The reduced resolvent satisfies the standard projector and inverse identities on the complementary invariant subspace

Every vector $v$ decomposes uniquely as $v=(y^{\ast }v)x+(v-(y^{\ast }v)x)$, with the second term in $\ker y^{\ast }$. If $z\in \ker y^{\ast }$ and $(A-\lambda I)z=0$, then $z$ is a right eigenvector for the simple eigenvalue $\lambda$, so $z=cx$ for some scalar $c$. Applying $y^{\ast }$ gives $0=y^{\ast }z=c{y}^{\ast }x=c$, hence $z=0$. Therefore the restriction of $A-\lambda I$ to $\ker y^{\ast }$ is injective, and since both domain and codomain have dimension $n-1$, it is bijective.

Define $S$ to be the inverse of that restriction on $\ker y^{\ast }$ and to vanish on $\mathrm{span}\{x\}$. Then $SP=PS=0$ by construction. For $v=\alpha x+z$ with $z\in \ker y^{\ast }$, [F1] gives $Pv=(y^{\ast }v)x=(\alpha +y^{\ast }z)x=\alpha x$, so $(I-P)v=z$. Therefore $S(A-\lambda I)v=S(A-\lambda I)z=z=(I-P)v$, and similarly $(A-\lambda I)Sv=(A-\lambda I)Sz=z=(I-P)v$. Thus all displayed identities hold.

If $S^{\mathrm{\prime }}$ is another linear map with the same identities, then $S^{\mathrm{\prime }}x=0$ because $S^{\mathrm{\prime }}P=0$ and $Px=x$. On $\ker y^{\ast }$ one has $(A-\lambda I)S^{\mathrm{\prime }}z=z=(A-\lambda I)Sz$, and the injectivity from step 1.1 gives $S^{\mathrm{\prime }}z=Sz$. Hence $S^{\mathrm{\prime }}=S$, so the reduced resolvent of The reduced resolvent, or group inverse, on the complementary invariant subspace of a simple eigenvalue is well defined and unique.