Iteration matrix and convergence Assuming $G_{JA}$ is an iteration matrix for Jacobi Algorithm:
$G_{JA}(A) = I -D^{-1}A$
$D$ is the diagonal of $A$.
The sufficient condition for convergence is spectral radius less than one($\rho(G) <1$). Now, what happen if the spectral radius is equal to 1? Is there a way to set the parameter(such as initial guess) that guarantee convergence  in this case?
I would appreciate any guidance on this matter.
Thanks in advance.
 A: $\rho(G_{JA}) < 1$ is a necessary (and sufficient) condition for the convergence.
With your notation for the iteration matrix (G), we can write a stationary, first order, linear iterative method as:
$$
\mathbf{x}^{k+1} = G \mathbf{x}^k + \mathbf{f}
$$
For an iterative method, (as Jacobi is) the theorem is:

Theorem
  An iterative method is convergent $\iff$  the spectral radius of the iterative matrix $\rho(G) < 1$

I will prove only the forward implication.
Suppose for contradiction that $\rho(G) \geq 1$, i.e. $\exists$ an eigenvalue $\lambda$ of G with $|\lambda| \geq 1$; let $\mathbf{v}$ be the corresponding 
eigenvector. We choose as starting point
$$\mathbf{x}^0 = \mathbf{x} + \mathbf{v}$$
With $\mathbf{x}$ the system solution. So the error is $\mathbf{e}^0 = \mathbf{v}$.
Thus:
$$
G\mathbf{e}^0 = \lambda \mathbf{e}^0 
\Rightarrow 
\mathbf{e}^k = G^k \mathbf{e}^0 = \lambda^k\mathbf{e}^0
$$
From the norm properties, 
$||\mathbf{e}^k|| = \lambda^k \cdot ||\mathbf{e}^0||$,
but by hypothesis $|\lambda| \geq 1$, so $\mathbf{e}^k$ cannot converge to zero. So if $\mathbf{x}^k \longrightarrow \mathbf{x}$ is necessary that $\rho(G) < 1$.

Note that the convergence is global and doesn't depend on the initial $\mathbf{x}^0$.
A: Take $A=\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$, then
$G = \begin{bmatrix} 0 &1 \\ 1 & 0  \end{bmatrix}$, and the spectral radius is $1$.
Starting with $x_0 = e_1$, we get the sequence $e_1, e_2, e_1, e_2,...$ which
never converges.
A: If the error/residual vector has components in the eigenspaces of the eigenvalues of absolute value $1$, then those components will never vanish, they will oscillate.
If an eigenvalue of absolute value $1$ has multiplicity greater 1, then the corresponding components can even grow like $$\text{(multiplicity)}^{\text{(iteration number)}}.$$
