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

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

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

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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).

\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.

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