Is the empty set a vector space?












4












$begingroup$


I think the empty set satisfies all of the axioms of a vector space except the one about the existence of an additive identity. Is this right?






vector-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    Wikipedia's definition and axioms suggest you need to specify a field so that you have a scalar multiplicative identity.
    $endgroup$
    – Henry
    Jan 22 '15 at 20:06






  • 3




    $begingroup$
    If the question is whether $(E,+,cdot)$ can be a vector space if $E=varnothing$, then I think the question answers itself: the additive identity is missing, so the answer is no.
    $endgroup$
    – David K
    Jan 22 '15 at 20:12
















4












$begingroup$


I think the empty set satisfies all of the axioms of a vector space except the one about the existence of an additive identity. Is this right?






vector-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    Wikipedia's definition and axioms suggest you need to specify a field so that you have a scalar multiplicative identity.
    $endgroup$
    – Henry
    Jan 22 '15 at 20:06






  • 3




    $begingroup$
    If the question is whether $(E,+,cdot)$ can be a vector space if $E=varnothing$, then I think the question answers itself: the additive identity is missing, so the answer is no.
    $endgroup$
    – David K
    Jan 22 '15 at 20:12














4












4








4


4



$begingroup$


I think the empty set satisfies all of the axioms of a vector space except the one about the existence of an additive identity. Is this right?






vector-spaces







share|cite|improve this question









$endgroup$




I think the empty set satisfies all of the axioms of a vector space except the one about the existence of an additive identity. Is this right?






vector-spaces




vector-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 22 '15 at 20:01









user209799user209799

3414




3414












  • $begingroup$
    Wikipedia's definition and axioms suggest you need to specify a field so that you have a scalar multiplicative identity.
    $endgroup$
    – Henry
    Jan 22 '15 at 20:06






  • 3




    $begingroup$
    If the question is whether $(E,+,cdot)$ can be a vector space if $E=varnothing$, then I think the question answers itself: the additive identity is missing, so the answer is no.
    $endgroup$
    – David K
    Jan 22 '15 at 20:12


















  • $begingroup$
    Wikipedia's definition and axioms suggest you need to specify a field so that you have a scalar multiplicative identity.
    $endgroup$
    – Henry
    Jan 22 '15 at 20:06






  • 3




    $begingroup$
    If the question is whether $(E,+,cdot)$ can be a vector space if $E=varnothing$, then I think the question answers itself: the additive identity is missing, so the answer is no.
    $endgroup$
    – David K
    Jan 22 '15 at 20:12
















$begingroup$
Wikipedia's definition and axioms suggest you need to specify a field so that you have a scalar multiplicative identity.
$endgroup$
– Henry
Jan 22 '15 at 20:06




$begingroup$
Wikipedia's definition and axioms suggest you need to specify a field so that you have a scalar multiplicative identity.
$endgroup$
– Henry
Jan 22 '15 at 20:06




3




3




$begingroup$
If the question is whether $(E,+,cdot)$ can be a vector space if $E=varnothing$, then I think the question answers itself: the additive identity is missing, so the answer is no.
$endgroup$
– David K
Jan 22 '15 at 20:12




$begingroup$
If the question is whether $(E,+,cdot)$ can be a vector space if $E=varnothing$, then I think the question answers itself: the additive identity is missing, so the answer is no.
$endgroup$
– David K
Jan 22 '15 at 20:12










3 Answers
3






active

oldest

votes


















7












$begingroup$

No! If $(E,+,cdot)$ is a vector space then $(E,+)$ is an abelian group so it contains a neutral element which is the zero vector hence $Enevarnothing$.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    That's exactly what I said, that the empty set has no zero vector!
    $endgroup$
    – user209799
    Jan 22 '15 at 20:11










  • $begingroup$
    It's obvious that the empty set doesn't contain a zero vector since it doesn't contain any element but my answer explains why the empty set isn't a vector space.
    $endgroup$
    – user63181
    Jan 22 '15 at 20:16










  • $begingroup$
    That's also what I said in the question, that since the empty set doesn't contain a 0 vector then it isn't a vector space; Thanks anyway, I like your answer since it's technical.
    $endgroup$
    – user209799
    Jan 22 '15 at 20:17












  • $begingroup$
    @user209799 Perhaps people are confused because you did not actually write "and therefore it is not a vector space" at the end of your first sentence, although evidently that's what you were thinking. Sometimes it's worth stating the obvious.
    $endgroup$
    – David K
    Jan 22 '15 at 20:19



















8












$begingroup$

The empty set is empty (no elements), hence it fails to have the zero vector as an element.



