Gauss–Seidel method — latest news, discussion & facts — IJR
About Gauss–Seidel method Also known as Gauss-Seidel method, Liebmann method, method of successive displacement
About Gauss–Seidel method In numerical linear algebra, the Gauss–Seidel method, also known as the Liebmann method or the method of successive displacement, is an iterative method used to solve a system of linear equations. It is named after the German mathematicians Carl Friedrich Gauss and Philipp Ludwig von Seidel. Though it can be applied to any matrix with non-zero elements on the diagonals, convergence is only guaranteed if the matrix is either strictly diagonally dominant, or symmetric and positive definite. It was only mentioned in a private letter from Gauss to his student Gerling in 1823. A publication was not delivered before 1874 by Seidel.
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In numerical linear algebra, the Gauss–Seidel method, also known as the Liebmann method or the method of successive displacement, is an iterative method used to solve a system of linear equations. It is named after the German mathematicians Carl Friedrich Gauss and Philipp Ludwig von Seidel. Though it can be applied to any matrix with non-zero elements on the diagonals, convergence is only guaranteed if the matrix is either strictly diagonally dominant, or symmetric and positive definite. It was only mentioned in a private letter from Gauss to his student Gerling in 1823. A publication was not delivered before 1874 by Seidel.
Article In numerical linear algebra, the Gauss–Seidel method, also known as the Liebmann method or the method of successive displacement, is an iterative method used to solve a system of linear equations. It is named after the German mathematicians Carl Friedrich Gauss and Philipp Ludwig von Seidel. Though it can be applied to any matrix with non-zero elements on the diagonals, convergence is only guaranteed if the matrix is either strictly diagonally dominant, or symmetric and positive definite. It was only mentioned in a private letter from Gauss to his student Gerling in 1823. A publication was not delivered before 1874 by Seidel.
Contents Description Let
A
x
=
b
{\textstyle \mathbf {A} \mathbf {x} =\mathbf {b} }
be a square system of n linear equations, where:
A
=
[
a
11
a
12
⋯
a
1
n
a
21
a
22
⋯
a
2
n
⋮
⋮
⋱
⋮
a
n
1
a
n
2
The solution is obtained iteratively via
L
x
(
k
+
1
)
=
b
−
U
x
(
k
)
,
{\displaystyle \mathbf {L} \mathbf {x} ^{(k+1)}=\mathbf {b} -\mathbf {U} \mathbf {x} ^{(k)},}
where the matrix
A
{\displaystyle \mathbf {A} }
is decomposed into a lower triangular component
L
{\displaystyle \mathbf {L} }
, and a strictly upper triangular component
U
{\displaystyle \mathbf {U} }
such that
A
=
L
However, by taking advantage of the triangular form of
L
{\displaystyle \mathbf {L} }
, the elements of
x
(
k
+
1
)
{\displaystyle \mathbf {x} ^{(k+1)}}
can be computed sequentially for each row
i
{\displaystyle i}
using forward substitution:
x
i
(
k
+
1
)
=
1
a
i
i
(
b
i
−
∑
j
=
Discussion The element-wise formula for the Gauss–Seidel method is related to that of the (iterative) Jacobi method, with an important difference:
In Gauss-Seidel, the computation of
x
(
k
+
1
)
{\displaystyle \mathbf {x} ^{(k+1)}}
uses the elements of
x
(
k
+
1
)
{\displaystyle \mathbf {x} ^{(k+1)}}
that have already been computed, and only the elements of
x
(
k
)
{\displaystyle \mathbf {x} ^{(k)}}
that have not been computed in the
(
k
+
1
)
{\displaystyle (k+1)}
Convergence The convergence properties of the Gauss–Seidel method are dependent on the matrix
A
{\displaystyle \mathbf {A} }
. Namely, the procedure is known to converge if either:
A
{\displaystyle \mathbf {A} }
is symmetric positive-definite, or
A
{\displaystyle \mathbf {A} }
is strictly or irreducibly diagonally dominant.
The Gauss–Seidel method may converge even if these conditions are not satisfied.
Golub and Van Loan give a theorem for an algorithm that splits
A
{\displaystyle \mathbf {A} }
into two parts. Suppose
A
=
M
−
N
{\displaystyle \mathbf {A} =\mathbf {M} -\mathbf {N} }
is nonsingular. Let
r
=
ρ
Algorithm Since elements can be overwritten as they are computed in this algorithm, only one storage vector is needed, and vector indexing is omitted. The algorithm goes as follows:
algorithm Gauss–Seidel method is
inputs: A, b
output: φ
Choose an initial guess φ to the solution
repeat until convergence
for i from 1 until n do
σ ← 0
for j from 1 until n do
if j ≠ i then
σ ← σ + aijφj
end if
end (j-loop)
φi ← (bi − σ) / aii
end (i-loop)
check if convergence is reached
end (repeat)
An example for the matrix version A linear system shown as
A
x
=
b
{\displaystyle \mathbf {A} \mathbf {x} =\mathbf {b} }
is given by:
A
=
[
16
3
7
−
11
]
and
b
=
[
11
13
]
.
