• Mar 17th 2009, 06:28 PM
DJDorianGray
Problem: $\displaystyle P_1$ and $\displaystyle P_2$ are two self-adjoint projections. Is it true that $\displaystyle \mbox{R} (P_1) \subseteq \mbox{R} (P_2) \Longrightarrow P_1 \leq P_2$? (Note that R() indicates the range).

I believe it is not true, but I cannot find any counterexamples or way to disprove.
One fact that might be useful to consider is that for a self-adjoint projection $\displaystyle \mbox{R}(P) = \mbox{N}(P)^\perp$
• Mar 18th 2009, 03:15 AM
Opalg
Quote:

Originally Posted by DJDorianGray
Problem: $\displaystyle P_1$ and $\displaystyle P_2$ are two self-adjoint projections. Is it true that $\displaystyle \mbox{R} (P_1) \subseteq \mbox{R} (P_2) \Longrightarrow P_1 \leq P_2$? (Note that R() indicates the range).

I believe it is not true, but I cannot find any counterexamples or way to disprove.
One fact that might be useful to consider is that for a self-adjoint projection $\displaystyle \mbox{R}(P) = \mbox{N}(P)^\perp$

If $\displaystyle \mbox{R} (P_1) \subseteq \mbox{R} (P_2)$ then $\displaystyle P_2P_1 = P_1$. Taking adjoints, you see that $\displaystyle P_1P_2= P_1$. Thus $\displaystyle P_1$ and $\displaystyle P_2$ commute, and it follows (as you can easily check) that $\displaystyle (P_2 - P_1)^2 = P_2 - P_1$. Thus $\displaystyle P_2 - P_1$ is also a selfadjoint projection and is therefore positive. Hence $\displaystyle P_1 \leqslant P_2$.