Prove that $ X_a={x_a; x_a=(a,a^2,a^3…), a in R }$ are linearly independent.












1












$begingroup$


I need to prove that all vectors in the set $X_a={x_a; x_a=(a,a^2,a^3...), a in mathbb{R} } subset mathbb{R}^{infty} $are linearly independent



I have tryed to prove it by induction but I don't know how to formulate it. I have tryed also to prove it by using polynomials (I know how to show that they are linearly independent) but these are just scalars of a specific form not $(x,x^2,x^3...)$ where $x$ is a variable (and then so and so hold for every $x$)



any Ideas?






linear-algebra polynomials vector-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    In what vector space? $ell^2$? $R^infty$? Does that ellipsis mean it is infinite dimensional?
    $endgroup$
    – Jeffery Opoku-Mensah
    Oct 22 '18 at 21:53












  • $begingroup$
    so $x_a$ is a set?
    $endgroup$
    – mathworker21
    Oct 22 '18 at 21:53










  • $begingroup$
    $X_a$ is a set (the notation was a bit confusing so I changed it now) the vector space is $R^{infty}$
    $endgroup$
    – Alexandar Solženjicin
    Oct 22 '18 at 21:55


















1












$begingroup$


I need to prove that all vectors in the set $X_a={x_a; x_a=(a,a^2,a^3...), a in mathbb{R} } subset mathbb{R}^{infty} $are linearly independent



I have tryed to prove it by induction but I don't know how to formulate it. I have tryed also to prove it by using polynomials (I know how to show that they are linearly independent) but these are just scalars of a specific form not $(x,x^2,x^3...)$ where $x$ is a variable (and then so and so hold for every $x$)



any Ideas?






linear-algebra polynomials vector-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    In what vector space? $ell^2$? $R^infty$? Does that ellipsis mean it is infinite dimensional?
    $endgroup$
    – Jeffery Opoku-Mensah
    Oct 22 '18 at 21:53












  • $begingroup$
    so $x_a$ is a set?
    $endgroup$
    – mathworker21
    Oct 22 '18 at 21:53










  • $begingroup$
    $X_a$ is a set (the notation was a bit confusing so I changed it now) the vector space is $R^{infty}$
    $endgroup$
    – Alexandar Solženjicin
    Oct 22 '18 at 21:55
















1












1








1





$begingroup$


I need to prove that all vectors in the set $X_a={x_a; x_a=(a,a^2,a^3...), a in mathbb{R} } subset mathbb{R}^{infty} $are linearly independent



I have tryed to prove it by induction but I don't know how to formulate it. I have tryed also to prove it by using polynomials (I know how to show that they are linearly independent) but these are just scalars of a specific form not $(x,x^2,x^3...)$ where $x$ is a variable (and then so and so hold for every $x$)



any Ideas?






linear-algebra polynomials vector-spaces







share|cite|improve this question











$endgroup$




I need to prove that all vectors in the set $X_a={x_a; x_a=(a,a^2,a^3...), a in mathbb{R} } subset mathbb{R}^{infty} $are linearly independent



I have tryed to prove it by induction but I don't know how to formulate it. I have tryed also to prove it by using polynomials (I know how to show that they are linearly independent) but these are just scalars of a specific form not $(x,x^2,x^3...)$ where $x$ is a variable (and then so and so hold for every $x$)



any Ideas?






linear-algebra polynomials vector-spaces




linear-algebra polynomials vector-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 12:17







Alexandar Solženjicin

















asked Oct 22 '18 at 21:48









Alexandar SolženjicinAlexandar Solženjicin

858




858












  • $begingroup$
    In what vector space? $ell^2$? $R^infty$? Does that ellipsis mean it is infinite dimensional?
    $endgroup$
    – Jeffery Opoku-Mensah
    Oct 22 '18 at 21:53












  • $begingroup$
    so $x_a$ is a set?
    $endgroup$
    – mathworker21
    Oct 22 '18 at 21:53










  • $begingroup$
    $X_a$ is a set (the notation was a bit confusing so I changed it now) the vector space is $R^{infty}$
    $endgroup$
    – Alexandar Solženjicin
    Oct 22 '18 at 21:55




















