How do I prove that 2 vector spaces are equal if they have different number of vectors that span them?












1












$begingroup$


For example, in 4-dimensions, vector space V is spanned by the linearly independent vectors x, y, and z
and vector space U is spanned by the linearly independent vectors r, and s. Would I try to prove that the vectors x, y, and z can span U and vectors r, and s can span V?






linear-algebra vector-spaces







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$endgroup$








  • 1




    $begingroup$
    Depends whether they are linearly independent. For example if the spanning vectors of $V$ are linearly independent it is impossible for two vectors to span $V$.
    $endgroup$
    – Jonas Lenz
    Jul 23 '18 at 6:09










  • $begingroup$
    That's one way! If the vectors are given e.g. "numerically" in $Bbb{R}^4$ (or with respect to a known basis of the 4D space), then you can also do the following. Form a 3x4 matrix with rows x,y,z. Calculate the reduced row echelon form. Do the same with the 2x4 matrix with rows r and s. The spanned subspaces are equal if and only if the non-zero rows of the two RREF-matrices are equal.
    $endgroup$
    – Jyrki Lahtonen
    Jul 23 '18 at 6:12












  • $begingroup$
    @JonasLenz oh, my bad I forgot to mention that they are linearly independent, ill edit that in
    $endgroup$
    – edmonda7
    Jul 23 '18 at 6:16






  • 2




    $begingroup$
    In that case, $dim V=3ne 2=dim U$, hence $Vne U$
    $endgroup$
    – Hagen von Eitzen
    Jul 23 '18 at 6:19
















1












$begingroup$


For example, in 4-dimensions, vector space V is spanned by the linearly independent vectors x, y, and z
and vector space U is spanned by the linearly independent vectors r, and s. Would I try to prove that the vectors x, y, and z can span U and vectors r, and s can span V?






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Depends whether they are linearly independent. For example if the spanning vectors of $V$ are linearly independent it is impossible for two vectors to span $V$.
    $endgroup$
    – Jonas Lenz
    Jul 23 '18 at 6:09










  • $begingroup$
    That's one way! If the vectors are given e.g. "numerically" in $Bbb{R}^4$ (or with respect to a known basis of the 4D space), then you can also do the following. Form a 3x4 matrix with rows x,y,z. Calculate the reduced row echelon form. Do the same with the 2x4 matrix with rows r and s. The spanned subspaces are equal if and only if the non-zero rows of the two RREF-matrices are equal.
    $endgroup$
    – Jyrki Lahtonen
    Jul 23 '18 at 6:12












  • $begingroup$
    @JonasLenz oh, my bad I forgot to mention that they are linearly independent, ill edit that in
    $endgroup$
    – edmonda7
    Jul 23 '18 at 6:16






  • 2




    $begingroup$
    In that case, $dim V=3ne 2=dim U$, hence $Vne U$
    $endgroup$
    – Hagen von Eitzen
    Jul 23 '18 at 6:19














1












1








1


0



$begingroup$


For example, in 4-dimensions, vector space V is spanned by the linearly independent vectors x, y, and z
and vector space U is spanned by the linearly independent vectors r, and s. Would I try to prove that the vectors x, y, and z can span U and vectors r, and s can span V?






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$




For example, in 4-dimensions, vector space V is spanned by the linearly independent vectors x, y, and z
and vector space U is spanned by the linearly independent vectors r, and s. Would I try to prove that the vectors x, y, and z can span U and vectors r, and s can span V?






linear-algebra vector-spaces




linear-algebra vector-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 10:51









José Carlos Santos

170k23132238




170k23132238










asked Jul 23 '18 at 5:59









edmonda7edmonda7

394




394








  • 1




    $begingroup$
    Depends whether they are linearly independent. For example if the spanning vectors of $V$ are linearly independent it is impossible for two vectors to span $V$.
    $endgroup$
    – Jonas Lenz
    Jul 23 '18 at 6:09










  • $begingroup$
    That's one way! If the vectors are given e.g. "numerically" in $Bbb{R}^4$ (or with respect to a known basis of the 4D space), then you can also do the following. Form a 3x4 matrix with rows x,y,z. Calculate the reduced row echelon form. Do the same with the 2x4 matrix with rows r and s. The spanned subspaces are equal if and only if the non-zero rows of the two RREF-matrices are equal.
    $endgroup$
    – Jyrki Lahtonen
    Jul 23 '18 at 6:12












  • $begingroup$
    @JonasLenz oh, my bad I forgot to mention that they are linearly independent, ill edit that in
    $endgroup$
    – edmonda7
    Jul 23 '18 at 6:16






