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Linearly independent vectors

  1. Feb 9, 2009 #1
    1. The problem statement, all variables and given/known data
    When are the vectors (1,x) and (1,y) linearly independent in R^2? When are the vectors (1,x,x^2), (1,y,y^2), and (1,z,z^2) linearly independent? Generalize to R^n.


    3. The attempt at a solution
    At first, I think I misconstrued the question and I ended up finding basis vectors. ie, this is what I did for R^2:

    In order for the vectors (1,x) and (1,y) to be linearly independent, there must exist some c1,c2 such that
    c1(1,x)+ c2(1,y)= (0,0)
    then we have c1=-c2
    c1x=-c2y

    Using the first equation you have C=(c1,c2)=c2(-1,1) and so the basis vector is (-1,1).

    I don't really know why i did this honestly. I just saw linearly independent, and from there I got to my last step. Then upon actually reading the question again, I realized that for them to be linearly independent their dot product should be 0. Also I guess my initial way of solving the problem was wrong since I should have gotten that c1=c2=0, right?

    So anyway if the dot product=0 then
    <(1,x),(1,y)> = 1+xy and we want this equal to 0
    --> xy=-1.

    Any help?
     
  2. jcsd
  3. Feb 9, 2009 #2

    Mark44

    Staff: Mentor

    If c1 = -c2, then -c2x = -c2y, which implies that x = y
    In that case, you have found a nontrivial solution to the equation c1(1, x) + c2(1, y) = (0,0).
    This means that these vectors are linearly dependent. In fact, they are the same vector.

    If x != y, then (1, x) and (1, y) point in different directions, and so must be linearly independent.

    Can you extend these ideas to the rest of your problem? Be aware that it's very simple to tell whether two vectors form a dependent set--each one is some constant multiple of the other. It's not as easy to tell when you have more than two vectors, but you can always go back to the definition of linear dependence; namely that c1*v1 + c2*v2 + ... + cn*vn = 0 has exactly one solution (linearly independent set of vectors) or has an infinite number of solutions (linearly dependent set).
     
  4. Feb 10, 2009 #3
    Does it have to do with being orthogonal?
     
  5. Feb 10, 2009 #4

    Mark44

    Staff: Mentor

    Not in this problem. Vectors that are othogonal have dot products that are zero, and I don't see any of that going on here.
     
  6. Feb 10, 2009 #5
    Well one vector cannot be a multiple of the other so we cannot have
    c1(1,x)=c2(1,y)
    ?
     
  7. Feb 10, 2009 #6

    Mark44

    Staff: Mentor

    Why can't a vector be a multiple of another? If x = y, then they are the same vector, hence either one is the 1 multiple of the other, hence the two are linearly dependent.

    If x != y, they are different vectors, hence form a linearly independent set.
     
  8. Feb 10, 2009 #7
    So x cannot equal y. (Is that what != means?)
     
  9. Feb 10, 2009 #8

    Mark44

    Staff: Mentor

    Yes. It's notation that comes from the "C" programming language.
     
  10. Feb 10, 2009 #9
    But is that the only answer? That x cannot = y?
     
  11. Feb 10, 2009 #10

    Mark44

    Staff: Mentor

    It's the only answer to this question:
    Think about it graphically. I.e., graph these two vectors for two cases: when x = y and when x != y.
     
  12. Feb 10, 2009 #11
    So then for R^3 it would be when x!=y!=z and x^2 != y^2 != z^2?
     
  13. Feb 10, 2009 #12

    Mark44

    Staff: Mentor

    Maybe, but maybe not. Do the same thing you did for the vectors (1, x) and (1, y); that is, solve the equation c1*(1, x, x^2) + c2*(1, y, y^2) + c3*(1, z, z^2) = 0 for the constants c1, c2, and c3. Clearly c1 = c2 = c3 = 0 is a solution, which is always the case whether the set of vectors is linearly dependent or linearly independent. This set of three vectors is linearly dependent if there is a nontrivial solution (one for which at least one of the ci's is not zero).

    It's very easy to tell if two vectors are linearly dependent/indepent. If dependent, one of them will be a multiple of the other. When you have more than two vectors, you can't tell as easily.
     
  14. Feb 10, 2009 #13
    Okay, when I solved
    c1(1,x,x^2) + c2(1,y,y^2) + c3(1,z,z^2)=0

    I got C=(c1,c2,c3)=c2(-1,1,0) + c3(-1,0,1).

    But aren't these the basis vectors?
     
    Last edited: Feb 10, 2009
  15. Feb 10, 2009 #14

    Mark44

    Staff: Mentor

    You can't solve that--it's not an equation! Look at the equation that I wrote. From it you should get three equations in c1, c2, and c3.
    This doesn't make any sense to me. You're trying to determine the conditions on x, y, and z for the three vectors to be linearly dependent. The goal is not to find a basis for <whatever>.
     
  16. Feb 10, 2009 #15
    For R^3 I got
    c1= [-c3(1,z,z^2) - c2(1,y,y^2)] / (1,x,x^2)
     
  17. Feb 10, 2009 #16

    Mark44

    Staff: Mentor

    And how do you propose to divide by (1, x, x^2)? Look at what I said in post 14.
     
  18. Feb 10, 2009 #17
    Okay, so
    (c1, c1x, c1x^2)+(c2,c2x,c2y^2)+(c3,c3z,c3z^2)=0
    (c1+c2+c3),(c1x+c2y+c3z),(c1x^2+c2y^2+c3z^2)=0
    c1=-c2-c3?

    Is that right at all? If not, I don't understand what you're asking.
     
  19. Feb 10, 2009 #18

    Mark44

    Staff: Mentor

    OK, I buy this (above).
    This I don't buy, and I don't even know what it means. Recall that I asked you to write three equations. Each of them will have c1, c2, and c3, so you should be able to solve this system for these numbers. Clearly, if c1=c2=c3=0, that works, but what we're interested in is other solutions where not all of the ci's are zero.

    The first equation above has to be true for all x, y, and z. Group together the constant terms, the first-degree terms, the second-degree terms. You should get three equations.
     
  20. Feb 10, 2009 #19
    Okay, doing what I did for c1 and c2, like you said to do, I got
    C=(c1,c2,c3)=c2(-1,1,0) + c3(-1,0,1)
     
  21. Feb 11, 2009 #20

    Mark44

    Staff: Mentor

    The three equations I was talking about are
    c1 + c2 + c3 = 0
    c1x + c2y + c3z = 0
    c1x^2 + c2y^2 + c3z^2 = 0

    If you put this system into an augmented matrix, you have:
    Code (Text):

    [1   1   1 | 0]
    [x   y   z  | 0]
    [x^2 y^2 z^2 | 0]
     
    You don't actually need the far-right column, since it will never change, so I will omit it in further work.
    This matrix can be row-reduced to echelon form, like so:
    Code (Text):

    [1   1      1     ]
    [0  y-x   z-x     ]
    [0   0  (z-x)(z-y)]
     
    (You should check my work.)

    The matrix just above tells us about the solutions c1, c2, and c3. Under what conditions will there be a solution for c1, c2, c3, where at least one of these numbers is not zero? Those are exactly the same conditions for the set of vectors {(1, x, x^2), (1, y, y^2), (1, z, z^2)} to be linearly dependent.
     
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