I know that the Gauss-Seidel method converges given that the matrix you want to solve is symmetric positive definite. However, I'm wondering if the "converse" of the statement is true. Namely, if $A$ is a symmetric matrix with positive diagonals and the Gauss-Seidel method converges for all initial guess $x\_0$, is $A$ guaranteed to be positive definite?



Is the matrix positive definite given the Gauss-Seidel method converges?
There is an interesting though partial answer to your question:



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> Under your assumptions, it is not possible that the signature of $A$ be $(n-1,1)$ (exactly one negative eigenvalue).
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Proof: With standard notations, $A=D-E-E^T$ where $E$ is strictly triangular and $D$ diagonal. By assumption, $D>0\_n$. Notice that the assumption of convergence implies that $A$ is invertible.


The iteration matrix is $G=(D-E)^{-1}E^T$. The quadratic form $q(x):=x^TAx$ satisfies the identity
$$q(Gx)+y^TDy=q(x),\qquad y:=x-Gx=(D-E)^{-1}Ax.$$
This implies that $G$ preserves the set $q<0$. If the signature of $A$ is $(n-1,1)$, this set is the union of two opposite convex cones $\pm K$. Thus $\pm G$ preserves $K$. One deduces from Brouwer fixed point theorem that $G$ admits an eigenvector $x\in K$, $Gx=\mu x$. Then $\mu^2q(x)<q(x)<0$ gives $|\mu|>1$, which contradicts the convergence of the Gauss-Seidel method.


*Nota*. The identity about $q$ is the one used to prove the convergence of the method when $A$ is positive definite. More generally, it is used to prove that any method with iteration matrix $L:=M^{-1}N$ where $A=M-N$, such that both $A$ and $M^T+N$ are positive definite, is convergent


