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Deriving the solutions of a system of linear equations
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$begingroup$
Let $Kin mathbb{N}$ with $Kgeq 2$ and consider the system of equations
$$
begin{cases}
sum_{j=1}^h y_j *x_{K+j-h}=sum_{j=1}^h x_j *y_{K+j-h} & text{ for }h=1,...,K-1\
sum_{j=1}^Ky_j=1\
end{cases}
$$
that is
$$
begin{cases}
y_1*x_K=x_1*y_K\
y_1*x_{K-1}+y_2*x_K=x_1*y_{K-1}+x_2*y_K\
y_1*x_{K-2}+y_2*x_{K-1}+y_3*x_K=x_1*y_{K-2}+x_2*y_{K-1}+x_3*y_K\
...\
y_1*x_2+y_2*x_3+...+y_{K-1}*x_K=x_1*y_2+x_2*y_3+...+x_{K-1}*y_K\
y_1+y_2+...+y_K=1\
end{cases}
$$
For example, when $K=3$, the system becomes
$$
begin{cases}
y_1*x_3=x_1*y_3\
y_1*x_2+y_2*x_3=x_1*y_2+x_2*y_3\
y_1+y_2+y_3=1\
end{cases}
$$
Note:
The unknowns of the system are $y_1,...,y_K$, while $x_1,...,x_K$ are treated as parameters.
I assume that $y_j>0,x_j>0$ for $j=1,...,K$.
We know that the parameters $x_1,...,x_K$ are such that $sum_{j=1}^K x_j=1$
Question: I want to show that the solutions of the system above are
$$
begin{cases}
(a) text{ }y_j=x_j & text{ for }j=1,...,K\
text{or }\
(b) text{ }y_j=y_{K+1-j} & text{ for }j=1,...,lfloorfrac{K+1}{2} rfloor
end{cases}
$$
where $lfloor A rfloor$ denotes the the largest integer less than or equal to $A$.
What I have tried (incomplete): I have the proof when $K=3$ (very simple).
$$
begin{cases}
y_1*x_3=x_1*y_3\
y_1*x_2+y_2*x_3=x_1*y_2+x_2*y_3\
y_1+y_2+y_3=1\
end{cases} Rightarrow begin{cases}
frac{x_3}{x_1}=frac{y_3}{y_1}\
frac{x_2}{x_1}+frac{y_2}{y_1}frac{x_3}{x_1}=frac{y_2}{y_1}+frac{y_3}{y_1}frac{x_2}{x_1}
end{cases}
$$
$$Rightarrow frac{x_2}{x_1}+frac{y_2}{y_1}frac{y_3}{y_1}=frac{y_2}{y_1}+frac{y_3}{y_1}frac{x_2}{x_1}Leftrightarrow Big( frac{y_2}{y_1}-frac{x_2}{x_1}Big)Big(frac{y_3}{y_1}-1 Big)=0
$$
$$
Leftrightarrow text{ }y_3=y_1 text{ or } frac{y_2}{y_1}=frac{x_2}{x_1}
$$
If $y_3=y_1$ then we are in case (b) of the general claim above.
If $ frac{y_2}{y_1}=frac{x_2}{x_1}$, then the system becomes
$$
begin{cases}
frac{y_3}{y_1}=frac{x_3}{x_1}\
frac{y_2}{y_1}=frac{x_2}{x_1}\
y_1+y_2+y_3=1\
end{cases} Rightarrow y_1+frac{x_2}{x_1}y_1+frac{x_3}{x_1}y_1=1
$$
$$
Leftrightarrow y_1(frac{overbrace{x_1+x_2+x_3}^{=1}}{x_1})=1 Leftrightarrow y_1=x_1 Rightarrow y_3=x_3, y_2=x_2
$$
that is case (a) of the general claim above. I am struggling to generalise this to any $K$ because I'm very weak in linear algebra. Any help from your side would be greatly appreciated.
