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Basis for a Vector space

  1. Mar 7, 2006 #1
    Hi

    Given a Vector Space V which has the basis [tex]\{ v_{1}, v_{2}, v_{3} \} [/tex] then I need to prove that the following set v = [tex]\{ v_{1}, v_{1}+ v_{2}, v_{1} + v_{2} + v_{3} \}[/tex] is also a basis for V.

    I know that in order for v to be a basis for V then [tex]V = span \{v_{1}, v_{1}+ v_{2}, v_{1} + v_{2} + v_{3} \}[/tex], and the vectors of v need to be linear independent.

    but by showing that the vectors of v are linear independent is the best way of show that v is a basis for V?

    Sincerely
    Fred
     
    Last edited: Mar 7, 2006
  2. jcsd
  3. Mar 7, 2006 #2

    0rthodontist

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    Probably.

    Also do you have the theorem that if {b1, ... ,bn} is a basis for a vector space V, then any linearly independent set of n vectors in V is also a basis?
     
  4. Mar 7, 2006 #3

    benorin

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    We know that every vector [tex]v\in V[/tex] can be expressed as a linear combination of the basis vectors [tex]\{ v_{1}, v_{2}, v_{3} \} , [/tex] that is there exists scalars [tex]c_{1}, c_{2}, c_{3} [/tex] such that [tex]v=c_1v_1+c_2v_2+c_3v_3 .[/tex] Is the same true of [tex]\{v_{1}, v_{1}+ v_{2}, v_{1} + v_{2} + v_{3} \}[/tex]?
     
  5. Mar 7, 2006 #4
    if I express [tex]\{v_{1}, v_{1}+ v_{2}, v_{1} + v_{2} + v_{3} \}[/tex] as a set of vectors [tex]\{x,y,z\}[/tex] which can be expressed

    [tex]v=c_1 x +c_2 y+c_3 z.[/tex]

    Then its the same.

    Sincerely
    Fred
     
  6. Mar 7, 2006 #5

    benorin

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    I want to be sure you understand:

    Suppose I take a linear combination of the vectors [tex]\{v_{1}, v_{1}+ v_{2}, v_{1} + v_{2} + v_{3} \} , [/tex] this is then a linear combination of linear combinations of the given basis vectors, which is again, a linear combination of the given basis vectors, which every vector in V may be expressed as. Follow?
     
    Last edited: Mar 7, 2006
  7. Mar 7, 2006 #6
    I get it ;)

    Then I show that the linear combination of the linear combination er linear independent ??

    like

    c1 * v1 = 0

    c2 * (v1+v2) = 0

    c3 * (v1 + v2 + v3) = 0

    Sincerly
    Fred

     
  8. Mar 7, 2006 #7

    benorin

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    Indeed, the existence of scalars [tex]c_{1}, c_{2}, c_{3} [/tex] such that [tex]v=c_1v_1+c_2v_2+c_3v_3 [/tex] implies it, then there also exists scalars a,b,c such that [tex]v=av_1+b(v_1+v_2)+c(v_1+v_2+v_3).[/tex]

    Proof: Fix v. Let the above hypothesis be so. Then

    [tex]v=av_1+b(v_1+v_2)+c(v_1+v_2+v_3)=(a+b+c)v_1 + (b+c)v_2+cv_3 [/tex]

    but we know there exist scalars [tex]c_{1}, c_{2}, c_{3} [/tex] such that [tex]v=c_1v_1+c_2v_2+c_3v_3 ,[/tex]

    so set these equal to obtain

    [tex]v=c_1v_1+c_2v_2+c_3v_3=(a+b+c)v_1 + (b+c)v_2+cv_3 [/tex]

    equate coefficients of the v_k's to get the systems of equations [tex]c_1=a+b+c,c_2=b+c,c_3=c[/tex] so that we may take [tex]c=c_3, b=c_2-c_3,a=c_1-c_2[/tex] and hence there must also exists scalars a,b,c such that [tex]v=av_1+b(v_1+v_2)+c(v_1+v_2+v_3),[/tex] as required.
     
    Last edited: Mar 7, 2006
  9. Mar 8, 2006 #8
    Here is my own idear for a proof that the set of vectors [tex]v = \{v_{1}, v_{1} + v_{2},v_{1} + v_{2} + v_{3} \} [/tex] is a basis for the vector space V.

    Definition: Basis for Vector Space

    Let V be a Vector Space. A set of vectors in V is a basis for V if the following conditions are meet:

    1. The set vectors spans V.
    2. The set of all vectors is linear independent.

    Proof:

    (2) The vectors in v is said to be linear independent iff there doesn't exists scalars [tex]C = (c_1,c_2,c_3) \neq 0[/tex] such that

    [tex]v = c_1 (v_1) + c_2 (v_1+v_2) + c_3 (v_1 + v_2 + v_3) = 0[/tex]

    By expression the above in matrix-equation form:

    [itex]\begin{array}{cc}\[ \left[ \begin{array}{ccc}
    v_{11}& (v_{11} + v_{12})& (v_{11} + v_{12} + v_{13}) \\
    v_{21}& (v_{21} + v_{22})& (v_{21} + v_{22} + v_{23})\\
    v_{31}& (v_{31} + v_{32})& (v_{31} + v_{32} + v_{33})\\
    \end{array} \right]\] \cdot \left[ \begin{array}{c} c_1 \\ c_2 \\ c_3 \end{array} \right] = 0 \] \end{array}[/itex]

    If v is fixed, it can be concluded that matrix columbs of v are linear independent cause the only solution for the equation vC = 0 is the trivial solution(zero-vector):

    [itex]\begin{array}{cc}\[ \left[ \begin{array}{ccc}
    c_{1} \\
    c_{2} \\
    c_{3} \\
    \end{array} \right]\] = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right] \] \end{array}[/itex]

    Which implies the dependence relation between v and C doesn't exist since C = 0.

    (1) Since the only solution for matrix equation is the trivial solution, then according to the definition for the Span of a vector subspace, then the set of v spans V.

    This proves that the set of is it fact a basis for the Vector Space V.

    Is my proof valid???

    Sincerely
    Fred
     
    Last edited: Mar 8, 2006
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