In the ordered <orthonormal basis>
$$
|b\rangle=|\psi_0\rangle|0\rangle,
\qquad
|g\rangle=|\psi_1\rangle|1\rangle
$$
of $\mathcal S$, the <Pauli Z gate> acts as
$$
I_{m-1}\otimes Z
=\begin{pmatrix}1&0\\0&-1\end{pmatrix}
=I_{\mathcal S}-2|g\rangle\langle g|.
$$
It fixes the line $|g\rangle^\perp\cap\mathcal S$ and negates $|g\rangle$, so it is the <reflection in a hyperplane> orthogonal to the good state.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 324 / 2 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-324/2/b/i/solution