Quadric (projective geometry)

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 linear system of equations. It is named after the German mathematicians Carl Friedrich Gauss and Philipp Ludwig von Seidel, and is similar to the Jacobi method. Though it can be applied to any matrix with non-zero elements on the diagonals, convergence is only guaranteed if the matrix is either 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.

Description

The Gauss–Seidel method is an iterative technique for solving a square system of n linear equations with unknown x:

$A\mathbf{x}=\mathbf{b}$.

It is defined by the iteration

$L_{*}{\mathbf{x}}^{(k+1)}=\mathbf{b}-U{\mathbf{x}}^{(k)},$

where the matrix A is decomposed into a lower triangular component $L_{*}$, and a strictly upper triangular component U: $A=L_{*}+U$.

In more detail, write out A, x and b in their components:

$A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right],\mathbf{x}=\left[\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right],\mathbf{b}=\left[\begin{array}{c}{b}_1\\ b_2\\ ⋮\\ b_n\end{array}\right].$

Then the decomposition of A into its lower triangular component and its strictly upper triangular component is given by:

$A=L_{*}+U\text{where}{L}_{*}=\left[\begin{array}{cccc}{a}_{11}& 0& \cdots & 0\\ a_{21}& a_{22}& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right],U=\left[\begin{array}{cccc}0& a_{12}& \cdots & a_{1n}\\ 0& 0& \cdots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & 0\end{array}\right].$

The system of linear equations may be rewritten as:

$L_{*}\mathbf{x}=\mathbf{b}-U\mathbf{x}$

The Gauss–Seidel method now solves the left hand side of this expression for x, using previous value for x on the right hand side. Analytically, this may be written as:

${\mathbf{x}}^{(k+1)}=L_{*}^{-1}(\mathbf{b}-U{\mathbf{x}}^{(k)}).$

However, by taking advantage of the triangular form of $L_{*}$, the elements of x^(k+1) can be computed sequentially using forward substitution:

$x_i^{(k+1)}=\frac{1}{a_{ii}}(b_i-\sum _{j<i}{a}_{ij}{x}_j^{(k+1)}-\sum _{j>i}{a}_{ij}{x}_j^{(k)}),i,j=1,2,\dots ,n.$

The procedure is generally continued until the changes made by an iteration are below some tolerance, such as a sufficiently small residual.

Discussion

The element-wise formula for the Gauss–Seidel method is extremely similar to that of the Jacobi method.

The computation of x_i^(k+1) uses only the elements of x^(k+1) that have already been computed, and only the elements of x^(k) that have not yet to be advanced to iteration k+1. 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 cannot be done in parallel. Furthermore, the values at each iteration are dependent on the order of the original equations.

Gauss-Seidel is the same as SOR (successive over-relaxation) with $\omega =1$.

Convergence

The convergence properties of the Gauss–Seidel method are dependent on the matrix A. Namely, the procedure is known to converge if either:

• A is symmetric positive-definite, or
• A is strictly or irreducibly diagonally dominant.

The Gauss–Seidel method sometimes converges even if these conditions are not satisfied.

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:

Inputs: Template:Var, Template:Var
Output: $\varphi$

Choose an initial guess $\varphi$ to the solution
repeat until convergence
    for Template:Var from 1 until Template:Var do
        $\sigma \leftarrow 0$
        for Template:Var from 1 until Template:Var do
            if Template:Var ≠ Template:Var then
                $\sigma \leftarrow \sigma +a_{ij}{\varphi }_j$
            end if
        end (Template:Var-loop)
        ${\varphi }_i\leftarrow \frac{1}{a_{ii}}(b_i-\sigma )$
    end (Template:Var-loop)
    check if convergence is reached
end (repeat)

Examples

An example for the matrix version

A linear system shown as $A\mathbf{x}=\mathbf{b}$ is given by:

$A=\left[\begin{array}{cc}16& 3\\ 7& -11\\ \end{array}\right]$ and $b=\left[\begin{array}{c}11\\ 13\end{array}\right].$

