> [!theorem] > > Let $\ch$ be a [[Hilbert Space|Hilbert space]] and $A \in K(\ch, \ch)$ be a [[Compact Operator|compact]], [[Self-Adjoint Operator|self-adjoint]] operator. If $\lambda \in \complex$ with $\lambda \ne 0$ and $y \in \ch$, then the equation > $ > (A - \lambda I)x = y > $ > has a solution if and only if $y \in \ker(A - \lambda I)^\perp$. > > *Proof*. Suppose that $\lambda \in \sigma(A)$ with $\lambda \ne 0$. Let $\seq{e_j}$ be an orthonormal eigenbasis for $A$ and $\seq{\lambda_j}$ be the corresponding eigenvalues. Write > $ > x = \sum_{j = 1}^\infty x_je_j \quad y = \sum_{j = 1}^\infty y_j e_j > $ > This induces an equation $(\lambda_j - \lambda)x_j = y_j$. Since $\lambda \in \sigma(A)$, there exists finitely many $n$s such that $\lambda_n = \lambda$. The span of the corresponding eigenvectors is the kernel of $(A - \lambda)$. As long as $y_n = 0$ for these $n$s, there is a uniform lower bound on $\abs{\lambda_j - \lambda}$ (because $0$ is the only accumulation point of $\sigma(A)$), which allows solving the coefficients directly.