[proofplan] We prove the four structural properties of $\sigma(K)$ in sequence. First, $0 \in \sigma(K)$ because invertibility of $K$ would force the identity to be compact, which is impossible in infinite dimensions. Second, every nonzero spectral value is an eigenvalue via the [Fredholm Alternative](/theorems/219). Third, each nonzero eigenvalue has finite-dimensional eigenspace, again by Fredholm theory. Fourth, nonzero eigenvalues cannot accumulate away from $0$, as shown by a Riesz lemma argument producing a sequence with no convergent subsequence, contradicting compactness. [/proofplan] [step:Show $0 \in \sigma(K)$ because $K$ cannot be invertible on an infinite-dimensional space] Since $X$ is infinite-dimensional and $K$ is compact, $K$ cannot be bijective. If $K$ were invertible, then the identity $I = K \cdot K^{-1}$ would be compact (as the composition of the compact operator $K$ with the bounded operator $K^{-1}$). But the identity operator $I: X \to X$ is compact if and only if $X$ is finite-dimensional (compactness of $I$ means the closed unit ball of $X$ is compact, which characterises finite-dimensional spaces by the Riesz theorem). Since $X$ is infinite-dimensional, this is a contradiction. Hence $K$ is not bijective and $0 \in \sigma(K)$. [/step] [step:Show every $\lambda \in \sigma(K) \setminus \{0\}$ is an eigenvalue via the Fredholm Alternative] Fix $\lambda \in \mathbb{C}$ with $\lambda \neq 0$ and suppose $K - \lambda I$ is injective. Factor: \begin{align*} K - \lambda I = -\lambda(I - \lambda^{-1}K). \end{align*} The operator $\lambda^{-1}K$ is compact (a scalar multiple of a compact operator). By the [Fredholm Alternative](/theorems/219) (Case A applied to the compact operator $\lambda^{-1}K$), injectivity of $I - \lambda^{-1}K$ implies surjectivity. Hence $K - \lambda I$ is bijective and $\lambda \in \rho(K)$. Taking the contrapositive: if $\lambda \in \sigma(K)$ and $\lambda \neq 0$, then $K - \lambda I$ is not injective, so there exists $v \neq 0$ with $Kv = \lambda v$, meaning $\lambda$ is an eigenvalue. [/step] [step:Show each nonzero eigenvalue has finite-dimensional eigenspace] For $\lambda \neq 0$, the eigenspace is \begin{align*} N(K - \lambda I) = N(I - \lambda^{-1}K). \end{align*} The operator $\lambda^{-1}K$ is compact. By the [Fredholm Alternative](/theorems/219) (specifically, the fact that $N(I - K')$ is finite-dimensional whenever $K'$ is compact, proved in Step 1 of the Fredholm Alternative proof), $\dim N(I - \lambda^{-1}K) < \infty$. [guided] The eigenspace $N(K - \lambda I) = N(I - \lambda^{-1}K)$ consists of vectors $u$ satisfying $u = \lambda^{-1}Ku$. The closed unit ball $B = \{u \in N(K - \lambda I) : \|u\|_X \leq 1\}$ satisfies \begin{align*} B = \lambda^{-1}K(B), \end{align*} which is precompact because $K$ is compact and $B$ is bounded. Since scalar multiplication by $\lambda^{-1} \neq 0$ preserves precompactness, $B$ itself is precompact. A normed space whose closed unit ball is precompact must be finite-dimensional (by the Riesz lemma characterisation), so $\dim N(K - \lambda I) < \infty$. This is in sharp contrast with bounded (but non-compact) operators, which can have infinite-dimensional eigenspaces. For example, the identity operator on an infinite-dimensional space has eigenvalue $1$ with eigenspace equal to the entire space. [/guided] [/step] [step:Show nonzero eigenvalues can only accumulate at $0$ via a Riesz lemma argument] Suppose for contradiction that there exist distinct eigenvalues $\lambda_k \in \sigma(K) \setminus \{0\}$ with $\lambda_k \to \lambda \neq 0$. For each $k \in \mathbb{N}$, let $e_k \in X$ be a unit eigenvector satisfying $Ke_k = \lambda_k e_k$. Eigenvectors corresponding to distinct eigenvalues of a linear operator are linearly independent, so the subspaces \begin{align*} X_k := \operatorname{span}\{e_1, \ldots, e_k\} \end{align*} form a strictly increasing chain $X_1 \subsetneq X_2 \subsetneq X_3 \subsetneq \cdots$. By Riesz's lemma, for each $k \geq 2$ there exists $u_k \in X_k$ with $\|u_k\|_X = 1$ and $\operatorname{dist}(u_k, X_{k-1}) \geq 1/2$. For $k > m$, the element $Ku_k \in X_k$ (since $K$ maps $X_k$ to itself: $K$ sends each basis vector $e_j$ to $\lambda_j e_j \in X_k$). Moreover, $Ku_k - Ku_m \in X_k$ because $Ku_m \in X_m \subseteq X_{k-1} \subset X_k$. Writing $Ku_k = \lambda_k u_k + (Ku_k - \lambda_k u_k)$, note that $Ku_k - \lambda_k u_k \in X_{k-1}$ (expanding $u_k = \sum_{j=1}^{k} \alpha_j e_j$, the term $Ku_k - \lambda_k u_k = \sum_{j=1}^{k} \alpha_j(\lambda_j - \lambda_k) e_j$, and the $j = k$ term vanishes). Therefore \begin{align*} \frac{Ku_k}{\lambda_k} = u_k + \frac{Ku_k - \lambda_k u_k}{\lambda_k}, \end{align*} where the second term lies in $X_{k-1}$. For $k > m$, the difference $\frac{Ku_k}{\lambda_k} - \frac{Ku_m}{\lambda_m}$ can be written as $u_k + w$ for some $w \in X_{k-1}$, giving \begin{align*} \left\|\frac{Ku_k}{\lambda_k} - \frac{Ku_m}{\lambda_m}\right\|_X \geq \operatorname{dist}(u_k, X_{k-1}) \geq \frac{1}{2}. \end{align*} Since $\lambda_k \to \lambda \neq 0$, the sequences $(Ku_k)$ and $(Ku_k / \lambda_k)$ differ by a vanishing correction, so $\|Ku_k - Ku_m\|_X$ is bounded away from zero for large $k \neq m$. But $\{u_k\}$ is bounded ($\|u_k\|_X = 1$) and $K$ is compact, so $\{Ku_k\}$ must have a convergent subsequence. A convergent sequence is Cauchy, contradicting the uniform lower bound $1/2$. Hence nonzero eigenvalues cannot accumulate at any $\lambda \neq 0$. Since $\sigma(K) \setminus \{0\}$ consists entirely of eigenvalues (by the second step) that can only accumulate at $0$, the set $\sigma(K) \setminus \{0\}$ is either finite or a countable sequence converging to $0$. [guided] The Riesz lemma argument constructs a sequence of unit vectors $u_k$ in the nested eigenspaces $X_k = \operatorname{span}\{e_1, \ldots, e_k\}$ that are uniformly separated after applying $K$. The key algebraic observation is that $Ku_k - \lambda_k u_k \in X_{k-1}$. Writing $u_k = \sum_{j=1}^{k}\alpha_j e_j$: \begin{align*} Ku_k - \lambda_k u_k = \sum_{j=1}^{k}\alpha_j(\lambda_j - \lambda_k)e_j, \end{align*} and the $j = k$ term vanishes. So $Ku_k/\lambda_k = u_k + (\text{element of } X_{k-1})$, meaning $Ku_k/\lambda_k$ is "essentially" $u_k$ modulo $X_{k-1}$. By Riesz's lemma, $\operatorname{dist}(u_k, X_{k-1}) \geq 1/2$, so \begin{align*} \left\|\frac{Ku_k}{\lambda_k} - \frac{Ku_m}{\lambda_m}\right\|_X \geq \operatorname{dist}(u_k, X_{k-1}) \geq \frac{1}{2} \quad \text{for } k > m. \end{align*} Since $\lambda_k \to \lambda \neq 0$, the rescaling by $1/\lambda_k$ is harmless ($1/\lambda_k$ stays bounded and bounded away from zero), so $\|Ku_k - Ku_m\|_X$ is bounded away from zero for large $k \neq m$. But $\{u_k\}$ is bounded and $K$ is compact, so $\{Ku_k\}$ must have a convergent (hence Cauchy) subsequence -- contradicting the uniform lower bound. What fails if $\lambda_k \to 0$? Then $Ku_k = \lambda_k u_k \to 0$, so the images do converge (to $0$), and there is no contradiction. This is why eigenvalues can accumulate at $0$ but nowhere else. [/guided] [/step]