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It is an exercise in Hunter's Applied Analysis:

Let A and B self adjoint operators,and let $AB=BA$,$A^2=B^2$.$P$ is the projection operator on the null space $L$ of $A-B$.I want to prove the following three :

1)Any bounded linear operator that commmutes with $A-B$ commutes with $P$

2)$Ax=0$ $\rightarrow$ $Px=x$

3)$A=(2P-1)B$

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what have you tried? –  Alex R. Dec 9 '12 at 21:53

1 Answer 1

up vote 2 down vote accepted

1) This part of the question holds even without requiring $AB=BA$ or $A^2=B^2$. Let $T$ be a selfadjoint operator that commutes with $A-B$. Then, for any vector $v$ we have $Pv=(A-B)w$ for some $w$, so $$ TPv=T(A-B)w=(A-B)Tw. $$ This implies that $TP=PTP$. As we were assuming $T=T^*$, we have $PT=(TP)^*=(PTP)^*=PT^*P=PTP=TP$. So $T$ commutes with $P$. If you now consider a general $T$, it is written as a linear combination of selfadjoints (i.e. real and imaginary parts), and the result follows.

2) This one is not true. Let $$ A=B=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ x=\begin{bmatrix}0\\1\end{bmatrix}. $$ Then $AB=BA$, $A^2=B^2$. We also have $A-B=0$, and so $P=0$. Then $$ Ax=0,\ Px=0\ne x. $$

3) Since $AB=BA$, the operators $A$ and $B$ can be simultaneously diagonalized, i.e. there exist spectral measure $E$ and Borel functions $f,g$, which are zero off the support of $E$, such that $$ A=\int\,f\,dE,\ \ B=\int\,g\,dE. $$ From $A^2=B^2$ we get that $f^2=g^2$. Then there exists measurable $h$, taking values in $\{-1,1\}$, such that $g=hf$. So $ A-B=\int\,(1-h)f, $ and $P=E\{h=1\}$. So $$ 2P-I=2\int_{h=1}\,1\,dE-\int\,1\,dE=\int_{h=1}],1\,dE-\int_{h=-1}\,1\,dE=\int h\,dE. $$ So, noting that $h=h^{-1}$, $$ (2P-I)B\int hg\,dE=\int\,f\,dE=A. $$

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it is the null space of A-B –  89085731 Dec 10 '12 at 20:34
    
@89085731: I have no idea what you are talking about. –  Martin Argerami Dec 11 '12 at 1:11

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