This is a question from numerical linear algebra. It originates from iteration method: Suppose $Ax = b$, we split $A = A_1+A_2$, then $A_1x = -A_2x+b$, if $A_1$ is invertible, then $x = -A_{1}^{-1}A_2 x+A_1^{-1}b$. Then we may begin iterations as $x^{k+1} = A_{1}^{-1}A_2 x^{k}+A_1^{-1}b$. It is known that iteration converges if and only if $\rho(A_1^{-1}A_2) <1$. We are now investigating conditions in which we can make the spectral radius less than 1.

Let $A$ be a symmetric positive definite matrix. Splitting $A$ into $A = A_1+A_2$. If $A_1,A_1 - A_2$ are all symmetric positive definite, show $\rho(A_{1}^{-1}A_2)<1$, where $\rho$ represents the spectral radius.

At first, I was about to use Spectral Decomposition of the symmetric matrix. I have successfully shown that $-\rho(A_1) < \lambda(A_2) < \rho(A_1)$, where $\lambda(A_2)$ represents eigenvalue of $A_2$. But I'm stuck here. I would be thankful for any help.


1 Answer 1


By assumptions $A_1\ge A_2$ and $A_1=A-A_2\ge -A_2.$ Therefore multiplying both inequalities by $A_1^{-1/2}$ from the right and the left side simultaneously we get $$I\ge A_1^{-1/2}A_2A_1^{-1/2},\quad I\ge -A_1^{-1/2}A_2A_1^{-1/2}$$ Thus $\|A_1^{-1/2}A_2A_1^{-1/2}\|\le 1.$ The spectral radius of $A_1^{-1}A_2$ coincides with the spectral radius of $A_1^{-1/2}A_2A_1^{-1/2},$ hence the conclusion follows.

Remarks We have used the following two facts. If $B$ is a symmetric matrix then $$\|B\|=\sup\{|\langle Bx,x\rangle |\,:\, \|x\|=1\}$$ Moreover for any matrices $B$ and $C$ the eigenvalues of $BC$ and $CB$ coincide, therefore $\rho(BC)=\rho(CB).$


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