[Math] Showing that the Jacobi method doesn’t converge with $A=\begin{bmatrix}2 & \pm2\sqrt2 & 0 \\ \pm2\sqrt2&8&\pm2\sqrt2 \\ 0&\pm2\sqrt2&2 \end{bmatrix}$

matricesnumerical linear algebranumerical methodssymmetric matrices

An exercise asks to find the values of $a\in\Bbb R$ such that, given a linear system with $A_a=\begin{bmatrix}2 & a & 0 \\ a&8&a \\ 0&a&2 \end{bmatrix}$, the Jacobi method converges.

I've found $\rho(A_a)=5-\sqrt{9+2a^2}$ and setting it to be smaller than $1$ I get $\lvert a\rvert <2\sqrt2$. However $\rho(A_a)<1$ is only sufficient for convergence, not necessary. Right?

So, I can rule it out for $\rho(A_a)>1$: in this case $A_a$ is tridiagonal and non-singular, so the Jacobi and Gauss-Seidel methods both either converge or diverge. Since it's symmetric and $a_{ii}>0$ for $i=1,2,3$, they both converge if and only if $A_a$ is positive definite, i.e. $\lvert a\rvert <2\sqrt2$.

What about $a=\pm2\sqrt2$? Does it hold in general that if the spectral radius is strictly larger than $1$, then the Jacobi and Gauss-Seidel methods don't converge?

Best Answer

The condition about the spectral radius < 1 is a sufficient and mandatory condition, for more detail see this my answer, but the matrix to check is not the matrix A of the linear system $Ax=b$. You must control the iteration matrix.

For the Jacobi method you use an additive splitting of the matrix $A$ with the form

$$ A = D + R $$

where, according with the nomenclature used by the wiki page linked,

  • $D$ is the diagonal matrix, $d_i = a_{ii}$
  • $R$ is the other entry, $r_{ii} = 0 \quad r_{ij} = a_{ij}$

In this method $$ \text{iteration matrix is } \quad D^{-1}R = D^{-1} (D-A) = I - D^{-1}A $$ and you must check $\rho(D^{-1}R)$. In this answer I report also another theorem for the convergence, in this case the condition is sufficient , and not mandatory.

At the end note that with the Gauss–Seidel method you have got a different iteration matrix and the theorem about the spectral radius hold for all the stationary iterative methods.