  • $begingroup$
    In what vector space? $ell^2$? $R^infty$? Does that ellipsis mean it is infinite dimensional?
    $endgroup$
    – Jeffery Opoku-Mensah
    Oct 22 '18 at 21:53












  • $begingroup$
    so $x_a$ is a set?
    $endgroup$
    – mathworker21
    Oct 22 '18 at 21:53










  • $begingroup$
    $X_a$ is a set (the notation was a bit confusing so I changed it now) the vector space is $R^{infty}$
    $endgroup$
    – Alexandar Solženjicin
    Oct 22 '18 at 21:55


















$begingroup$
In what vector space? $ell^2$? $R^infty$? Does that ellipsis mean it is infinite dimensional?
$endgroup$
– Jeffery Opoku-Mensah
Oct 22 '18 at 21:53






$begingroup$
In what vector space? $ell^2$? $R^infty$? Does that ellipsis mean it is infinite dimensional?
$endgroup$
– Jeffery Opoku-Mensah
Oct 22 '18 at 21:53














$begingroup$
so $x_a$ is a set?
$endgroup$
– mathworker21
Oct 22 '18 at 21:53




$begingroup$
so $x_a$ is a set?
$endgroup$
– mathworker21
Oct 22 '18 at 21:53












$begingroup$
$X_a$ is a set (the notation was a bit confusing so I changed it now) the vector space is $R^{infty}$
$endgroup$
– Alexandar Solženjicin
Oct 22 '18 at 21:55






$begingroup$
$X_a$ is a set (the notation was a bit confusing so I changed it now) the vector space is $R^{infty}$
$endgroup$
– Alexandar Solženjicin
Oct 22 '18 at 21:55












2 Answers
2






active

oldest

votes


















3












$begingroup$

I assume you mean $ane 0$, otherwise it is of course not true. Well, let $a_1,a_2,...,a_n$ be different non zero real numbers and write a linear combination $lambda_1x_{a_1}+lambda_2x_{a_2}+...+lambda_nx_{a_n}=0$. Here the zero vector is $x_0$, all its coordinates are zeros, including the first $n$ coordinates of course. So we get a system of linear equations:



$lambda_1a_1+lambda_2a_2+...+lambda_na_n=0$



$lambda_1a_1^2+lambda_2a_2^2+...+lambda_na_n^2=0$



.................................................



$lambda_1a_1^n+lambda_2a_n^2+...+lambda_na_n^n=0$



This is a system of linear equations with $lambda_1,...,lambda_n$ being the variables. To show $lambda_1=...=lambda_n=0$ it is enough to prove that the determinant of the matrix which represents the system is not zero. Write the matrix and see that its determinant is $a_1a_2...a_n$ times the determinant of a Vandermonde matrix which I assume you know is not zero. (of course we are using the fact that $a_1,a_2,...,a_n$ are all different)






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Assume that there exist a nonempty minimally sized set of nonzero elements



    $$mathcal{S} = left{(x_0, x_0^2, x_0^3ldots), ldots , (x_n, x_n^2, x_n^3 ldots)right}
    $$



    that are linearly dependent. Then



    $$x_0^k = sum_{i=1}^{n} c_i x_i^k$$



    for all $k in mathbb{N}_0$. Then



    $$sum_{i=1}^{n} (c_ix_i) x_i^{k-1} = x_0^k = x_0(x_0)^{k-1} = sum_{i=1}^{n} (x_0c_i) x_i^{k-1},$$



    which implies that for all $k in mathbb{N}$,



    $$sum_{i=1}^{n} (x_0c_i - x_ic_i)x_i^{k-1} = 0$$



    Since $mathcal{S} setminus {(x_0, x_0^2, x_0^3, ldots )}$ is by assumption linearly independent, it must be that for each $i$, $x_0c_i - x_ic_i = 0$, thus $x_0 = x_i$, or $c_i = 0$. In other words, either
    $$x_0 = x_1 = ldots = x_n,$$
    or $mathbf{0} in mathcal{S}$.
    Contradiction.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      is $x_0^k=(x_0,x_0^2..) $or je k-th element of that list, or something else?
      $endgroup$
      – Alexandar Solženjicin
      Oct 22 '18 at 22:17