We want to use the equation

${\mathbf{x}}^{(k+1)}=L_{*}^{-1}(\mathbf{b}-U{\mathbf{x}}^{(k)})$

in the form

${\mathbf{x}}^{(k+1)}=T{\mathbf{x}}^{(k)}+C$

where:

$T=-L_{*}^{-1}U$ and $C=L_{*}^{-1}\mathbf{b}.$

We must decompose $A_{}^{}$ into the sum of a lower triangular component $L_{*}^{}$ and a strict upper triangular component $U_{}^{}$:

$L_{*}=\left[\begin{array}{cc}16& 0\\ 7& -11\\ \end{array}\right]$ and $U=\left[\begin{array}{cc}0& 3\\ 0& 0\end{array}\right].$

The inverse of $L_{*}^{}$ is:

$L_{*}^{-1}={\left[\begin{array}{cc}16& 0\\ 7& -11\end{array}\right]}^{-1}=\left[\begin{array}{cc}0.0625& 0.0000\\ 0.0398& -0.0909\\ \end{array}\right]$.

Now we can find:

$T=-\left[\begin{array}{cc}0.0625& 0.0000\\ 0.0398& -0.0909\end{array}\right]\times \left[\begin{array}{cc}0& 3\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0.000& -0.1875\\ 0.000& -0.1193\end{array}\right],$

$C=\left[\begin{array}{cc}0.0625& 0.0000\\ 0.0398& -0.0909\end{array}\right]\times \left[\begin{array}{c}11\\ 13\end{array}\right]=\left[\begin{array}{c}0.6875\\ -0.7439\end{array}\right].$

Now we have $T_{}^{}$ and $C_{}^{}$ and we can use them to obtain the vectors $\mathbf{x}$ iteratively.

First of all, we have to choose ${\mathbf{x}}^{(0)}$: we can only guess. The better the guess, the quicker the algorithm will perform.

We suppose:

$x^{(0)}=\left[\begin{array}{c}1.0\\ 1.0\end{array}\right].$

We can then calculate:

$x^{(1)}=\left[\begin{array}{cc}0.000& -0.1875\\ 0.000& -0.1193\end{array}\right]\times \left[\begin{array}{c}1.0\\ 1.0\end{array}\right]+\left[\begin{array}{c}0.6875\\ -0.7443\end{array}\right]=\left[\begin{array}{c}0.5000\\ -0.8636\end{array}\right].$

$x^{(2)}=\left[\begin{array}{cc}0.000& -0.1875\\ 0.000& -0.1193\end{array}\right]\times \left[\begin{array}{c}0.5000\\ -0.8636\end{array}\right]+\left[\begin{array}{c}0.6875\\ -0.7443\end{array}\right]=\left[\begin{array}{c}0.8494\\ -0.6413\end{array}\right].$

$x^{(3)}=\left[\begin{array}{cc}0.000& -0.1875\\ 0.000& -0.1193\end{array}\right]\times \left[\begin{array}{c}0.8494\\ -0.6413\\ \end{array}\right]+\left[\begin{array}{c}0.6875\\ -0.7443\end{array}\right]=\left[\begin{array}{c}0.8077\\ -0.6678\end{array}\right].$

$x^{(4)}=\left[\begin{array}{cc}0.000& -0.1875\\ 0.000& -0.1193\end{array}\right]\times \left[\begin{array}{c}0.8077\\ -0.6678\end{array}\right]+\left[\begin{array}{c}0.6875\\ -0.7443\end{array}\right]=\left[\begin{array}{c}0.8127\\ -0.6646\end{array}\right].$

$x^{(5)}=\left[\begin{array}{cc}0.000& -0.1875\\ 0.000& -0.1193\end{array}\right]\times \left[\begin{array}{c}0.8127\\ -0.6646\end{array}\right]+\left[\begin{array}{c}0.6875\\ -0.7443\end{array}\right]=\left[\begin{array}{c}0.8121\\ -0.6650\end{array}\right].$

