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U,v,w a spanning set for V?

  1. Mar 13, 2014 #1
    1. The problem statement, all variables and given/known data


    In R2 let

    u = (4, -2), v = (8, 5), w = (4, 1).

    a)Is the set {u, v, w} a spanning set for R2?
    b) Are the vectors u, v linearly independent?
    c) Are the vectors u, v, w linearly independent?


    3. The attempt at a solution

    a) u, v and w is a spanning set for the vector space R2 iff every vector in R2 can be expressed as a linear combination of the vectors in {u,v,w}. In other words, and if my defintion is correct, if a vector space V spans set A of vectors, then the set A of vector is a spanning for the vector space V.

    (x,y) = λ1(4,-2) + λ2(8,5) + λ3(4,1)
    4λ1 + 8λ2 + 4λ3 = x
    -2λ1 +5λ2 + λ3 = y

    In augmented matrix:

    4 8 4 | x
    -2 5 1 | y

    R2<- R2 +0.5R1

    4 8 4 | x
    0 9 3 | y + x/2

    There is one free variable λ3-so, 1 parameter.
    There are 2 pivots and, therefore, dimension, 2.
    u, v, and, w are spanning sets for R2.


    b) u, v are linearly independent iff span {u,v} = 0 where λ1=λ2= 0.


    4 8 | 0
    0 9 | 0

    9λ2 = 0
    λ2 = 0
    4λ1 + 8(0) = 0
    λ1 = 0

    since λ1 = λ2 = 0, u and v are linearly independent.

    c) (0,0) = λ1 ( 4,-2) + λ2 ( 8,5) + λ3 (4,1)

    4 8 4 | 0
    -2 5 1 | 0

    ....RREF

    4 8 4 | 0
    0 9 3 | 0

    λ3 = free variables
    The corollary from this is that there are infinite solutions. From this, it can be deduced that the solution is non-trivial, and therefore, u,v and w is not linearly independent. I'm making this conclusion based on the definition that for a system of linear equation to be linearly independent, it must be the case that the only solution is such that λ1=λ2=...=λn=0 but infinite solution implies that the solutions are either non-trivial or that a trivial solution is only one of the many infinite possible solutions. (PLEASE correct me if I am wrong)

    The computer system states my answer to be correct. However, I would appreciate if my conclusion in part (c) is correct or a fluke.
     
    Last edited: Mar 13, 2014
  2. jcsd
  3. Mar 13, 2014 #2

    PeroK

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    Gold Member

    I didn't look closely at what you did, but there is a much simpler way to do this. Note that:

    Two vectors are linearly dependent iff one is a multiple of the other

    Any two linearly independent vectors will span R2

    No three vectors can be linearly independent in R2.
     
  4. Mar 13, 2014 #3
    Hi, I have reedited my entire OP.
     
  5. Mar 13, 2014 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, your definition is correct.
    More fundamentally, if we multiply -2λ1+ 4λ2+ λ3= y by 2 and add it to 4λ1+ 8λ2+ 4λ3= x, we get 16λ2+ 6λ3= x+ 2y. Take λ2= 0 and we have 6λ3= x+ 2y so λ3= (x+ 2y)/6 and then -2λ1+ (x+ 2y)/6= y or λ1= (4y- x)/6. Since we can find λ1, λ2, λ3 for any x, y, yes, it spans R2.

    u, v, and w are NOT sets. {u, v, w} is a spanning set for R2.


    That is not well stated. The "span {u,v}" consists of all λ1u+ λ2v. That span is NOT "0".
    You mean "if λ1u+ λ2v= 0 then λ1= λ2= 0"
    Yes, that is correct.

    Your answer to (c) is completely correct. If λ1u+ λ2v+ λ3w= 0 then
    λ1(4,-2)+ λ2(8, 5)+ λ3(4, 1)= (0, 0)
    (4λ1+ 8λ2+ 3λ3, -2λ1+ 5λ2+ λ3)= (0, 0)
    4λ1+ 8λ2+ 3λ3= 0, -2λ1+ 5λ2+ λ3= 0

    Multiply the second equation by 2 and add to the first equation: 18λ2+ 5λ3= 0 so if you take λ1= 0, λ2= 5 and λ3= -18 you have non-zero coefficients that make the linear combination 0. The set of vectors is NOT independent.
     
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