I have to prove that a symmetric positive definite matrix $A \in \mathbb{R^{2 \times 2}}$ converges for the Jacobi method. Any ideas?

The matrix $A$ is said to be positive definite if $x^t A x > 0 \ (\forall x \neq 0$ )

Thus, we can consider matrix $A$ with the form $\begin{pmatrix} a & c \\ c & b \end{pmatrix}$ with the matrix $D = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$ and $R = \begin{pmatrix} 0 & c \\ c & 0 \end{pmatrix}$ for the iterative Jacobi method which is $\mathbf{x}^{(k+1)} = D^{-1} (\mathbf{b} - R \mathbf{x}^{(k)})$.

The standard convergence condition (for any iterative method) is when the spectral radius (the matrix eigenvalue with supremum absolute value) of the iteration matrix is less than 1: $\rho(D^{-1}R) < 1. $

SOLUTION: The SPD matrix has $a, b > 0$ and $det(A) > 0$. Thus we consider $det(A) = ab -c^2 > 0 \rightarrow \frac{c^2}{ab} < 1$. The eigenvalues of $D^{-1}R$ are obtained from the roots of $\lambda^2 - \frac{c^2}{ab} $ and along with the first formula this implies $|\lambda| < 1$ for all eigenvalues of $D^{-1}R$.


1 Answer 1


One characterization for $A$ to be positive definite is that $a > 0$ and $\det(A) > 0$. Now compute the eigenvalues of $D^{-1} R$, and see that they both have absolute value less than $1$.

  • 1
    $\begingroup$ I'm not allowed to give you a point but I solved it with your help. Thanks! $\endgroup$
    – marlanbar
    Dec 6, 2013 at 15:09

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