5
$\begingroup$

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 $D^{-1}R$.

$\endgroup$
3
$\begingroup$

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 $1$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ I'm not allowed to give you a point but I solved it with your help. Thanks! $\endgroup$ – marlanbar Dec 6 '13 at 15:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.