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Math Help - Tricky first year proof

  1. #1
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    Tricky first year proof

    let V be a vector space and suppose B is a basis for V. show that any vector in V be written uniquely as a linear combination of elements from B.

    Starters/pointers/hints would be great

    Nic
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  2. #2
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    Basis B is the largest set of linearly independent vectors in V. Number of elements of that set represents the dimension of V. Take any vector v in V. Then v together with the elements of B forms a set of linearly dependent vectors. Form a linear combination and it can be zero with not all coefficients equal to zero.

    See where that gets you.
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  3. #3
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    Suppose otherwise, and you’ll end up enlarging your basis.
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    Don't forget to prove uniqueness. Suppose there are two ways to express a single vector v and look for contradiction.
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  5. #5
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    Quote Originally Posted by MathoMan View Post
    Basis B is the largest set of linearly independent vectors in V. Number of elements of that set represents the dimension of V. Take any vector v in V. Then v together with the elements of B forms a set of linearly dependent vectors. Form a linear combination and it can be zero with not all coefficients equal to zero.

    See where that gets you.
    Ok, so I understand the description, but in terms of actually defining the vectors. For instance, you say 'take any vector v in V, would you define it is [A1+A2....An]? and then how would you define the elements of B?

    I understand what I have to do, just can't get it down on paper so to say.
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  6. #6
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    First, what is the definition of "basis" you are using? In some texts, it is exactly what you are trying to prove. Probably your definition is what Mathoman said- a largest independent set of vectors. Although, be careful about saying the basis is the largest set- there exist an infinite number of bases, all having the same size.

    Since this is "first year", can we assume that the vector space is finite- dimensional? In that case you can write your basis, B, as, say, \{v_1, v_2, \cdot\cdot\cdot, v_n\}.

    Let v be any vector in the vector space. Since B is a largest set of independent vectors, \{v_1, v_2, \cdot\cdot\cdot, v_n\} is NOT independent. Use the definiton of "dependent" to show that v can be written as a linear combination of vectors in B.

    Now uniqueness. Suppose there were some vector, v, which could be written as two different linear combinations of vectors in B. That is, v= a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n and v= b_1v_1+ b_2v_2+ \cdot\cdot\cdot+ b_nv_n where, for at least some index i, a_i\ne b_i. Subtract the two equations to get a contradiction to the fact that the set is independent.
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  7. #7
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    Quote Originally Posted by Tikoloshe View Post
    Suppose otherwise, and you’ll end up enlarging your basis.
    Tikoloshe stated the most elegant way of doing this.

    B is basis for V. Then B is largest possible set of vectors all being linearly independent of each other. If arbitrarily chosen vector v in V cannot be expressed as a linear combination of elements in B, than it follows it is linearly independent of all elements in B. Hence v together with elements in B forms another set B' of linearly independent vectors that is larger then B (obviously B is contained in B'). That contradicts with the fact that B IS basis for V and as such B is largest posible set of linearly independent. Hence one must reject the assumption that v cannot be expressed as a linear combination of elements in B, thus leaving the oposite as true: any v in V can be expressed as a linear combination of elements from B.
    Last edited by MathoMan; September 22nd 2010 at 03:40 AM. Reason: removed THE in front of attribute 'largest' :)
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  8. #8
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    Hey. By the definition of basis, any vector in the space can be expressed as a linear combination of the elements of V. Suppose that this representation was not unique:

    ie c_1v_1+...+c_nv_n=d_1v_1+...+d_nv_n

    Then we have  0=(c_1-d_1)v_1+...+(c_n-d_n)v_n. Then, kick in linear independence.
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