 Solution

Put $E=\mathrm{im}S$, a finite-dimensional subspace, and decompose $H=E\oplus E^{\perp }$. The restrictionis injective, because a nonzero vector in its kernel would be a $\lambda$-eigenvector of $S$. By finite-dimensionality it is therefore surjective.

Given $y=y_E+y_{\perp }$, first put $x_{\perp }=-y_{\perp }/\lambda$. Then $S{x}_{\perp }\in E$, so choose $x_E\in E$ such thatFor $x=x_E+x_{\perp }$ we obtain $(S-\lambda I)x=y$. Hence $S-\lambda I$ is surjective. This is the finite-rank Fredholm alternative.