Gauss–Seidel method
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 = and b =. {\displaystyle \mathbf {A} ={\begin{bmatrix}16&3\\7&-11\\\end{bmatrix}}\quad {\text{and}}\quad \mathbf {b} ={\begin{bmatrix}11\\13\end{bmatrix}}.}

We want to use the equation x (k + 1) = L − 1 (b − U x (k)) {\displaystyle \mathbf {x} ^{(k+1)}=\mathbf {L} ^{-1}(\mathbf {b} -\mathbf {U} \mathbf {x} ^{(k)})} in the form x (k + 1) = T x (k) + c {\displaystyle \mathbf {x} ^{(k+1)}=\mathbf {T} \mathbf {x} ^{(k)}+\mathbf {c} } where:

We must decompose A {\displaystyle \mathbf {A} } into the sum of a lower triangular component L {\displaystyle \mathbf {L} } and a strict upper triangular component U {\displaystyle U}: L = and U =. {\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}}.}

The inverse of L {\displaystyle \mathbf {L} } is: L − 1 = − 1 =. {\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}}.}

Now we can find: T = − =, c = =. {\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}},\\\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}}}

Now we have T {\displaystyle \mathbf {T} } and c {\displaystyle \mathbf {c} } and we can use them to obtain the vectors x {\displaystyle \mathbf {x} } iteratively.

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

We choose a starting point: x (0) =. {\displaystyle \mathbf {x} ^{(0)}={\begin{bmatrix}1.0\\1.0\end{bmatrix}}.}

We can then calculate: x (1) = + =. x (2) = + =. x (3) = + =. x (4) = + =. x (5) = + =. x (6) = + =. x (7) = + =. {\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}}.\\\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}}.\\\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}}.\\\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}}.\\\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}}.\\\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}}.\\\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: x = A − 1 b ≈. {\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).
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