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Linear algebra problem check

  1. Apr 3, 2013 #1
    1. The problem statement, all variables and given/known data

    Let x and y be linearly independent vectors in R2.

    If ||x||=2 and ||y||=3, what if anything can we conclude about the possible values of |xTy|

    I know that ||x||*||y||cos(∅)=|xTy|

    If ||x||=2 and ||y||=3, then we can get either 6 or -6, i'm not sure if this is right. It is the only logical thing I can come up with.





    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 3, 2013 #2

    Mark44

    Staff: Mentor

    If cos(θ) were equal to 1, then ||x|| ||y|| cos(θ) would be 6, but that can't happen, given what we know about x and y. Do you see why?

    Note that |xTy| is the absolute value of the dot product of x and y, so it can't be less than 0.
     
  4. Apr 3, 2013 #3
    Which means if it cant be less then zero the only possible outcome is 6?
     
  5. Apr 3, 2013 #4
    No. [tex] \|\vec{x}\cdot \vec{y}\| = 6 |\cos \theta| [/tex] as you have shown. What are all the possible values of [tex] 6 |\cos \theta| [/tex] if [itex] \theta [/itex] is not zero or π?
     
  6. Apr 3, 2013 #5
    Theta can be 90 pretty much only 90 or 270?
     
  7. Apr 3, 2013 #6
    Why can not [itex] \theta [/itex] take on other values? Linearly independent does not mean orthogonal.
     
  8. Apr 3, 2013 #7
    Actually using the cauchy-schwarz inequality

    if x and y are vectors in either R.^2 or R.^3 then

    |xTy|≤||x||*||y||

    or simply

    |xTy|≤6
     
  9. Apr 3, 2013 #8
    Not quite. Since the two vectors are linearly independent, it can not be the case that [itex] \cos\theta=1 [/itex] so the inequality is strict.
     
  10. Apr 3, 2013 #9
    Well if cosθ=1 can not equal 1 then I see only one option. For it to be either zero or -1 but the absolute value takes care of the negative case.
     
  11. Apr 3, 2013 #10
    Why can't [itex] |cos\theta | [/itex] equal 1/10, .38, or any other value in [itex] [0,1) [/itex]?
     
  12. Apr 3, 2013 #11
    Tbh I'm still quite unsure why cant cosθ be zero. It is the dot product that has a limitation on it. Not cosθ.
     
  13. Apr 3, 2013 #12
    [itex] cos\theta [/itex] can be zero. Just not 1 from the condition of linear independence.
     
  14. Apr 3, 2013 #13
    Ok. Well if cosθ can be zero then we know that it can in fact be 6. and cosθ can be anything less then 1?
     
  15. Apr 3, 2013 #14
    Um, if [itex] cos\theta [/itex] is zero, then the expression is zero. [itex] cos\theta [/itex] can be anything between zero and one, including zero but not one. Explain why this is the case.
     
  16. Apr 3, 2013 #15
    Sorry all this time I thought we were referring to theta.

    If it can't be 1 because that would mean that orthogonality would no longer be there.
     
  17. Apr 3, 2013 #16
    Correct. So now that you understand the constraint on the cosine, can you finish off the problem?
     
  18. Apr 3, 2013 #17
    Which means the answer would be

    6cos(∅)=|xTy| such that 0≤cos(theta)∠1
     
  19. Apr 3, 2013 #18

    Mark44

    Staff: Mentor

    There's nothing about orthogonality in this problem. It's given that the vectors are linearly independent, which means they can't be parallel. The vectors CAN be orthogonal, but don't have to be.
     
  20. Apr 3, 2013 #19
    Yes, it should have said there would no longer be linear independence. I did not catch that. Do you understand why having a cosine equal to one violates linear independence?
     
  21. Apr 3, 2013 #20
    So is my above answer incorrect?
     
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