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 $U$

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.

Updated April 17, 2017 18:20 PM

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