 Solution

Let ${\Pi }_g$ project onto the good-output subspace and let $|g⟩=|{\Psi }_g⟩$, $|b⟩=|{\Psi }_b⟩$. They are orthonormal because their supporting computational basis strings are disjoint. The verification unitary operator is $V=I-2{\Pi }_g$, so $V|g⟩=-|g⟩$ and $V|b⟩=|b⟩$.

Also $V_0=I-2|0⟩⟨0|$. Since $A|0⟩=|\Psi ⟩$, conjugating gives $A{V}_0{A}^{†}=I-2|\Psi ⟩⟨\Psi |$. ThusBoth reflections preserve the span of $|g⟩$ and $|b⟩$: $V$ is diagonal there, and the other reflection only adds a multiple of $|\Psi ⟩$, already in that span. Direct substitution of $|\Psi ⟩=\sqrt{p}|g⟩+\sqrt{1-p}|b⟩$ yieldsThe minus sign in the first image follows from applying $V$ before the state reflection. In the ordered basis $(|b⟩,|g⟩)$ the matrix isThis exhibits the invariant plane without assumptions about the action of $A$ on the other input basis vectors.