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 $D^{-1}R$.
Best Answer
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 $1$.