Jacobi's Method on $A^{T}$ Is there a theorem, a Lemma, or something, about Jacobi's method on $A^{T}$?
Especially, is it known that if Jacobi's method works correctly on $A\in \mathbb{R}^{n\times n}$ ($x^{(m)}$ converges on $x$), then it does on $A^{T}$ as well? Thanks a lot!
 A: Not in general, but in the following standard case, yes.
Let $A = D + L + U$ for $D$ diagonal, $L$ and $U$ lower and upper triangular. A standard criterion is that the Jacobi method converges when $A$ is strictly diagonally dominant. The reason is that then $D^{-1}(L+U)$ has spectral radius $<1$.
Suppose more generally then that you know the Jacobi method converges for $A$ since $D^{-1}(L+U)$ has spectral radius $<1$. The characteristic polynomial of $D^{-1}(L+U)$ is the same as for $(L+U)D^{-1}$ and also the same for the transpose $D^{-1}(U^\top+L^\top)$, so the spectral radius of $D^{-1}(U^\top+L^\top)$ is $<1$. But that says the Jacobi method for $A^\top = D + U^\top + L^\top$ converges.
This argument guides the search for a counterexample. The Jacobi method simply finds the fixed point $u$ of $u = -D^{-1}(L+U)u + D^{-1}b$. So, first pick $A$ where the spectral radius of $D^{-1}(L+U)$ is $>1$ but where there is an eigenvector $v$ with eigenvalue less than $1$. Now pick $b$ so that $D^{-1}b \in \mathrm{span}\{v\}$ and pick your initial guess $u_0 \in \mathrm{span}\{v\}$. Then this fixed point iteration will have a solution, so the Jacobi method will converge for $(A, b, u_0)$. If moreover $-D^{-1}(U^\top+L^\top)u_0$ has components with eigenvalue $>1$, you would not expect $u=-D^{-1}(U^\top+L^\top)u + D^{-1}b$ to converge, so you would not expect the Jacobi method to converge for $(A^\top, b, u_0)$. I leave it to you to find an explicit counterexample--look at $3 \times 3$ matrices.