{\displaystyle \mathbf {A} ={\begin{bmatrix}16&3\\7&-11\\\end{bmatrix}}\quad {\text{and}}\quad \mathbf {b} ={\begin{bmatrix}11\\13\end{bmatrix}}.}
Use the equation
x
(
k
+
1
)
=
Another example for the matrix version Another linear system shown as
A
x
=
b
{\displaystyle \mathbf {A} \mathbf {x} =\mathbf {b} }
is given by:
A
=
[
2
3
5
7
]
and
b
=
[
11
13
]
.
{\displaystyle \mathbf {A} ={\begin{bmatrix}2&3\\5&7\\\end{bmatrix}}\quad {\text{and}}\quad \mathbf {b} ={\begin{bmatrix}11\\13\\\end{bmatrix}}.}
Use the equation
x
(
k
+
1
)
=
L
An example for the equation version Suppose given
n
{\displaystyle n}
equations and a starting point
x
0
{\displaystyle \mathbf {x} _{0}}
.
At any step in a Gauss-Seidel iteration, solve the first equation for
x
1
{\displaystyle x_{1}}
in terms of
x
2
,
…
,
x
n
{\displaystyle x_{2},\dots ,x_{n}}
; then solve the second equation for
x
2
{\displaystyle x_{2}}
in terms of
x
1
{\displaystyle x_{1}}
just found and the remaining
An example using Python and NumPy The following iterative procedure produces the solution vector of a linear system of equations:
Produces the output:
Program to solve arbitrary number of equations using Matlab The following code uses the formula
x
i
(
k
+
1
)
=
1
a
i
i
(
b
i
−
∑
j
<
i
a
i
j
x
j
(
k
+
1
)
−
∑
j
>
i
a
Article from Wikipedia (CC BY-SA 4.0), where it is maintained by volunteer editors.
{\displaystyle \mathbf {A} ={\begin{bmatrix}a_{11}&a_{12}&\cdots &a_{1n}\\a_{21}&a_{22}&\cdots &a_{2n}\\\vdots &\vdots &\ddots &\vdots \\a_{n1}&a_{n2}&\cdots &a_{nn}\end{bmatrix}},\qquad \mathbf {x} ={\begin{bmatrix}x_{1}\\x_{2}\\\vdots \\x_{n}\end{bmatrix}},\qquad \mathbf {b} ={\begin{bmatrix}b_{1}\\b_{2}\\\vdots \\b_{n}\end{bmatrix}}.}
{\displaystyle \mathbf {A} }
{\displaystyle \mathbf {b} }
{\displaystyle \mathbf {x} }
is unknown, the Gauss–Seidel method can be used to iteratively approximate
{\displaystyle \mathbf {x} }
{\displaystyle \mathbf {x} ^{(0)}}
denotes the initial guess for
{\displaystyle \mathbf {x} }
{\displaystyle \mathbf {x} _{i}^{(0)}=0}
{\displaystyle i=1,2,...,n}
{\displaystyle \mathbf {x} ^{(k)}}
-th approximation or iteration of
{\displaystyle \mathbf {x} }
{\displaystyle \mathbf {x} ^{(k+1)}}
{\displaystyle \mathbf {x} }
{\displaystyle \mathbf {A} =\mathbf {L} +\mathbf {U} }
. More specifically, the decomposition of
{\displaystyle \mathbf {A} }
{\displaystyle \mathbf {L} }
{\displaystyle \mathbf {U} }
{\displaystyle \mathbf {A} =\underbrace {\begin{bmatrix}a_{11}&0&\cdots &0\\a_{21}&a_{22}&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\a_{n1}&a_{n2}&\cdots &a_{nn}\end{bmatrix}} _{\textstyle \mathbf {L} }+\underbrace {\begin{bmatrix}0&a_{12}&\cdots &a_{1n}\\0&0&\cdots &a_{2n}\\\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &0\end{bmatrix}} _{\textstyle \mathbf {U} }.}
The system of linear equations may be rewritten as:
{\displaystyle {\begin{alignedat}{1}\mathbf {A} \mathbf {x} &=\mathbf {b} \\(\mathbf {L} +\mathbf {U} )\mathbf {x} &=\mathbf {b} \\\mathbf {L} \mathbf {x} +\mathbf {U} \mathbf {x} &=\mathbf {b} \\\mathbf {L} \mathbf {x} &=\mathbf {b} -\mathbf {U} \mathbf {x} \end{alignedat}}}