linear-algebra systems-of-equations linear-programming
asked Mar 25 at 17:08
STFSTF
671422
$endgroup$
add a comment |
$begingroup$
Let $Kin mathbb{N}$ with $Kgeq 2$ and consider the system of equations
$$
begin{cases}
sum_{j=1}^h y_j *x_{K+j-h}=sum_{j=1}^h x_j *y_{K+j-h} & text{ for }h=1,...,K-1\
sum_{j=1}^Ky_j=1\
end{cases}
$$
that is
$$
begin{cases}
y_1*x_K=x_1*y_K\
y_1*x_{K-1}+y_2*x_K=x_1*y_{K-1}+x_2*y_K\
y_1*x_{K-2}+y_2*x_{K-1}+y_3*x_K=x_1*y_{K-2}+x_2*y_{K-1}+x_3*y_K\
...\
y_1*x_2+y_2*x_3+...+y_{K-1}*x_K=x_1*y_2+x_2*y_3+...+x_{K-1}*y_K\
y_1+y_2+...+y_K=1\
end{cases}
$$
For example, when $K=3$, the system becomes
$$
begin{cases}
y_1*x_3=x_1*y_3\
y_1*x_2+y_2*x_3=x_1*y_2+x_2*y_3\
y_1+y_2+y_3=1\
end{cases}
$$
Note:
The unknowns of the system are $y_1,...,y_K$, while $x_1,...,x_K$ are treated as parameters.
I assume that $y_j>0,x_j>0$ for $j=1,...,K$.
We know that the parameters $x_1,...,x_K$ are such that $sum_{j=1}^K x_j=1$
Question: I want to show that the solutions of the system above are
$$
begin{cases}
(a) text{ }y_j=x_j & text{ for }j=1,...,K\
text{or }\
(b) text{ }y_j=y_{K+1-j} & text{ for }j=1,...,lfloorfrac{K+1}{2} rfloor
end{cases}
$$
where $lfloor A rfloor$ denotes the the largest integer less than or equal to $A$.
What I have tried (incomplete): I have the proof when $K=3$ (very simple).
$$
begin{cases}
y_1*x_3=x_1*y_3\
y_1*x_2+y_2*x_3=x_1*y_2+x_2*y_3\
y_1+y_2+y_3=1\
end{cases} Rightarrow begin{cases}
frac{x_3}{x_1}=frac{y_3}{y_1}\
frac{x_2}{x_1}+frac{y_2}{y_1}frac{x_3}{x_1}=frac{y_2}{y_1}+frac{y_3}{y_1}frac{x_2}{x_1}
end{cases}
$$
$$Rightarrow frac{x_2}{x_1}+frac{y_2}{y_1}frac{y_3}{y_1}=frac{y_2}{y_1}+frac{y_3}{y_1}frac{x_2}{x_1}Leftrightarrow Big( frac{y_2}{y_1}-frac{x_2}{x_1}Big)Big(frac{y_3}{y_1}-1 Big)=0
$$
$$
Leftrightarrow text{ }y_3=y_1 text{ or } frac{y_2}{y_1}=frac{x_2}{x_1}
$$
If $y_3=y_1$ then we are in case (b) of the general claim above.
If $ frac{y_2}{y_1}=frac{x_2}{x_1}$, then the system becomes
$$
begin{cases}
frac{y_3}{y_1}=frac{x_3}{x_1}\
frac{y_2}{y_1}=frac{x_2}{x_1}\
y_1+y_2+y_3=1\
end{cases} Rightarrow y_1+frac{x_2}{x_1}y_1+frac{x_3}{x_1}y_1=1
$$
$$
Leftrightarrow y_1(frac{overbrace{x_1+x_2+x_3}^{=1}}{x_1})=1 Leftrightarrow y_1=x_1 Rightarrow y_3=x_3, y_2=x_2
$$
that is case (a) of the general claim above. I am struggling to generalise this to any $K$ because I'm very weak in linear algebra. Any help from your side would be greatly appreciated.
linear-algebra systems-of-equations linear-programming
asked Mar 25 at 17:08
STFSTF
671422
$endgroup$
$begingroup$
Why are you interested in this question?
$endgroup$
– James
Mar 25 at 17:31
$begingroup$
It is part of a proof I have to develop.
