A level shift preserves a fixed occupied projector exactly under block ordering

Assume $PF=FP$. Then both $\mathrm{ran}P$ and $\mathrm{ran}(I-P)$ are invariant under $F$: if $x=Px$, then $P(Fx)=F(Px)=Fx$, and the analogous identity holds when $Px=0$. Relative to their orthogonal direct sum, $F$ is block diagonal and $F_s$ leaves the occupied block unchanged while adding $sI$ on the virtual block.

By [F2], the eigenvalues of $F_s$ are exactly those of $F_{occ}$ together with those of $F_{vir}$ shifted by $s$. If $\max \sigma (F_{occ})<\min \sigma (F_{vir})+s$, the lowest $r$ eigenvalues are precisely the occupied block and the gap is strict; [L1] gives ${\mathrm{\Pi }}_r(F_s)=P$.

Conversely, if ${\mathrm{\Pi }}_r(F_s)=P$, its defining strict gap makes $P$ the sum of the eigenspace projectors of $F_s$ for the lowest $r$ eigenvalues, so $P$ commutes with $F_s$. Since $[s(I-P),P]=0$, this gives $[F,P]=0$, and the two restrictions are the occupied and virtual blocks of $F$. Their eigenvalues in $F_s$ are respectively $\sigma (F_{occ})$ and $\sigma (F_{vir})+s$. Because $P$ is exactly the lowest-$r$ spectral projector with a strict gap, the largest occupied eigenvalue is strictly smaller than the smallest shifted virtual eigenvalue; hence the displayed ordering holds.