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Homework Help: Spanning Sets

  1. Oct 24, 2004 #1
    I don't totally understand spanning sets...

    Can anyone explain this problem to me:

    let V = set of all polynomials with degree of 2 or less (a vector space_
    let S = {t + 1, t^2 + 1, t^2 - t}

    Does S span V?

    I know that (t^2 + 1) - (t + 1) = t^2 - t

    But I just don't see what that tells me....
  2. jcsd
  3. Oct 24, 2004 #2


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    This tells you that your vectors are linearly dependant. Do you see why?

    V has dimension 3. You have 3 vectors. If they spanned V, then they would be linearly independant as well (it's crucial here that you have 3 vectors and dimV=3, you probably have a theorem to this effect). You've shown they aren't linearly independant, so can they span?
  4. Oct 24, 2004 #3
    Thanks for the response....so basically since I reduced the system to 2 equations, it cannot span since it needs 3 equation (vectors) to span it?

    Another related question...

    let V = vector space of all continous functions on the interval of the real numbers

    let S = {sin(t), sin(2t), sin(3t)}

    is S of V linearly dependent or lineraly independent?

    Thanks any help
  5. Oct 24, 2004 #4
    Ok, I get the first question now....since they are linearly dependant, they cannot span a 3d vector space....that makes sense now. I still need some help on that second question though please.
  6. Oct 24, 2004 #5


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    nvm, you got it.

    Have you tried the usual test for linear independance? Do you know what it is?
    Last edited: Oct 24, 2004
  7. Oct 24, 2004 #6
    well, I do know what it is...but this is all I get when using it:

    (c1)sin(t) + (c2)sin(2t) + (c3)sin(3t) = 0

    I dont know where to go from there
  8. Oct 25, 2004 #7
    A set is linearly independant if no vector in the set can be written as a linear combination of of the other vectors
  9. Oct 25, 2004 #8


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    Good. That has to be true for all values of t, so try substituting some in and see what you can say about the coefficients.

    Ideally you want to put in a value of t that will make 2 of the terms vanish so you can elimante one coefficient.

    E.g. to shaow t and cos(t) are independant, look at

    (c1)t+(c2)cos(t)=0 (*)

    Put in t=0 and get:


    so c2=0 and equation (*) becomes


    Put in t=1, and we see c1=0, therefore t and cos(t) are independant.
  10. Oct 25, 2004 #9
    well, I cannot think of any values that would cancel out just two of the constants...which leads me to believe it is linearly independent....but how do I know that there are no values....how do I know I am not missing some value?? Is there a better way of doing it?
  11. Oct 25, 2004 #10


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    It's not necessary to have 2 of them cancel, that would make life easier, but it's not necessary and not always possible.

    Putting in any value of t (where they don't all vanish), will allow you to simplify your equation by getting a relation among c1, c2, and c3. Can you find any values of t to make one of them vanish and not the other 2?
  12. Oct 25, 2004 #11
    how about pi/2

    then we have (c1) - (c3) = 0
    so (c1) = (c3)
  13. Oct 25, 2004 #12


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    Good. You've simplified down to 2 variables now. Try substituting in some more values of t to see what you can get.

    Another thing you can do is differentiate your equation. You'll get a linear combination of cos's. Try putting in convenient values for t and see what you can say about c1 and c2. You could differentiate this again if need be.
  14. Oct 25, 2004 #13
    differentiate the equation, and you get all of it in terms of cos....then plug in pi/2 again and only the c2 term is left.

    so (c2)*2*cost(2(pi/2)) = 0

    So (c2) = 0

    And then back to the original eq..

    (c1)[sin(t) + sint(3t)] = 0

    WE can ignore the (c2) term since it is always 0, but now if t = pi/2, then the equation is true for any value of (c1) and thus any value of (c3)
    Last edited: Oct 25, 2004
  15. Oct 25, 2004 #14


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    Ok, c2=0 like you showed, that's good, and you had c1=c3 from earlier post, so you're down to (c1)[sin(t) + sint(3t)] = 0, which must be true for all t.

    What value of c1 makes (c1)[sin(t) + sint(3t)] = 0 for all values of t? What if t=pi/6, what can you say about c1?
  16. Oct 25, 2004 #15
    then c1 is 0 and c3 must also be 0....ok

    It doesn't matter that c1 could be anything when t is pi/2 becuase the equation must hold for all values of t, and that is only possible when c1 is zero.

    Therefore, the set is linearly independent.

    Thanks for all your help!
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