$endgroup$
– STF
Mar 25 at 18:21
$begingroup$
(b) is not a solution to your system, but just a symmetry that your system exhibits. If you know half the $y$ values, it will tell you the other half, but it doesn't tell you that first half.
$endgroup$
– Paul Sinclair
Mar 26 at 4:21
$begingroup$
You can show easily that $y_i = x_i$ for all $i$ is a solution. If you can prove that your $K$ equations in the $K$ unknowns are all independent (none is a linear combination of the others), then it is the only solution, so you are done.
$endgroup$
– Paul Sinclair
Mar 26 at 4:30
add a comment |
$begingroup$
Let $Kin mathbb{N}$ with $Kgeq 2$ and consider the system of equations
$$
begin{cases}
sum_{j=1}^h y_j *x_{K+j-h}=sum_{j=1}^h x_j *y_{K+j-h} & text{ for }h=1,...,K-1\
sum_{j=1}^Ky_j=1\
end{cases}
$$
that is
$$
begin{cases}
y_1*x_K=x_1*y_K\
y_1*x_{K-1}+y_2*x_K=x_1*y_{K-1}+x_2*y_K\
y_1*x_{K-2}+y_2*x_{K-1}+y_3*x_K=x_1*y_{K-2}+x_2*y_{K-1}+x_3*y_K\
...\
y_1*x_2+y_2*x_3+...+y_{K-1}*x_K=x_1*y_2+x_2*y_3+...+x_{K-1}*y_K\
y_1+y_2+...+y_K=1\
end{cases}
$$
For example, when $K=3$, the system becomes
$$
begin{cases}
y_1*x_3=x_1*y_3\
y_1*x_2+y_2*x_3=x_1*y_2+x_2*y_3\
y_1+y_2+y_3=1\
end{cases}
$$
Note:
The unknowns of the system are $y_1,...,y_K$, while $x_1,...,x_K$ are treated as parameters.
I assume that $y_j>0,x_j>0$ for $j=1,...,K$.
We know that the parameters $x_1,...,x_K$ are such that $sum_{j=1}^K x_j=1$
Question: I want to show that the solutions of the system above are
$$
begin{cases}
(a) text{ }y_j=x_j & text{ for }j=1,...,K\
text{or }\
(b) text{ }y_j=y_{K+1-j} & text{ for }j=1,...,lfloorfrac{K+1}{2} rfloor
end{cases}
$$
where $lfloor A rfloor$ denotes the the largest integer less than or equal to $A$.
What I have tried (incomplete): I have the proof when $K=3$ (very simple).
$$
begin{cases}
y_1*x_3=x_1*y_3\
y_1*x_2+y_2*x_3=x_1*y_2+x_2*y_3\
y_1+y_2+y_3=1\
end{cases} Rightarrow begin{cases}
frac{x_3}{x_1}=frac{y_3}{y_1}\
frac{x_2}{x_1}+frac{y_2}{y_1}frac{x_3}{x_1}=frac{y_2}{y_1}+frac{y_3}{y_1}frac{x_2}{x_1}
end{cases}
$$
$$Rightarrow frac{x_2}{x_1}+frac{y_2}{y_1}frac{y_3}{y_1}=frac{y_2}{y_1}+frac{y_3}{y_1}frac{x_2}{x_1}Leftrightarrow Big( frac{y_2}{y_1}-frac{x_2}{x_1}Big)Big(frac{y_3}{y_1}-1 Big)=0
$$
$$
Leftrightarrow text{ }y_3=y_1 text{ or } frac{y_2}{y_1}=frac{x_2}{x_1}
$$
If $y_3=y_1$ then we are in case (b) of the general claim above.
