"eigenvector" meaning in Anglais

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      "kind": "other",
      "name": "Calques en anglais issus d’un mot en allemand",
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      "source": "w"
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    },
    {
      "kind": "other",
      "name": "Anglais",
      "orig": "anglais",
      "parents": [],
      "source": "w"
    }
  ],
  "etymology_texts": [
    "Calque de l’allemand Eigenvektor composé de eigen (« propre, personnel ») et de Vektor (« vecteur »)."
  ],
  "forms": [
    {
      "form": "eigenvectors",
      "tags": [
        "plural"
      ]
    }
  ],
  "lang": "Anglais",
  "lang_code": "en",
  "pos": "noun",
  "pos_title": "Nom commun",
  "related": [
    {
      "word": "eigenvalue"
    },
    {
      "word": "characteristic polynomial"
    }
  ],
  "senses": [
    {
      "categories": [
        {
          "kind": "other",
          "name": "Exemples en anglais",
          "parents": [],
          "source": "w"
        },
        {
          "kind": "other",
          "name": "Exemples en anglais à traduire",
          "parents": [],
          "source": "w"
        },
        {
          "kind": "other",
          "name": "Lexique en anglais des mathématiques",
          "parents": [],
          "source": "w"
        }
      ],
      "examples": [
        {
          "bold_text_offsets": [
            [
              75,
              87
            ]
          ],
          "ref": "Serge Lang, Algebra, 1965",
          "text": "We say that S is diagonalizable if there exists a basis of E consisting of eigenvectors of S."
        }
      ],
      "glosses": [
        "Vecteur propre."
      ],
      "id": "fr-eigenvector-en-noun-rY~DFpLK",
      "topics": [
        "mathematics"
      ]
    }
  ],
  "sounds": [
    {
      "ipa": "\\ˌaɪɡnˈvɛk.tɜː\\"
    },
    {
      "ipa": "\\ˌaɪɡnˈvɛk.tɜː\\"
    },
    {
      "audio": "En-us-eigenvector.ogg",
      "ipa": "[ˌaɪɡnˈvɛk.tɜː]",
      "mp3_url": "https://upload.wikimedia.org/wikipedia/commons/transcoded/e/e8/En-us-eigenvector.ogg/En-us-eigenvector.ogg.mp3",
      "ogg_url": "https://commons.wikimedia.org/wiki/Special:FilePath/En-us-eigenvector.ogg",
      "raw_tags": [
        "États-Unis"
      ]
    }
  ],
  "word": "eigenvector"
}

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    "anglais"
  ],
  "etymology_texts": [
    "Calque de l’allemand Eigenvektor composé de eigen (« propre, personnel ») et de Vektor (« vecteur »)."
  ],
  "forms": [
    {
      "form": "eigenvectors",
      "tags": [
        "plural"
      ]
    }
  ],
  "lang": "Anglais",
  "lang_code": "en",
  "pos": "noun",
  "pos_title": "Nom commun",
  "related": [
    {
      "word": "eigenvalue"
    },
    {
      "word": "characteristic polynomial"
    }
  ],
  "senses": [
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      "examples": [
        {
          "bold_text_offsets": [
            [
              75,
              87
            ]
          ],
          "ref": "Serge Lang, Algebra, 1965",
          "text": "We say that S is diagonalizable if there exists a basis of E consisting of eigenvectors of S."
        }
      ],
      "glosses": [
        "Vecteur propre."
      ],
      "topics": [
        "mathematics"
      ]
    }
  ],
  "sounds": [
    {
      "ipa": "\\ˌaɪɡnˈvɛk.tɜː\\"
    },
    {
      "ipa": "\\ˌaɪɡnˈvɛk.tɜː\\"
    },
    {
      "audio": "En-us-eigenvector.ogg",
      "ipa": "[ˌaɪɡnˈvɛk.tɜː]",
      "mp3_url": "https://upload.wikimedia.org/wikipedia/commons/transcoded/e/e8/En-us-eigenvector.ogg/En-us-eigenvector.ogg.mp3",
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  "word": "eigenvector"
}