Question #143190
show that ( 1, 1, 1), ( 0, 1, 1) and ( 0, 1, -1) generate or spanned R^3
1
Expert's answer
2020-11-12T19:08:46-0500

a1=(1,1,1),a2=(0,1,1),a3=(0,1,1)\vec a _1 =(1,1,1), \vec a_2=(0,1,1), \vec a_3=(0,1,-1)

We show that the vectors form the basis.

Consider a linear combination of vectors

αa1+βa2+γa3=0\alpha \vec a_1+\beta\vec a_2+\gamma\vec a_3=\vec 0

and show that the numbers α,β,γ\alpha,\beta,\gamma are equal 0.

αa1+βa2+γa3=0α(1,1,1)+β(0,1,1)+γ(0,1,1)=(0,0,0)α=0α+β+γ=0α+βγ=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=0

Then vectors a1=(1,1,1),a2=(0,1,1),a3=(0,1,1)\vec a _1 =(1,1,1), \vec a_2=(0,1,1), \vec a_3=(0,1,-1) form the basis in R3R^3


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