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    I understand that a basis is a linearly independent set of vectors which spans a subspace of  \mathbb{R}^n ... but I really have no idea where to even begin with this question. Help please?


    If  V = {v_1, v_2,....., v_n} is a basis for  \mathbb{R}^n then prove that every vector  v in  \mathbb{R}^n can be expressed in the form:  v = a_1 v_1 + a_2 v_2 + ... + a_n v_n in exactly one way where  a_i \in \mathbb{R} for all  i .
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    since V spans \mathbb{R}^{n} every vector in \mathbb{R}^{n} is expressible as a linear combination of the vectors in V

    suppose you take an arbitrary vector in V which can be written as:

    a_{1}v_{1}+a_{2} v_{2}+...a_{n} v_{n}

    but also as:

    k_{1}v_{1}+k_{2}v_{2}+...k_{n}v_  {n}

    subtract one from the other and you get equality...
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    (Original post by Hasufel)
    since V spans \mathbb{R}^{n} every vector in \mathbb{R}^{n} is expressible as a linear combination of the vectors in V

    suppose you take an arbitrary vector in V which can be written as:

    a_{1}v_{1}+a_{2} v_{2}+...a_{n} v_{n}

    but also as:

    k_{1}v_{1}+k_{2}v_{2}+...k_{n}v_  {n}

    subtract one from the other and you get equality...

    Oh gosh... this makes so much sense!!
    I think I must have just been having a mental block. Thank you so much
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    (Original post by Hasufel)
    since V spans \mathbb{R}^{n} every vector in \mathbb{R}^{n} is expressible as a linear combination of the vectors in V

    suppose you take an arbitrary vector in V which can be written as:

    a_{1}v_{1}+a_{2} v_{2}+...a_{n} v_{n}

    but also as:

    k_{1}v_{1}+k_{2}v_{2}+...k_{n}v_  {n}

    subtract one from the other and you get equality...

    After going through this I've actually become a bit stuck.

    I have:

     v = a_1 v_1 + a_2 v_2 + ... + a_n v_n

    and

     v = k_1 v_1 + k_2 v_2 + ... + k_n v_n

    I subtract them:

     0 = (a_1 - k_1 ) v_1 + (a_2 - k_2 ) v_2 + ... + (a_n - k_n ) v_n ... and I get confused about the coefficients here. :confused:

    Are a_i and k_i the same number? ... that would make sense
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    (Original post by SummerPi)
    After going through this I've actually become a bit stuck.

    I have:

     v = a_1 v_1 + a_2 v_2 + ... + a_n v_n

    and

     v = k_1 v_1 + k_2 v_2 + ... + k_n v_n

    I subtract them:

     0 = (a_1 - k_1 ) v_1 + (a_2 - k_2 ) v_2 + ... + (a_n - k_n ) v_n ... and I get confused about the coefficients here. :confused:

    Are a_i and k_i the same number? ... that would make sense
    since you have a zero vector, all the coefficients must be zero (linear independence) which means:

    a_{1}=k_{1},  a_{2}=k_{2}... etc - i.e. the coeffs are equal, so you`ve demonstrated that the vector is unique
 
 
 
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