If $ frac{y_2}{y_1}=frac{x_2}{x_1}$, then the system becomes
$$
begin{cases}
frac{y_3}{y_1}=frac{x_3}{x_1}\
frac{y_2}{y_1}=frac{x_2}{x_1}\
y_1+y_2+y_3=1\
end{cases} Rightarrow y_1+frac{x_2}{x_1}y_1+frac{x_3}{x_1}y_1=1
$$
$$
Leftrightarrow y_1(frac{overbrace{x_1+x_2+x_3}^{=1}}{x_1})=1 Leftrightarrow y_1=x_1 Rightarrow y_3=x_3, y_2=x_2
$$
that is case (a) of the general claim above. I am struggling to generalise this to any $K$ because I'm very weak in linear algebra. Any help from your side would be greatly appreciated.
linear-algebra systems-of-equations linear-programming
asked Mar 25 at 17:08
STFSTF
671422
$endgroup$
Let $Kin mathbb{N}$ with $Kgeq 2$ and consider the system of equations
$$
begin{cases}
sum_{j=1}^h y_j *x_{K+j-h}=sum_{j=1}^h x_j *y_{K+j-h} & text{ for }h=1,...,K-1\
sum_{j=1}^Ky_j=1\
end{cases}
$$
that is
$$
begin{cases}
y_1*x_K=x_1*y_K\
y_1*x_{K-1}+y_2*x_K=x_1*y_{K-1}+x_2*y_K\
y_1*x_{K-2}+y_2*x_{K-1}+y_3*x_K=x_1*y_{K-2}+x_2*y_{K-1}+x_3*y_K\
...\
y_1*x_2+y_2*x_3+...+y_{K-1}*x_K=x_1*y_2+x_2*y_3+...+x_{K-1}*y_K\
y_1+y_2+...+y_K=1\
end{cases}
$$
For example, when $K=3$, the system becomes
$$
begin{cases}
y_1*x_3=x_1*y_3\
y_1*x_2+y_2*x_3=x_1*y_2+x_2*y_3\
y_1+y_2+y_3=1\
end{cases}
$$
Note:
The unknowns of the system are $y_1,...,y_K$, while $x_1,...,x_K$ are treated as parameters.
I assume that $y_j>0,x_j>0$ for $j=1,...,K$.
We know that the parameters $x_1,...,x_K$ are such that $sum_{j=1}^K x_j=1$
Question: I want to show that the solutions of the system above are
$$
begin{cases}
(a) text{ }y_j=x_j & text{ for }j=1,...,K\
text{or }\
(b) text{ }y_j=y_{K+1-j} & text{ for }j=1,...,lfloorfrac{K+1}{2} rfloor
end{cases}
$$
where $lfloor A rfloor$ denotes the the largest integer less than or equal to $A$.
What I have tried (incomplete): I have the proof when $K=3$ (very simple).
$$
begin{cases}
y_1*x_3=x_1*y_3\
y_1*x_2+y_2*x_3=x_1*y_2+x_2*y_3\
y_1+y_2+y_3=1\
end{cases} Rightarrow begin{cases}
frac{x_3}{x_1}=frac{y_3}{y_1}\
frac{x_2}{x_1}+frac{y_2}{y_1}frac{x_3}{x_1}=frac{y_2}{y_1}+frac{y_3}{y_1}frac{x_2}{x_1}
end{cases}
$$
$$Rightarrow frac{x_2}{x_1}+frac{y_2}{y_1}frac{y_3}{y_1}=frac{y_2}{y_1}+frac{y_3}{y_1}frac{x_2}{x_1}Leftrightarrow Big( frac{y_2}{y_1}-frac{x_2}{x_1}Big)Big(frac{y_3}{y_1}-1 Big)=0
$$
$$
Leftrightarrow text{ }y_3=y_1 text{ or } frac{y_2}{y_1}=frac{x_2}{x_1}
$$
If $y_3=y_1$ then we are in case (b) of the general claim above.
If $ frac{y_2}{y_1}=frac{x_2}{x_1}$, then the system becomes
$$
begin{cases}
frac{y_3}{y_1}=frac{x_3}{x_1}\
frac{y_2}{y_1}=frac{x_2}{x_1}\
y_1+y_2+y_3=1\
end{cases} Rightarrow y_1+frac{x_2}{x_1}y_1+frac{x_3}{x_1}y_1=1
$$
$$
Leftrightarrow y_1(frac{overbrace{x_1+x_2+x_3}^{=1}}{x_1})=1 Leftrightarrow y_1=x_1 Rightarrow y_3=x_3, y_2=x_2
$$
that is case (a) of the general claim above. I am struggling to generalise this to any $K$ because I'm very weak in linear algebra. Any help from your side would be greatly appreciated.
linear-algebra systems-of-equations linear-programming
linear-algebra systems-of-equations linear-programming
asked Mar 25 at 17:08
STFSTF
671422
asked Mar 25 at 17:08
STFSTF
671422
asked Mar 25 at 17:08
STFSTF
671422
asked Mar 25 at 17:08
STFSTF
671422
asked Mar 25 at 17:08
STFSTF
671422
671422
$begingroup$
Why are you interested in this question?
