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





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    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


















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