$x^{(6)}=\left[\begin{array}{cc}0.000& -0.1875\\ 0.000& -0.1193\end{array}\right]\times \left[\begin{array}{c}0.8121\\ -0.6650\end{array}\right]+\left[\begin{array}{c}0.6875\\ -0.7443\end{array}\right]=\left[\begin{array}{c}0.8122\\ -0.6650\end{array}\right].$

$x^{(7)}=\left[\begin{array}{cc}0.000& -0.1875\\ 0.000& -0.1193\end{array}\right]\times \left[\begin{array}{c}0.8122\\ -0.6650\end{array}\right]+\left[\begin{array}{c}0.6875\\ -0.7443\end{array}\right]=\left[\begin{array}{c}0.8122\\ -0.6650\end{array}\right].$

As expected, the algorithm converges to the exact solution:

$\mathbf{x}=A^{-1}\mathbf{b}=\left[\begin{array}{c}0.8122\\ -0.6650\end{array}\right].$

In fact, the matrix A is strictly diagonally dominant (but not positive definite).

Another example for the matrix version

Another linear system shown as $A\mathbf{x}=\mathbf{b}$ is given by:

$A=\left[\begin{array}{cc}2& 3\\ 5& 7\\ \end{array}\right]$ and $b=\left[\begin{array}{c}11\\ 13\\ \end{array}\right].$

We want to use the equation

${\mathbf{x}}^{(k+1)}=L_{*}^{-1}(\mathbf{b}-U{\mathbf{x}}^{(k)})$

in the form

${\mathbf{x}}^{(k+1)}=T{\mathbf{x}}^{(k)}+C$

where:

$T=-L_{*}^{-1}U$ and $C=L_{*}^{-1}\mathbf{b}.$

We must decompose $A_{}^{}$ into the sum of a lower triangular component $L_{*}^{}$ and a strict upper triangular component $U_{}^{}$:

$L_{*}=\left[\begin{array}{cc}2& 0\\ 5& 7\\ \end{array}\right]$ and $U=\left[\begin{array}{cc}0& 3\\ 0& 0\\ \end{array}\right].$

The inverse of $L_{*}^{}$ is:

$L_{*}^{-1}={\left[\begin{array}{cc}2& 0\\ 5& 7\\ \end{array}\right]}^{-1}=\left[\begin{array}{cc}0.500& 0.000\\ -0.357& 0.143\\ \end{array}\right]$.

Now we can find:

$T=-\left[\begin{array}{cc}0.500& 0.000\\ -0.357& 0.143\\ \end{array}\right]\times \left[\begin{array}{cc}0& 3\\ 0& 0\\ \end{array}\right]=\left[\begin{array}{cc}0.000& -1.500\\ 0.000& 1.071\\ \end{array}\right],$

$C=\left[\begin{array}{cc}0.500& 0.000\\ -0.357& 0.143\\ \end{array}\right]\times \left[\begin{array}{c}11\\ 13\\ \end{array}\right]=\left[\begin{array}{c}5.500\\ -2.071\\ \end{array}\right].$

Now we have $T_{}^{}$ and $C_{}^{}$ and we can use them to obtain the vectors $\mathbf{x}$ iteratively.

First of all, we have to choose ${\mathbf{x}}^{(0)}$: we can only guess. The better the guess, the quicker will perform the algorithm.

We suppose:

$x^{(0)}=\left[\begin{array}{c}1.1\\ 2.3\\ \end{array}\right].$

We can then calculate:

$x^{(1)}=\left[\begin{array}{cc}0& -1.500\\ 0& 1.071\\ \end{array}\right]\times \left[\begin{array}{c}1.1\\ 2.3\\ \end{array}\right]+\left[\begin{array}{c}5.500\\ -2.071\\ \end{array}\right]=\left[\begin{array}{c}2.050\\ 0.393\\ \end{array}\right].$

$x^{(2)}=\left[\begin{array}{cc}0& -1.500\\ 0& 1.071\\ \end{array}\right]\times \left[\begin{array}{c}2.050\\ 0.393\\ \end{array}\right]+\left[\begin{array}{c}5.500\\ -2.071\\ \end{array}\right]=\left[\begin{array}{c}4.911\\ -1.651\\ \end{array}\right].$

$x^{(3)}=\cdots .$

If we test for convergence we'll find that the algorithm diverges. In fact, the matrix A is neither diagonally dominant nor positive definite. Then, convergence to the exact solution

$\mathbf{x}=A^{-1}\mathbf{b}=\left[\begin{array}{c}-38\\ 29\end{array}\right]$

is not guaranteed and, in this case, will not occur.

