements is also linearly independent.