  • 2




    $begingroup$
    In that case, $dim V=3ne 2=dim U$, hence $Vne U$
    $endgroup$
    – Hagen von Eitzen
    Jul 23 '18 at 6:19














  • 1




    $begingroup$
    Depends whether they are linearly independent. For example if the spanning vectors of $V$ are linearly independent it is impossible for two vectors to span $V$.
    $endgroup$
    – Jonas Lenz
    Jul 23 '18 at 6:09










  • $begingroup$
    That's one way! If the vectors are given e.g. "numerically" in $Bbb{R}^4$ (or with respect to a known basis of the 4D space), then you can also do the following. Form a 3x4 matrix with rows x,y,z. Calculate the reduced row echelon form. Do the same with the 2x4 matrix with rows r and s. The spanned subspaces are equal if and only if the non-zero rows of the two RREF-matrices are equal.
    $endgroup$
    – Jyrki Lahtonen
    Jul 23 '18 at 6:12












  • $begingroup$
    @JonasLenz oh, my bad I forgot to mention that they are linearly independent, ill edit that in
    $endgroup$
    – edmonda7
    Jul 23 '18 at 6:16






  • 2




    $begingroup$
    In that case, $dim V=3ne 2=dim U$, hence $Vne U$
    $endgroup$
    – Hagen von Eitzen
    Jul 23 '18 at 6:19








1




1




$begingroup$
Depends whether they are linearly independent. For example if the spanning vectors of $V$ are linearly independent it is impossible for two vectors to span $V$.
$endgroup$
– Jonas Lenz
Jul 23 '18 at 6:09




$begingroup$
Depends whether they are linearly independent. For example if the spanning vectors of $V$ are linearly independent it is impossible for two vectors to span $V$.
$endgroup$
– Jonas Lenz
Jul 23 '18 at 6:09












$begingroup$
That's one way! If the vectors are given e.g. "numerically" in $Bbb{R}^4$ (or with respect to a known basis of the 4D space), then you can also do the following. Form a 3x4 matrix with rows x,y,z. Calculate the reduced row echelon form. Do the same with the 2x4 matrix with rows r and s. The spanned subspaces are equal if and only if the non-zero rows of the two RREF-matrices are equal.
$endgroup$
– Jyrki Lahtonen
Jul 23 '18 at 6:12






$begingroup$
That's one way! If the vectors are given e.g. "numerically" in $Bbb{R}^4$ (or with respect to a known basis of the 4D space), then you can also do the following. Form a 3x4 matrix with rows x,y,z. Calculate the reduced row echelon form. Do the same with the 2x4 matrix with rows r and s. The spanned subspaces are equal if and only if the non-zero rows of the two RREF-matrices are equal.
$endgroup$
– Jyrki Lahtonen
Jul 23 '18 at 6:12














$begingroup$
@JonasLenz oh, my bad I forgot to mention that they are linearly independent, ill edit that in
$endgroup$
– edmonda7
Jul 23 '18 at 6:16




$begingroup$
@JonasLenz oh, my bad I forgot to mention that they are linearly independent, ill edit that in
$endgroup$
– edmonda7
Jul 23 '18 at 6:16




2




2




$begingroup$
In that case, $dim V=3ne 2=dim U$, hence $Vne U$
$endgroup$
– Hagen von Eitzen
Jul 23 '18 at 6:19




$begingroup$
In that case, $dim V=3ne 2=dim U$, hence $Vne U$
$endgroup$
– Hagen von Eitzen
Jul 23 '18 at 6:19










1 Answer
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$begingroup$

If $S$ and $S'$ are subsets of a vector space $V$, then $operatorname{span}(S)=operatorname{span}(S')$ if and only if each vector of $S$ is a linear combination of elements of $S'$ and vice-versa.



Consider, for instance, the case in which:




  • $V=mathbb{R}^3$;

  • $S=bigl{(1,-1,0),(1,0,-1),(0,1,-1)bigr}$;

  • $S'=bigl{(2,-1,-1),(-1,-1,2)bigr}$.


I claim that $operatorname{span}(S)=operatorname{span}(S')$. In order to prove it, all I have to do is to note that:




  • $displaystyle(1,-1,0)=frac23(2,-1,-1)+frac13(-1,-1,2)$;

  • $displaystyle(1,0,-1)=frac13(2,-1,-1)-frac13(-1,-1,2)$;

  • $displaystyle(0,1,-1)=-frac13(2,-1,-1)-frac23(-1,-1,2)$;

  • $displaystyle(2,-1,-1)=(1,-1,0)+(1,0,-1)$;

  • $displaystyle(-1,-1,2)=-(1,0,-1)-(0,1,-1)$.