An example for the equation version

Suppose given k equations where x_n are vectors of these equations and starting point x₀. From the first equation solve for x₁ in terms of $x_{n+1},x_{n+2},\dots ,x_n.$ For the next equations substitute the previous values of xs.

To make it clear let's consider an example.

$\begin{array}{rl}10x_1-x_2+2x_3& =6,\\ -x_1+11x_2-x_3+3x_4& =25,\\ 2x_1-x_2+10x_3-x_4& =-11,\\ 3x_2-x_3+8x_4& =15.\end{array}$

Solving for $x_1$, $x_2$, $x_3$ and $x_4$ gives:

$\begin{array}{rl}{x}_1& =x_2/10-x_3/5+3/5,\\ x_2& =x_1/11+x_3/11-3x_4/11+25/11,\\ x_3& =-x_1/5+x_2/10+x_4/10-11/10,\\ x_4& =-3x_2/8+x_3/8+15/8.\end{array}$

Suppose we choose (0, 0, 0, 0) as the initial approximation, then the first approximate solution is given by

$\begin{array}{rl}{x}_1& =3/5=0.6,\\ x_2& =(3/5)/11+25/11=3/55+25/11=2.3272,\\ x_3& =-(3/5)/5+(2.3272)/10-11/10=-3/25+0.23272-1.1=-0.9873,\\ x_4& =-3(2.3272)/8+(-0.9873)/8+15/8=0.8789.\end{array}$

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.

$x_1$	$x_2$	$x_3$	$x_4$
$0.6$	$2.32727$	$-0.987273$	$0.878864$
$1.03018$	$2.03694$	$-1.01446$	$0.984341$
$1.00659$	$2.00356$	$-1.00253$	$0.998351$
$1.00086$	$2.0003$	$-1.00031$	$0.99985$

The exact solution of the system is (1, 2, −1, 1).

An example using Python

The following numerical procedure simply iterates through to produce the solution vector.

#initialize the matrix
mat = [ [3/5.0,    0.0,    1/10.0, -1/5.0, 0.0    ], \
        [25/11.0,  1/11.0, 0.0,    1/11.0, -3/11.0], \
        [-11/10.0, -1/5.0, 1/10.0, 0,      1/10.0 ], \
        [15/8.0,   0.0,    -3/8.0, 1/8.0,  0.0    ] ]

x = [0,0,0,0] #initial guess

for i in xrange(6):
     x[0] = mat[0][0] + mat[0][1]*0  + mat[0][2]*x[1] + mat[0][3]*x[2] + mat[0][4]*x[3]
     x[1] = mat[1][0] + mat[1][1]*x[0] + mat[1][2]*0  + mat[1][3]*x[2] + mat[1][4]*x[3]
     x[2] = mat[2][0] + mat[2][1]*x[0] + mat[2][2]*x[1] + mat[2][3]*0  + mat[2][4]*x[3]
     x[3] = mat[3][0] + mat[3][1]*x[0] + mat[3][2]*x[1] + mat[3][3]*x[2] + mat[3][4]*0
     print '%f %f %f %f' %(x[0],x[1],x[2],x[3]) #display the iterations to the user

Produces the output,

0.600000 2.327273 -0.987273 0.878864
1.030182 2.036938 -1.014456 0.984341
1.006585 2.003555 -1.002527 0.998351
1.000861 2.000298 -1.000307 0.999850
1.000091 2.000021 -1.000031 0.999988
1.000008 2.000001 -1.000003 0.999999

