Show that spectral radius is $\rho(M^{-1}K)\ge1$

Given a splitting of symmetric singular matrix by A=M-K where M is nonsingular. Show that spectral radius of $\rho(M^{-1}K)\ge1$. Recall spectral radius is $\rho(A)=\max_i|\lambda_i|$ Hint: any zero eigenvector of A, Au=0, is also an eigenvector of $M^{-1}K$.

My thoughts on the problem are to show that the iteration $x_{n+1}=M^{-1}Kx_n+c$ will not converge, where $c=M^{-1}b.$

Solution 1:

Your idea is good.

Let $Ax=b\neq 0,$ and take your initial guess to be your eigenvector $x_0$ with eigenvalue $0$ so that $Ax_0=0.$ Then, $$M^{-1}Kx_0=M^{-1}(M-A)x_0=x_0-M^{-1}Ax_0=x_0.$$

Thus, for any $n\geq 0,$ $$x_{n+1}=x_0.$$ This converges to $x=x_0$, but $Ax=Ax_0=0\neq b.$ This implies that the iteration does not converge to the solution for the given $x_0$ and $b$, which implies that $\rho(M^{-1}K)\geq 1.$