# If $P$ is a projection operator, is $1-P$ also a projection operator?

by priyanka   Last Updated October 20, 2019 05:20 AM - source

Show that if $P$ is a (hermitian) projection operator, so are (a) $1-P$ and (b) $$U^{+}PU$$ for any operator $U$

Tags :

Definition of projection operator is $P\circ P = P$, for (a) you can simply expand $(1-P)\circ (1-P)$ and find it holds true. You have to define the notation $U^+$ to get an answer for (b).

P Vanchinathan
October 11, 2014 00:50 AM

\begin{align} (1-P)^2(x) &=(1-P) \circ ((1-P)(x)) \\ \implies (1-P)^2(x) &=(1-P) \circ (x-P(x)) \\ \implies (1-P)^2(x) &=(x-P(x))-(P(x)-P^2(x)) \\ \end{align} Now using the fact that P is a projection operator, i.e. $$P^2=P$$, we get: \begin{align} \implies (1-P)^2(x)&=(1-P)(x)-(P(x)-P(x)) \\ \implies (1-P)^2(x)&=(1-P)(x)\\ \end{align}

Hence we get $$(1-P)^2=(1-P)$$, which implies $$(1-P)$$ is a projection operator.

Anirban Saha
October 20, 2019 05:11 AM