    • $begingroup$
      $x_0$ is just the first element of the vector $(x_0, x_0^2, x_0^3, ldots)$. I didn't use your notation since it would be a bit clunky. Then $x_0^k$ is that raised to the $kth$ power.
      $endgroup$
      – Jeffery Opoku-Mensah
      Oct 22 '18 at 22:18












    • $begingroup$
      what if $S$ is not minimaly sized? and then $ S/{(x_0,x_0^2..)}$ is not linearly independent?
      $endgroup$
      – Alexandar Solženjicin
      Oct 22 '18 at 22:37












    • $begingroup$
      If there exists a linearly dependent set of vectors, then there must be a finite minimally sized set of linearly dependent vectors. Thus if there is no minimally sized linearly dependent set of vectors, there is no linearly dependent set of vectors at all.
      $endgroup$
      – Jeffery Opoku-Mensah
      Oct 22 '18 at 22:55













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    2 Answers
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    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    I assume you mean $ane 0$, otherwise it is of course not true. Well, let $a_1,a_2,...,a_n$ be different non zero real numbers and write a linear combination $lambda_1x_{a_1}+lambda_2x_{a_2}+...+lambda_nx_{a_n}=0$. Here the zero vector is $x_0$, all its coordinates are zeros, including the first $n$ coordinates of course. So we get a system of linear equations:



    $lambda_1a_1+lambda_2a_2+...+lambda_na_n=0$



    $lambda_1a_1^2+lambda_2a_2^2+...+lambda_na_n^2=0$



    .................................................



    $lambda_1a_1^n+lambda_2a_n^2+...+lambda_na_n^n=0$



    This is a system of linear equations with $lambda_1,...,lambda_n$ being the variables. To show $lambda_1=...=lambda_n=0$ it is enough to prove that the determinant of the matrix which represents the system is not zero. Write the matrix and see that its determinant is $a_1a_2...a_n$ times the determinant of a Vandermonde matrix which I assume you know is not zero. (of course we are using the fact that $a_1,a_2,...,a_n$ are all different)






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      I assume you mean $ane 0$, otherwise it is of course not true. Well, let $a_1,a_2,...,a_n$ be different non zero real numbers and write a linear combination $lambda_1x_{a_1}+lambda_2x_{a_2}+...+lambda_nx_{a_n}=0$. Here the zero vector is $x_0$, all its coordinates are zeros, including the first $n$ coordinates of course. So we get a system of linear equations:



      $lambda_1a_1+lambda_2a_2+...+lambda_na_n=0$



      $lambda_1a_1^2+lambda_2a_2^2+...+lambda_na_n^2=0$



      .................................................



      $lambda_1a_1^n+lambda_2a_n^2+...+lambda_na_n^n=0$



      This is a system of linear equations with $lambda_1,...,lambda_n$ being the variables. To show $lambda_1=...=lambda_n=0$ it is enough to prove that the determinant of the matrix which represents the system is not zero. Write the matrix and see that its determinant is $a_1a_2...a_n$ times the determinant of a Vandermonde matrix which I assume you know is not zero. (of course we are using the fact that $a_1,a_2,...,a_n$ are all different)






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        I assume you mean $ane 0$, otherwise it is of course not true. Well, let $a_1,a_2,...,a_n$ be different non zero real numbers and write a linear combination $lambda_1x_{a_1}+lambda_2x_{a_2}+...+lambda_nx_{a_n}=0$. Here the zero vector is $x_0$, all its coordinates are zeros, including the first $n$ coordinates of course. So we get a system of linear equations:



        $lambda_1a_1+lambda_2a_2+...+lambda_na_n=0$



        $lambda_1a_1^2+lambda_2a_2^2+...+lambda_na_n^2=0$



        .................................................



        $lambda_1a_1^n+lambda_2a_n^2+...+lambda_na_n^n=0$



        This is a system of linear equations with $lambda_1,...,lambda_n$ being the variables. To show $lambda_1=...=lambda_n=0$ it is enough to prove that the determinant of the matrix which represents the system is not zero. Write the matrix and see that its determinant is $a_1a_2...a_n$ times the determinant of a Vandermonde matrix which I assume you know is not zero. (of course we are using the fact that $a_1,a_2,...,a_n$ are all different)






        share|cite|improve this answer









        $endgroup$



        I assume you mean $ane 0$, otherwise it is of course not true. Well, let $a_1,a_2,...,a_n$ be different non zero real numbers and write a linear combination $lambda_1x_{a_1}+lambda_2x_{a_2}+...+lambda_nx_{a_n}=0$. Here the zero vector is $x_0$, all its coordinates are zeros, including the first $n$ coordinates of course. So we get a system of linear equations:



        $lambda_1a_1+lambda_2a_2+...+lambda_na_n=0$



        $lambda_1a_1^2+lambda_2a_2^2+...+lambda_na_n^2=0$



        .................................................