Since it fails to contain zero vector, it cannot be a vector space.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    according to the definition of a vector space:
    one of those requirements is that there exists an element 0 belongs to V(a set) such that v + 0 =v for all v belongs to V, above which denotes the additive identity of a vector space.
    and a empty set has no element even a 0, so that it does not fit the requirements of a vector space.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Welcome to MSE. Your answer adds nothing new to the already existing answers.
      $endgroup$
      – José Carlos Santos
      Jan 2 at 13:59












    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1115505%2fis-the-empty-set-a-vector-space%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7












    $begingroup$

    No! If $(E,+,cdot)$ is a vector space then $(E,+)$ is an abelian group so it contains a neutral element which is the zero vector hence $Enevarnothing$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      That's exactly what I said, that the empty set has no zero vector!
      $endgroup$
      – user209799
      Jan 22 '15 at 20:11










    • $begingroup$
      It's obvious that the empty set doesn't contain a zero vector since it doesn't contain any element but my answer explains why the empty set isn't a vector space.
      $endgroup$
      – user63181
      Jan 22 '15 at 20:16










    • $begingroup$
      That's also what I said in the question, that since the empty set doesn't contain a 0 vector then it isn't a vector space; Thanks anyway, I like your answer since it's technical.
      $endgroup$
      – user209799
      Jan 22 '15 at 20:17












    • $begingroup$
      @user209799 Perhaps people are confused because you did not actually write "and therefore it is not a vector space" at the end of your first sentence, although evidently that's what you were thinking. Sometimes it's worth stating the obvious.
      $endgroup$
      – David K
      Jan 22 '15 at 20:19
















    7












    $begingroup$

    No! If $(E,+,cdot)$ is a vector space then $(E,+)$ is an abelian group so it contains a neutral element which is the zero vector hence $Enevarnothing$.






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      That's exactly what I said, that the empty set has no zero vector!
      $endgroup$
      – user209799
      Jan 22 '15 at 20:11










    • $begingroup$
      It's obvious that the empty set doesn't contain a zero vector since it doesn't contain any element but my answer explains why the empty set isn't a vector space.
      $endgroup$
      – user63181
      Jan 22 '15 at 20:16










    • $begingroup$
      That's also what I said in the question, that since the empty set doesn't contain a 0 vector then it isn't a vector space; Thanks anyway, I like your answer since it's technical.
      $endgroup$
      – user209799
      Jan 22 '15 at 20:17












    • $begingroup$
      @user209799 Perhaps people are confused because you did not actually write "and therefore it is not a vector space" at the end of your first sentence, although evidently that's what you were thinking. Sometimes it's worth stating the obvious.
      $endgroup$
      – David K
      Jan 22 '15 at 20:19














    7












    7








    7





    $begingroup$

    No! If $(E,+,cdot)$ is a vector space then $(E,+)$ is an abelian group so it contains a neutral element which is the zero vector hence $Enevarnothing$.






    share|cite|improve this answer









    $endgroup$



    No! If $(E,+,cdot)$ is a vector space then $(E,+)$ is an abelian group so it contains a neutral element which is the zero vector hence $Enevarnothing$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 22 '15 at 20:03







    user63181















    • 1




      $begingroup$
      That's exactly what I said, that the empty set has no zero vector!
      $endgroup$
      – user209799
      Jan 22 '15 at 20:11










    • $begingroup$
      It's obvious that the empty set doesn't contain a zero vector since it doesn't contain any element but my answer explains why the empty set isn't a vector space.
      $endgroup$
      – user63181
      Jan 22 '15 at 20:16










    • $begingroup$
      That's also what I said in the question, that since the empty set doesn't contain a 0 vector then it isn't a vector space; Thanks anyway, I like your answer since it's technical.
      $endgroup$
      – user209799
      Jan 22 '15 at 20:17












    • $begingroup$
      @user209799 Perhaps people are confused because you did not actually write "and therefore it is not a vector space" at the end of your first sentence, although evidently that's what you were thinking. Sometimes it's worth stating the obvious.
      $endgroup$
      – David K
      Jan 22 '15 at 20:19














    • 1




      $begingroup$
      That's exactly what I said, that the empty set has no zero vector!
      $endgroup$
      – user209799
      Jan 22 '15 at 20:11










    • $begingroup$
      It's obvious that the empty set doesn't contain a zero vector since it doesn't contain any element but my answer explains why the empty set isn't a vector space.
      $endgroup$
      – user63181
      Jan 22 '15 at 20:16










    • $begingroup$
      That's also what I said in the question, that since the empty set doesn't contain a 0 vector then it isn't a vector space; Thanks anyway, I like your answer since it's technical.
      $endgroup$
      – user209799
      Jan 22 '15 at 20:17












    • $begingroup$
      @user209799 Perhaps people are confused because you did not actually write "and therefore it is not a vector space" at the end of your first sentence, although evidently that's what you were thinking. Sometimes it's worth stating the obvious.
      $endgroup$
      – David K
      Jan 22 '15 at 20:19








    1




    1




    $begingroup$
    That's exactly what I said, that the empty set has no zero vector!
    $endgroup$
    – user209799
    Jan 22 '15 at 20:11




    $begingroup$
    That's exactly what I said, that the empty set has no zero vector!
    $endgroup$
    – user209799
    Jan 22 '15 at 20:11












