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I am trying to prove why Ax=b where $$ A= \begin{bmatrix} 1 & 0 & 1\\ -1 & 1 & 0 \\ 1 & 2 & -3 \\ \end{bmatrix} b= \begin{bmatrix} 0\\0\\0\\ \end{bmatrix} $$ converges for Jacobi and does not converge for Gauss-Seidel. So far, I found that the spectral radius for the Jacobi iteration matrix is less than 1 and that the matrix is diagonally dominant. However, for the Gauss-Seidel method I am having a harder time understanding where to start. I was thinking of finding the spectral radius of the iteration matrix $-(D+L)^{-1}*U$, but I don't know if this the correct formula as it was one I found in a random online source and I do not see it in my class textbook. Any help would be appreciated

user1106787
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    I would imagine you have to have a solution vector (RHS) values for each row. Unless you are after just a matrix inverse. – NoChance Mar 07 '23 at 19:54
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    The formula is correct. For convergence, you have to look at the spectral radius. If the spectral radius is less than $1$, you get convergence. If it is larger than $1$, you get divergence. If it is exactly one, you have to find some argument yourself, whether it converges, diverges. This might be different for different first iterates. – Hyperbolic PDE friend Mar 07 '23 at 20:01
  • yes I believe so, It seems very similar to what I did. Thank you – user1106787 Mar 08 '23 at 00:35
  • @user1106787 Your answer needs some clarification... I suppose you want to establish convergence, regardless of the initial approximation, but you do not explicitly say so. Note that if you take as initial approximation $x^{(0)} = 0$, both methods converge. – PierreCarre Mar 08 '23 at 11:46
  • @PierreCarre, the initial guess is (1,1,1), I saw that the solutions bounced around between -1,-1,-1 and 1,1,1. Is there a different way without the spectral radius to show why this diverges or do we use the spectral radius in conjunction with the initial guess – user1106787 Mar 08 '23 at 19:27

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