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Basis for a Vector space

  • Thread starter Mathman23
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  • #1
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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
 
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  • #2
0rthodontist
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Mathman23 said:
but by showing that the vectors of v are linear independent is the best way of show that v is a basis for V?
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?
 
  • #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]?
 
  • #4
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benorin said:
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]?
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
 
  • #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?
 
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  • #6
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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

benorin said:
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?
 
  • #7
benorin
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benorin said:
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]?
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.
 
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  • #8
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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
 
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