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Math Help - Why row vectors equal to linear combination of basis vectors of rowspc(in this case)?

  1. #1
    Senior Member x3bnm's Avatar
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    Why row vectors equal to linear combination of basis vectors of rowspc(in this case)?

    Let v_1, v_2, ...,\,\,\, and \,\,\,v_m be the row vectors of a matrix:

      A = \begin{bmatrix} a_{11} & a_{12} & ... &a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ ... & ... & ... & ...\\a_{m1} & a_{m2} & ... & a_{mn} \end{bmatrix}

    Suppose the row space of A has dimension r and basis S = \{b_1, b_2, ..., b_r\} where b_i = (b_{i1}, b_{i2}, ..., b_{in}).


    Using this basis, you can write the row vectors of A as(the book wrote this)

     v_1 = c_{11}b_1 + c_{12}b_2  + ... + c_{1r}b_r

     v_2 = c_{21}b_1 + c_{22}b_2  + ... + c_{2r}b_r

     ...\,\,\,    ... \,\,\,    ....\,\,\,

     v_m = c_{m1}b_1 + c_{m2}b_2  + ... + c_{mr}b_r

    My question: why v_1 = c_{11}b_1 + c_{12}b_2  + ... + c_{1r}b_r and v_2 = c_{21}b_1 + c_{22}b_2  + ... + c_{2r}b_r and v_m = c_{m1}b_1 + c_{m2}b_2  + ... + c_{mr}b_r ?

    Why the book states that the linear combination of basis vectors equal to v_1, v_2, ...\,\,\, and\,\,\, v_m?

    What method they used to find this? I can't see it. Sorry.
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  2. #2
    Senior Member x3bnm's Avatar
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    Re: Why row vectors equal to linear combination of basis vectors of rowspc(in this ca

    We know that a set of vectors S = \{g_1, g_2, ...,g_n\} in a vector space V is called a basis for V if the following conditions are true.
    1) S spans V
    2) S is linearly independent


    I can't find the answer of my above mentioned question. All I can think of is this.


    So in the above post \{v_1, v_2, ..., v_r\} are the row vectors of matrix A and so spans the row space of A which is row\,\,space\,\,of\,\,A \,\,\, = d_1 v_1 + d_2 v_2 + ... + d_m v_m:
    Shouldn't d_1 v_1 + d_2 v_2 + ... + d_m v_m = e_1 b_1 + e_2 b_2 +... + e_r b_r because \{b_1, b_2, ..., b_r\} is the basis of row space of A and for the fact that row\,\,space\,\,of\,\,A \,\,\, = e_1 b_1 + e_2 b_2 +... + e_r b_r?

    So how and why
     v_1 = c_{11}b_1 + c_{12}b_2  + ... + c_{1r}b_r

     v_2 = c_{21}b_1 + c_{22}b_2  + ... + c_{2r}b_r

     ...\,\,\,    ... \,\,\,    ....

     v_m = c_{m1}b_1 + c_{m2}b_2  + ... + c_{mr}b_r
    ?

    Can anyone kindly help me understand please?
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    Re: Why row vectors equal to linear combination of basis vectors of rowspc(in this ca

    Quote Originally Posted by x3bnm View Post
    Let [tex]
    Why the book states that the linear combination of basis vectors equal to v_1, v_2, ...\,\,\, and\,\,\, v_m?
    b_{1}, b_{2},...,b_{r} are a basis for the rowspace of A, so every combination of the rows of A can be written in terms of this basis.
    Each of v_1, v_2, ...\,\,\, and\,\,\, v_m is in the row space of A, and so can be written in terms of b_{1}, b_{2},...,b_{r}.

    Did you understand column spaces? Row space works the same way, except obviously related to rows rather than columns. If you did understand column space,
    note that the row space of A is equal to the column space of A^{T}, and a basis for the row space of A is a basis for the column space of A^{T}
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  4. #4
    Senior Member x3bnm's Avatar
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    Re: Why row vectors equal to linear combination of basis vectors of rowspc(in this ca

    Quote Originally Posted by Joanna View Post
    b_{1}, b_{2},...,b_{r} are a basis for the rowspace of A, so every combination of the rows of A can be written in terms of this basis.
    Each of v_1, v_2, ...\,\,\, and\,\,\, v_m is in the row space of A, and so can be written in terms of b_{1}, b_{2},...,b_{r}.

    Did you understand column spaces? Row space works the same way, except obviously related to rows rather than columns. If you did understand column space,
    note that the row space of A is equal to the column space of A^{T}, and a basis for the row space of A is a basis for the column space of A^{T}
    Thanks a lot for clarifying this.
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