Answer to Question #143190 in Linear Algebra for Ojugbele Daniel

Question

show that ( 1, 1, 1), ( 0, 1, 1) and ( 0, 1, -1) generate or spanned R^3

Answer 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

$\alpha \vec a_1+\beta\vec a_2+\gamma\vec a_3=\vec 0$

and show that the numbers $\alpha,\beta,\gamma$ are equal 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 $\vec a _1 =(1,1,1), \vec a_2=(0,1,1), \vec a_3=(0,1,-1)$ form the basis in $R^3$