Program to solve arbitrary no. of equations using Matlab

disp('Give the input to solve the set of equations AX=B')
a=input('Input the square matrix A : \n');
b=input('Input the column matrix B : \n');
m=length(a);
%z is a two dimensional array in which row corresponds to values of X in a
%specific iteration and the column corresponds to values of specific
%element of X in different iterations
c=0;%random assignment
e=1;%'e' represents the maximum error
d=0;%random assignment
for u=1:m
    x(u)=b(u,1)/a(u,u);
    z(1,u)=0;%initializing the values for matrix X(x1;x2;...xm)
end
l=2;%'l' represents the iteration no.
%loop for finding the convergence factor (C.F)
for r = 1:m
    for s = 1:m
        if r~=s
           p(r)=abs(a(r,s)/a(r,r))+d;%p(r) is the C.F for equation no. r
           d=p(r);
        end
    end
    d=0;
end
if min(p)>=1 %at least one equation must satisfy the condition p<1
   fprintf('Roots will not converge for this set of equations')
else
    while(e>=1e-4)
        j1=1;%while calculating elements in first column we consider only the old values of X
        for i1=2:m
            q(j1)=(a(j1,i1)/a(j1,j1))*z(l-1,i1)+c;
            c=q(j1);
        end
        c=0;
        z(l,j1)=x(j1)-q(j1);%elements of z in the iteration no. l
        x(j1)=z(l,j1);
        for u=1:m
            x(u)=b(u,1)/a(u,u);
            z(1,u)=0;
        end
        for j1=2:m-1%for intermediate columns between 1 and m, we use the updated values of X 
            for i1=1:j1-1
                q(j1)=(a(j1,i1)/a(j1,j1))*z(l,i1)+c;
                c=q(j1);
            end
            for i1=j1+1:m
                q(j1)=(a(j1,i1)/a(j1,j1))*z(l-1,i1)+c;
                c=q(j1);
            end
            c=0;
            z(l,j1)=x(j1)-q(j1);
            x(j1)=z(l,j1);
            for u=1:m
                x(u)=b(u,1)/a(u,u);
                z(1,u)=0;
            end
        end
        j1=m;%for the last column, we use only the updated values of X
        for i1=1:m-1
            q(j1)=(a(j1,i1)/a(j1,j1))*z(l,i1)+c;
            c=q(j1);
        end
        c=0;
        z(l,j1)=x(j1)-q(j1);
        for v=1:m
            t=abs(z(l,v)-z(l-1,v));%calculates the error 
        end 
        e=max(t);%evaluates the maximum error out of errors of all elements of X 
        l=l+1;%iteration no. gets updated
        for i=1:m 
            X(1,i)=z(l-1,i);%the final solution X 
        end
    end
    %loop to show iteration number along with the values of z 
    for i=1:l-1    
        for j=1:m        
            w(i,j+1)=z(i,j);    
        end
        w(i,1)=i;
    end
    disp('   It. no.      x1        x2       x3        x4 ') 
    disp(w) 
    disp('The final solution is ') 
    disp(X) 
    fprintf('The total number of iterations is %d',l-1)
end

Program output is

Give the input to solve the set of equations AX=B
Input the square matrix A : 
[10 -2 -1 -1;-2 10 -1 -1;-1 -1 10 -2;-1 -1 -2 10]
Input the column matrix B : 
[3;15;27;-9]
   It. no.      x1        x2       x3        x4 

    1.0000         0         0         0         0
    2.0000    0.3000    1.5600    2.8860   -0.1368
    3.0000    0.8869    1.9523    2.9566   -0.0248
    4.0000    0.9836    1.9899    2.9924   -0.0042
    5.0000    0.9968    1.9982    2.9987   -0.0008
    6.0000    0.9994    1.9997    2.9998   -0.0001
    7.0000    0.9999    1.9999    3.0000   -0.0000
    8.0000    1.0000    2.0000    3.0000   -0.0000

The final solution is 

    1.0000    2.0000    3.0000   -0.0000

The total number of iterations is 8

See also

• Jacobi method
• Successive over-relaxation
• Iterative method. Linear systems
• Gaussian belief propagation
• Matrix splitting

Notes

43 year old Petroleum Engineer Harry from Deep River, usually spends time with hobbies and interests like renting movies, property developers in singapore new condominium and vehicle racing. Constantly enjoys going to destinations like Camino Real de Tierra Adentro.