The Gauss–Seidel method now solves the left hand side of this expression for
{\displaystyle \mathbf {x} }
, using the previous value for
{\displaystyle \mathbf {x} }
on the right hand side. Analytically, this may be written as
{\displaystyle \mathbf {x} ^{(k+1)}=\mathbf {L} ^{-1}\left(\mathbf {b} -\mathbf {U} \mathbf {x} ^{(k)}\right).}
{\displaystyle x_{i}^{(k+1)}={\frac {1}{a_{ii}}}\left(b_{i}-\sum _{j=1}^{i-1}a_{ij}x_{j}^{(k+1)}-\sum _{j=i+1}^{n}a_{ij}x_{j}^{(k)}\right),\quad i=1,2,\dots ,n.}
Notice that the formula uses two summations per iteration which can be expressed as one summation
{\displaystyle \sum _{j\neq i}a_{ij}x_{j}}
that uses the most recently calculated iteration of
. The procedure is generally continued until the changes made by an iteration are below some tolerance, such as a sufficiently small residual.
-th iteration. This means that, unlike the Jacobi method, only one storage vector is required as elements can be overwritten as they are computed, which can be advantageous for very large problems.
However, unlike the Jacobi method, the computations for each element are generally much harder to implement in parallel, since they can have a very long critical path, and are thus most feasible for sparse matrices. Furthermore, the values at each iteration are dependent on the order of the original equations.
Gauss-Seidel is the same as successive over-relaxation with
{\displaystyle \omega =1}
{\displaystyle r=\rho (\mathbf {M} ^{-1}\mathbf {N} )}
be the spectral radius of
{\displaystyle \mathbf {M} ^{-1}\mathbf {N} }
{\displaystyle \mathbf {x} ^{(k)}}
{\displaystyle \mathbf {M} \mathbf {x} ^{(k+1)}=\mathbf {N} \mathbf {x} ^{(k)}+\mathbf {b} }
{\displaystyle \mathbf {x} =\mathbf {A} ^{-1}\mathbf {b} }
{\displaystyle \mathbf {x} ^{(0)}}
{\displaystyle \mathbf {M} }
{\displaystyle \mathbf {x} ^{(k+1)}=\mathbf {L} ^{-1}(\mathbf {b} -\mathbf {U} \mathbf {x} ^{(k)})}
{\displaystyle \mathbf {x} ^{(k+1)}=\mathbf {T} \mathbf {x} ^{(k)}+\mathbf {c} }
{\displaystyle \mathbf {T} =-\mathbf {L} ^{-1}\mathbf {U} \quad {\text{and}}\quad \mathbf {c} =\mathbf {L} ^{-1}\mathbf {b} .}
{\displaystyle \mathbf {A} }
into the sum of a lower triangular component
{\displaystyle \mathbf {L} }
and a strict upper triangular component
{\displaystyle \mathbf {L} ={\begin{bmatrix}16&0\\7&-11\\\end{bmatrix}}\quad {\text{and}}\quad \mathbf {U} ={\begin{bmatrix}0&3\\0&0\end{bmatrix}}.}
{\displaystyle \mathbf {L} }
{\displaystyle \mathbf {L} ^{-1}={\begin{bmatrix}16&0\\7&-11\end{bmatrix}}^{-1}={\begin{bmatrix}0.0625&0.0000\\0.0398&-0.0909\\\end{bmatrix}}.}
{\displaystyle {\begin{aligned}\mathbf {T} &=-{\begin{bmatrix}0.0625&0.0000\\0.0398&-0.0909\end{bmatrix}}{\begin{bmatrix}0&3\\0&0\end{bmatrix}}={\begin{bmatrix}0.000&-0.1875\\0.000&-0.1194\end{bmatrix}},\\[1ex]\mathbf {c} &={\begin{bmatrix}0.0625&0.0000\\0.0398&-0.0909\end{bmatrix}}{\begin{bmatrix}11\\13\end{bmatrix}}={\begin{bmatrix}0.6875\\-0.7439\end{bmatrix}}.\end{aligned}}}
{\displaystyle \mathbf {T} }
{\displaystyle \mathbf {c} }
{\displaystyle \mathbf {x} }
can be obtained iteratively.
