For which $a \in \mathbb{R}$ Jacobi converge? I tried to solve the following problem and I don't know if it's correct and I have a few questions:
Let
\begin{align}
A =
\left[ {\begin{array}{cc}
a & 1 & 0 \\
1 & a & 1 \\
0 & 1 & a
\end{array} } \right]
\end{align}
Question: For which $a \in \mathbb{R}$ does Jacobi converge?
First of all: We know that Jacobi and Gauss Seidel converge if $A$ ist diagonal dominant. So $A$ converges at least for all $a \in \mathbb{R} \setminus [-2,2]$. But the only way to find all those $a \in \mathbb{R}$ is to calculate the spectral radius of the iteration matrix (right?).
Let $A = D + R$ with
\begin{align}
D =
\left[ {\begin{array}{cc}
a & 0 & 0 \\
0 & a & 0 \\
0 & 0 & a
\end{array} } \right] &&&
R =
\left[ {\begin{array}{cc}
0 & 1 & 0 \\
1 & 0 & 1 \\
0 & 1 & 0
\end{array} } \right]
\end{align}
So $Ax = b \Leftrightarrow (D+R)x=b \Leftrightarrow x = -D^{-1}Rx+D^{-1}b$, so $T = D^{-1}R$ is the iteration matrix is $T$ and now we have to find all $a \in \mathbb{R}$ for which $\rho(T)<1$. 
We find that $D^{-1} = \mathrm{diag}(\frac{1}{a}, \frac{1}{a}, \frac{1}{a})$ and 
\begin{align}
D^{-1}R =
\left[ {\begin{array}{cc}
0 & \frac{1}{a} & 0 \\
\frac{1}{a} & 0 & \frac{1}{a} \\
0 & \frac{1}{a} & 0
\end{array} } \right] &&&
\chi (D^{-1}R-\lambda I) = \det
\left[ {\begin{array}{cc}
-\lambda & \frac{1}{a} & 0 \\
\frac{1}{a} & -\lambda & \frac{1}{a} \\
0 & \frac{1}{a} & -\lambda
\end{array} } \right]
\end{align}
And 
\begin{align}
\det(D^{-1}R - \lambda I)= -\lambda^3 + 2\lambda^2 \cdot \frac{1}{a} = -\lambda(\lambda^2 - 2\lambda \cdot \frac{1}{a})
\end{align}
 and so we find two eigenvalues $\lambda_1 = 0$ and $\lambda_2 = 2 \cdot \frac{1}{a}$. For $|\lambda_2|<1$ we need $a \in \mathbb{R} \setminus [-2,2]$.
So the Jacobi method converges if and only if $a \in \mathbb{R} \setminus [-2,2]$


*

*Is this correct?

*Is there any easier way? This procedure takes a while, and I need to do this in an exam next week. I will have only 15 minutes to do this with Jacobi, Gauss Seidel and cg method

*Does it make a difference for the spectral radius if I do the calculations without the minus, e.g. $T=-D^{-1}R$ and $T=+D^{-1}R$

 A: 1) The general idea is correct, but you've made a mistake calculating the characteristic polynomial of $T$.
In fact, taking the Laplace expansion along the first row we get:
$$\begin{array}{c}p\left( \lambda  \right) = \left| {\begin{array}{*{20}{c}}{ - \lambda }&{\frac{1}{a}}&0\\{\frac{1}{a}}&{ - \lambda }&{\frac{1}{a}}\\0&{\frac{1}{a}}&{ - \lambda }\end{array}} \right| =  - \lambda \left| {\begin{array}{*{20}{c}}{ - \lambda }&{\frac{1}{a}}\\{\frac{1}{a}}&{ - \lambda }\end{array}} \right| - \frac{1}{a}\left| {\begin{array}{*{20}{c}}{\frac{1}{a}}&{\frac{1}{a}}\\0&{ - \lambda }\end{array}} \right|\\ =  - \lambda \left( {{\lambda ^2} - \frac{1}{{{a^2}}}} \right) - \frac{1}{a}\left( { - \frac{\lambda }{a}} \right) =  - {\lambda ^3} + \frac{2\lambda }{{{a^2}}} =  - \lambda \left( {{\lambda ^2} + \frac{2}{a}} \right)\end{array}$$
Hence, the eigenvalues are ${\lambda _1} = 0$, ${\lambda _2} = \frac{{\sqrt 2 }}{a}$ and ${\lambda _3} =  - \frac{{\sqrt 2 }}{a}$. So, in order to assure that $\rho \left( T \right) < 1$, we must have:
$$\left| { \pm \frac{{\sqrt 2 }}{a}} \right| < 1 \Leftrightarrow \left| a \right| > \sqrt 2  \Leftrightarrow a \in \mathbb{R}\backslash  [ { - \sqrt 2 ,\sqrt 2 } ]$$
2) If you really have to find all the values of $a$ for which the method is convergent, then I don't see any other way; you'll have to perform the whole process. However, if it happens that the question only asks for some values of $a$ for which the method is convergent, then you can just check the values of $a$ for which $A$ is strictly diagonal dominant.
3) Remember that if $\lambda$ is an eigenvalue of a matrix $A$, then $-\lambda$ will be an eigenvalue of $-A$. The spectral radius of a matrix $A$ is defined as $\rho \left( A \right) = \mathop {\max }\limits_i \left( {\left| {{\lambda _i}} \right|} \right)$, so what matters is the absolute value of the eigenvalues. Thus, the answer to your last question is: no, it doesn't make any difference; but I don't see any reason not to find the eigenvalues of the correct iteration matrix. Also, note that the iteration matrix for the Jacobi method is $T = - {D^{ - 1}}R$, without the $x$. 
