Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Two linearly dependent vectors

  1. Mar 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that two vectors are linearly dependent if and only if one is a scalar multiple of the other.

    2. Relevant equations



    3. The attempt at a solution This seems at glance to be a fairly easy proof:

    Part I Assume that vectors u and v are linearly dependent.

    Then c1u + c2v = 0 where c1 and c2 are not both 0

    then u = -c2/c1 * v
    and v = -c1/c2 * u But this doesn't make sense to me because what if one of c1 or c2 does equal zero?

    Part II Assume that u =av and v=bu , where a and b are constants

    then u - av = 0 where the coefficient of u is 1 and v - bu = 0 where the coefficient of v is 1.

    Therefore u and v are linearly dependent.

    I'm struggling a bit with linear algebra proofs, so any critique or suggestions that anyone could offer would be greatly appreciated.
     
  2. jcsd
  3. Mar 6, 2009 #2
    Either c1 or c2 is nonzero. Without loss of generality, let c1 be nonzero. Then since c1u + c2v = 0, dividing by c1, we get u + (c2/c1)v = 0. What does this tell you?
     
  4. Mar 6, 2009 #3

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Look at two different cases:

    Case1: c1=0 and c2≠0

    Case 2: c2=0 and c1≠0

    ...your method in partII is not valid, since it neglects the c1=0 case.
     
  5. Mar 6, 2009 #4
    Ok so v = c1u

    v -c1u = 0 and suppose that c1 = 0.
    v must be the zero vector

    now suppose that v-c1=0 and c1 is not equal to zero.
    then v-c1u=0 where the coefficients of v and u are both nonzero

    Am I going on the right track?

    and then I would do a similar thing for u = c2v
     
  6. Mar 6, 2009 #5

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    No, start with c1u+c2v=0...as you did before

    If c1=0, what can you say about c2? (Remember, the vectors are by assumption not zero vectors)

    If c2=0, what can you say about c1?

    The case where neither of them are zero was already covered in your 1st post.
     
  7. Mar 7, 2009 #6
    If c1=0, what can you say about c2? : c2 is nonzero

    If c2=0, what can you say about c1? c1 is nonzero

    so assume c1u + c2v = 0

    if c1 is nonzero u = -c2/c1 * v
    if c2 is nonzero v= -c1/c2 * u


    part II

    assume that u and v are scalar multiples of each other:

    u = av and v = bu where both a and b are nonzero scalars and u and v are nonzero vectors

    u - av = 0 and v - bu = 0 therefore u and v are linearly dependent


    the problem also gives a hint that I should separately consider the case where one of the vectors is the zero vector

    so assume u = av and v=bu

    let u be the zero vector 1u-av = 0 therefore u and v are linearly dependent

    let v be the zero vector 1v - bu= 0 therefore u and v are linearly dependent


    I don't know if I'm getting this right at all.
     
  8. Mar 7, 2009 #7

    HallsofIvy

    User Avatar
    Science Advisor

    Yes, that is correct.


    You are making the same mistake you did before. If u and v are scalar multiples of one another, then u= av and v= bu but it does NOT FOLLOW THAT "both a and b are nonzero scalars and u and v are nonzero vectors". If neither u nor v are zero, then you can say u= av for a non zero and so u- av= 0 with coefficients 1 and -a. If u= 0 then au+ 0v= 0 for any non-zero a and if v= 0, then 0u+ bv= 0 for any non-zero b.

     
  9. Mar 7, 2009 #8
    part i. If c1 is zero and c2 is not zero then for c1u + c2v = 0 to hold, either u = o(vector) or v = 0 and vice versa since it has been said that u and v are linearly dependent
     
  10. Mar 7, 2009 #9
    Is there a linear space V in which the union of any subspaces of V is a subspace except the trivial subspaces V and {0}? pls help
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook