text: ### Welcome to our community
  
  #### Yankel
  
  ##### Active member
  which one of the next statements is the correct one ?
  
  Let v,u,w be linearly dependent vectors in a vector space over R.
  
  u1 = 2u
  v1 = -3u+4v
  w1 = u+2v-aw (a scalar from R)
  
  (1) the vectors u1, v1 and w1 are linearly dependent for every value of a
  (2) the vectors u1, v1 and w1 are linearly independent for every value of a
  (3) the vectors u1, v1 and w1 are linearly independent for every value of a apart from 0
  (4) the vectors u1, v1 and w1 are linearly independent for every positive value of a
  (5) there exists a value of a for which the vectors u1, v1 and w1 are linearly independent
  
  Thanks a lot !!
  
  #### Chris L T521
  
  ##### Well-known member
  Staff member
  which one of the next statements is the correct one ?
  
  Let v,u,w be linearly dependent vectors in a vector space over R.
  
  u1 = 2u
  v1 = -3u+4v
  w1 = u+2v-aw (a scalar from R)
  
  (1) the vectors u1, v1 and w1 are linearly dependent for every value of a
  (2) the vectors u1, v1 and w1 are linearly independent for every value of a
  (3) the vectors u1, v1 and w1 are linearly independent for every value of a apart from 0
  (4) the vectors u1, v1 and w1 are linearly independent for every positive value of a
  (5) there exists a value of a for which the vectors u1, v1 and w1 are linearly independent
  
  Thanks a lot !!
  This is somewhat similar to the last one. Again, we know that $u,v,w$ are dependent, implying that there are constants $c_1,c_2,c_3$ not all zero such that $c_1u+c_2v+c_3w=0$. Now, we want to analyze when the following is true:
  
  $d_1u_1+d_2v_1+d_3w_1=0$
  
  where $d_1,d_2,d_3\in\mathbb{R}$ are arbitrary constants. The idea now is to express the above equation in terms of a linear combination of just $u,v,w$, then use the fact that $u,v,w$ are linearly dependent to come up with the appropriate conclusion.
  
  I hope this helps!
  
  #### Yankel
  
  ##### Active member
  right, so if I am not mistaken I get:
  
  (2d1-3d2+d3)u + (-4d2+2d3)v + (-ad3)w = 0
  
  what does it tells me ? I know that at least one of the coefficients is not zero, because u,v and w are dependent...what can I say about u1,v1,w1 and what about a ?
  
  #### Deveno
  
  ##### Well-known member
  MHB Math Scholar
  i think focusing on the coefficients overmuch is a mistake.
  
  it is clear that:
  
  $\{u_1,v_1,w_1\} \subset \text{Span}(\{u,v,w\})$.
  
  since {u,v,w} is linearly dependent, this has dimension ≤ 2.
  
  therefore $u_1,v_1,w_1$ cannot be linearly independent, else we have a subspace of greater dimension than a space which contains it.
  
  ("a" is a red herring).
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