Question

Show the vector v1=(1,2,3),v2=(0,1,2),v3=(0.0.1)generate v3(R)

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO PROVE

$\sf{The \: vector \: \: v_1=(1,2,3),v_2=(0,1,2),v_3=(0, 0, 1)}$

$\sf{generates \: \: \: { \mathbb{R}}^{3} }$

PROOF

$\sf{Let \: \: S = \: \{ \: (1,2,3),(0,1,2),(0, 0, 1) \}\: }$

$\sf{Let \: \alpha = (a, b, c) \: be \: an \: arbitrary \: vector \: of \: { \mathbb{R}}^{3}}$

$\sf{We \: have \: to \: show \: that \: \: \alpha \in L(S) } \:$

If possible let

$\sf{ \alpha = \: r_1v_1+r_2v_2+r_3v_3 \: \: for \: real \: \: r_1,r_2,r_3}$

$\implies \sf{(a,b,c) = \: r_1(1,2,3)+r_2(0,1,2)+r_3(0,0,1) }$

$\sf{ \implies (a, b, c) =(r_1,2r_1+r_2,3r_1+2r_2+r_3)}$

Comparing

$\sf{ \implies r_1= a, 2r_1+r_2=b,c=3r_1+2r_2+r_3}$

Which gives

$\sf{r_1= a}$

$\sf{r_2=b - 2 a}$

$\sf{r_3=c - 2b - 3 a}$

$\sf{ This \: proves \: that \: \: \alpha \in L(S) } \:$

$\sf{ \therefore \: \: \mathbb{{R}^{3} } \: \subset \: \: L(S)}$

$\sf{Again \: \: \: S \: \subset { \mathbb{R}}^{3} }$

Also L(S) being the smallest subspace containing S

$\sf{ \therefore \: \: \: L(S) \: \subset { \mathbb{R}}^{3} }$

$\sf{ Consequently \: \: \: L(S) \: = { \mathbb{R}}^{3} }$

Hence

$\sf{The \: vector \: \: v_1=(1,2,3),v_2=(0,1,2),v_3=(0, 0, 1)}$

$\sf{generates \: \: \: { \mathbb{R}}^{3} }$

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