{ "schema_version": 1, "kind": "ConceptView", "producer": "concept-view-1", "concept_id": "urn:uuid:cb1aa3aa-d213-4d5c-9061-9e805aea073a", "labels": { "zh-CN": "可逆的", "en": "invertible" }, "alternative_labels": [], "scope_note": "", "identity_status": "active", "identity_revision": "f20fd70ae76e266e12bcfc096736ef1aa779f20d01cfd4a756d9c235fdd67e7e", "occurrences": [ { "concept_id": "urn:uuid:cb1aa3aa-d213-4d5c-9061-9e805aea073a", "book_id": "szekeres", "edition_id": "source-snapshot-bc3404156f6e", "lang": "zh-CN", "route": "zh/chapter-13/04-bounded-linear-operators", "anchor": "concept-13-invertible", "chapter": "13", "section": "13.4", "offset": 3873, "range": [ 9751, 9977 ], "source_revision": "194ca7b68e093551807e302c52f29521e73d76a77f31f6fd5b10ca6c3ee642c4", "source_category": "translation", "review_status": "unreviewed", "role": "occurrence", "role_basis": { "kind": "explicit-source-uuid-marker" }, "label": "可逆的(invertible)", "context": { "range": [ 9652, 10582 ], "quote": "对一个有界算符$A:V\\rightarrow V$而言，如果存在$A^{-1}:V\\rightarrow V$，使得$A A^{-1}=A^{-1} A=\\mathrm{id}_V$，那么$A$称为可逆的(invertible)，$A^{-1}$称为$A$的逆(inverse)，显而易见是唯一的，因为如果$BA=CA$，那么$B=B I=B A A^{-1}=C A A^{-1}=C$。我们需要指明的是，$A^{-1}$既是一个左逆又是一个右逆。例如在$\\ell^2$中，在例13.8中的平移算符$S$就只有左逆$S'$，因为$S'S=I$，但是这不是其右逆因为$SS'(x_1,x_2,x_3,\\ldots)=(0,x_2,x_3,\\ldots)$，那么$S$就不是一个可逆算符，尽管它是一个等距的单射，$\\|Sx\\|=\\|x\\|$，对于有限维空间这已经足够保证可逆性了。", "quote_sha256": "d32ebf2f93657326547a46a4d46f60b503866b0866b8a8c9d00f4da5a5b5255a", "environment": null, "excerpt_omitted": false }, "parallel_readings": [ { "route": "en/chapter-13/04-bounded-linear-operators", "lang": "en", "source_category": "original_transcription", "basis": "registered-book-edition-section-language-context; not-concept-equivalence" } ], "id": "occ-dbaa91cdc722051081948cd9" }, { "concept_id": "urn:uuid:cb1aa3aa-d213-4d5c-9061-9e805aea073a", "book_id": "szekeres", "edition_id": "source-snapshot-bc3404156f6e", "lang": "en", "route": "en/chapter-13/04-bounded-linear-operators", "anchor": "first-1b9652eb4e8887748082-en", "chapter": "13", "section": "13.4", "offset": 5049, "range": [ 10418, 10647 ], "source_revision": "eb0725f6ab7fc44fa4ee614b878f2ec644f55a6b84b31a4afac5555a1c5df299", "source_category": "original_transcription", "review_status": "unreviewed", "role": "occurrence", "role_basis": { "kind": "explicit-source-uuid-marker" }, "label": "invertible", "context": { "range": [ 10374, 11926 ], "quote": "A bounded operator $A:V\\to V$ is said to be invertible if there exists a bounded operator $A ^ { - 1 } : V \\to V$ such that $A A ^ { - 1 } = A ^ { - 1 } A = I \\equiv \\mathrm { i d } _ { V } . A ^ { - 1 }$ is called the inverse of $A$. It is clearly unique, for if $B A = C A$ then $B = B I = B A A ^ { - 1 } = C A A ^ { - 1 } = C$ . It is important that we specify $A ^ { - 1 }$ to be both a right and left inverse. For example, in $\\ell ^ { 2 }$ , the shift operator $S$ defined in Example 13.8 has left inverse $S ^ { \\prime } { } _ { ; }$ , since $S ^ { \\prime } S = I$ , but it is not a right inverse for $S S ^ { \\prime } ( x _ { 1 } , x _ { 2 } , \\dots ) = ( 0 , x _ { 2 } , x _ { 3 } , \\dots )$ . Thus $S$ is not an invertible operator, despite the fact that it is injective and an isometry, $\\| S x \\| = \\| x \\|$ . 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