a⃗1=(1,1,1),a⃗2=(0,1,1),a⃗3=(0,1,−1)\vec a _1 =(1,1,1), \vec a_2=(0,1,1), \vec a_3=(0,1,-1)a1=(1,1,1),a2=(0,1,1),a3=(0,1,−1)
We show that the vectors form the basis.
Consider a linear combination of vectors
αa⃗1+βa⃗2+γa⃗3=0⃗\alpha \vec a_1+\beta\vec a_2+\gamma\vec a_3=\vec 0αa1+βa2+γa3=0
and show that the numbers α,β,γ\alpha,\beta,\gammaα,β,γ are equal 0.
αa⃗1+βa⃗2+γa⃗3=0⃗α(1,1,1)+β(0,1,1)+γ(0,1,−1)=(0,0,0)α=0α+β+γ=0α+β−γ=0α=β=γ=0\alpha \vec a_1+\beta\vec a_2+\gamma\vec a_3=\vec 0\\ \alpha (1,1,1)+\beta(0,1,1)+\gamma(0,1,-1)=(0,0,0)\\ \begin{matrix} \alpha=0\\ \alpha+\beta+\gamma=0\\ \alpha+\beta-\gamma=0 \\\end{matrix}\\ \alpha=\beta=\gamma=0αa1+βa2+γa3=0α(1,1,1)+β(0,1,1)+γ(0,1,−1)=(0,0,0)α=0α+β+γ=0α+β−γ=0α=β=γ=0
Then vectors a⃗1=(1,1,1),a⃗2=(0,1,1),a⃗3=(0,1,−1)\vec a _1 =(1,1,1), \vec a_2=(0,1,1), \vec a_3=(0,1,-1)a1=(1,1,1),a2=(0,1,1),a3=(0,1,−1) form the basis in R3R^3R3
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