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  "body": "\n## Definition\n\nLet $X$ and $Y$ be [[normed spaces|03J5RAKB30]] over $\\mathbb{K} = \\R$ or $\\C$, and let $T \\colon X \\to Y$ be a [[bounded linear map|03TM135G0N]].\nThe map $T$ is a *finite-rank operator* if its [[image|0WSFKKT6FF]] $T(X)$ is a finite-dimensional subspace of $Y$.\nIts *rank* is the dimension of $T(X)$.\n\n## Remarks\n\nEquivalently, there are $n \\ge 0$, bounded linear functionals $f_1, \\ldots, f_n$ on $X$ and vectors $y_1, \\ldots, y_n \\in Y$ with\n$$T x = \\sum_{k=1}^{n} f_k(x) \\, y_k \\quad \\text{for all } x \\in X.$$\nExample: for a nonzero bounded linear functional $f$ on $X$ and a nonzero $y \\in Y$, the map $x \\mapsto f(x) y$ has image $\\mathbb{K} y$, so it has rank $1$.\nNon-example: the identity map of $\\ell^2$ is bounded, but its image $\\ell^2$ is infinite-dimensional, so it is not of finite rank.\nFor linear maps between finite-dimensional spaces, the rank is the usual [[rank of a linear map|0ZZXSBTR4J]].\nA bounded finite-rank operator is compact, because bounded subsets of a finite-dimensional normed space are relatively compact.\nOn Hilbert spaces, every compact operator is a norm limit of finite-rank operators.\nOn general Banach spaces this can fail; this is the question of the approximation property.\n",
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