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Math Help - n dimensional vectors -linear dependence

  1. #1
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    n dimensional vectors -linear dependence

    i read a proof that there do no exist n+1 linearly independent vectors in a n-dimensional space but i dont quite get some bits
    (1)Let \bf{v}_{1}, {v}_{2},{v}_{3},...,{v}_{n + 1} be linearly independent vectors in n dimensional space
    (2)then \bf{v}_{1}, {v}_{2},{v}_{3},...,{v}_{n} must be linearly independent
    (3)Hence the matrix A whose columns are \bf{v}_{1}, {v}_{2},{v}_{3},...,{v}_{n} is nonsingular ie the only solution to A\bf {c} = \bf0 is \bf{c} = \bf0 , where c is a column vector THIS I GET
    (4)then the books says therefore the equation A\bf c = \bf{v}_{n + 1} has unique soln for c ie. {\bf{v}_{n + 1}} = {c}_{1}{\bf{v}_{1}} + {c}_{2}{\bf {v}_{2}} + {c}_{3}{\bf{v}_{3}} + ... + {c}_{n}{\bf {v}_{n}} CAN SOMEONE PLEASE EXPLAIN HOW THIS IS, as in how does A\bf c = \bf{v}_{n + 1} have a soln and how it's unique??
    (5)then it just follows that \bf{v}_{1}, {v}_{2},{v}_{3},...,{v}_{n + 1} would be linearly dependent and contary to assumption which i get...
    so can someone please please please explain how line (4) came about
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  2. #2
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    c = A^-1*v_n+1
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  3. #3
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    that means in a n-dimensional space
    you can assume that v_1,...v_n are linearly independent
    then v_1,...v_n-1 are linearly independent
    then you can find a soln for A{\bf c}={\bf {v}_{n}} as you said similiary {\bf c}={A}^{-1}{\bf {v}_{n}}
    which means  {\bf{v}_{n}}= {c}_{1}{\bf{v}_{1}}+{c}_{2}{\bf{v}_{2}}+...+{c}_{n-1}{\bf{v}_{n-1}}
    and that means the original assumption is wrong as v_1,...v_n will not be linearly independent ???
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  4. #4
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    that means in a n-dimensional space
    you can assume that v_1,...v_n are linearly independent
    then v_1,...v_n-1 are linearly independent
    then you can find a soln for as you said similiary
    which means
    and that means the original assumption is wrong as v_1,...v_n will not be linearly independent ???
    No. The matrix A will have only n-1 columns and c will have n elements so the product Ac is not defined.

    c = A^-1*v_n+1
    Skalkaz used the result, which you are expected to know, that A is invertible if and only if it's columns are linearly independent.
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