Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 4 / 22F / iv / Solution

If $w\perp X$, then $⟨w,(I-U)g⟩=0$ for all $g$. The adjoint identity gives $(I-U^{\ast })w=0$, so $U^{\ast }w=w$. Multiplying by $U$ givesSince $X\subseteq V=\overline{X}$, $w\perp V$ implies $w\perp X$ and yields the same result. Conversely a fixed vector is perpendicular to $X$ by this calculation, identifying $V^{\perp }$ as the fixed subspace.