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Show that v, ø(v), ø(v)^2 are independent

  1. Feb 8, 2014 #1
    Please see attached question.
    I can finish part (a)

    For part b, how can I find ø(v) ?
    Although I can find ø(v1) and ø(v2) but I think it is unrelated to ø(v)....
     

    Attached Files:

  2. jcsd
  3. Feb 8, 2014 #2

    Dick

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    Part b isn't really related to the first part. If you assume ##v, \phi(v), \phi^2(v)## are linearly DEPENDENT what does that tell you about them? What's the definition of linear dependence? Try to use that to reach a contradiction to what you know about the vectors. If you can show one of them must be zero then you've shown they are linearly independent.
     
  4. Feb 8, 2014 #3
    (b)
    Thank you for your reply

    This is my solution. I do not know if it is correct.
    Assume v, ø(v) and ø(v)^2 are linearly dependent

    [tex] => a_{1}v+a_{2}\phi (v)+a_{3}\phi (v)^{2} = 0 [/tex]
    for [tex] a_{1}, a_{2}, a_{3} [/tex] not all zero

    since [tex] \phi (v)^{2} \neq 0 [/tex] we have [tex]a_{3}=0[/tex]

    Contradiction, so v, ø(v) and ø(v)^2 are linearly dependent
     
    Last edited: Feb 8, 2014
  5. Feb 8, 2014 #4
    I have another question about part c.
    Why column 1 is M^2*v? How can we know?

    Please see attached. Many thanks.
     

    Attached Files:

  6. Feb 8, 2014 #5

    Dick

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    That's not correct at all. You did get the definition of linear dependence right. So you have ##a_{1}v+a_{2}\phi (v)+a_{3}\phi^2 (v) = 0## (notice I wrote ##\phi^2 (v)## not ##\phi (v)^2## that's what you really want). It's worth trying to take it slowly and get this right. It's pretty basic. ##\phi^2 (v) \ne 0## doesn't imply anything about ##a_3##. I'll give you a hint. What happens if you apply ##\phi^2## to that equation? What does that tell you about ##a_1##?
     
    Last edited: Feb 8, 2014
  7. Feb 8, 2014 #6

    Dick

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    It's much better to post new questions in new threads.
     
  8. Feb 11, 2014 #7
    I am a beginner in this topic...I know the trick is to add phi to the equation
    is this a correct proof?

    Assume v, ø(v) and ø(v)^2 are linearly dependent

    [tex] => a_{1}v+a_{2}\phi (v)+a_{3}\phi ^{2}(v) = 0 [/tex]
    for [tex] a_{1}, a_{2}, a_{3} [/tex] not all zero

    [tex] => \phi^2( a_{1}v+a_{2}\phi (v)+a_{3}\phi ^{2}(v)) = 0 [/tex]

    since [tex] \phi^2 \neq 0 [/tex]

    it leads to contradiction
     
  9. Feb 11, 2014 #8

    Dick

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    Nope, not correct. ##\phi^2( a_{1}v+a_{2}\phi (v)+a_{3}\phi ^{2}(v)) = a_1 \phi^2(v) +a_2 \phi^3(v) + a_3 \phi^4(v)##, what can you say about ##\phi^3(v)## and ##\phi^4(v)##??
     
  10. Feb 11, 2014 #9
    In my opinion, ##\phi^3(v)## = 0 => ##\phi^4(v)## = 0

    so, ##a_1 \phi^2(v) +a_2 \phi^3(v) + a_3 \phi^4(v)##=0 implies ##a_1 \phi^2(v)## = 0

    which gives ##a_1=0##

    and it leads to contradiction??
     
  11. Feb 11, 2014 #10

    Dick

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    It's not a contradiction yet. Now put ##a_1=0## into ##a_{1}v+a_{2}\phi (v)+a_{3}\phi ^{2}(v) = 0##. Any ideas what to do next?
     
  12. Feb 11, 2014 #11
    Thanks dick, i know the trick now.
    So this is my complete proof of part (b), please have a look.

    Assume v, ##\phi (v)## and ##\phi ^{2}(v)## are linearly dependent

    i.e ##a_{1}v+a_{2}\phi (v)+a_{3}\phi^2 (v) = 0## for ##a_{1}, a_{2}, a_{3}## not all zero

    ## \phi^2( a_{1}v+a_{2}\phi (v)+a_{3}\phi ^{2}(v)) = 0 ##

    => ## a_1 \phi^2(v) +a_2 \phi^3(v) + a_3 \phi^4(v) = 0##

    Since ##\phi^3(v)## = 0 => ##\phi^4(v)## = 0, we have ##a_{1}=0##

    So the equation ##a_{1}v+a_{2}\phi (v)+a_{3}\phi ^{2}(v) = 0 ## gives ##a_{2}\phi (v)+a_{3}\phi ^{2}(v) = 0##

    ## \phi (a_{2}\phi (v)+a_{3}\phi ^{2}(v)) = 0 ##

    => ##a_{2}\phi ^2(v)+a_{3}\phi ^{3}(v) = 0##

    and once again, since ##\phi^3(v)= 0## and ##\phi^2 \neq 0 ##

    we have ##a_{2}=0 ##

    so the equation ##a_{1}v+a_{2}\phi (v)+a_{3}\phi ^{2}(v) = 0 ## gives ##a_{3}\phi ^{2}(v) = 0##

    since ##\phi^2 \neq 0 ## we have ##a_{3}=0 ##

    This leads to contradiction, since we assume ##a_{1}, a_{2}, a_{3}## not all zero

    so v, ##\phi (v)## and ##\phi ^{2}(v)## are linearly INDEPENDENT
     
  13. Feb 11, 2014 #12

    Dick

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    That's perfect.
     
  14. Feb 11, 2014 #13

    LCKurtz

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    Almost. Correction in red. It isn't ##\phi^2## that isn't zero, it's ##\phi^2(v)## that isn't.
     
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