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?

Your claim seems to be not true. Let $A=\begin{pmatrix}1&0\0&0\end{pmatrix}$ and $E=\begin{pmatrix}0&0\0&-1\end{pmatrix}$ and $\lambda=1$. — Michael Hoppe, Apr 11 '18 at 11:51
@MichaelHoppe: In this case $(A-\lambda E)^{-1} = I$ and so the question is simply whether $[A,E]=0$ which is clearly true because $AE=EA=0$. What were you driving at? (The result does hold in general, as follows from the line of reasoning I give.) — not all wrong, Apr 11 '18 at 11:58

score 2 · Answer 1 · answered Apr 11 '18 at 11:47

2

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 $XE$?

answered Apr 11 '18 at 11:47

not all wrong

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I dont understand ...........It seems you have done nothing beyond my progress. – Myshkin Apr 11 '18 at 11:54
@Wow Try to compute both sides of [XE, XA-\lambda XE] = [XE,I]. What do you get? (PS: Can the downvoter please explain?) – not all wrong Apr 11 '18 at 12:00
$XEXA-\lambda XEXE-XAXE+\lambda XEXE$ – Myshkin Apr 11 '18 at 12:07
The $XEXA-XAXE = [XE,XA]$ terms are what you are trying to evaluate, right? And you can save time with the other terms: $[XE, XE] = 0$ automatically! – not all wrong Apr 11 '18 at 12:10
Since $E$ is not given invertible how can you consider commutator of $XE$ ? – user371231 Oct 01 '18 at 20:29
@user371231 "Commutator" in the sense of rings, not in the sense of groups! That does not require inverting anything. (https://en.wikipedia.org/wiki/Commutator#Ring_theory) – not all wrong Oct 02 '18 at 08:16

score 1 · Answer 2 · answered Apr 11 '18 at 12:09

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

how to show these two matrices commutes

2 Answers2