# How do I prove that $P_\pm=\frac12(1\pm U)$ if $U^2=I$?

Suppose I have an $$n$$-qubit Hermitian operator $$U$$ such that $$U^2=I$$.

The projection operators with eigenvalue $$+1$$ and $$−1$$ are $$P_+$$ and $$P_-$$.

How can I prove that $$P_+=\frac{1}{2}(1+U)$$ and $$P_-=\frac{1}{2}(1-U)$$?

I think $$\sum P=1$$ but I have no idea to get this conclusion.

First, we can start with $$U = P_+ - P_-$$, since the Hermitian is the sum of the projection operators of the eigenspaces scaled by their eigenvalues. If $$U^2 = I$$, that means $$I = (P_+ - P_-)(P_+ - P_-) = P_+P_+ - P_+P_- - P_-P_+ + P_-P_-$$. Since the two projectors correspond to orthogonal eigenspaces, operating on themselves doesn't change them and one operating on the other gets a zero, so we get $$P_+ + P_- = I$$. With $$P_+ + P_- = I$$ and $$P_+ - P_- = U$$, we can add the two equations to get $$P_+ = \frac{1}{2}(I + U)$$ or subtract the $$U$$ from the $$I$$ to get $$P_- = \frac{1}{2}(I - U)$$.
• In the first line you're assuming that the projectors are rank one but the OP says that $U$ is an $n$-qubit operator. In particular if $n>1$ then they cannot be rank $1$ as $U$ needs full rank to square to the identity. Dec 17, 2020 at 12:47