how to show these two matrices commutes

Question

could any one tell me how to show $(A-\lambda E)^{-1}E$ and $(A-\lambda E)^{-1}A$ commutes where $\lambda$ is chosen in way such that $(A-\lambda E)$ is invertible?

I tried $I=(A-\lambda E)^{-1}(A-\lambda E)=(A-\lambda E)(A-\lambda E)^{-1}$

comparing the equation we get $$E(A-\lambda E)^{-1} =(A-\lambda E)^{-1} E\dots(1)$$ and $$A(A-\lambda E)^{-1} =(A-\lambda E)^{-1} A\dots(2)$$

now I am thinking of multiplyinf ridesides of both equation:

$(A-\lambda E)^{-1} E (A-\lambda E)^{-1}A=E(A-\lambda E)^{-1}A(A-\lambda E)^{-1}=$

but i am not sure it will help much?

Answer 1

Let $X = (A- \lambda E)^{-1}$. You have $$X(A-\lambda E) = XA-\lambda XE = I$$ What can you say about the commutator of both sides with $XE$?

Answer 2

$$\begin{align}(A-\lambda E)^{-1}E(A-\lambda E)^{-1}A&=(A-\lambda E)^{-1}E(A-\lambda E)^{-1}(A-\lambda E)+(A-\lambda E)^{-1}E(A-\lambda E)^{-1}\lambda E\\ &=(A-\lambda E)^{-1}E+(A-\lambda E)^{-1}(\lambda E)(A-\lambda E)^{-1}E\\ &=[I+(A-\lambda E)^{-1}\lambda E](A-\lambda E)^{-1}E\\ &=[(A-\lambda E)^{-1}(A-\lambda E)+(A-\lambda E)^{-1}\lambda E](A-\lambda E)^{-1}E\\ &=(A-\lambda E)^{-1}A(A-\lambda E)^{-1}E\end{align}$$