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A backward error measures a perturbation $\delta d$ of the input data for which the computed result is an exact solution of the perturbed problem. Specify which data may change and which norm measures the perturbation. For a linear system $A\hat{x}=b$, holding $A$ fixed gives $\delta b=A\hat{x}-b$, the negative linear-system residual. The corresponding forward error also depends on the condition number.

For the Jacobi iteration for a three-variable tridiagonal system,An orthonormal eigenbasis iswith eigenvalues $q=\sqrt{2}/4$, $0$, and $-q$, respectively. These are error modes: their shapes persist under the iteration matrix, while their coefficients change. Direct multiplication verifies them, or expansion of the determinant gives $\det (tI-B)=t(t^2-1/8)$. Thus $\rho (B)=q<1$ and convergence holds for every starting vector.

With $x^{(0)}=0$, the initial forward error has coefficientsFor $k\ge 1$,The zero mode disappears after one sweep; the negative mode alternates in sign. The Euclidean norm drops by exactly $q$ per sweep from $k=1$ onward. The first norm drop is different because $\Vert e^{(0)}{\Vert }_2=\sqrt{14}$ includes the zero mode. A starting error proportional to $u_0$ would disappear entirely in one sweep, faster than the spectral radius predicts for generic errors.

The curves are computed iterates and exact formulas for this linear system. Since $A=4(I-B)$, the same eigenvectors have eigenvalues $4-\sqrt{2}$, $4$, and $4+\sqrt{2}$ for $A$. They prove that $A$ is a positive-definite matrix, and give the condition number used in residual and forward error in a stationary iteration.

For the Gauss-Seidel iteration for a two-variable positive-definite system, define the forward error by $e_x^{(k)}=x^{(k)}-x_{\ast }$ and $e_y^{(k)}=y^{(k)}-y_{\ast }$. Subtract the two exact solution equations from the two update equations:The fresh first coordinate passes the old second-coordinate error through two successive factors. The iteration matrix is thereforeIts eigenvalues are $0$ and $1/12$, as seen from the determinant. The zero eigenvalue explains why the old first-coordinate error disappears completely: the first row overwrites it. After one sweep, every remaining forward error lies on the eigenvector line spanned by $(-3,1{)}^T$.

For this zero start, $e_y^{(0)}=-7/11$, so the exact vectors for $k\ge 1$ areThe Euclidean norm of the forward error decreases by exactly $1/12$ per sweep from $k=1$ onward. The first decrease is different because the initial error has not yet reached that eigenvector line. Each row update moves horizontally or vertically in the coordinate plane; sweep endpoints then approach the solution along one straight line.

The linear-system residual is an independent way to check the calculation. Completing a sweep satisfies the second row exactly, but changing $y$ disturbs the first row:Thus a zero residual in the last updated row alone does not certify a solution; inspect the complete linear-system residual.

For the same matrix, the stale schedule of the Jacobi method gives the iteration matrixHere its eigenvalues are $\pm 1/\sqrt{12}$, whereas the fresh schedule gives $\rho (H)=\rho (B{)}^2$. This comparison belongs to this particular linear system; it is not a universal relation between the two methods' spectral radii.

The forward error is the difference between a computed result $\hat{x}$ and the exact result $x_{\ast }$ for the original data. Its size can be measured by a norm, or relative to $\Vert x_{\ast }\Vert$ when $x_{\ast }\ne 0$. A linear-system residual measures the equation imbalance instead; the condition number relates that imbalance to the forward error.

An invertible matrix and nonzero diagonal entries make the update well defined, but do not guarantee convergence. For the original linear systemthe Gauss-Seidel method updates $x^{(k+1)}=1-2y^{(k)}$ and $y^{(k+1)}=1-2x^{(k+1)}=-1+4y^{(k)}$. Starting at zero gives $(1,-1)$, then $(3,-5)$. The second-coordinate forward error is $e_y^{(k)}=-4^k/3$, and the iteration matrix $H=\left(\begin{array}{cc}0& -2\\ 0& 4\end{array}\right)$ has spectral radius $4$. The linear system has a unique solution while this particular stationary iterative method for a linear system diverges.

For a stationary iteration $x^{(k+1)}=H{x}^{(k)}+v$, the matrix $H$ propagates the error by $e^{(k+1)}=H{e}^{(k)}$. Convergence from every initial error is equivalent to $\rho (H)<1$.

For a matrix splitting $A=M-N$, the iteration matrix is $H=M^{-1}N=I-M^{-1}A$, where $I$ is the identity matrix. If $A{x}_{\ast }=b$, subtracting its fixed point equation from the update givesThus $b$ determines the fixed point, while $H$ determines how the forward error moves. An eigenvector $v$ with eigenvalue $\lambda$ is an error direction whose coefficient is multiplied by $\lambda$ on every step. A negative real eigenvalue reverses its sign; a zero eigenvalue removes that component in one step.

The linear-system residual $r^{(k)}=b-A{x}^{(k)}$ is computable without knowing $x_{\ast }$. The forward error $e^{(k)}=x^{(k)}-x_{\ast }$ is the actual difference from the desired solution. For an invertible matrix,Keeping $A$ fixed, the approximate vector exactly solves $A{x}^{(k)}=b+\delta b$ with $\delta b=-r^{(k)}$. Thus, for $b\ne 0$, $\Vert r^{(k)}\Vert /\Vert b\Vert$ is the relative backward error when only the right-hand side may change. For a vector norm and its induced operator norm,A small relative linear-system residual ensures a small relative forward error when the condition number is moderate; an ill-conditioned matrix can amplify it. The bound follows directly from $e=-A^{-1}r$ and $\Vert b\Vert =\Vert A{x}_{\ast }\Vert \le \Vert A\Vert \Vert x_{\ast }\Vert$.^[ref]

For the Jacobi iteration for a three-variable tridiagonal system, the first linear-system residual is $(1,3,1{)}^T$, while the first forward error is $(-1/2,-1,-1/2{)}^T$. In the Euclidean norm, the relative forward error is $\sqrt{3/28}\approx 0.3273$ and the relative linear-system residual is $\sqrt{11/120}\approx 0.3028$. Its spectral condition number of a positive-definite matrix is $(4+\sqrt{2})/(4-\sqrt{2})\approx 2.094$, so the bound is about $0.6340$. The bound is a guarantee, not an equality. In floating-point arithmetic, recompute the linear-system residual from the current vector when checking termination; a small update alone need not certify a small forward error.