        $lambda_1a_1^n+lambda_2a_n^2+...+lambda_na_n^n=0$



        This is a system of linear equations with $lambda_1,...,lambda_n$ being the variables. To show $lambda_1=...=lambda_n=0$ it is enough to prove that the determinant of the matrix which represents the system is not zero. Write the matrix and see that its determinant is $a_1a_2...a_n$ times the determinant of a Vandermonde matrix which I assume you know is not zero. (of course we are using the fact that $a_1,a_2,...,a_n$ are all different)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Oct 22 '18 at 22:03









        MarkMark

        6,488416




        6,488416























            1












            $begingroup$

            Assume that there exist a nonempty minimally sized set of nonzero elements



            $$mathcal{S} = left{(x_0, x_0^2, x_0^3ldots), ldots , (x_n, x_n^2, x_n^3 ldots)right}
            $$



            that are linearly dependent. Then



            $$x_0^k = sum_{i=1}^{n} c_i x_i^k$$



            for all $k in mathbb{N}_0$. Then



            $$sum_{i=1}^{n} (c_ix_i) x_i^{k-1} = x_0^k = x_0(x_0)^{k-1} = sum_{i=1}^{n} (x_0c_i) x_i^{k-1},$$



            which implies that for all $k in mathbb{N}$,



            $$sum_{i=1}^{n} (x_0c_i - x_ic_i)x_i^{k-1} = 0$$



            Since $mathcal{S} setminus {(x_0, x_0^2, x_0^3, ldots )}$ is by assumption linearly independent, it must be that for each $i$, $x_0c_i - x_ic_i = 0$, thus $x_0 = x_i$, or $c_i = 0$. In other words, either
            $$x_0 = x_1 = ldots = x_n,$$
            or $mathbf{0} in mathcal{S}$.
            Contradiction.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              is $x_0^k=(x_0,x_0^2..) $or je k-th element of that list, or something else?
              $endgroup$
              – Alexandar Solženjicin
              Oct 22 '18 at 22:17












            • $begingroup$
              $x_0$ is just the first element of the vector $(x_0, x_0^2, x_0^3, ldots)$. I didn't use your notation since it would be a bit clunky. Then $x_0^k$ is that raised to the $kth$ power.
              $endgroup$
              – Jeffery Opoku-Mensah
              Oct 22 '18 at 22:18












            • $begingroup$
              what if $S$ is not minimaly sized? and then $ S/{(x_0,x_0^2..)}$ is not linearly independent?
              $endgroup$
              – Alexandar Solženjicin
              Oct 22 '18 at 22:37












            • $begingroup$
              If there exists a linearly dependent set of vectors, then there must be a finite minimally sized set of linearly dependent vectors. Thus if there is no minimally sized linearly dependent set of vectors, there is no linearly dependent set of vectors at all.
              $endgroup$
              – Jeffery Opoku-Mensah
              Oct 22 '18 at 22:55


















            1












            $begingroup$

            Assume that there exist a nonempty minimally sized set of nonzero elements



            $$mathcal{S} = left{(x_0, x_0^2, x_0^3ldots), ldots , (x_n, x_n^2, x_n^3 ldots)right}
            $$



            that are linearly dependent. Then



            $$x_0^k = sum_{i=1}^{n} c_i x_i^k$$



            for all $k in mathbb{N}_0$. Then



            $$sum_{i=1}^{n} (c_ix_i) x_i^{k-1} = x_0^k = x_0(x_0)^{k-1} = sum_{i=1}^{n} (x_0c_i) x_i^{k-1},$$



            which implies that for all $k in mathbb{N}$,



            $$sum_{i=1}^{n} (x_0c_i - x_ic_i)x_i^{k-1} = 0$$



            Since $mathcal{S} setminus {(x_0, x_0^2, x_0^3, ldots )}$ is by assumption linearly independent, it must be that for each $i$, $x_0c_i - x_ic_i = 0$, thus $x_0 = x_i$, or $c_i = 0$. In other words, either
            $$x_0 = x_1 = ldots = x_n,$$
            or $mathbf{0} in mathcal{S}$.
            Contradiction.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              is $x_0^k=(x_0,x_0^2..) $or je k-th element of that list, or something else?
              $endgroup$
              – Alexandar Solženjicin
              Oct 22 '18 at 22:17












