Rewrite a system of linear equations












3












$begingroup$


Fix $r$ and $dequiv r+binom{r}{2}$ and consider $d$ real numbers $b_1,b_2,..., b_d$.



I have the following system of linear equations:
$$
begin{cases}
text{$(diamond)$ Differences between the elements ${b_1,b_h}$ for $h=2,...,r$ $hspace{1cm}$ ($r-1$ elements)}\
b_{r+1}=b_1-b_2\
b_{r+2}=b_1-b_3\
...\
b_{2r-1}=b_1-b_r\
-------\
text{$(diamond)$ Differences between the elements ${b_2,b_h}$ for $h=3,...,r$ $hspace{1cm}$ ($r-2$ elements)}\
b_{2r}=b_2-b_3\
...\
b_{3r-3}=b_2-b_r\
-------\
text{$(diamond)$ Etc. }\
....\
-------\
text{$(diamond)$ Differences between the elements ${b_{r-1},b_h}$ for $h=r$ $hspace{1cm}$ ($1$ element)} \
b_d=b_{r-1}-b_r
end{cases}
$$



Question: For any $r$, I want to re-write in an EQUIVALENT way such that it




  • has only relations with the $+$ sign on the RHS of each equation


  • has only 1 variable on the LHS


  • has all the variables on the LHS different between each other


  • has the least possible amount of equations



I'm unable to find a generic expression. Below I report my attempts which should also clarify what I'm trying to do.




My attempts:
For $r=2$, the original system is
$$
begin{cases}
b_3=b_1-b_2
end{cases}
$$
which simply can be written as
$$
begin{cases}
b_1=b_2+b_3
end{cases}
$$
which satisfies my requirements.




For $r=3$, the original system is
$$
begin{cases}
b_4=b_1-b_2\
b_5=b_1-b_3\
b_6=b_2-b_3\
end{cases} Leftrightarrow begin{cases}
b_1=b_2+b_4\
b_1=b_3+b_5\
b_2=b_3+b_6
end{cases} Leftrightarrow begin{cases}
b_1=b_3+b_6+b_4\
b_1=b_3+b_5\
b_2=b_3+b_6
end{cases} Leftrightarrow begin{cases}
b_1=b_2+b_4\
b_2=b_3+b_6\
b_5=b_6+b_4\
end{cases}
$$
which satisfies my requirements.




For $r=4$, the original system is
$$
begin{cases}
b_5=b_1-b_2\
b_6=b_1-b_3\
b_7=b_1-b_4\
b_8=b_2-b_3\
b_9=b_2-b_4\
b_{10}=b_3-b_4
end{cases} Leftrightarrow begin{cases}
b_1=b_2+b_5\
b_1=b_3+b_6\
b_1=b_4+b_7\
b_2=b_3+b_8\
b_2=b_4+b_9\
b_3=b_4+b_{10}
end{cases}Leftrightarrow begin{cases}
b_1=b_2+b_5\
b_1=b_4+b_{10}+b_6\
b_1=b_4+b_7\
b_2=b_4+b_{10}+b_8\
b_2=b_4+b_9\
b_3=b_4+b_{10}
end{cases}Leftrightarrow begin{cases}
b_1=b_2+b_5\
b_2=b_3+b_8\
b_3=b_4+b_{10}\
b_7=b_{10}+b_6\
b_9=b_{10}+b_8\
end{cases}
$$
which satisfies my requirements. In principle, as I keep increasing $r$, I should be able to find some path, but I don't see anything!






linear-algebra combinatorics linear-transformations systems-of-equations







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Fix $r$ and $dequiv r+binom{r}{2}$ and consider $d$ real numbers $b_1,b_2,..., b_d$.



    I have the following system of linear equations:
    $$
    begin{cases}
    text{$(diamond)$ Differences between the elements ${b_1,b_h}$ for $h=2,...,r$ $hspace{1cm}$ ($r-1$ elements)}\
    b_{r+1}=b_1-b_2\
    b_{r+2}=b_1-b_3\
    ...\
    b_{2r-1}=b_1-b_r\
    -------\
    text{$(diamond)$ Differences between the elements ${b_2,b_h}$ for $h=3,...,r$ $hspace{1cm}$ ($r-2$ elements)}\
    b_{2r}=b_2-b_3\
    ...\
    b_{3r-3}=b_2-b_r\
    -------\
    text{$(diamond)$ Etc. }\
    ....\
    -------\
    text{$(diamond)$ Differences between the elements ${b_{r-1},b_h}$ for $h=r$ $hspace{1cm}$ ($1$ element)} \
    b_d=b_{r-1}-b_r
    end{cases}
    $$



    Question: For any $r$, I want to re-write in an EQUIVALENT way such that it




    • has only relations with the $+$ sign on the RHS of each equation


    • has only 1 variable on the LHS


    • has all the variables on the LHS different between each other


    • has the least possible amount of equations



    I'm unable to find a generic expression. Below I report my attempts which should also clarify what I'm trying to do.




