Span of linear dependent set












0












$begingroup$


Let ${v_1 ldots v_n }$ be a linear dependent set of vectors from a vector space $V$. We can show then that:



$$ v_j = span{v_1 ldots v_{j-1}, v_{j+1}, ldots v_n } $$



Because since it is linearly dependent, we can always choose a non-null (all together) set of elements $x_i$ of a field $mathbf{F}$ such that:



$$ v_j = - frac{sum_{ineq j}^{n} x_i v_i}{x_j}$$



Yet I have to prove a corollary which states that under the same conditions:



$$ span(v_1 ldots v_n) = span(v_1 ldots v_{j-1}, v_{j+1}, ,ldots v_n) $$



and I am stuck because my results show that they are not the same. Any tips or advices to prove it? Thanks!






linear-algebra vector-spaces finite-fields







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  • $begingroup$
    How is it that a vector ($v_j$) is equal to a set of vectors (the span on the right-hand side)?
    $endgroup$
    – amd
    Jan 13 at 20:27
















0












$begingroup$


Let ${v_1 ldots v_n }$ be a linear dependent set of vectors from a vector space $V$. We can show then that:



$$ v_j = span{v_1 ldots v_{j-1}, v_{j+1}, ldots v_n } $$



Because since it is linearly dependent, we can always choose a non-null (all together) set of elements $x_i$ of a field $mathbf{F}$ such that:



$$ v_j = - frac{sum_{ineq j}^{n} x_i v_i}{x_j}$$



Yet I have to prove a corollary which states that under the same conditions:



$$ span(v_1 ldots v_n) = span(v_1 ldots v_{j-1}, v_{j+1}, ,ldots v_n) $$



and I am stuck because my results show that they are not the same. Any tips or advices to prove it? Thanks!






linear-algebra vector-spaces finite-fields







share|cite|improve this question









$endgroup$












  • $begingroup$
    How is it that a vector ($v_j$) is equal to a set of vectors (the span on the right-hand side)?
    $endgroup$
    – amd
    Jan 13 at 20:27














0












0








0





$begingroup$


Let ${v_1 ldots v_n }$ be a linear dependent set of vectors from a vector space $V$. We can show then that:



$$ v_j = span{v_1 ldots v_{j-1}, v_{j+1}, ldots v_n } $$



Because since it is linearly dependent, we can always choose a non-null (all together) set of elements $x_i$ of a field $mathbf{F}$ such that:



$$ v_j = - frac{sum_{ineq j}^{n} x_i v_i}{x_j}$$



Yet I have to prove a corollary which states that under the same conditions:



$$ span(v_1 ldots v_n) = span(v_1 ldots v_{j-1}, v_{j+1}, ,ldots v_n) $$



and I am stuck because my results show that they are not the same. Any tips or advices to prove it? Thanks!






linear-algebra vector-spaces finite-fields







share|cite|improve this question









$endgroup$




Let ${v_1 ldots v_n }$ be a linear dependent set of vectors from a vector space $V$. We can show then that:



$$ v_j = span{v_1 ldots v_{j-1}, v_{j+1}, ldots v_n } $$



Because since it is linearly dependent, we can always choose a non-null (all together) set of elements $x_i$ of a field $mathbf{F}$ such that:



$$ v_j = - frac{sum_{ineq j}^{n} x_i v_i}{x_j}$$



Yet I have to prove a corollary which states that under the same conditions:



$$ span(v_1 ldots v_n) = span(v_1 ldots v_{j-1}, v_{j+1}, ,ldots v_n) $$



and I am stuck because my results show that they are not the same. Any tips or advices to prove it? Thanks!






linear-algebra vector-spaces finite-fields




linear-algebra vector-spaces finite-fields






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share|cite|improve this question











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share|cite|improve this question










asked Jan 13 at 19:39









M.GonzalezM.Gonzalez

1146




1146












  • $begingroup$
    How is it that a vector ($v_j$) is equal to a set of vectors (the span on the right-hand side)?
    $endgroup$
    – amd
    Jan 13 at 20:27


















  • $begingroup$
    How is it that a vector ($v_j$) is equal to a set of vectors (the span on the right-hand side)?
    $endgroup$
    – amd
    Jan 13 at 20:27
















