Subspace basis of $mathbb{R}^n$ only with positive values












4












$begingroup$


It seems obvious but I didn't find a proof yet:
Let $U$ be an arbitrary subspace of $mathbb{R}^n$. Set $m:=dim{U}$.



Can $U$ be written as $U=mathrm{span}{b_1,dotsc,b_m}$, $b_j=begin{pmatrix}b_{i,1}\vdots\b_{i,n}end{pmatrix}$ with $b_{i,j}ge0;forall iin{1,dotsc,m};forall jin{1,dotsc,n}$?



An equivalent formulation for this question:
$exists B=begin{pmatrix}b_{1,1}&cdots &b_{1,n}\vdots&&vdots\b_{m,1}&cdots &b_{m,n}end{pmatrix}inmathbb{R}^{mtimes n}: U=mathrm{img}(B)$ with $b_{i,j}ge0;forall iin{1,dotsc,m};forall jin{1,dotsc,n}$?



If yes, can you please give a construction algorithm for a given base ${u_1,dotsc,u_m}$ of $U$?










share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    It seems obvious but I didn't find a proof yet:
    Let $U$ be an arbitrary subspace of $mathbb{R}^n$. Set $m:=dim{U}$.



    Can $U$ be written as $U=mathrm{span}{b_1,dotsc,b_m}$, $b_j=begin{pmatrix}b_{i,1}\vdots\b_{i,n}end{pmatrix}$ with $b_{i,j}ge0;forall iin{1,dotsc,m};forall jin{1,dotsc,n}$?



    An equivalent formulation for this question:
    $exists B=begin{pmatrix}b_{1,1}&cdots &b_{1,n}\vdots&&vdots\b_{m,1}&cdots &b_{m,n}end{pmatrix}inmathbb{R}^{mtimes n}: U=mathrm{img}(B)$ with $b_{i,j}ge0;forall iin{1,dotsc,m};forall jin{1,dotsc,n}$?



    If yes, can you please give a construction algorithm for a given base ${u_1,dotsc,u_m}$ of $U$?










    share|cite|improve this question









    $endgroup$















      4












      4








      4


      1



      $begingroup$


      It seems obvious but I didn't find a proof yet:
      Let $U$ be an arbitrary subspace of $mathbb{R}^n$. Set $m:=dim{U}$.



      Can $U$ be written as $U=mathrm{span}{b_1,dotsc,b_m}$, $b_j=begin{pmatrix}b_{i,1}\vdots\b_{i,n}end{pmatrix}$ with $b_{i,j}ge0;forall iin{1,dotsc,m};forall jin{1,dotsc,n}$?



      An equivalent formulation for this question:
      $exists B=begin{pmatrix}b_{1,1}&cdots &b_{1,n}\vdots&&vdots\b_{m,1}&cdots &b_{m,n}end{pmatrix}inmathbb{R}^{mtimes n}: U=mathrm{img}(B)$ with $b_{i,j}ge0;forall iin{1,dotsc,m};forall jin{1,dotsc,n}$?



      If yes, can you please give a construction algorithm for a given base ${u_1,dotsc,u_m}$ of $U$?










      share|cite|improve this question









      $endgroup$




      It seems obvious but I didn't find a proof yet:
      Let $U$ be an arbitrary subspace of $mathbb{R}^n$. Set $m:=dim{U}$.



      Can $U$ be written as $U=mathrm{span}{b_1,dotsc,b_m}$, $b_j=begin{pmatrix}b_{i,1}\vdots\b_{i,n}end{pmatrix}$ with $b_{i,j}ge0;forall iin{1,dotsc,m};forall jin{1,dotsc,n}$?



      An equivalent formulation for this question:
      $exists B=begin{pmatrix}b_{1,1}&cdots &b_{1,n}\vdots&&vdots\b_{m,1}&cdots &b_{m,n}end{pmatrix}inmathbb{R}^{mtimes n}: U=mathrm{img}(B)$ with $b_{i,j}ge0;forall iin{1,dotsc,m};forall jin{1,dotsc,n}$?



      If yes, can you please give a construction algorithm for a given base ${u_1,dotsc,u_m}$ of $U$?







      matrices vector-spaces






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      asked Dec 15 '18 at 15:20









      quiliupquiliup

      1799




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          2 Answers
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          $begingroup$

          Consider the line $y=-x$ as a subspace of $Bbb R^2$. Any basis consists of one vector, which must be of the form $(alpha,-alpha)$ for some non-zero $alphainBbb R$.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            It's not true. For a counter example, consider $Bbb R^3$ and the 2D subspace $U$ whose normal vector is $(1,1,1)$. Any two non-negative non-zero vectors in $Bbb R^3_{ge 0}$ will span a subspace part of which is out of $U$ because $UcapBbb R^3_{ge 0} = {0}$, hence they cannot span $U$.






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              2 Answers
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              2 Answers
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              active

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              2












              $begingroup$

              Consider the line $y=-x$ as a subspace of $Bbb R^2$. Any basis consists of one vector, which must be of the form $(alpha,-alpha)$ for some non-zero $alphainBbb R$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Consider the line $y=-x$ as a subspace of $Bbb R^2$. Any basis consists of one vector, which must be of the form $(alpha,-alpha)$ for some non-zero $alphainBbb R$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Consider the line $y=-x$ as a subspace of $Bbb R^2$. Any basis consists of one vector, which must be of the form $(alpha,-alpha)$ for some non-zero $alphainBbb R$.






                  share|cite|improve this answer









                  $endgroup$



                  Consider the line $y=-x$ as a subspace of $Bbb R^2$. Any basis consists of one vector, which must be of the form $(alpha,-alpha)$ for some non-zero $alphainBbb R$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 15 '18 at 15:53









                  TonyKTonyK

                  42.6k355134




                  42.6k355134























                      2












                      $begingroup$

                      It's not true. For a counter example, consider $Bbb R^3$ and the 2D subspace $U$ whose normal vector is $(1,1,1)$. Any two non-negative non-zero vectors in $Bbb R^3_{ge 0}$ will span a subspace part of which is out of $U$ because $UcapBbb R^3_{ge 0} = {0}$, hence they cannot span $U$.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        It's not true. For a counter example, consider $Bbb R^3$ and the 2D subspace $U$ whose normal vector is $(1,1,1)$. Any two non-negative non-zero vectors in $Bbb R^3_{ge 0}$ will span a subspace part of which is out of $U$ because $UcapBbb R^3_{ge 0} = {0}$, hence they cannot span $U$.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          It's not true. For a counter example, consider $Bbb R^3$ and the 2D subspace $U$ whose normal vector is $(1,1,1)$. Any two non-negative non-zero vectors in $Bbb R^3_{ge 0}$ will span a subspace part of which is out of $U$ because $UcapBbb R^3_{ge 0} = {0}$, hence they cannot span $U$.






                          share|cite|improve this answer









                          $endgroup$



                          It's not true. For a counter example, consider $Bbb R^3$ and the 2D subspace $U$ whose normal vector is $(1,1,1)$. Any two non-negative non-zero vectors in $Bbb R^3_{ge 0}$ will span a subspace part of which is out of $U$ because $UcapBbb R^3_{ge 0} = {0}$, hence they cannot span $U$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 15 '18 at 15:37









                          VimVim

                          8,12031348




                          8,12031348






























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