Spectral radius for Jacobi and Gauss Seidel iterative methods Let 
$
L=
\left[
\begin{matrix}
0 & \ldots & & &  &0 \\
a & 0 & \ldots &  &  &0 \\
0 & a & 0 & \ldots &  &0\\
0 & 0& a & 0 & \ldots & 0 \\
\vdots 
\end{matrix}
\right]
$
that is $L$ has its sub-diagonal equal to $a$, the other elements are zero.
Show that
$$
\rho(L+L^T)^2 = \rho((\text{Id}+L)^{-1} L^T) 
$$
where $\rho(M)$ is the spectral radius of the matrix $M$.
 A: I can only remember the following proof. So there might be an easier way.
Definiton 1:
 Let $A∈ℝ^{n×n}$ be a matrix with $$A=L+D+R,$$
with $D$ invertible.
If the eigenvalues of the matrix $$J(α)=-D^{-1}\{αL+α^{-1}R\}, α∈ℂ\setminus\{0\}$$
are independent of $α$, then we call $A$ consistently ordered.

Theorem 1: Tri-diagonal matrices of the following form are consistently ordered.
$$\begin{pmatrix}
d_1 & a_{12} & \\
a_{21}& d_2 & \ddots \\
& \ddots & \ddots & a_{n-1,n} \\
& & a_{n-1,n} & d_n
\end{pmatrix}$$
with $d_i\neq0$.
Why? 
Let $T=\text{diag}(1, α, …, α^{n-1})$, then we can do a similarity transformation: 
$$-J(α) = αD^{-1}L+α^{-1}D^{-1}R = T(D^{-1}L+D^{-1}R)T^{-1}$$
And we know that eigenvalues are preserved under similarity transformation. 

In your case it is $A=L+I+L^\top$, which is a tri-diagonal matrix. Hence your matrix is consistently ordered. And the two matrices where you want to know the spectral radii are simply the Jacobi-matrix $J=D^{-1}(L+R)$ and the Gauß-Seidel-matrix $H=-(D+L)^{-1}L^\top$ of that $A$.

Theorem 2: 
Let $A∈ℝ^{n×n}$ be a consistently ordered matrix. Then the eigenvalues $μ∈σ(J)$ and $λ∈σ(H_ω)$ fulfil: 
$$\sqrt{λ}ωμ = λ+ω-1.$$ 
Probably $ω∈(0,2)$, so that the SOR-method converges. I don't remember the proof. You should be able to find it somewhere in these lecture notes. (I was an exercise group leader to that lecture, that is why I remember this answer.)

The Gauß-Seidel-matrix is the special case of the SOR-matrix $H_ω$ with $ω=1$. Hence: 
$$ρ(H_1)=ρ(J)^2$$
