# Is Gauss-Seidel guaranteed to converge on *semi* positive definite matrices?

I know that the Gauss-Seidel method is guaranteed to converge given that the matrix you want to solve is positive definite. I've looked at the proofs of convergence, and it appears that one cannot change the assumption so that the smallest eigenvalue is zero, without coming up with a totally new proof. This makes sense, since of course a PSD matrix is singular.

However, assuming that $b$, the right hand side of the equation $A \cdot x = b$ is in $A$'s image (and that $A$ is PSD) is Gauss-Seidel guaranteed to converge?

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This is an interesting question, to which the answer is positive.

Here is the proof. Of course, for the method to make sense, we must assume that the diagonal $D>0$. The notations below are borrowed from Section 12.3.2 of my book Matrices (2nd edition, Springer-Verlag, GTM 216). Let $G=(D-E)^{-1}E^T$ the iteration matrix. One checks easily that $\ker A$ is the eigenspace of $G$, associated with the eigenvalue $\lambda=1$. The corresponding eigenspace for $G^T$ is $(D-E)^T\ker A$, fro which we find that $G$ has the invariant subspace $(D-E)^{-1}R(A)$. This can be verified directly with the help of the formula $G(D-E)^{-1}A=(D-E)^{-1}AG$. It turns out that $1$ is semi-simple: if $Gw=w+V$ and $Gv=0$, we obtain $v^T(D+A)v=0$, hence $v=0$. Therefore $${\mathbb R}^n=\ker A\oplus (D-E)^{-1}R(A)$$ is a decomposition into $G$-invariant subspaces.

There remains to prove that for every vector $x^0$, the sequence $x^m:=G^mx^0$ is convergent. Let us decompose $x^m=y^m+z^m$, according to the invariant subspaces above. We have $y^m=y^0$, so this part is obviously convergent. There remains to prove that the spectral radius of the restriction $g$ of $G$ to $(D-E)^{-1}R(A)$ is smaller than $1$. This is proved exactly the same way as in Lemma 20 of the reference. We have to prove that if $x\in(D-E)^{-1}R(A)$, then $(Gx^T)A(Gx)\le x^TAx$, and equality implies $x=0$. With $y=(D-E)^{-1}Ax$, we have $$(Gx^T)A(Gx)-x^TAx=-y^TDy\le0.$$ If the right-hand side vanishes, then $y=0$, which means $x\in\ker A$, hence $x=0$. Q.E.D

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Here are some general comments (drawn from the nice book: Accuracy and stability of numerical algorithms by N. J. Higham) that might prove helpful.

Suppose $A$ is singular. Split $A=M-N$, where $M$ is nonsingular. Iterate

$$Mx_{k+1}=Nx_k+b.$$

Define the "action" matrix $H=M^{-1}N$, using which the above iteration becomes $$x_{k+1} = Hx_k + M^{-1}b.$$

Now, unroll this loop starting at $x_1$, to get

$$x_{k+1} = H^kx_1 + \sum_{j=0}^{k-1} H^jM^{-1}b.$$

For the sequence $(x_k)$ to be convergent, a necessary condition is that the sequence $(H^k)$ converges. This, in turn can be ensured if $rank(I-H) = rank( (I-H)^2 )$. Eventually, though, the limit of $(x_k)$ will depend on where we started, i.e., $x_1$.

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According to section 17.6 of "Solving PDEs in C++: numerical methods in a unified object-oriented approach" Gauss-Seidel does not necessarily converge when the matrix is indefinite.

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But SPD's aren't considered strictly indefinite. Usually they are yield the same results as SD, e.g conjugate gradient will converge to the correct solution for SPD as well, as long as the solution is in the matrix's image. – noam Nov 13 '11 at 5:39

What do you think about the projected gauss Seidel method? The projected Gauss-Seidel method for a mixed linear complemetarity problem (MLCP) converges if the quadratic matrix of the quadratic optimization problem related to MLCP is definite positive. If this matrix is only semidefinite positive, is Gauss-Seidel guaranteed to converge? Or is there a condition for the convergence?

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Please to not post new questions as answers. Create a question of your own instead! – Alex Degtyarev Mar 14 '14 at 8:41