 Solution

Subtract the exact solution from the Jacobi method to get $e_n=H{e}_{n-1}$, hence $e_n=H^n{e}_0$. Convergence from every initial vector is equivalent to $H^n\to 0$. If every eigenvalue has modulus below one, a Jordan block $\lambda I+N$ has powers that are finite sums $\left(\genfrac{}{}{0ex}{}{n}{k}\right){\lambda }^{n-k}{N}^k$. The polynomial factor in $n$ is dominated by geometric decay; for $\lambda =0$, the block is nilpotent. Consequently $H^n\to 0$.

Conversely, for a complex eigenvector $v$ with eigenvalue $\lambda$, $H^{n}v={\lambda }^{n}v$. Convergence for real errors also implies convergence for their complexification, so this tends to zero only if $|\lambda |<1$. Therefore the iteration converges from every starting vector if and only if $\rho (H)<1$, where $\rho$ is the spectral radius.

For the specified matrix,Its eigenvalues are $0,\mu ,-\mu$, so $\rho (H)=|\mu |$. Hence $-1<\mu <1$ is the convergence range. The original coefficient matrix has determinant $12(1-{\mu }^2)$, so the endpoints also violate its assumed nonsingularity.