Simplify the matrix of a linear system knowing that some of the solutions are equal
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In order to improve the efficiency of my python program I'm trying to take advantage of some properties of a linear system I need to solve.
I have a linear system $Ax = b$ and I know beforehand that some of the solutions are equal, though I don't know their values before solving the system (i.e. $x_1 = x_{10}$, $x_3 = x_5$, ...).
Is there any way I can simplify my system ($A$ and b) for faster solving? My first thought was to eliminate one of the two unknowns for each pair and to do the sum of both columns associated with the elements of the pair, weighted by 1/2, but I'm not sure at all that this is mathematically correct.
linear-algebra matrices systems-of-equations
asked Dec 15 '18 at 17:08
FredVFredV
876
$endgroup$
add a comment |
$begingroup$
In order to improve the efficiency of my python program I'm trying to take advantage of some properties of a linear system I need to solve.
I have a linear system $Ax = b$ and I know beforehand that some of the solutions are equal, though I don't know their values before solving the system (i.e. $x_1 = x_{10}$, $x_3 = x_5$, ...).
Is there any way I can simplify my system ($A$ and b) for faster solving? My first thought was to eliminate one of the two unknowns for each pair and to do the sum of both columns associated with the elements of the pair, weighted by 1/2, but I'm not sure at all that this is mathematically correct.
linear-algebra matrices systems-of-equations
asked Dec 15 '18 at 17:08
FredVFredV
876
$endgroup$
1
$begingroup$
Don't weight by $1/2$ !
$endgroup$
– Damien
Dec 15 '18 at 18:38
1
$begingroup$
You don't have to modify $b$
$endgroup$
– Damien
Dec 15 '18 at 18:45
add a comment |
$begingroup$
In order to improve the efficiency of my python program I'm trying to take advantage of some properties of a linear system I need to solve.
I have a linear system $Ax = b$ and I know beforehand that some of the solutions are equal, though I don't know their values before solving the system (i.e. $x_1 = x_{10}$, $x_3 = x_5$, ...).
Is there any way I can simplify my system ($A$ and b) for faster solving? My first thought was to eliminate one of the two unknowns for each pair and to do the sum of both columns associated with the elements of the pair, weighted by 1/2, but I'm not sure at all that this is mathematically correct.
linear-algebra matrices systems-of-equations
asked Dec 15 '18 at 17:08
FredVFredV
876
$endgroup$
In order to improve the efficiency of my python program I'm trying to take advantage of some properties of a linear system I need to solve.
I have a linear system $Ax = b$ and I know beforehand that some of the solutions are equal, though I don't know their values before solving the system (i.e. $x_1 = x_{10}$, $x_3 = x_5$, ...).
Is there any way I can simplify my system ($A$ and b) for faster solving? My first thought was to eliminate one of the two unknowns for each pair and to do the sum of both columns associated with the elements of the pair, weighted by 1/2, but I'm not sure at all that this is mathematically correct.
linear-algebra matrices systems-of-equations
linear-algebra matrices systems-of-equations
asked Dec 15 '18 at 17:08
FredVFredV
876
asked Dec 15 '18 at 17:08
FredVFredV
876
asked Dec 15 '18 at 17:08
FredVFredV
876
asked Dec 15 '18 at 17:08
FredVFredV
876
asked Dec 15 '18 at 17:08
FredVFredV
876
876
1
$begingroup$
Don't weight by $1/2$ !
$endgroup$
– Damien
Dec 15 '18 at 18:38
1
$begingroup$
You don't have to modify $b$
$endgroup$
– Damien
Dec 15 '18 at 18:45
add a comment |
1
$begingroup$
Don't weight by $1/2$ !
$endgroup$
– Damien
Dec 15 '18 at 18:38
1
$begingroup$
You don't have to modify $b$
$endgroup$
– Damien
Dec 15 '18 at 18:45
1
1
$begingroup$
Don't weight by $1/2$ !
$endgroup$
– Damien
Dec 15 '18 at 18:38
$begingroup$
Don't weight by $1/2$ !
$endgroup$
– Damien
Dec 15 '18 at 18:38
1
1
$begingroup$
You don't have to modify $b$
$endgroup$
– Damien
Dec 15 '18 at 18:45
$begingroup$
You don't have to modify $b$
$endgroup$
– Damien
Dec 15 '18 at 18:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let us call $(c_i)_i$ the columns of$A$.
The equation can be written as:
$$b = sum_{i=1}^{n} x_ic_i $$
If for example we know that $x_1 = x_2$, the equation becomes:
$$b = x_2(c_1 + c_2) + sum_{i=3}^{n} x_ic_i $$
The procedure follows:
- Replace $c_2$ by $c_1 + c_2$
- Suppress $c_1$
- Remove $x_1$ from the set of unknown values
- Don't modify $b$
answered Dec 15 '18 at 19:50
DamienDamien
59714
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
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active
oldest
votes
$begingroup$
Let us call $(c_i)_i$ the columns of$A$.
The equation can be written as:
$$b = sum_{i=1}^{n} x_ic_i $$
If for example we know that $x_1 = x_2$, the equation becomes:
$$b = x_2(c_1 + c_2) + sum_{i=3}^{n} x_ic_i $$
The procedure follows:
- Replace $c_2$ by $c_1 + c_2$
- Suppress $c_1$
- Remove $x_1$ from the set of unknown values
- Don't modify $b$
answered Dec 15 '18 at 19:50
DamienDamien
59714
$endgroup$
add a comment |
$begingroup$
Let us call $(c_i)_i$ the columns of$A$.
The equation can be written as:
$$b = sum_{i=1}^{n} x_ic_i $$
If for example we know that $x_1 = x_2$, the equation becomes:
$$b = x_2(c_1 + c_2) + sum_{i=3}^{n} x_ic_i $$
The procedure follows:
- Replace $c_2$ by $c_1 + c_2$
- Suppress $c_1$
- Remove $x_1$ from the set of unknown values
- Don't modify $b$
answered Dec 15 '18 at 19:50
DamienDamien
59714
$endgroup$
add a comment |
$begingroup$
Let us call $(c_i)_i$ the columns of$A$.
The equation can be written as:
$$b = sum_{i=1}^{n} x_ic_i $$
If for example we know that $x_1 = x_2$, the equation becomes:
$$b = x_2(c_1 + c_2) + sum_{i=3}^{n} x_ic_i $$
The procedure follows:
- Replace $c_2$ by $c_1 + c_2$
- Suppress $c_1$
- Remove $x_1$ from the set of unknown values
- Don't modify $b$
answered Dec 15 '18 at 19:50
DamienDamien
59714
$endgroup$
Let us call $(c_i)_i$ the columns of$A$.
The equation can be written as:
$$b = sum_{i=1}^{n} x_ic_i $$
If for example we know that $x_1 = x_2$, the equation becomes:
$$b = x_2(c_1 + c_2) + sum_{i=3}^{n} x_ic_i $$
The procedure follows:
- Replace $c_2$ by $c_1 + c_2$
- Suppress $c_1$
- Remove $x_1$ from the set of unknown values
- Don't modify $b$
answered Dec 15 '18 at 19:50
DamienDamien
59714
answered Dec 15 '18 at 19:50
DamienDamien
59714
answered Dec 15 '18 at 19:50
DamienDamien
59714
answered Dec 15 '18 at 19:50
DamienDamien
59714
59714
add a comment |
add a comment |
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1
$begingroup$
Don't weight by $1/2$ !
$endgroup$
– Damien
Dec 15 '18 at 18:38
1
$begingroup$
You don't have to modify $b$
$endgroup$
– Damien
Dec 15 '18 at 18:45