            • $begingroup$
              $x_0$ is just the first element of the vector $(x_0, x_0^2, x_0^3, ldots)$. I didn't use your notation since it would be a bit clunky. Then $x_0^k$ is that raised to the $kth$ power.
              $endgroup$
              – Jeffery Opoku-Mensah
              Oct 22 '18 at 22:18












            • $begingroup$
              what if $S$ is not minimaly sized? and then $ S/{(x_0,x_0^2..)}$ is not linearly independent?
              $endgroup$
              – Alexandar Solženjicin
              Oct 22 '18 at 22:37












            • $begingroup$
              If there exists a linearly dependent set of vectors, then there must be a finite minimally sized set of linearly dependent vectors. Thus if there is no minimally sized linearly dependent set of vectors, there is no linearly dependent set of vectors at all.
              $endgroup$
              – Jeffery Opoku-Mensah
              Oct 22 '18 at 22:55
















            1












            1








            1





            $begingroup$

            Assume that there exist a nonempty minimally sized set of nonzero elements



            $$mathcal{S} = left{(x_0, x_0^2, x_0^3ldots), ldots , (x_n, x_n^2, x_n^3 ldots)right}
            $$



            that are linearly dependent. Then



            $$x_0^k = sum_{i=1}^{n} c_i x_i^k$$



            for all $k in mathbb{N}_0$. Then



            $$sum_{i=1}^{n} (c_ix_i) x_i^{k-1} = x_0^k = x_0(x_0)^{k-1} = sum_{i=1}^{n} (x_0c_i) x_i^{k-1},$$



            which implies that for all $k in mathbb{N}$,



            $$sum_{i=1}^{n} (x_0c_i - x_ic_i)x_i^{k-1} = 0$$



            Since $mathcal{S} setminus {(x_0, x_0^2, x_0^3, ldots )}$ is by assumption linearly independent, it must be that for each $i$, $x_0c_i - x_ic_i = 0$, thus $x_0 = x_i$, or $c_i = 0$. In other words, either
            $$x_0 = x_1 = ldots = x_n,$$
            or $mathbf{0} in mathcal{S}$.
            Contradiction.






            share|cite|improve this answer









            $endgroup$



            Assume that there exist a nonempty minimally sized set of nonzero elements



            $$mathcal{S} = left{(x_0, x_0^2, x_0^3ldots), ldots , (x_n, x_n^2, x_n^3 ldots)right}
            $$



            that are linearly dependent. Then



            $$x_0^k = sum_{i=1}^{n} c_i x_i^k$$



            for all $k in mathbb{N}_0$. Then



            $$sum_{i=1}^{n} (c_ix_i) x_i^{k-1} = x_0^k = x_0(x_0)^{k-1} = sum_{i=1}^{n} (x_0c_i) x_i^{k-1},$$



            which implies that for all $k in mathbb{N}$,



            $$sum_{i=1}^{n} (x_0c_i - x_ic_i)x_i^{k-1} = 0$$



            Since $mathcal{S} setminus {(x_0, x_0^2, x_0^3, ldots )}$ is by assumption linearly independent, it must be that for each $i$, $x_0c_i - x_ic_i = 0$, thus $x_0 = x_i$, or $c_i = 0$. In other words, either
            $$x_0 = x_1 = ldots = x_n,$$
            or $mathbf{0} in mathcal{S}$.
            Contradiction.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Oct 22 '18 at 22:14









            Jeffery Opoku-MensahJeffery Opoku-Mensah

            3,528720




            3,528720












            • $begingroup$
              is $x_0^k=(x_0,x_0^2..) $or je k-th element of that list, or something else?
              $endgroup$
              – Alexandar Solženjicin
              Oct 22 '18 at 22:17












