Basis and Vector Spaces: Proof of Linear Independence and Spanning Property

[Mathman23]

(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

$$v = \{v_{1}, v_{1} + v_{2},v_{1} + v_{2} + v_{3} \}$$

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 $$C = (c_1,c_2,c_3) \neq 0$$ such that

$$v = c_1 (v_1) + c_2 (v_1+v_2) + c_3 (v_1 + v_2 + v_3) = 0$$

By expression the above in matrix-equation form:

$\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}$

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):

$\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}$

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

 

[JFo]

Your "theorem" is not well stated. You're trying to prove that the set

$$S = \{v_{1}, v_{1} + v_{2},v_{1} + v_{2} + v_{3} \}$$

is what? A basis for a vector space V?

If this is what your trying to prove you need to add constrictions on the vectors v1, v2, etc... for example v1 can not equal zero because any set containing the zero vector is linearly dependent.

 

[Mathman23]

Your "theorem" is not well stated. You're trying to prove that the set

$$S = \{v_{1}, v_{1} + v_{2},v_{1} + v_{2} + v_{3} \}$$

is what? A basis for a vector space V?

If this is what your trying to prove you need to add constrictions on the vectors v1, v2, etc... for example v1 can not equal zero because any set containing the zero vector is linearly dependent.

Hello


I'm indeed trying to prove that

$$S = \{v_{1}, v_{1} + v_{2},v_{1} + v_{2} + v_{3} \}$$

is a basis for V.

If I add the constiction that$$S \neq 0$$ is my proof valid?

Sincerely
Fre

 

[shmoe]

How do $$v_1,\ v_2,\ v_3$$ relate to V?

 

[Mathman23]

How do $$v_1,\ v_2,\ v_3$$ relate to V?

Its know in my assigment that the set T = $$\{v_1,\ v_2,\ v_3 \}$$ is a basis for the Vector Space V.

I'm tasked with showing that the set S mentioned in my previous post is also a basis for V.

I add that detail shmoe is my proof then valid?

Sincerly
Fred

 

[shmoe]

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):

I don't see how this follows from what you've written. Could you explain more why you think the only solution is the trivial one?

 

[Mathman23]

Yes,

If v is nonzero and fixed then the only solution for the matrix equation vC = 0 is the trivial solution.

/Fred

I don't see how this follows from what you've written. Could you explain more why you think the only solution is the trivial one?