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.
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 $$\rho(BC)=\rho(CB).$$