Does $ PX_1P+(1-P)X_1(1-P) =PX_2 P+ (1-P)X_2 (1-P)$ imply $X_1=X_2$？

Question

Let $H$ be a Hilbert space and $X_1,X_2\in B(H)$. If $P$ is a projection on $B(H)$ and $$ PX_1P+(1-P)X_1(1-P) =PX_2 P+ (1-P)X_2 (1-P),$$ then do we have $X_1=X_2$？

Answer 1

No, look at $\Bbb C^2$ and let $X_1=\begin{pmatrix}0& 1\\ 0&0\end{pmatrix}$, $P=\begin{pmatrix}1&0\\0 &0\end{pmatrix}$, $X_2=0$.

Then $X_1P=0$ and $(1-P)X_1=0$, so $PX_1P+(1-P)X_1(1-P)=0=P X_2P +(1-P)X_2(1-P)$.

Answer 2

You already have an example in s.harp's answer. More conceptually, what you are asking is whether equal (block) trace implies equality.

Given a projection $P$, you get to split $H=PH+(I-P)H$. This allows you to write operators $T$ as $2\times 2$ block matrices; the "diagonal entries" are $PTP$ and $(I-P)T(I-P)$. So you have $$ X=\begin{bmatrix} X_{11}&X_{12}\\ X_{21}&X_{22}\end{bmatrix}, \ \ \ Y=\begin{bmatrix} Y_{11}& Y_{12} \\ Y_{21} & Y_{22}\end{bmatrix}. $$ So your question comes down to asking whether $X_{11}+X_{22}=Y_{11}+Y_{22}$ implies $X=Y$.