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  "body": "\n## Definition\n\nLet $X$ and $Y$ be [[Banach spaces|06GZ956657]], and let $T \\colon X \\to Y$ be a [[bounded linear map|03TM135G0N]].\nThe map $T$ is *strictly singular* if for every infinite-dimensional closed subspace $M$ of $X$,\n$$\\inf \\set{\\norm{Tx} : x \\in M, \\ \\norm{x} = 1} = 0.$$\nEquivalently, no restriction $T|_M$ to an infinite-dimensional closed subspace $M$ is an isomorphism onto its image.\n\n## Remarks\n\nExample: every [[compact operator|M2NEP1K5AE]] is strictly singular.\nIndeed, suppose that $T|_M$ had a bounded inverse on $T(M)$.\nThen the closed unit ball of $M$ would be the image of a relatively compact set under a bounded map, so it would be compact, and $M$ would be finite-dimensional.\nFor $1 \\le p \u003c q \u003c \\infty$, the inclusion $\\ell^p \\to \\ell^q$ is strictly singular but not compact.\nNon-example: the identity map of an infinite-dimensional Banach space $X$ is not strictly singular, since it is an isomorphism on $M = X$.\nOn Hilbert spaces, the strictly singular operators are exactly the compact operators.\nFor a Banach space $X$, the strictly singular operators from $X$ to $X$ form a closed two-sided ideal of the algebra of bounded operators.\nIf $F$ is a [[Fredholm operator|8HWGKWD6ZM]] and $S$ is strictly singular, then $F + S$ is Fredholm with the same index.\n",
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