Let $X$ be an infinite-dimensional complex [Banach space](/page/Banach%20Space) and let $T \in \mathcal{L}(X)$ be compact. If $\lambda \in \sigma(T)$ and $\lambda \ne 0$, then $\lambda$ is an eigenvalue of $T$. Moreover, the eigenspace $\ker(T-\lambda I)$ is finite-dimensional.

Let $(X,\|\cdot\|_X)$ be a [normed vector space](/page/Normed%20Vector%20Space), let $(Y,\|\cdot\|_Y)$ be a [Banach space](/page/Banach%20Space), and let $T:X\to Y$ be a [linear map](/page/Linear%20Map). Then $T$ is compact, meaning that $\overline{T(A)}^{\,Y}$ is compact in $Y$ for every bounded subset $A\subset X$, if and only if for every bounded sequence $(x_k)_{k\in\mathbb N}$ in $X$, the sequence $(Tx_k)_{k\in\mathbb N}$ has a subsequence converging in $Y$.

Let $X$ be a [normed vector space](/page/Normed%20Vector%20Space), let $Y$ be a normed [vector space](/page/Vector%20Space), and let $T: X \to Y$ be a bounded finite-rank linear operator. Then $T$ is compact.

Let $X$ and $Y$ be normed vector spaces over the same field $\mathbb{F} \in \{\mathbb{R}, \mathbb{C}\}$, and let $T: X \to Y$ be a [linear map](/page/Linear%20Map). Define the closed unit ball of $X$ by \begin{align*} B_X = \{x \in X : \|x\|_X \leq 1\}. \end{align*} Then $T$ is a compact linear operator if and only if $T(B_X)$ is relatively compact in $Y$, meaning that $\overline{T(B_X)}$ is compact in $Y$.

Let $(a_k)$ be a bounded scalar sequence, and define $T: \ell^2 \to \ell^2$ by $(Tx)_k = a_k x_k$. Then $T$ is compact if and only if $a_k \to 0$.

Let $X$ and $Y$ be normed vector spaces. If $S,T: X \to Y$ are compact and $\alpha,\beta$ are scalars, then $\alpha S + \beta T$ is compact.

Let $U \subset \mathbb{R}^n$ be measurable with $\mathcal{L}^n(U)<\infty$. If $K \in L^2(U \times U)$, then the associated operator $T_K: L^2(U) \to L^2(U)$ is compact.

Let $X$, $Y$, and $Z$ be normed vector spaces. If $T: X \to Y$ is compact, $A: Z \to X$ is bounded and linear, and $B: Y \to Z$ is bounded and linear, then $T A: Z \to Y$ and $B T: X \to Z$ are compact.

Let $H$ be a [Hilbert space](/page/Hilbert%20Space). An orthonormal system in $H$ is maximal if and only if it is an [orthonormal basis](/page/Orthonormal%20Basis) of $H$.

Let $H$ be a [Hilbert space](/page/Hilbert%20Space) over $\mathbb{R}$ or $\mathbb{C}$, and let $(e_i)_{i \in I}$ be an [orthonormal basis](/page/Orthonormal%20Basis) of $H$, meaning that $(e_i)_{i \in I}$ is an orthonormal family and \begin{align*} \overline{\operatorname{span}\{e_i : i \in I\}} = H. \end{align*} Define $\ell^2(I)$ to be the Hilbert space of scalar families $a = (a_i)_{i \in I}$ such that \begin{align*} \|a\|_{\ell^2(I)}^2 := \sup\left\{\sum_{i \in F} |a_i|^2 : F \subset I \text{ finite}\right\} < \infty. \end{align*} Then the coordinate map \begin{align*} U: H &\to \ell^2(I) \\ x &\mapsto \bigl((x,e_i)_H\bigr)_{i \in I} \end{align*} is a linear isometric isomorphism. Its inverse is the map \begin{align*} V: \ell^2(I) &\to H \\ (a_i)_{i \in I} &\mapsto \lim_{F} \sum_{i \in F} a_i e_i, \end{align*}…