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Math Help - vector spaces

  1. #1
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    vector spaces

    show that every set with more than n vectors is necessarily linearly dependent in a n-dimensional vector space.


    The following shows how i give it a try. am i getting things right. is there any way of tackling the same problem?


    proof:


    suppose B = {e1,e2,...,en} is a basis for V. i.e dim(V)= n.


    Let S = {s1,s2,...sN} n < N.
    then 1 of the elements in S can be expressed as a linear combination of other elements. say s1.

    s1
    = k1*s2 + k2*s2 + ...+kN*sN

    = k1(a1*e1 +... + an*en) + k2(b1*e1 +...+ bn*en) +...+ KN(x1*e1 +...+xN*en)

    = ...(some steps)

    = y1*e1 + y2*e2 + ... + yn*en
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  2. #2
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    Quote Originally Posted by mtmath View Post
    show that every set with more than n vectors is necessarily linearly dependent in a n-dimensional vector space.


    The following shows how i give it a try. am i getting things right. is there any way of tackling the same problem?


    proof:


    suppose B = {e1,e2,...,en} is a basis for V. i.e dim(V)= n.


    Let S = {s1,s2,...sN} n < N.
    then 1 of the elements in S can be expressed as a linear combination of other elements. say s1.

    s1
    = k1*s2 + k2*s2 + ...+kN*sN

    = k1(a1*e1 +... + an*en) + k2(b1*e1 +...+ bn*en) +...+ KN(x1*e1 +...+xN*en)

    = ...(some steps)

    = y1*e1 + y2*e2 + ... + yn*en
    I have proven that if n vectors span \mathbb{R}^n then they are lin ind. in the lin alg sticky proofs. Since n vectors in dim=n form a basis, they must be lin ind.

    Now looking at your problem. If we have at least n+1 vectors in \mathbb{R}^n, then one of them has to be lin. dep. Since n vectors from a basis, then any additional vector beyond the basis must be a lin. combination of the basis.
    Last edited by dwsmith; May 17th 2010 at 11:54 AM.
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    I have proven this in the lin alg sticky proofs
    i've gone through the proofs.i can figure out which one is it.
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  4. #4
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    Quote Originally Posted by mtmath View Post
    i've gone through the proofs.i can figure out which one is it.
    I thought that may be the case so I edited my post. There really isn't much to show since we have n+k vectors and n vectors in n dimension form a basis.

    A basis is the minimum spanning set. If we have a minimum spanning set, then any other vector must be dependent.
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  5. #5
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    Here is a general proof for a Vector Space that shows when k+1 will be lin. ind. and lin. dep.

    Let \mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_k be lin. ind. vectors in V. If we add a vector \mathbf{x}_{k+1}, do we still have a set of lin. ind. vectors?

    (i) Assume \mathbf{x}_{k+1}\in Span (\mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_k)

    \mathbf{x}_{k+1}=c_1\mathbf{x}_{1}+...+c_k\mathbf{  x}_{k}

    c_1\mathbf{x}_{1}+...+c_k\mathbf{x}_{k}+c_{k+1}\ma  thbf{x}_{k+1}=0

    c_{k+1}=-1

    -1\neq 0; therefore, (\mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_k, \mathbf{x}_{k+1}) are lin. dep.

    (ii) Assume \mathbf{x}_{k+1}\notin Span (\mathbf{x}_1, \mathbf{x}_2,  ..., \mathbf{x}_k)

    c_1\mathbf{x}_{1}+...+c_k\mathbf{x}_{k}+c_{k+1}\ma  thbf{x}_{k+1}=0

    c_{k+1}=0 otherwise \mathbf{x}_{k+1}=\frac{-c_1}{c_{k+1}}\mathbf{x}+....+\frac{-c_k}{c_{k+1}}\mathbf{x}_k which is a contradiction.

    c_1\mathbf{x}_{1}+...+c_k\mathbf{x}_{k}+c_{k+1}\ma  thbf{x}_{k+1}=0 since (\mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_k, \mathbf{x}_{k+1}) are lin ind., c_1=...c_{k+1}=0.
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