$begingroup$
How is it that a vector ($v_j$) is equal to a set of vectors (the span on the right-hand side)?
$endgroup$
– amd
Jan 13 at 20:27




$begingroup$
How is it that a vector ($v_j$) is equal to a set of vectors (the span on the right-hand side)?
$endgroup$
– amd
Jan 13 at 20:27










2 Answers
2






active

oldest

votes


















1












$begingroup$

Without loss of generality, we may assume $j=n$. Let $X=text{span}{x_1,x_2,ldots,x_n}$ and $Y=text{span}{x_1,x_2,ldots,x_{n-1}}$. It is easy to see that
$$
Xgeq Y
$$from the definition of the spanned subspace. Now, to prove that
$$
Xleq Y,
$$ it is sufficient to show that
$$
{x_1,x_2,ldots,x_{n}}subset Y.
$$ But this is obvious from the fact that $x_nin Y$ and ${x_1,x_2,ldots,x_{n-1}}subset Y$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you! It has surely helped me :)
    $endgroup$
    – M.Gonzalez
    Jan 14 at 16:45



















0












$begingroup$

Yes, they are the same:




  • since ${v_1,ldots v_{j-1},v_{j+1},ldots,v_n}subset{v_1,ldots,v_n}$,$$operatorname{span}bigl({v_1,ldots v_{j-1},v_{j+1},ldots,v_n}bigr)subsetoperatorname{span}bigl({v_1,ldots,v_n}bigr);$$

  • each element of ${v_1,ldots,v_n}$ is a linear combination of elements of ${v_1,ldots v_{j-1},v_{j+1},ldots,v_n}$,$$operatorname{span}bigl({v_1,ldots v_{j-1},v_{j+1},ldots,v_n}bigr)supsetoperatorname{span}bigl({v_1,ldots,v_n}bigr).$$






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Without loss of generality, we may assume $j=n$. Let $X=text{span}{x_1,x_2,ldots,x_n}$ and $Y=text{span}{x_1,x_2,ldots,x_{n-1}}$. It is easy to see that
    $$
    Xgeq Y
    $$    from the definition of the spanned subspace. Now, to prove that
    $$
    Xleq Y,
    $$     it is sufficient to show that
    $$
    {x_1,x_2,ldots,x_{n}}subset Y.
    $$     But this is obvious from the fact that $x_nin Y$ and ${x_1,x_2,ldots,x_{n-1}}subset Y$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you! It has surely helped me :)
      $endgroup$
      – M.Gonzalez
      Jan 14 at 16:45
















    1












    $begingroup$

    Without loss of generality, we may assume $j=n$. Let $X=text{span}{x_1,x_2,ldots,x_n}$ and $Y=text{span}{x_1,x_2,ldots,x_{n-1}}$. It is easy to see that
    $$
    Xgeq Y
    $$    from the definition of the spanned subspace. Now, to prove that
    $$
    Xleq Y,
    $$     it is sufficient to show that
    $$
    {x_1,x_2,ldots,x_{n}}subset Y.
    $$     But this is obvious from the fact that $x_nin Y$ and ${x_1,x_2,ldots,x_{n-1}}subset Y$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you! It has surely helped me :)
      $endgroup$
      – M.Gonzalez
      Jan 14 at 16:45














    1












    1








    1





    $begingroup$

    Without loss of generality, we may assume $j=n$. Let $X=text{span}{x_1,x_2,ldots,x_n}$ and $Y=text{span}{x_1,x_2,ldots,x_{n-1}}$. It is easy to see that
    $$
    Xgeq Y
    $$    from the definition of the spanned subspace. Now, to prove that
    $$
    Xleq Y,
    $$     it is sufficient to show that
    $$
    {x_1,x_2,ldots,x_{n}}subset Y.
    $$     But this is obvious from the fact that $x_nin Y$ and ${x_1,x_2,ldots,x_{n-1}}subset Y$.






