I need to show that if $A$ is an operator defined on a euclidean space so that it commutes with every other operator $B$ defined on the same euclidean space, then $A=\lambda I$.
Attempt at a solution
If $[A,B]=0$ for every operator on $E$ (euclidean space) then, in particular, $[A,P_x]=0$, where $P_x$ is a projector operator on $x\in E$ define as: $$P_x y = \frac{x(x,y)}{||x||^2}$$
Then, given an element $y\in E$ we can calculate $[A,P_x]y=0$: $$AP_x y = P_x Ay$$ $$AP_x y = \frac{x(x,Ay)}{||x||^2}$$ $$Ax = \frac{(x,Ay)}{||x||^2}x$$
What we have calculated above gives a explicit expression for $\lambda$. But, this quantity $\frac{(x,Ay)}{||x||^2}$ is dependent , in principle, on $y\in E$.
Nevertheless, we can show that $\lambda_x$ is independent of the choice of $x,y$. Taking two L.I vectors on $E$ $u,v$ we want to show the independence of $\lambda_x$:
Is evident that $Ax = \lambda_x x$ for every $x\in E$, so in particular , $$A(u+v)=\lambda_{u+v}(u+v)$$ Using the linearity of $A$: $$Au+Av=\lambda_{u+v}(u+v)$$ $$\lambda_u u+\lambda_v v=\lambda_{u+v}(u+v)$$ $$(\lambda_{u+v}-\lambda_u)u+(\lambda_{u+v}-\lambda_v)v=0$$ Because of $u,v$ are L.I then $\lambda_{u+v}-\lambda_u=0 \ \wedge \ \lambda_{u+v}-\lambda_v=0$. This implies $\lambda_u = \lambda_v$. This also holds in the case $u,v$ are not L.I. Consider $u=\alpha v$ then $Au=\alpha Au$, this implies $\lambda_u u=\alpha \lambda_v v=\lambda_v u$, then $\lambda_u=\lambda_v$.
This completes the proof.
My question is: Is there a proof that treats the general case for $A$ and $B$ arbitrary?
Any help is appreciated.