TITLE: Proof involving Gauss–Seidel method.
QUESTION [1 upvotes]: I've got a symmetric positive definite matrix $A$ that I decompose into $A=U+R$ where $U$ is the upper triangular portion of $A$ including the diagonal and $R$ is $A-U$. I've shown that $$x^TAx = (1+\lambda)x^TUx$$ with $x$ an eigen-vector of $U^{-1}R$ with eigen-value $\lambda$ and from the previous I know $\lambda \neq -1$. Now I'm supposed to prove $$x^TAx \leq 2x^TUx$$ to show that $|\lambda| < 1$ and the method converges. I'm having a bit of trouble showing this inequality holds $$x^TAx \leq 2x^TUx$$
Can anybody point me in the right direction?
Thanks in advance.

REPLY [4 votes]: Hints.  Since $A$ is symmetric, we have $U = R^T + D$, where $R^T$ is the portion of $A$ above the diagonal, and $D$ is the diagonal part.  Since $A$ is positive definite, the diagonal entries of $D$ are positive.
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