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Proving Vector space

  1. May 17, 2012 #1
    1. The problem statement, all variables and given/known data

    V1 is defined as the span of the vectors b1,b2,b3,b4

    Prove that V1 is not a vector space.


    2. Relevant equations

    A set of axioms :
    If u and v are tow vectors in the span of b1,b2,b3,b4 then:
    1. u + v belongs to V1
    2. ku also belongs to V1

    3. The attempt at a solution
    Don't know how to prove 1. wrong !
     

    Attached Files:

  2. jcsd
  3. May 18, 2012 #2

    tiny-tim

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    hi hms.tech! :smile:
    i'm confused :redface:

    i thought span is defined as the smallest vector subspace that (etc) :confused:

    what definition of span have you been given?​
     
  4. May 18, 2012 #3

    sharks

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    To prove that [itex]V_1[/itex] is a vector space, the two closure axioms must be satisfied, as stated in your relevant equations.
    [tex]V_1= \left ( b_1,b_2,b_3,b_4 \right )[/tex]
    [tex]V_1= \left ( \begin{bmatrix}1 \\ 0 \\ 0\\0 \end{bmatrix}, \begin{bmatrix}1 \\ 1 \\ 0\\0 \end{bmatrix},\begin{bmatrix}1 \\ 1 \\1\\0 \end{bmatrix},\begin{bmatrix}1 \\ 1 \\ 1\\1 \end{bmatrix} \right )[/tex]
    According to the two properties:
    [tex]1.\;b_1+b_2\not=b_3
    \\2.\;k\times b_1\not=b_2[/tex]Also, if you write the 4 vectors as a matrix and reduce to its row echelon form, you will see that it has no free variables, meaning, all the 4 vectors are linearly independent.
     
  5. May 18, 2012 #4
    I think i might have to rephrase the question. Becuase i think i interpreted it wrongly.
    so again :

    Here is the complete question :
     

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  6. May 18, 2012 #5

    sharks

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    That's a completely different question from your post #1.[tex]V_1= \left ( \begin{bmatrix}1 \\ 0 \\ 0\\0 \end{bmatrix}, \begin{bmatrix}1 \\ 1 \\ 0\\0 \end{bmatrix},\begin{bmatrix}1 \\ 1 \\ 1\\0 \end{bmatrix} \right )= \begin{bmatrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\\ 0 & 0 & 0 \end{bmatrix}[/tex]
    [tex]V_2= \left ( \begin{bmatrix}1 \\ 0 \\ 0\\0 \end{bmatrix}, \begin{bmatrix}1 \\ 1 \\ 0\\0 \end{bmatrix},\begin{bmatrix}1 \\ 1 \\ 1\\1 \end{bmatrix} \right )= \begin{bmatrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\\ 0 & 0 & 1 \end{bmatrix}[/tex]Note: vector spaces are sometimes called linear spaces.
     
    Last edited: May 18, 2012
  7. May 18, 2012 #6
    yes, i know that vector space is the same thing as a linear space, but again, how to we prove that V1 union V2 is not a vector space ?
     
  8. May 18, 2012 #7

    sharks

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    [tex]V_1+V_2\not=kV_1\;or\;V_1+V_2\not=kV_2[/tex]where k is a scalar multiple.
     
  9. May 18, 2012 #8
    what axiom is that, i can't recall any such axiom of a vector space ?
     
  10. May 18, 2012 #9

    sharks

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    [itex]V_1\cup V_2[/itex] is the sum of all corresponding elements of [itex]V_1[/itex] and [itex]V_2[/itex]. Therefore, it is [itex]V_1 + V_2[/itex].

    The result of the vector addition, let's call it, [itex]V_3[/itex], is to be tested against the closure axioms to verify if it is a vector space of either [itex]V_1[/itex] or [itex]V_2[/itex].

    Now, according to the closure axiom of scalar multiplication, if [itex]V_3[/itex] is a scalar multiple of [itex]V_1[/itex] or [itex]V_2[/itex], then the former is a vector space. You can get the answer easily from here.
     
  11. May 18, 2012 #10

    tiny-tim

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    the simplest way is to find a in V1 and b in V2 such that a+b is not in V1 and is not in V2 :wink:

    (and the simplest choices for a and b would be one 1 and three 0s)
     
  12. May 19, 2012 #11
    yeah i got it, thanks for the explanation. I took the vectors b3 and b4, added them together and proved that the resultant vector does not lie in v3.
     
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