References

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  Naturally, you will have to pay a safety deposit and that is usually one month rent for annually of the settlement. That is the place your good religion deposit will likely be taken into account and will kind part or all of your security deposit. Anticipate to have a proportionate amount deducted out of your deposit if something is discovered to be damaged if you move out. It's best to you'll want to test the inventory drawn up by the owner, which can detail all objects in the property and their condition. If you happen to fail to notice any harm not already mentioned within the inventory before transferring in, you danger having to pay for it yourself.
  
  In case you are in search of an actual estate or Singapore property agent on-line, you simply should belief your intuition. It's because you do not know which agent is nice and which agent will not be. Carry out research on several brokers by looking out the internet. As soon as if you end up positive that a selected agent is dependable and reliable, you can choose to utilize his partnerise in finding you a home in Singapore. Most of the time, a property agent is taken into account to be good if he or she locations the contact data on his website. This may mean that the agent does not mind you calling them and asking them any questions relating to new properties in singapore in Singapore. After chatting with them you too can see them in their office after taking an appointment.
  
  Have handed an trade examination i.e Widespread Examination for House Brokers (CEHA) or Actual Property Agency (REA) examination, or equal; Exclusive brokers are extra keen to share listing information thus making certain the widest doable coverage inside the real estate community via Multiple Listings and Networking. Accepting a severe provide is simpler since your agent is totally conscious of all advertising activity related with your property. This reduces your having to check with a number of agents for some other offers. Price control is easily achieved. Paint work in good restore-discuss with your Property Marketing consultant if main works are still to be done. Softening in residential property prices proceed, led by 2.8 per cent decline within the index for Remainder of Central Region
  
  Once you place down the one per cent choice price to carry down a non-public property, it's important to accept its situation as it is whenever you move in – faulty air-con, choked rest room and all. Get round this by asking your agent to incorporate a ultimate inspection clause within the possibility-to-buy letter. HDB flat patrons routinely take pleasure in this security net. "There's a ultimate inspection of the property two days before the completion of all HDB transactions. If the air-con is defective, you can request the seller to repair it," says Kelvin.
  
  15.6.1 As the agent is an intermediary, generally, as soon as the principal and third party are introduced right into a contractual relationship, the agent drops out of the image, subject to any problems with remuneration or indemnification that he could have against the principal, and extra exceptionally, against the third occasion. Generally, agents are entitled to be indemnified for all liabilities reasonably incurred within the execution of the brokers´ authority.
  
  To achieve the very best outcomes, you must be always updated on market situations, including past transaction information and reliable projections. You could review and examine comparable homes that are currently available in the market, especially these which have been sold or not bought up to now six months. You'll be able to see a pattern of such report by clicking here It's essential to defend yourself in opposition to unscrupulous patrons. They are often very skilled in using highly unethical and manipulative techniques to try and lure you into a lure. That you must also protect your self, your loved ones, and personal belongings as you'll be serving many strangers in your home. Sign a listing itemizing of all of the objects provided by the proprietor, together with their situation. HSR Prime Recruiter 2010.
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• Other Sports Official Kull from Drumheller, has hobbies such as telescopes, property developers in singapore and crocheting. Identified some interesting places having spent 4 months at Saloum Delta.
  
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• Gauss–Seidel from www.math-linux.com
• Module for Gauss–Seidel Iteration
• Gauss–Seidel From Holistic Numerical Methods Institute
• Gauss Siedel Iteration from www.geocities.com
• The Gauss-Seidel Method
• Bickson
• Matlab code
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