Question

Determine whether or not the following vectors span R^3

u1 = ( 1, 1, 2), u2 = ( 1, -1, 2 ) and u3 = ( 1, 0, 1 ).

Answer 1

$\mathbf{Solution:} let ~u=(a,b,c)~\in ~R^3 ~be~ any~ element. \\ Given~vectors: u_{1}=(1,1,2),u_{2}=(1,-1,2),u_{3}=(1,0,1) \\Let ~if~ possible~u=k_{1}u_{1}+ k_{2}u_{2}+k_{3}u_{3},~where~k_{1},k_{2},k_{3}\in R^3 \\(a,b,c)=k_{1}(1,1,2)+ k_{2}(1,-1,2)+k_{3}(1,0,1) \\(a,b,c)=(k_{1},k_{1},2k_{1})+ (k_{2},-k_{2},2k_{2})+(k_{3},0,k_{3}) \\(a,b,c)=(k_{1}+k_{2}+k_{3},k_{1}-k_{2},2k_{1}+2k_{2}+k_{3}) \\ Comapring~ \\~~~~~~~~~~k_{1}+k_{2}+k_{3}=a \\~~~~~~~~~~k_{1}-k_{2}=b \\~and~2k_{1}+2k_{2}+k_{3}=c$

$Hence,~~~~~~number ~of~ variables=number~of~equations \\ Matrix ~A~in~corresponding~matrix~equation~ \\AX=~B~for ~system~is~"Square~Matrix" .$

$Where ~~~~~~A=\begin{bmatrix}1 & 1 & 1\\1 & -1 &0 \\2 & 2 & 1\end{bmatrix}$

$\therefore |A|=1(-1)-1(1)+1(2-(-2))=-1-1+4=-2+4=2\neq0$

$\therefore A~is~non~singular. ~i.e.~A~is~invertible. \\ \therefore linear~system~AX=B~has \,a \, solution. \\ \therefore~Every~Element~of~R^3 ~can ~be~ expressed~as \,a~linear~combination~of~u_{1},u_{2},u_{3}.$

$Hence ~u_{1},u_{2},u_{3}~span~R^3.$