[proofplan] We establish that $T - \mu I$ is Fredholm of index zero for every $\mu \neq 0$. The key tool is compactness: it forces finite-dimensional kernels (any bounded sequence in the eigenspace has a convergent image), closed ranges (via a normalisation-and-compactness argument), finite ascent (the Riesz lemma produces sequences contradicting compactness if the chain is infinite), and isolation of nonzero eigenvalues (accumulation would again contradict compactness). The Fredholm alternative follows from index zero. [/proofplan] [step:Show the kernel is finite-dimensional for every $\mu \neq 0$] [claim:Finite-Dimensional Kernel] For every $\mu \neq 0$, $\dim \ker(T - \mu I) < \infty$. [/claim] [proof] If $\ker(T - \mu I)$ were infinite-dimensional, the Riesz lemma provides $\{x_k\}$ with $\|x_k\| = 1$ and $\|x_j - x_k\| > 1/2$. Since $Tx_k = \mu x_k$, $\|Tx_j - Tx_k\| = |\mu|\|x_j - x_k\| > |\mu|/2$, contradicting compactness. [/proof] [/step] [step:Show the range is closed for every $\mu \neq 0$] [claim:Closed Range] $\operatorname{Range}(T - \mu I)$ is closed for every $\mu \neq 0$. [/claim] [proof] Let $y_k = (T - \mu I)x_k \to y$ with $x_k \perp \ker(T - \mu I)$. If $\|x_k\|$ is unbounded, set $z_k = x_k/\|x_k\|$; then $(T - \mu I)z_k \to 0$ and compactness gives a subsequence with $Tz_{k_j} \to w$, so $z_{k_j} \to w/\mu$. But $w/\mu \in \ker(T - \mu I)$ and $z_{k_j} \perp \ker(T - \mu I)$, a contradiction. So $\|x_k\|$ is bounded, compactness extracts a convergent subsequence, and $y \in \operatorname{Range}(T - \mu I)$. [/proof] [/step] [step:Show the ascending chain of kernels stabilises (finite ascent)] [claim:Finite Ascent] The chain $\ker(T - \mu I) \subseteq \ker(T - \mu I)^2 \subseteq \cdots$ stabilises after finitely many steps. [/claim] [proof] If the chain is strictly increasing, the Riesz lemma gives $x_k$ with $\|x_k\| = 1$ and $\operatorname{dist}(x_k, \ker(T - \mu I)^k) > 1/2$. For $j < k$, $Tx_k - Tx_j = \mu x_k - w$ with $w \in \ker(T - \mu I)^k$, so $\|Tx_k - Tx_j\| \geq |\mu|/2$, contradicting compactness. [/proof] The generalized eigenspace $\ker(T - \mu I)^m$ is finite-dimensional (by induction from finite-dimensionality of each kernel). [/step] [step:Show nonzero eigenvalues cannot accumulate away from zero] [claim:Isolation of Nonzero Eigenvalues] The set of nonzero eigenvalues has no accumulation point in $\mathbb{C} \setminus \{0\}$. [/claim] [proof] Suppose $\mu_k \to \mu \neq 0$ with distinct eigenvalues and unit eigenvectors $x_k$. The spaces $E_k = \operatorname{span}\{x_1, \ldots, x_k\}$ form a strictly increasing chain. The Riesz lemma gives $y_k \in E_k$ with $\|y_k\| = 1$ and $\operatorname{dist}(y_k, E_{k-1}) > 1/2$. Since $(T - \mu_k I)$ maps $E_k$ into $E_{k-1}$, we get $\|Ty_k - Ty_j\| \geq |\mu_k|/2 \geq |\mu|/4$ for large $k$, contradicting compactness. [/proof] [/step] [step:State the Fredholm alternative] For each $\mu \neq 0$, the ascent equals the descent (by an analogous descending-chain argument), giving a direct sum $X = \ker(T - \mu I)^m \oplus \operatorname{Range}(T - \mu I)^m$. The index is zero, so exactly one alternative holds: (a) $\ker(T - \mu I) = \{0\}$ and $T - \mu I$ is bijective (hence $\mu \notin \sigma(T)$); (b) $\ker(T - \mu I) \neq \{0\}$ with $\dim \ker(T - \mu I) = \dim \ker(T^* - \bar{\mu} I)$, and $(T - \mu I)x = y$ is solvable iff $y$ annihilates $\ker(T^* - \bar{\mu} I)$. [/step]