    share|cite|improve this answer











    $endgroup$



    Without loss of generality, we may assume $j=n$. Let $X=text{span}{x_1,x_2,ldots,x_n}$ and $Y=text{span}{x_1,x_2,ldots,x_{n-1}}$. It is easy to see that
    $$
    Xgeq Y
    $$    from the definition of the spanned subspace. Now, to prove that
    $$
    Xleq Y,
    $$     it is sufficient to show that
    $$
    {x_1,x_2,ldots,x_{n}}subset Y.
    $$     But this is obvious from the fact that $x_nin Y$ and ${x_1,x_2,ldots,x_{n-1}}subset Y$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 13 at 22:37

























    answered Jan 13 at 19:46









    SongSong

    11.2k628




    11.2k628












    • $begingroup$
      Thank you! It has surely helped me :)
      $endgroup$
      – M.Gonzalez
      Jan 14 at 16:45


















    • $begingroup$
      Thank you! It has surely helped me :)
      $endgroup$
      – M.Gonzalez
      Jan 14 at 16:45
















    $begingroup$
    Thank you! It has surely helped me :)
    $endgroup$
    – M.Gonzalez
    Jan 14 at 16:45




    $begingroup$
    Thank you! It has surely helped me :)
    $endgroup$
    – M.Gonzalez
    Jan 14 at 16:45











    0












    $begingroup$

    Yes, they are the same:




    • since ${v_1,ldots v_{j-1},v_{j+1},ldots,v_n}subset{v_1,ldots,v_n}$,$$operatorname{span}bigl({v_1,ldots v_{j-1},v_{j+1},ldots,v_n}bigr)subsetoperatorname{span}bigl({v_1,ldots,v_n}bigr);$$

    • each element of ${v_1,ldots,v_n}$ is a linear combination of elements of ${v_1,ldots v_{j-1},v_{j+1},ldots,v_n}$,$$operatorname{span}bigl({v_1,ldots v_{j-1},v_{j+1},ldots,v_n}bigr)supsetoperatorname{span}bigl({v_1,ldots,v_n}bigr).$$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Yes, they are the same:




      • since ${v_1,ldots v_{j-1},v_{j+1},ldots,v_n}subset{v_1,ldots,v_n}$,$$operatorname{span}bigl({v_1,ldots v_{j-1},v_{j+1},ldots,v_n}bigr)subsetoperatorname{span}bigl({v_1,ldots,v_n}bigr);$$

      • each element of ${v_1,ldots,v_n}$ is a linear combination of elements of ${v_1,ldots v_{j-1},v_{j+1},ldots,v_n}$,$$operatorname{span}bigl({v_1,ldots v_{j-1},v_{j+1},ldots,v_n}bigr)supsetoperatorname{span}bigl({v_1,ldots,v_n}bigr).$$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Yes, they are the same:




        • since ${v_1,ldots v_{j-1},v_{j+1},ldots,v_n}subset{v_1,ldots,v_n}$,$$operatorname{span}bigl({v_1,ldots v_{j-1},v_{j+1},ldots,v_n}bigr)subsetoperatorname{span}bigl({v_1,ldots,v_n}bigr);$$

        • each element of ${v_1,ldots,v_n}$ is a linear combination of elements of ${v_1,ldots v_{j-1},v_{j+1},ldots,v_n}$,$$operatorname{span}bigl({v_1,ldots v_{j-1},v_{j+1},ldots,v_n}bigr)supsetoperatorname{span}bigl({v_1,ldots,v_n}bigr).$$






        share|cite|improve this answer











        $endgroup$



        Yes, they are the same:




        • since ${v_1,ldots v_{j-1},v_{j+1},ldots,v_n}subset{v_1,ldots,v_n}$,$$operatorname{span}bigl({v_1,ldots v_{j-1},v_{j+1},ldots,v_n}bigr)subsetoperatorname{span}bigl({v_1,ldots,v_n}bigr);$$

        • each element of ${v_1,ldots,v_n}$ is a linear combination of elements of ${v_1,ldots v_{j-1},v_{j+1},ldots,v_n}$,$$operatorname{span}bigl({v_1,ldots v_{j-1},v_{j+1},ldots,v_n}bigr)supsetoperatorname{span}bigl({v_1,ldots,v_n}bigr).$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 13 at 22:04

























        answered Jan 13 at 19:45









        José Carlos SantosJosé Carlos Santos

        157k22126227




        157k22126227






























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