Showing measurement of a Hermitian Unitary operator gives final states as eigenvectors

This is related to exercise 4.34,

The operation described can be written as $$(H \otimes I)C^1(U)(H \otimes I)(|0\rangle \otimes |\psi\rangle)$$

I can get to the point where the state of the system is given by:

$$|0\rangle \otimes(I+U)|\psi\rangle + |1\rangle \otimes(I-U)|\psi\rangle$$,

where $$U$$ is a Hermitian unitary with eigenvalues $$-1$$ and $$+1$$ with corresponding eigenvectors $$|\lambda_-\rangle$$ and $$|\lambda_+\rangle$$ respectively.

however I am stuck on the final part that when measuring $$q_0$$ the post-measurement state is given by the corresponding eigenvector of $$U$$. This reduces down to to showing that

$$(I-U)|\psi\rangle = |\lambda_-\rangle$$

$$(I+U)|\psi\rangle = |\lambda_+\rangle$$

I have tried using the spectral decomp of $$U$$ however I can't seem to get it to lead anywhere. My current trail of thought (not sure if correct) is if taking the density of the system, $$(I\pm U)$$ reduces down to projectors for $$|\lambda_\pm\rangle$$, s.t. $$P_{\pm}|\psi\rangle = c_{\pm}\lambda_\pm$$.

-- Update --

An answer using projectors (as suspected) is using the fact that for $$U$$ to be unitary and Hermitian then $$U = (2P - I)$$ for an orthogonal projector $$P$$ (https://math.stackexchange.com/questions/57148/matrices-which-are-both-unitary-and-hermitian), hence $$(I+U)$$ and $$(I-U)$$ reduce down to a projector $$P$$ and its orthogonal complement $$2(I-P)$$, thus projecting $$\psi$$ onto the eigenvectors.

Just need to write it out: \begin{align} (I-U)|\psi\rangle &= I\big(c_+|\lambda_{-}\rangle + c_{-}|\lambda_{-}\rangle\big) - U\big(c_+|\lambda_{-}\rangle + c_{-}|\lambda_{-}\rangle\big) \\ &= c_+|\lambda_{+}\rangle + c_{-}|\lambda_{-}\rangle - \big(c_+|\lambda_{+}\rangle - c_{-}|\lambda_{-}\rangle\big) \\ &= c_+|\lambda_{+}\rangle + c_{-}|\lambda_{-}\rangle - c_+|\lambda_{+}\rangle + c_{-}|\lambda_{-}\rangle \\ &= c_{-}|\lambda_{-}\rangle \end{align}
I've used $$U(c_{-}|\lambda_{-}\rangle) = -c_{-}|\lambda_{-}\rangle$$.
• I'm a little confused as to how/where you decomposed $\psi$, unless you're saying that the eigenvectors of $U$ are $|\lambda_+\rangle = |0\rangle$ and $|\lambda_-\rangle = |1\rangle$ Apr 28 '20 at 19:26