    My attempts:
    For $r=2$, the original system is
    $$
    begin{cases}
    b_3=b_1-b_2
    end{cases}
    $$
    which simply can be written as
    $$
    begin{cases}
    b_1=b_2+b_3
    end{cases}
    $$
    which satisfies my requirements.




    For $r=3$, the original system is
    $$
    begin{cases}
    b_4=b_1-b_2\
    b_5=b_1-b_3\
    b_6=b_2-b_3\
    end{cases} Leftrightarrow begin{cases}
    b_1=b_2+b_4\
    b_1=b_3+b_5\
    b_2=b_3+b_6
    end{cases} Leftrightarrow begin{cases}
    b_1=b_3+b_6+b_4\
    b_1=b_3+b_5\
    b_2=b_3+b_6
    end{cases} Leftrightarrow begin{cases}
    b_1=b_2+b_4\
    b_2=b_3+b_6\
    b_5=b_6+b_4\
    end{cases}
    $$
    which satisfies my requirements.




    For $r=4$, the original system is
    $$
    begin{cases}
    b_5=b_1-b_2\
    b_6=b_1-b_3\
    b_7=b_1-b_4\
    b_8=b_2-b_3\
    b_9=b_2-b_4\
    b_{10}=b_3-b_4
    end{cases} Leftrightarrow begin{cases}
    b_1=b_2+b_5\
    b_1=b_3+b_6\
    b_1=b_4+b_7\
    b_2=b_3+b_8\
    b_2=b_4+b_9\
    b_3=b_4+b_{10}
    end{cases}Leftrightarrow begin{cases}
    b_1=b_2+b_5\
    b_1=b_4+b_{10}+b_6\
    b_1=b_4+b_7\
    b_2=b_4+b_{10}+b_8\
    b_2=b_4+b_9\
    b_3=b_4+b_{10}
    end{cases}Leftrightarrow begin{cases}
    b_1=b_2+b_5\
    b_2=b_3+b_8\
    b_3=b_4+b_{10}\
    b_7=b_{10}+b_6\
    b_9=b_{10}+b_8\
    end{cases}
    $$
    which satisfies my requirements. In principle, as I keep increasing $r$, I should be able to find some path, but I don't see anything!






    linear-algebra combinatorics linear-transformations systems-of-equations







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Fix $r$ and $dequiv r+binom{r}{2}$ and consider $d$ real numbers $b_1,b_2,..., b_d$.



      I have the following system of linear equations:
      $$
      begin{cases}
      text{$(diamond)$ Differences between the elements ${b_1,b_h}$ for $h=2,...,r$ $hspace{1cm}$ ($r-1$ elements)}\
      b_{r+1}=b_1-b_2\
      b_{r+2}=b_1-b_3\
      ...\
      b_{2r-1}=b_1-b_r\
      -------\
      text{$(diamond)$ Differences between the elements ${b_2,b_h}$ for $h=3,...,r$ $hspace{1cm}$ ($r-2$ elements)}\
      b_{2r}=b_2-b_3\
      ...\
      b_{3r-3}=b_2-b_r\
      -------\
      text{$(diamond)$ Etc. }\
      ....\
      -------\
      text{$(diamond)$ Differences between the elements ${b_{r-1},b_h}$ for $h=r$ $hspace{1cm}$ ($1$ element)} \
      b_d=b_{r-1}-b_r
      end{cases}
      $$



      Question: For any $r$, I want to re-write in an EQUIVALENT way such that it




      • has only relations with the $+$ sign on the RHS of each equation


      • has only 1 variable on the LHS


      • has all the variables on the LHS different between each other


      • has the least possible amount of equations



      I'm unable to find a generic expression. Below I report my attempts which should also clarify what I'm trying to do.




      My attempts:
      For $r=2$, the original system is
      $$
      begin{cases}
      b_3=b_1-b_2
      end{cases}
      $$
      which simply can be written as
      $$
      begin{cases}
      b_1=b_2+b_3
      end{cases}
      $$
      which satisfies my requirements.




      For $r=3$, the original system is
      $$
      begin{cases}
      b_4=b_1-b_2\
      b_5=b_1-b_3\
      b_6=b_2-b_3\
      end{cases} Leftrightarrow begin{cases}
      b_1=b_2+b_4\
      b_1=b_3+b_5\
      b_2=b_3+b_6
      end{cases} Leftrightarrow begin{cases}
      b_1=b_3+b_6+b_4\
      b_1=b_3+b_5\
      b_2=b_3+b_6
      end{cases} Leftrightarrow begin{cases}
      b_1=b_2+b_4\
      b_2=b_3+b_6\
      b_5=b_6+b_4\
      end{cases}
      $$
      which satisfies my requirements.