{\displaystyle \mathbf {x} ^{(0)}}
{\displaystyle \mathbf {x} ^{(0)}={\begin{bmatrix}1.0\\1.0\end{bmatrix}}.}
The closer the guess to the final solution, the fewer iterations the algorithm will need.
{\displaystyle {\begin{aligned}\mathbf {x} ^{(1)}&={\begin{bmatrix}0.000&-0.1875\\0.000&-0.1193\end{bmatrix}}{\begin{bmatrix}1.0\\1.0\end{bmatrix}}+{\begin{bmatrix}0.6875\\-0.7443\end{bmatrix}}={\begin{bmatrix}0.5000\\-0.8636\end{bmatrix}}.\\[1ex]\mathbf {x} ^{(2)}&={\begin{bmatrix}0.000&-0.1875\\0.000&-0.1193\end{bmatrix}}{\begin{bmatrix}0.5000\\-0.8636\end{bmatrix}}+{\begin{bmatrix}0.6875\\-0.7443\end{bmatrix}}={\begin{bmatrix}0.8494\\-0.6413\end{bmatrix}}.\\[1ex]\mathbf {x} ^{(3)}&={\begin{bmatrix}0.000&-0.1875\\0.000&-0.1193\end{bmatrix}}{\begin{bmatrix}0.8494\\-0.6413\\\end{bmatrix}}+{\begin{bmatrix}0.6875\\-0.7443\end{bmatrix}}={\begin{bmatrix}0.8077\\-0.6678\end{bmatrix}}.\\[1ex]\mathbf {x} ^{(4)}&={\begin{bmatrix}0.000&-0.1875\\0.000&-0.1193\end{bmatrix}}{\begin{bmatrix}0.8077\\-0.6678\end{bmatrix}}+{\begin{bmatrix}0.6875\\-0.7443\end{bmatrix}}={\begin{bmatrix}0.8127\\-0.6646\end{bmatrix}}.\\[1ex]\mathbf {x} ^{(5)}&={\begin{bmatrix}0.000&-0.1875\\0.000&-0.1193\end{bmatrix}}{\begin{bmatrix}0.8127\\-0.6646\end{bmatrix}}+{\begin{bmatrix}0.6875\\-0.7443\end{bmatrix}}={\begin{bmatrix}0.8121\\-0.6650\end{bmatrix}}.\\[1ex]\mathbf {x} ^{(6)}&={\begin{bmatrix}0.000&-0.1875\\0.000&-0.1193\end{bmatrix}}{\begin{bmatrix}0.8121\\-0.6650\end{bmatrix}}+{\begin{bmatrix}0.6875\\-0.7443\end{bmatrix}}={\begin{bmatrix}0.8122\\-0.6650\end{bmatrix}}.\\[1ex]\mathbf {x} ^{(7)}&={\begin{bmatrix}0.000&-0.1875\\0.000&-0.1193\end{bmatrix}}{\begin{bmatrix}0.8122\\-0.6650\end{bmatrix}}+{\begin{bmatrix}0.6875\\-0.7443\end{bmatrix}}={\begin{bmatrix}0.8122\\-0.6650\end{bmatrix}}.\end{aligned}}}
As expected, the algorithm converges to the solution:
{\displaystyle \mathbf {x} =\mathbf {A} ^{-1}\mathbf {b} \approx {\begin{bmatrix}0.8122\\-0.6650\end{bmatrix}}.}
In fact, the matrix A is strictly diagonally dominant, but not positive definite.
{\displaystyle \mathbf {x} ^{(k+1)}=\mathbf {L} ^{-1}(\mathbf {b} -\mathbf {U} \mathbf {x} ^{(k)})}
{\displaystyle \mathbf {x} ^{(k+1)}=\mathbf {T} \mathbf {x} ^{(k)}+\mathbf {c} }
{\displaystyle \mathbf {T} =-\mathbf {L} ^{-1}\mathbf {U} \quad {\text{and}}\quad \mathbf {c} =\mathbf {L} ^{-1}\mathbf {b} .}
{\displaystyle \mathbf {A} }
into the sum of a lower triangular component
{\displaystyle \mathbf {L} }
and a strict upper triangular component
{\displaystyle \mathbf {U} }
{\displaystyle \mathbf {L} ={\begin{bmatrix}2&0\\5&7\\\end{bmatrix}}\quad {\text{and}}\quad \mathbf {U} ={\begin{bmatrix}0&3\\0&0\\\end{bmatrix}}.}
{\displaystyle \mathbf {L} }
{\displaystyle \mathbf {L} ^{-1}={\begin{bmatrix}2&0\\5&7\\\end{bmatrix}}^{-1}={\begin{bmatrix}0.500&0.000\\-0.357&0.143\\\end{bmatrix}}.}
{\displaystyle {\begin{aligned}\mathbf {T} &=-{\begin{bmatrix}0.500&0.000\\-0.357&0.143\\\end{bmatrix}}{\begin{bmatrix}0&3\\0&0\\\end{bmatrix}}={\begin{bmatrix}0.000&-1.500\\0.000&1.071\\\end{bmatrix}},\\[1ex]\mathbf {c} &={\begin{bmatrix}0.500&0.000\\-0.357&0.143\\\end{bmatrix}}{\begin{bmatrix}11\\13\\\end{bmatrix}}={\begin{bmatrix}5.500\\-2.071\\\end{bmatrix}}.\end{aligned}}}
{\displaystyle \mathbf {T} }
{\displaystyle \mathbf {c} }
{\displaystyle \mathbf {x} }
can be obtained iteratively.