    $begingroup$
    It's obvious that the empty set doesn't contain a zero vector since it doesn't contain any element but my answer explains why the empty set isn't a vector space.
    $endgroup$
    – user63181
    Jan 22 '15 at 20:16




    $begingroup$
    It's obvious that the empty set doesn't contain a zero vector since it doesn't contain any element but my answer explains why the empty set isn't a vector space.
    $endgroup$
    – user63181
    Jan 22 '15 at 20:16












    $begingroup$
    That's also what I said in the question, that since the empty set doesn't contain a 0 vector then it isn't a vector space; Thanks anyway, I like your answer since it's technical.
    $endgroup$
    – user209799
    Jan 22 '15 at 20:17






    $begingroup$
    That's also what I said in the question, that since the empty set doesn't contain a 0 vector then it isn't a vector space; Thanks anyway, I like your answer since it's technical.
    $endgroup$
    – user209799
    Jan 22 '15 at 20:17














    $begingroup$
    @user209799 Perhaps people are confused because you did not actually write "and therefore it is not a vector space" at the end of your first sentence, although evidently that's what you were thinking. Sometimes it's worth stating the obvious.
    $endgroup$
    – David K
    Jan 22 '15 at 20:19




    $begingroup$
    @user209799 Perhaps people are confused because you did not actually write "and therefore it is not a vector space" at the end of your first sentence, although evidently that's what you were thinking. Sometimes it's worth stating the obvious.
    $endgroup$
    – David K
    Jan 22 '15 at 20:19











    8












    $begingroup$

    The empty set is empty (no elements), hence it fails to have the zero vector as an element.



    Since it fails to contain zero vector, it cannot be a vector space.






    share|cite|improve this answer









    $endgroup$


















      8












      $begingroup$

      The empty set is empty (no elements), hence it fails to have the zero vector as an element.



      Since it fails to contain zero vector, it cannot be a vector space.






      share|cite|improve this answer









      $endgroup$
















        8












        8








        8





        $begingroup$

        The empty set is empty (no elements), hence it fails to have the zero vector as an element.



        Since it fails to contain zero vector, it cannot be a vector space.






        share|cite|improve this answer









        $endgroup$



        The empty set is empty (no elements), hence it fails to have the zero vector as an element.



        Since it fails to contain zero vector, it cannot be a vector space.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 22 '15 at 20:04









        NamasteNamaste

        1




        1























            1












            $begingroup$

            according to the definition of a vector space:
            one of those requirements is that there exists an element 0 belongs to V(a set) such that v + 0 =v for all v belongs to V, above which denotes the additive identity of a vector space.
            and a empty set has no element even a 0, so that it does not fit the requirements of a vector space.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Welcome to MSE. Your answer adds nothing new to the already existing answers.
              $endgroup$
              – José Carlos Santos
              Jan 2 at 13:59
















            1












            $begingroup$

            according to the definition of a vector space:
            one of those requirements is that there exists an element 0 belongs to V(a set) such that v + 0 =v for all v belongs to V, above which denotes the additive identity of a vector space.
            and a empty set has no element even a 0, so that it does not fit the requirements of a vector space.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Welcome to MSE. Your answer adds nothing new to the already existing answers.
              $endgroup$
              – José Carlos Santos
              Jan 2 at 13:59














            1












            1








            1





            $begingroup$

            according to the definition of a vector space:
            one of those requirements is that there exists an element 0 belongs to V(a set) such that v + 0 =v for all v belongs to V, above which denotes the additive identity of a vector space.
            and a empty set has no element even a 0, so that it does not fit the requirements of a vector space.






            share|cite|improve this answer









            $endgroup$



            according to the definition of a vector space:
            one of those requirements is that there exists an element 0 belongs to V(a set) such that v + 0 =v for all v belongs to V, above which denotes the additive identity of a vector space.
            and a empty set has no element even a 0, so that it does not fit the requirements of a vector space.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 2 at 13:35









            user631095user631095

            111




            111












            • $begingroup$
              Welcome to MSE. Your answer adds nothing new to the already existing answers.
              $endgroup$
              – José Carlos Santos
              Jan 2 at 13:59


















            • $begingroup$
              Welcome to MSE. Your answer adds nothing new to the already existing answers.
              $endgroup$
              – José Carlos Santos
              Jan 2 at 13:59
















            $begingroup$
            Welcome to MSE. Your answer adds nothing new to the already existing answers.
            $endgroup$
            – José Carlos Santos
            Jan 2 at 13:59




            $begingroup$
            Welcome to MSE. Your answer adds nothing new to the already existing answers.
            $endgroup$
            – José Carlos Santos
            Jan 2 at 13:59


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1115505%2fis-the-empty-set-a-vector-space%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Wiesbaden

            Image
            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
            Read more

            Marschland

            Image
            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
            Read more

            Dieringhausen

            Image
            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
            Read more