      For $r=4$, the original system is
      $$
      begin{cases}
      b_5=b_1-b_2\
      b_6=b_1-b_3\
      b_7=b_1-b_4\
      b_8=b_2-b_3\
      b_9=b_2-b_4\
      b_{10}=b_3-b_4
      end{cases} Leftrightarrow begin{cases}
      b_1=b_2+b_5\
      b_1=b_3+b_6\
      b_1=b_4+b_7\
      b_2=b_3+b_8\
      b_2=b_4+b_9\
      b_3=b_4+b_{10}
      end{cases}Leftrightarrow begin{cases}
      b_1=b_2+b_5\
      b_1=b_4+b_{10}+b_6\
      b_1=b_4+b_7\
      b_2=b_4+b_{10}+b_8\
      b_2=b_4+b_9\
      b_3=b_4+b_{10}
      end{cases}Leftrightarrow begin{cases}
      b_1=b_2+b_5\
      b_2=b_3+b_8\
      b_3=b_4+b_{10}\
      b_7=b_{10}+b_6\
      b_9=b_{10}+b_8\
      end{cases}
      $$
      which satisfies my requirements. In principle, as I keep increasing $r$, I should be able to find some path, but I don't see anything!






      linear-algebra combinatorics linear-transformations systems-of-equations







      share|cite|improve this question











      $endgroup$




      Fix $r$ and $dequiv r+binom{r}{2}$ and consider $d$ real numbers $b_1,b_2,..., b_d$.



      I have the following system of linear equations:
      $$
      begin{cases}
      text{$(diamond)$ Differences between the elements ${b_1,b_h}$ for $h=2,...,r$ $hspace{1cm}$ ($r-1$ elements)}\
      b_{r+1}=b_1-b_2\
      b_{r+2}=b_1-b_3\
      ...\
      b_{2r-1}=b_1-b_r\
      -------\
      text{$(diamond)$ Differences between the elements ${b_2,b_h}$ for $h=3,...,r$ $hspace{1cm}$ ($r-2$ elements)}\
      b_{2r}=b_2-b_3\
      ...\
      b_{3r-3}=b_2-b_r\
      -------\
      text{$(diamond)$ Etc. }\
      ....\
      -------\
      text{$(diamond)$ Differences between the elements ${b_{r-1},b_h}$ for $h=r$ $hspace{1cm}$ ($1$ element)} \
      b_d=b_{r-1}-b_r
      end{cases}
      $$



      Question: For any $r$, I want to re-write in an EQUIVALENT way such that it




      • has only relations with the $+$ sign on the RHS of each equation


      • has only 1 variable on the LHS


      • has all the variables on the LHS different between each other


      • has the least possible amount of equations



      I'm unable to find a generic expression. Below I report my attempts which should also clarify what I'm trying to do.




      My attempts:
      For $r=2$, the original system is
      $$
      begin{cases}
      b_3=b_1-b_2
      end{cases}
      $$
      which simply can be written as
      $$
      begin{cases}
      b_1=b_2+b_3
      end{cases}
      $$
      which satisfies my requirements.




      For $r=3$, the original system is
      $$
      begin{cases}
      b_4=b_1-b_2\
      b_5=b_1-b_3\
      b_6=b_2-b_3\
      end{cases} Leftrightarrow begin{cases}
      b_1=b_2+b_4\
      b_1=b_3+b_5\
      b_2=b_3+b_6
      end{cases} Leftrightarrow begin{cases}
      b_1=b_3+b_6+b_4\
      b_1=b_3+b_5\
      b_2=b_3+b_6
      end{cases} Leftrightarrow begin{cases}
      b_1=b_2+b_4\
      b_2=b_3+b_6\
      b_5=b_6+b_4\
      end{cases}
      $$
      which satisfies my requirements.




      For $r=4$, the original system is
      $$
      begin{cases}
      b_5=b_1-b_2\
      b_6=b_1-b_3\
      b_7=b_1-b_4\
      b_8=b_2-b_3\
      b_9=b_2-b_4\
      b_{10}=b_3-b_4
      end{cases} Leftrightarrow begin{cases}
      b_1=b_2+b_5\
      b_1=b_3+b_6\
      b_1=b_4+b_7\
      b_2=b_3+b_8\
      b_2=b_4+b_9\
      b_3=b_4+b_{10}
      end{cases}Leftrightarrow begin{cases}
      b_1=b_2+b_5\
      b_1=b_4+b_{10}+b_6\
      b_1=b_4+b_7\
      b_2=b_4+b_{10}+b_8\
      b_2=b_4+b_9\
      b_3=b_4+b_{10}
      end{cases}Leftrightarrow begin{cases}
      b_1=b_2+b_5\
      b_2=b_3+b_8\
      b_3=b_4+b_{10}\
      b_7=b_{10}+b_6\
      b_9=b_{10}+b_8\
      end{cases}
      $$
      which satisfies my requirements. In principle, as I keep increasing $r$, I should be able to find some path, but I don't see anything!






      linear-algebra combinatorics linear-transformations systems-of-equations




      linear-algebra combinatorics linear-transformations systems-of-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 17 at 16:14







      STF

















      asked Jan 16 at 17:01









      STFSTF

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