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    $begingroup$

    If $S$ and $S'$ are subsets of a vector space $V$, then $operatorname{span}(S)=operatorname{span}(S')$ if and only if each vector of $S$ is a linear combination of elements of $S'$ and vice-versa.



    Consider, for instance, the case in which:




    • $V=mathbb{R}^3$;

    • $S=bigl{(1,-1,0),(1,0,-1),(0,1,-1)bigr}$;

    • $S'=bigl{(2,-1,-1),(-1,-1,2)bigr}$.


    I claim that $operatorname{span}(S)=operatorname{span}(S')$. In order to prove it, all I have to do is to note that:




    • $displaystyle(1,-1,0)=frac23(2,-1,-1)+frac13(-1,-1,2)$;

    • $displaystyle(1,0,-1)=frac13(2,-1,-1)-frac13(-1,-1,2)$;

    • $displaystyle(0,1,-1)=-frac13(2,-1,-1)-frac23(-1,-1,2)$;

    • $displaystyle(2,-1,-1)=(1,-1,0)+(1,0,-1)$;

    • $displaystyle(-1,-1,2)=-(1,0,-1)-(0,1,-1)$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      If $S$ and $S'$ are subsets of a vector space $V$, then $operatorname{span}(S)=operatorname{span}(S')$ if and only if each vector of $S$ is a linear combination of elements of $S'$ and vice-versa.



      Consider, for instance, the case in which:




      • $V=mathbb{R}^3$;

      • $S=bigl{(1,-1,0),(1,0,-1),(0,1,-1)bigr}$;

      • $S'=bigl{(2,-1,-1),(-1,-1,2)bigr}$.


      I claim that $operatorname{span}(S)=operatorname{span}(S')$. In order to prove it, all I have to do is to note that:




      • $displaystyle(1,-1,0)=frac23(2,-1,-1)+frac13(-1,-1,2)$;

      • $displaystyle(1,0,-1)=frac13(2,-1,-1)-frac13(-1,-1,2)$;

      • $displaystyle(0,1,-1)=-frac13(2,-1,-1)-frac23(-1,-1,2)$;

      • $displaystyle(2,-1,-1)=(1,-1,0)+(1,0,-1)$;

      • $displaystyle(-1,-1,2)=-(1,0,-1)-(0,1,-1)$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        If $S$ and $S'$ are subsets of a vector space $V$, then $operatorname{span}(S)=operatorname{span}(S')$ if and only if each vector of $S$ is a linear combination of elements of $S'$ and vice-versa.



        Consider, for instance, the case in which:




        • $V=mathbb{R}^3$;

        • $S=bigl{(1,-1,0),(1,0,-1),(0,1,-1)bigr}$;

        • $S'=bigl{(2,-1,-1),(-1,-1,2)bigr}$.


        I claim that $operatorname{span}(S)=operatorname{span}(S')$. In order to prove it, all I have to do is to note that:




        • $displaystyle(1,-1,0)=frac23(2,-1,-1)+frac13(-1,-1,2)$;

        • $displaystyle(1,0,-1)=frac13(2,-1,-1)-frac13(-1,-1,2)$;

        • $displaystyle(0,1,-1)=-frac13(2,-1,-1)-frac23(-1,-1,2)$;

        • $displaystyle(2,-1,-1)=(1,-1,0)+(1,0,-1)$;

        • $displaystyle(-1,-1,2)=-(1,0,-1)-(0,1,-1)$.






        share|cite|improve this answer











        $endgroup$



        If $S$ and $S'$ are subsets of a vector space $V$, then $operatorname{span}(S)=operatorname{span}(S')$ if and only if each vector of $S$ is a linear combination of elements of $S'$ and vice-versa.



        Consider, for instance, the case in which:




        • $V=mathbb{R}^3$;

        • $S=bigl{(1,-1,0),(1,0,-1),(0,1,-1)bigr}$;

        • $S'=bigl{(2,-1,-1),(-1,-1,2)bigr}$.


        I claim that $operatorname{span}(S)=operatorname{span}(S')$. In order to prove it, all I have to do is to note that:




        • $displaystyle(1,-1,0)=frac23(2,-1,-1)+frac13(-1,-1,2)$;

        • $displaystyle(1,0,-1)=frac13(2,-1,-1)-frac13(-1,-1,2)$;

        • $displaystyle(0,1,-1)=-frac13(2,-1,-1)-frac23(-1,-1,2)$;

        • $displaystyle(2,-1,-1)=(1,-1,0)+(1,0,-1)$;

        • $displaystyle(-1,-1,2)=-(1,0,-1)-(0,1,-1)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jul 23 '18 at 8:49

























        answered Jul 23 '18 at 6:41









        José Carlos SantosJosé Carlos Santos

        170k23132238




        170k23132238






























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