Rewrite a system of linear equations
$begingroup$
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
asked Jan 16 at 17:01
STFSTF
11420
$endgroup$
add a comment |
$begingroup$
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
asked Jan 16 at 17:01
STFSTF
11420
$endgroup$
add a comment |
$begingroup$
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
asked Jan 16 at 17:01
STFSTF
11420
$endgroup$
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
asked Jan 16 at 17:01
STFSTF
11420
asked Jan 16 at 17:01
STFSTF
11420
edited Jan 17 at 16:14
STF
asked Jan 16 at 17:01
STFSTF
11420
asked Jan 16 at 17:01
STFSTF
11420
asked Jan 16 at 17:01
STFSTF
11420
11420
add a comment |
add a comment |
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