[proofplan] We establish the four structural properties of the spectrum of a compact operator on an infinite-dimensional Banach space. First, $0 \in \sigma(K)$ because bijectivity of $K$ would force the identity to be compact, contradicting infinite-dimensionality. Second, every nonzero spectral value is an eigenvalue via the Fredholm alternative applied to $I - \lambda^{-1}K$. Third, each nonzero eigenspace is finite-dimensional because compactness forces its unit ball to be precompact. Fourth, the nonzero eigenvalues cannot accumulate away from zero, proved by a Riesz lemma argument showing that accumulation would produce a bounded sequence whose image under $K$ has no convergent subsequence. [/proofplan] [step:Show $0 \in \sigma(K)$ by contradicting compactness of the identity] Suppose for contradiction that $0 \in \rho(K)$, so $K$ is bijective with bounded inverse $K^{-1} \in \mathcal{L}(X)$. Then \begin{align*} I = K^{-1} \circ K. \end{align*} Since $K$ is compact and $K^{-1}$ is bounded, the composition $K^{-1} \circ K$ is compact (a bounded operator composed with a compact operator is compact). But the identity operator on an infinite-dimensional [Banach space](/page/Banach%20Space) is not compact: any infinite orthonormal-type sequence (or, more generally, any sequence produced by Riesz's lemma applied to a chain of finite-dimensional subspaces) is bounded with no convergent subsequence. This contradicts compactness of $I$, so $0 \in \sigma(K)$. [/step] [step:Prove every $\lambda \in \sigma(K) \setminus \{0\}$ is an eigenvalue via the Fredholm alternative] Let $\lambda \in \mathbb{C}$ with $\lambda \ne 0$ and $\lambda \in \sigma(K)$. Write $K - \lambda I = -\lambda(I - \lambda^{-1} K)$. Define $K_\lambda := \lambda^{-1} K$, which is compact since $K$ is compact and scalar multiplication preserves compactness. Then $\lambda \in \sigma(K)$ means $K - \lambda I$ is not bijective, equivalently $I - K_\lambda$ is not bijective. By [the Fredholm alternative](/page/The%20Fredholm%20Alternative) applied to the compact operator $K_\lambda$: either $N(I - K_\lambda) = \{0\}$ and $I - K_\lambda$ is bijective, or $N(I - K_\lambda) \ne \{0\}$. Since $I - K_\lambda$ is not bijective, we must have $N(I - K_\lambda) \ne \{0\}$. That is, there exists $u \ne 0$ with $(I - K_\lambda)u = 0$, giving $u = \lambda^{-1}Ku$, hence $Ku = \lambda u$. Therefore $\lambda$ is an eigenvalue: $\lambda \in \sigma_p(K)$. [guided] The key mechanism is the [Fredholm alternative](/page/The%20Fredholm%20Alternative), which provides a sharp dichotomy for operators of the form $I - T$ when $T$ is compact: either $N(I - T) = \{0\}$ (in which case $I - T$ is bijective), or $N(I - T) \neq \{0\}$. The contrapositive states: if $I - T$ fails to be bijective, it must fail to be injective. We factor $K - \lambda I = -\lambda(I - K_\lambda)$ where $K_\lambda := \lambda^{-1}K$. The operator $K_\lambda$ is compact because $K$ is compact and scalar multiplication preserves compactness. Since $\lambda \in \sigma(K)$, the operator $K - \lambda I$ is not invertible, so $I - K_\lambda$ is not bijective. The Fredholm alternative then forces \begin{align*} N(I - K_\lambda) \neq \{0\}, \end{align*} meaning there exists $u \neq 0$ with $K_\lambda u = u$, i.e., $\lambda^{-1}Ku = u$, hence $Ku = \lambda u$. Therefore $\lambda$ is an eigenvalue. This argument fails at $\lambda = 0$: we cannot form $K_0 = 0^{-1}K$. And indeed $0$ may or may not be an eigenvalue of $K$ -- it is always in the spectrum (by the previous step), but $\ker K$ could be trivial. [/guided] [/step] [step:Show every nonzero eigenspace is finite-dimensional] Let $\lambda \ne 0$ be an eigenvalue of $K$ and let $E_\lambda := N(K - \lambda I)$ be the corresponding eigenspace. The closed unit ball $B_\lambda := \{u \in E_\lambda : \|u\|_X \le 1\}$ satisfies $K(B_\lambda) = \lambda B_\lambda$, since $Ku = \lambda u$ for every $u \in E_\lambda$. Therefore \begin{align*} B_\lambda = \lambda^{-1} K(B_\lambda), \end{align*} which is precompact because $K(B_\lambda)$ is precompact (as $K$ is compact and $B_\lambda$ is bounded) and scalar multiplication by $\lambda^{-1}$ preserves precompactness. A normed space whose closed unit ball is precompact is finite-dimensional (by the Riesz lemma characterisation), so $\dim E_\lambda < \infty$. [/step] [step:Prove the nonzero eigenvalues cannot accumulate away from zero via the Riesz lemma] Suppose for contradiction that there exists a sequence of distinct eigenvalues $\{\lambda_k\}_{k=1}^\infty \subseteq \sigma_p(K) \setminus \{0\}$ with $\lambda_k \to \mu \ne 0$. For each $k \in \mathbb{N}$, let $u_k \in X$ be a unit eigenvector with $Ku_k = \lambda_k u_k$. Define the finite-dimensional subspaces \begin{align*} E_k := \mathrm{span}\{u_1, u_2, \ldots, u_k\}. \end{align*} Eigenvectors corresponding to distinct eigenvalues of a linear operator are linearly independent, so $\dim E_k = k$ and $E_k \subsetneq E_{k+1}$ for every $k \in \mathbb{N}$. By the Riesz lemma, for each $k \ge 2$ there exists $v_k \in E_k$ with $\|v_k\|_X = 1$ and $\mathrm{dist}(v_k, E_{k-1}) \ge \frac{1}{2}$. For $k > m$, we show that $\|Kv_k - Kv_m\|_X$ is bounded below. Write $v_k = \sum_{j=1}^k \alpha_j^{(k)} u_j$. Since $K$ acts on $E_k$ by $Ku_j = \lambda_j u_j$, the operator $K$ maps $E_k$ into itself. Moreover, $Kv_k - \lambda_k v_k = \sum_{j=1}^k \alpha_j^{(k)}(\lambda_j - \lambda_k)u_j \in E_{k-1}$ (the $j = k$ term vanishes). Therefore \begin{align*} Kv_k - Kv_m = \lambda_k v_k + (Kv_k - \lambda_k v_k) - Kv_m, \end{align*} where $(Kv_k - \lambda_k v_k) \in E_{k-1}$ and $Kv_m \in E_m \subseteq E_{k-1}$. Hence $Kv_k - Kv_m - \lambda_k v_k \in E_{k-1}$, giving \begin{align*} \|Kv_k - Kv_m\|_X = |\lambda_k| \cdot \left\|v_k - \frac{Kv_k - Kv_m - \lambda_k v_k}{-\lambda_k}\right\|_X \ge |\lambda_k| \cdot \mathrm{dist}(v_k, E_{k-1}) \ge \frac{|\lambda_k|}{2}. \end{align*} Since $\lambda_k \to \mu \ne 0$, we have $|\lambda_k| \ge |\mu|/2 > 0$ for all sufficiently large $k$. The sequence $\{v_k\}$ is bounded ($\|v_k\|_X = 1$), yet $\{Kv_k\}$ has no convergent subsequence (successive terms are separated by at least $|\mu|/4$). This contradicts compactness of $K$. [guided] The Riesz lemma argument is a standard technique for extracting contradictions from compactness. The idea is to build a sequence $\{v_k\}$ in the expanding chain of eigenspaces $E_1 \subsetneq E_2 \subsetneq \cdots$ such that each $v_k$ is "far" from the previous subspace $E_{k-1}$. Why does $K$ separate these vectors? The key algebraic observation is that $Kv_k - \lambda_k v_k \in E_{k-1}$: when we apply $K$ to $v_k = \sum_{j=1}^k \alpha_j^{(k)} u_j$, each $u_j$ gets scaled by $\lambda_j$, and subtracting $\lambda_k v_k$ kills the top component (the $u_k$ term), leaving a vector in $E_{k-1}$. So $Kv_k = \lambda_k v_k + (\text{something in } E_{k-1})$. When we compute $Kv_k - Kv_m$ for $k > m$, the term $Kv_m$ lies entirely in $E_m \subseteq E_{k-1}$, so $Kv_k - Kv_m$ differs from $\lambda_k v_k$ by an element of $E_{k-1}$. Since $\mathrm{dist}(v_k, E_{k-1}) \ge 1/2$ and $|\lambda_k|$ is bounded away from zero (because $\lambda_k \to \mu \ne 0$), we get \begin{align*} \|Kv_k - Kv_m\|_X \ge |\lambda_k| \cdot \mathrm{dist}(v_k, E_{k-1}) \ge \frac{|\lambda_k|}{2} \ge \frac{|\mu|}{4} \end{align*} for large $k$. This means $\{Kv_k\}$ has no Cauchy subsequence, contradicting the compactness of $K$ applied to the bounded sequence $\{v_k\}$. [/guided] [/step] [step:Conclude the spectral structure of $\sigma(K)$] Combining the previous steps: $0 \in \sigma(K)$ (first step), every element of $\sigma(K) \setminus \{0\}$ is an eigenvalue (second step) with finite-dimensional eigenspace (third step), and the nonzero eigenvalues cannot accumulate at any point other than $0$ (fourth step). A subset of $\mathbb{C}$ whose only possible accumulation point is $0$ is either finite or forms a sequence converging to $0$. [/step]