First of all, we have to choose
{\displaystyle \mathbf {x} ^{(0)}}
{\displaystyle \mathbf {x} ^{(0)}={\begin{bmatrix}1.1\\2.3\end{bmatrix}}}
{\displaystyle {\begin{aligned}\mathbf {x} ^{(1)}&={\begin{bmatrix}0&-1.500\\0&1.071\\\end{bmatrix}}{\begin{bmatrix}1.1\\2.3\\\end{bmatrix}}+{\begin{bmatrix}5.500\\-2.071\\\end{bmatrix}}={\begin{bmatrix}2.050\\0.393\\\end{bmatrix}}.\\[1ex]\mathbf {x} ^{(2)}&={\begin{bmatrix}0&-1.500\\0&1.071\\\end{bmatrix}}{\begin{bmatrix}2.050\\0.393\\\end{bmatrix}}+{\begin{bmatrix}5.500\\-2.071\\\end{bmatrix}}={\begin{bmatrix}4.911\\-1.651\end{bmatrix}}.\\[1ex]\mathbf {x} ^{(3)}&=\cdots .\end{aligned}}}
In a test for convergence we find that the algorithm diverges. In fact, the matrix
{\displaystyle \mathbf {A} }
is neither diagonally dominant nor positive definite.
Then, convergence to the exact solution
{\displaystyle \mathbf {x} =\mathbf {A} ^{-1}\mathbf {b} ={\begin{bmatrix}-38\\29\end{bmatrix}}}
is not guaranteed and, in this case, will not occur.
{\displaystyle x_{3},\dots ,x_{n}}
. Then, repeat iterations until convergence is achieved, or break if the divergence in the solutions start to diverge beyond a predefined level.
{\displaystyle {\begin{array}{rrrrl}10x_{1}&-x_{2}&+2x_{3}&&=6,\\-x_{1}&+11x_{2}&-x_{3}&+3x_{4}&=25,\\2x_{1}&-x_{2}&+10x_{3}&-x_{4}&=-11,\\&3x_{2}&-x_{3}&+8x_{4}&=15.\end{array}}}
{\displaystyle x_{1},x_{2},x_{3}}
{\displaystyle {\begin{aligned}x_{1}&=x_{2}/10-x_{3}/5+3/5,\\x_{2}&=x_{1}/11+x_{3}/11-3x_{4}/11+25/11,\\x_{3}&=-x_{1}/5+x_{2}/10+x_{4}/10-11/10,\\x_{4}&=-3x_{2}/8+x_{3}/8+15/8.\end{aligned}}}
Suppose (0, 0, 0, 0) is the initial approximation, then the first approximate solution is given by:
{\displaystyle {\begin{aligned}x_{1}&=3/5=0.6,\\x_{2}&=(3/5)/11+25/11=3/55+25/11=2.3272,\\x_{3}&=-(3/5)/5+(2.3272)/10-11/10=-3/25+0.23272-1.1=-0.9873,\\x_{4}&=-3(2.3272)/8+(-0.9873)/8+15/8=0.8789.\end{aligned}}}
Using the approximations obtained, the iterative procedure is repeated until the desired accuracy has been reached. The following are the approximated solutions after four iterations.
The exact solution of the system is (1, 2, −1, 1).
{\displaystyle x_{i}^{(k+1)}={\frac {1}{a_{ii}}}\left(b_{i}-\sum _{j<i}a_{ij}x_{j}^{(k+1)}-\sum _{j>i}a_{ij}x_{j}^{(k)}\right),\quad {\begin{array}{l}i=1,2,\ldots ,n\\k=0,1,2,\ldots \end{array}}}
