Iterative methods: What happens when the spectral radius of a matrix is exactly 1? I know that an iterative method (I'm using Jacobi and Gauss-Seidel in this case) will converge iff the spectral radius (max absolute value of eigenvalues) of its iterative matrix is strictly less than 1. 
I've encountered a situation where the spectral radius of the iterative matrix is exactly 1. I'm pretty sure that it will converge for certain initial "guesses", but not others. Why is this the case? Is there a way to determine which initial guesses result in convergence?
 A: Consider $A=M-N,$ where in Gauss-Seidel method $M=D-L,$ and $N=U,$ then the iteration matrix $B_{G}=$
$M^{-1} N=(D-L)^{-1} U .$ For the Gauss-Seidel method to converge, the spectral radius of $B_{G}$ should be less than $1,$ i.e.
$$
\rho\left(B_{G}\right)=\max _{\lambda \in \Lambda\left(B_{G}\right)}|\lambda|<1
$$
Let $(\lambda, v)$ be a pair of eigenvalue and eigenvector of $B_{G},\left(\lambda \in \Lambda\left(B_{G}\right)\right)$ then
$$
\begin{aligned}
B_{G} v &=(D-L)^{-1} U v=\lambda v \\
U v &=\lambda(D-L) v \\
-\sum_{j>i} A(i, j) * v(j) &=\lambda \sum_{j \leqslant i} A(i, j) * v(j)=\lambda A(i, i) * v(i)+\lambda \sum_{j<i} A(i, j) * v(j) \quad \text { for } i=1,2, \cdots, n \\
\Longrightarrow \lambda A(i, i) * v(i) &=-\sum_{j>i} A(i, j) * v(j)-\lambda \sum_{j<i} A(i, j) * v(j) \quad \text { for } i=1,2, \cdots, n \\
\Longrightarrow|\lambda||A(i, i)||v(i)| &=\left|-\sum_{j>i} A(i, j) * v(j)-\lambda \sum_{j<i} A(i, j) * v(j)\right| \quad \text { for } i=1,2, \cdots, n \\
& \leqslant \sum_{j>i}|A(i, j)||v(j)|+|\lambda| \sum_{j<i}|A(i, j)||v(j)| \\
& \leqslant \max \{1,|\lambda|\}\left(\sum_{j>i}|A(i, j)||v(j)|+\sum_{j<i}|A(i, j)||v(j)|\right) \quad\\
&=\max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i}^{n}|A(i, j)||v(j)|
\end{aligned}
$$
But $v(j) \leqslant\|v\|_{\infty},$ thus
$$
|\lambda||v(i)|\leqslant \max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i}^{n}\left|\frac{A(i, j)}{A(i, i)} \right| \|v_{\infty}\|=\|v\|_{\infty} \max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i}^{n}\left|\frac{A(i, j)}{A(i, i)}\right| \quad \text { for } i=1,2, \cdots, n
$$
Let $i_{0}$ be such that $\left|v\left(i_{0}\right)\right|=\|v\|_{\infty},$ then we have that
$$
\begin{aligned}
|\lambda|\left|v\left(i_{0}\right)\right| \leqslant &\|v\|_{\infty} \max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i_{0}}^{n}\left|\frac{A\left(i_{0}, j\right)}{A\left(i_{0}, i_{0}\right)}\right| \\
\therefore & |\lambda|  \leqslant \max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i_{0}}^{n}\left|\frac{A\left(i_{0}, j\right)}{A\left(i_{0}, i_{0}\right)}\right|
\;\llap{\mathrel{\boxed{\phantom{\therefore  |\lambda|  \leqslant \max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i_{0}}^{n}\left|\frac{A\left(i_{0}, j\right)}{A\left(i_{0}, i_{0}\right)}\right|}}}}
\end{aligned}
$$
Considering $A$ to be strictly diagonally dominant, that is to say :
$$
|A(i, i)|>\sum_{j=1 \atop j \neq i}^{n}|A(i, j)|, \quad \forall i=1,2, \cdots, n
$$
We will have that :
$$
\begin{aligned}
|\lambda|& \leqslant \max \{1,|\lambda|\} \sum_{j=1 \atop j \neq i_{0}}^{n}\left|\frac{A\left(i_{0}, j\right)}{A\left(i_{0}, i_{0}\right)}\right| \\
&<\max \{1,|\lambda|\} \frac{1}{\left|A\left(i_{0}, i_{0}\right)\right|}\left|A\left(i_{0}, i_{0}\right)\right|=\max \{1,|\lambda|\}
\end{aligned}
\tag{1}
$$
\begin{aligned}
&\text { If }|\lambda| \color{red}{\geqslant} 1, \text { then this leads to a contradiction since equation (1) becomes }|\lambda|\color{red}{<}|\lambda| \text { . }\\
&\text { Thus, }|\lambda|<1 \text { and } :
\end{aligned}
$$
\boxed{\rho\left(B_{G}\right)=\max _{\lambda \in \Lambda\left(B_{G}\right)}|\lambda|<1}
$$