$endgroup$
– James
Mar 25 at 17:31
$begingroup$
It is part of a proof I have to develop.
$endgroup$
– STF
Mar 25 at 18:21
$begingroup$
(b) is not a solution to your system, but just a symmetry that your system exhibits. If you know half the $y$ values, it will tell you the other half, but it doesn't tell you that first half.
$endgroup$
– Paul Sinclair
Mar 26 at 4:21
$begingroup$
You can show easily that $y_i = x_i$ for all $i$ is a solution. If you can prove that your $K$ equations in the $K$ unknowns are all independent (none is a linear combination of the others), then it is the only solution, so you are done.
$endgroup$
– Paul Sinclair
Mar 26 at 4:30
add a comment |
$begingroup$
Why are you interested in this question?
$endgroup$
– James
Mar 25 at 17:31
$begingroup$
It is part of a proof I have to develop.
$endgroup$
– STF
Mar 25 at 18:21
$begingroup$
(b) is not a solution to your system, but just a symmetry that your system exhibits. If you know half the $y$ values, it will tell you the other half, but it doesn't tell you that first half.
$endgroup$
– Paul Sinclair
Mar 26 at 4:21
$begingroup$
You can show easily that $y_i = x_i$ for all $i$ is a solution. If you can prove that your $K$ equations in the $K$ unknowns are all independent (none is a linear combination of the others), then it is the only solution, so you are done.
$endgroup$
– Paul Sinclair
Mar 26 at 4:30
$begingroup$
Why are you interested in this question?
$endgroup$
– James
Mar 25 at 17:31
$begingroup$
Why are you interested in this question?
$endgroup$
– James
Mar 25 at 17:31
$begingroup$
It is part of a proof I have to develop.
$endgroup$
– STF
Mar 25 at 18:21
$begingroup$
It is part of a proof I have to develop.
$endgroup$
– STF
Mar 25 at 18:21
$begingroup$
(b) is not a solution to your system, but just a symmetry that your system exhibits. If you know half the $y$ values, it will tell you the other half, but it doesn't tell you that first half.
$endgroup$
– Paul Sinclair
Mar 26 at 4:21
$begingroup$
(b) is not a solution to your system, but just a symmetry that your system exhibits. If you know half the $y$ values, it will tell you the other half, but it doesn't tell you that first half.
$endgroup$
– Paul Sinclair
Mar 26 at 4:21
$begingroup$
You can show easily that $y_i = x_i$ for all $i$ is a solution. If you can prove that your $K$ equations in the $K$ unknowns are all independent (none is a linear combination of the others), then it is the only solution, so you are done.
$endgroup$
– Paul Sinclair
Mar 26 at 4:30
$begingroup$
You can show easily that $y_i = x_i$ for all $i$ is a solution. If you can prove that your $K$ equations in the $K$ unknowns are all independent (none is a linear combination of the others), then it is the only solution, so you are done.
$endgroup$
– Paul Sinclair
Mar 26 at 4:30
add a comment |
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$begingroup$
Why are you interested in this question?
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– James
Mar 25 at 17:31
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It is part of a proof I have to develop.
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– STF
Mar 25 at 18:21
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(b) is not a solution to your system, but just a symmetry that your system exhibits. If you know half the $y$ values, it will tell you the other half, but it doesn't tell you that first half.
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– Paul Sinclair
Mar 26 at 4:21
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You can show easily that $y_i = x_i$ for all $i$ is a solution. If you can prove that your $K$ equations in the $K$ unknowns are all independent (none is a linear combination of the others), then it is the only solution, so you are done.
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– Paul Sinclair
Mar 26 at 4:30