            • $begingroup$
              $x_0$ is just the first element of the vector $(x_0, x_0^2, x_0^3, ldots)$. I didn't use your notation since it would be a bit clunky. Then $x_0^k$ is that raised to the $kth$ power.
              $endgroup$
              – Jeffery Opoku-Mensah
              Oct 22 '18 at 22:18












            • $begingroup$
              what if $S$ is not minimaly sized? and then $ S/{(x_0,x_0^2..)}$ is not linearly independent?
              $endgroup$
              – Alexandar Solženjicin
              Oct 22 '18 at 22:37












            • $begingroup$
              If there exists a linearly dependent set of vectors, then there must be a finite minimally sized set of linearly dependent vectors. Thus if there is no minimally sized linearly dependent set of vectors, there is no linearly dependent set of vectors at all.
              $endgroup$
              – Jeffery Opoku-Mensah
              Oct 22 '18 at 22:55




















            • $begingroup$
              is $x_0^k=(x_0,x_0^2..) $or je k-th element of that list, or something else?
              $endgroup$
              – Alexandar Solženjicin
              Oct 22 '18 at 22:17












            • $begingroup$
              $x_0$ is just the first element of the vector $(x_0, x_0^2, x_0^3, ldots)$. I didn't use your notation since it would be a bit clunky. Then $x_0^k$ is that raised to the $kth$ power.
              $endgroup$
              – Jeffery Opoku-Mensah
              Oct 22 '18 at 22:18












            • $begingroup$
              what if $S$ is not minimaly sized? and then $ S/{(x_0,x_0^2..)}$ is not linearly independent?
              $endgroup$
              – Alexandar Solženjicin
              Oct 22 '18 at 22:37












            • $begingroup$
              If there exists a linearly dependent set of vectors, then there must be a finite minimally sized set of linearly dependent vectors. Thus if there is no minimally sized linearly dependent set of vectors, there is no linearly dependent set of vectors at all.
              $endgroup$
              – Jeffery Opoku-Mensah
              Oct 22 '18 at 22:55


















            $begingroup$
            is $x_0^k=(x_0,x_0^2..) $or je k-th element of that list, or something else?
            $endgroup$
            – Alexandar Solženjicin
            Oct 22 '18 at 22:17






            $begingroup$
            is $x_0^k=(x_0,x_0^2..) $or je k-th element of that list, or something else?
            $endgroup$
            – Alexandar Solženjicin
            Oct 22 '18 at 22:17














            $begingroup$
            $x_0$ is just the first element of the vector $(x_0, x_0^2, x_0^3, ldots)$. I didn't use your notation since it would be a bit clunky. Then $x_0^k$ is that raised to the $kth$ power.
            $endgroup$
            – Jeffery Opoku-Mensah
            Oct 22 '18 at 22:18






            $begingroup$
            $x_0$ is just the first element of the vector $(x_0, x_0^2, x_0^3, ldots)$. I didn't use your notation since it would be a bit clunky. Then $x_0^k$ is that raised to the $kth$ power.
            $endgroup$
            – Jeffery Opoku-Mensah
            Oct 22 '18 at 22:18














            $begingroup$
            what if $S$ is not minimaly sized? and then $ S/{(x_0,x_0^2..)}$ is not linearly independent?
            $endgroup$
            – Alexandar Solženjicin
            Oct 22 '18 at 22:37






            $begingroup$
            what if $S$ is not minimaly sized? and then $ S/{(x_0,x_0^2..)}$ is not linearly independent?
            $endgroup$
            – Alexandar Solženjicin
            Oct 22 '18 at 22:37














            $begingroup$
            If there exists a linearly dependent set of vectors, then there must be a finite minimally sized set of linearly dependent vectors. Thus if there is no minimally sized linearly dependent set of vectors, there is no linearly dependent set of vectors at all.
            $endgroup$
            – Jeffery Opoku-Mensah
            Oct 22 '18 at 22:55






            $begingroup$
            If there exists a linearly dependent set of vectors, then there must be a finite minimally sized set of linearly dependent vectors. Thus if there is no minimally sized linearly dependent set of vectors, there is no linearly dependent set of vectors at all.
            $endgroup$
            – Jeffery Opoku-Mensah
            Oct 22 '18 at 22:55




















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