(adsbygoogle = window.adsbygoogle || []).push({}); (Urgend)Basis and Vector Spaces (Need review of my proof)

Hi I'm trying to proof the following statement:

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]

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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# Basis and Vector Spaces

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