Prove that if $A$ is diagonally dominant and if $Q$ is chosen as in the Jacobi method, then $\rho(I-Q^{-1}A)<1$

Question

Prove that if $A$ is diagonally dominant and if $Q$ is chosen as in the Jacobi method, then $\rho(I-Q^{-1}A)<1$

I know that $\rho(A)=\inf_{\|.\|}\|A\|$ and \begin{equation} ||I-Q^{-1}A||_\infty = \max_{1\leq i\leq n}\sum\limits_{j=1,j\neq i}^n \left|\frac{a_{ij}}{a_{ii}}\right| \end{equation}, but I do not know what else I can do to show what I want, could someone help me please?

Answer 1

This property can be easily shown with the help of this property of the spectral radius: For the spectral radius the following holds for any matrix $A \in \mathbb R^{n \times n} $ and any (matrix) norm $\Vert \cdot \Vert$: $$\rho(A) \leq \Vert A \Vert. $$

Thus, for your chosen norm and a strictly diagonally dominant matrix $A$ you get convergence since $$\sum\limits_{j=1,j\neq i}^n \left|\frac{a_{ij}}{a_{ii}}\right| < 1 \: \forall \: i$$ and thus $$\rho \big(I-Q^{-1}A\big) \leq \big\Vert I-Q^{-1}A \big\Vert_\infty < 1 \checkmark$$