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Show that {u+v,v-w,w-u} is a basis of R^3 if {u,v,w} is

  1. May 19, 2007 #1
    1. The problem statement, all variables and given/known data

    Suppose {u,v,w} is a basis for (R3). Show that {u+v,v-w,w-u} is also a basis for (R3).


    3. The attempt at a solution

    By definition, u,v and w are linearly independent and they span (R3).
    I really don't know where to go from here though.

    Thanks
     
    Last edited: May 20, 2007
  2. jcsd
  3. May 20, 2007 #2

    daniel_i_l

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    Can you prove that u+v , v-w , w-v is also linearly independent? If so, do you know why that's enough to make it a base?
     
  4. May 20, 2007 #3

    matt grime

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    I think there is an error in what is written. That set of three vectors is obvisouly linearly dependent: w-v = -(v-w).
     
  5. May 20, 2007 #4
    My bad, its fixed now
     
  6. May 20, 2007 #5

    matt grime

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    Was post 2 sufficient to help you? There is an easier way, in this case. It is perfectly possible to do it by inspection. Just adding some of those vectors together yields that u,v,w are all in the span of the three new ones, hence they form a basis (assuming char=/=2).
     
  7. May 20, 2007 #6
    How would I show that they are Linearly independent.

    But part (ii) is :
    Suppose {u,v,w,x} is a basis for R4. Show that {u+v,v+w,w+x,x+u} is NOT a basis for R4 So I can't really do it by inspection.

    Thanks guys
     
  8. May 20, 2007 #7

    matt grime

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    Yes, you can. Because it is easy to see by inspection that those 4 elements are linearly dependent. It is also good to spot easy things. And it will work in general, since that new set of vectors is a basis if and only if you can expess u,v,w as combinations of the three new ones, and it is easy to see how to do that by eye. In general, you need to set up and solve some simultaneous equations. In this case you can see the solution without doing and linear algebra.

    As for your other question: you know how to solve simultaneous equations, so do it. Find the a,b,c such that a(u+v)+b(v-w)+c(w-u)=0, and show that the only solution is for a=b=c=0. Note you have to rearrange to write things in terms of u,v,w which are known to be a basis. More succintly, note this implies (a-c)u+(a+b)v+(c-b)w=0, so since, u,v,w are a basis, this implies a-c=a+b=c-b=0, whence a=b=c=0.
     
    Last edited: May 20, 2007
  9. May 20, 2007 #8
    How do you show things are linearly independent in general? Remember that lin. independence comes from

    [tex]c_1 \mathbf{v_1} + c_2 \mathbf{v_2} + ... + c_n \mathbf{v_n} = 0[/tex]

    Is there a way to pick the constants in the linear combinations such that you get one of the other vectors?
     
  10. May 20, 2007 #9
    thanks everyone and especially matt grime for your patience
    I've done it now.
     
    Last edited: May 20, 2007
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