Linearly independent set in a vector space

  • Thread starter AkilMAI
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  • #1
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Homework Statement


I need to prove that, if {u;v;w} is a linearly independent set in a
vector space, then the set
{2u + v + w; u + 2v + w; u + v + 2w}
is also linearly independent.



Homework Equations


......


The Attempt at a Solution


if {u;v;w} is a linearly independent set=> c1*u +c2*v+c3*w=0 and c1=c2=c3=0 and also u,v,w are distinct and different from 0.
=>c1(2u + v + w)+ c2(u + 2v + w)+ c3(u + v + 2w)=0,but if I create a system of equations based on this i get u=v=w=0....any help?
 

Answers and Replies

  • #2
radou
Homework Helper
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From this upper equation you get a system of equations involving the coefficients c1, c2, c3. This system yields a solution c1 = c2 = c3 = 0. This, in turn, implies what you need to show.

Edit: but note that the c's are not the same as in the upper equation - your equation contains coefficients which are sums of the c's.
 
  • #3
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writing u(2c1+c2+c3)+v(c1+2c2+c3)+w(c1+c2+2c3)=0...and by definition u v and w have to be distinct and different from 0.=>2c1+c2+c3=0,c1+2c2+c3=0,c1+c2+2c3=0....calculating this system of equations and i get c1=c2=c3=0....is this correct?
 
  • #4
radou
Homework Helper
3,115
6
writing u(2c1+c2+c3)+v(c1+2c2+c3)+w(c1+c2+2c3)=0...and by definition u v and w have to be distinct and different from 0.=>2c1+c2+c3=0,c1+2c2+c3=0,c1+c2+2c3=0....calculating this system of equations and i get c1=c2=c3=0....is this correct?
Yes. And the important thing is that c1 = c2 = c3 = 0 implies 2c1 + c2 + c3 = 0, c1 + 2c2 + c3 = 0 and c1 + c2 + 2c3 = 0, which is exactly what you need to show.
 

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