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Let-vector-set-u-1-u-2-u-3-u-4-in-C-n-free-linear-So-that-u-1-u-2-u-2-u-3-u-3-u-4-u-4-u-1-too-free-linear-scalar-have-to-

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Question Number 54955 by gunawan last updated on 15/Feb/19
Let vector set {u_1 , u_2 , u_3 , u_4 } in C^n free linear. So that {u_1 +αu_2 , u_2 +αu_3 , u_3 +αu_4 , u_4 +αu_1 } too free linear , scalar α have to...
$$\mathrm{Let}\:\mathrm{vector}\:\mathrm{set}\:\left\{{u}_{\mathrm{1}} ,\:{u}_{\mathrm{2}} ,\:{u}_{\mathrm{3}} ,\:{u}_{\mathrm{4}} \right\}\:\mathrm{in}\:\mathbb{C}^{{n}} \\ $$$$\mathrm{free}\:\mathrm{linear}.\:\mathrm{So}\:\mathrm{that}\: \\ $$$$\left\{{u}_{\mathrm{1}} +\alpha{u}_{\mathrm{2}} ,\:{u}_{\mathrm{2}} +\alpha{u}_{\mathrm{3}} ,\:{u}_{\mathrm{3}} +\alpha{u}_{\mathrm{4}} ,\:{u}_{\mathrm{4}} +\alpha{u}_{\mathrm{1}} \right\} \\ $$$$\mathrm{too}\:\mathrm{free}\:\mathrm{linear}\:,\:\mathrm{scalar}\:\alpha\:\mathrm{have}\:\mathrm{to}… \\ $$
Answered by kaivan.ahmadi last updated on 15/Feb/19
c_1 (u_1 +αu_2 )+c_2 (u_2 +αu_3 )+c_3 (u_3 +αu_4 )+c_4 (u_4 +αu_1 )=0⇒ (c_1 +c_4 α)u_1 +(c_1 α+c_2 )u_2 +(c_2 α+c_3 )u_3 +(c_3 α+c_4 )u_4 =0⇒since u_i are free linear { ((c_1 +c_4 α=0 ⇒c_1 =−c_4 α)),((c_1 α+c_2 =0 ⇒c_4 α^2 =c_2 )),((c_2 α+c_3 =0 ⇒−c_4 α^3 =c_3 )),((c_3 α+c_4 =0 ⇒−c_4 α^4 +c_4 =0⇒(1−α^4 )c_4 =0⇒c_4 =0)) :} ⇒c_3 =c_2 =c_1 =0 we should have 1−α^4 ≠0⇒α^4 ≠1
$${c}_{\mathrm{1}} \left({u}_{\mathrm{1}} +\alpha{u}_{\mathrm{2}} \right)+{c}_{\mathrm{2}} \left({u}_{\mathrm{2}} +\alpha{u}_{\mathrm{3}} \right)+{c}_{\mathrm{3}} \left({u}_{\mathrm{3}} +\alpha{u}_{\mathrm{4}} \right)+{c}_{\mathrm{4}} \left({u}_{\mathrm{4}} +\alpha{u}_{\mathrm{1}} \right)=\mathrm{0}\Rightarrow \\ $$$$\left({c}_{\mathrm{1}} +{c}_{\mathrm{4}} \alpha\right){u}_{\mathrm{1}} +\left({c}_{\mathrm{1}} \alpha+{c}_{\mathrm{2}} \right){u}_{\mathrm{2}} +\left({c}_{\mathrm{2}} \alpha+{c}_{\mathrm{3}} \right){u}_{\mathrm{3}} +\left({c}_{\mathrm{3}} \alpha+{c}_{\mathrm{4}} \right){u}_{\mathrm{4}} =\mathrm{0}\Rightarrow{since}\:{u}_{{i}} \:{are}\:{free}\:{linear} \\ $$$$\begin{cases}{{c}_{\mathrm{1}} +{c}_{\mathrm{4}} \alpha=\mathrm{0}\:\:\:\:\:\Rightarrow{c}_{\mathrm{1}} =−{c}_{\mathrm{4}} \alpha}\\{{c}_{\mathrm{1}} \alpha+{c}_{\mathrm{2}} =\mathrm{0}\:\:\:\:\:\Rightarrow{c}_{\mathrm{4}} \alpha^{\mathrm{2}} ={c}_{\mathrm{2}} }\\{{c}_{\mathrm{2}} \alpha+{c}_{\mathrm{3}} =\mathrm{0}\:\:\:\:\:\Rightarrow−{c}_{\mathrm{4}} \alpha^{\mathrm{3}} ={c}_{\mathrm{3}} }\\{{c}_{\mathrm{3}} \alpha+{c}_{\mathrm{4}} =\mathrm{0}\:\:\:\:\Rightarrow−{c}_{\mathrm{4}} \alpha^{\mathrm{4}} +{c}_{\mathrm{4}} =\mathrm{0}\Rightarrow\left(\mathrm{1}−\alpha^{\mathrm{4}} \right){c}_{\mathrm{4}} =\mathrm{0}\Rightarrow{c}_{\mathrm{4}} =\mathrm{0}}\end{cases} \\ $$$$\Rightarrow{c}_{\mathrm{3}} ={c}_{\mathrm{2}} ={c}_{\mathrm{1}} =\mathrm{0} \\ $$$${we}\:{should}\:{have}\:\mathrm{1}−\alpha^{\mathrm{4}} \neq\mathrm{0}\Rightarrow\alpha^{\mathrm{4}} \neq\mathrm{1} \\ $$
Commented by gunawan last updated on 15/Feb/19
nice thank you Sir
$$\mathrm{nice} \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$

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