Is a vector space a subspace of itself?












2












$begingroup$


We know that a subspace is a vector space that follows the same addition and multiplication rules as $Bbb V$, but is a vector space a subspace of itself?
Also, I'm getting confused doing the practice questions, on when we prove that something is a vector space by using the subspace test and when we prove V1 - V10.






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    How do you define a subspace of a vector space?
    $endgroup$
    – Brian
    10 hours ago






  • 3




    $begingroup$
    Is a set a subset of itself?? What’s V1-V10?
    $endgroup$
    – J. W. Tanner
    10 hours ago








  • 2




    $begingroup$
    The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
    $endgroup$
    – Theo Bendit
    10 hours ago






  • 4




    $begingroup$
    As other commenters have noted, your question lacks context. Please edit your question to include more context, lest your question be closed. Please give a definition of a subspace. Please explain what V1 - V10 means. If you are working from a particular text, a citation to that text would be helpful, too.
    $endgroup$
    – Xander Henderson
    9 hours ago






  • 3




    $begingroup$
    @CompuChip That's not a usage of "proper set" that I'm accustomed to, but I can certainly imagine it being used by certain fields of maths. I would instead say "proper, non-trivial" to mean not the whole set/space, and not the empty set/$0$ subspace.
    $endgroup$
    – Theo Bendit
    3 hours ago
















2












$begingroup$


We know that a subspace is a vector space that follows the same addition and multiplication rules as $Bbb V$, but is a vector space a subspace of itself?
Also, I'm getting confused doing the practice questions, on when we prove that something is a vector space by using the subspace test and when we prove V1 - V10.






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    How do you define a subspace of a vector space?
    $endgroup$
    – Brian
    10 hours ago






  • 3




    $begingroup$
    Is a set a subset of itself?? What’s V1-V10?
    $endgroup$
    – J. W. Tanner
    10 hours ago








  • 2




    $begingroup$
    The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
    $endgroup$
    – Theo Bendit
    10 hours ago






  • 4




    $begingroup$
    As other commenters have noted, your question lacks context. Please edit your question to include more context, lest your question be closed. Please give a definition of a subspace. Please explain what V1 - V10 means. If you are working from a particular text, a citation to that text would be helpful, too.
    $endgroup$
    – Xander Henderson
    9 hours ago






  • 3




    $begingroup$
    @CompuChip That's not a usage of "proper set" that I'm accustomed to, but I can certainly imagine it being used by certain fields of maths. I would instead say "proper, non-trivial" to mean not the whole set/space, and not the empty set/$0$ subspace.
    $endgroup$
    – Theo Bendit
    3 hours ago














2












2








2


1



$begingroup$


We know that a subspace is a vector space that follows the same addition and multiplication rules as $Bbb V$, but is a vector space a subspace of itself?
Also, I'm getting confused doing the practice questions, on when we prove that something is a vector space by using the subspace test and when we prove V1 - V10.






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$




We know that a subspace is a vector space that follows the same addition and multiplication rules as $Bbb V$, but is a vector space a subspace of itself?
Also, I'm getting confused doing the practice questions, on when we prove that something is a vector space by using the subspace test and when we prove V1 - V10.






linear-algebra vector-spaces




linear-algebra vector-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 8 hours ago









Eric Wofsey

192k14220352




192k14220352










asked 10 hours ago









mingming

4536




4536








  • 1




    $begingroup$
    How do you define a subspace of a vector space?
    $endgroup$
    – Brian
    10 hours ago






  • 3




    $begingroup$
    Is a set a subset of itself?? What’s V1-V10?
    $endgroup$
    – J. W. Tanner
    10 hours ago








  • 2




    $begingroup$
    The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
    $endgroup$
    – Theo Bendit
    10 hours ago






  • 4




    $begingroup$
    As other commenters have noted, your question lacks context. Please edit your question to include more context, lest your question be closed. Please give a definition of a subspace. Please explain what V1 - V10 means. If you are working from a particular text, a citation to that text would be helpful, too.
    $endgroup$
    – Xander Henderson
    9 hours ago






  • 3




    $begingroup$
    @CompuChip That's not a usage of "proper set" that I'm accustomed to, but I can certainly imagine it being used by certain fields of maths. I would instead say "proper, non-trivial" to mean not the whole set/space, and not the empty set/$0$ subspace.
    $endgroup$
    – Theo Bendit
    3 hours ago














  • 1




    $begingroup$
    How do you define a subspace of a vector space?
    $endgroup$
    – Brian
    10 hours ago






  • 3




    $begingroup$
    Is a set a subset of itself?? What’s V1-V10?
    $endgroup$
    – J. W. Tanner
    10 hours ago








  • 2




    $begingroup$
    The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
    $endgroup$
    – Theo Bendit
    10 hours ago






  • 4




    $begingroup$
    As other commenters have noted, your question lacks context. Please edit your question to include more context, lest your question be closed. Please give a definition of a subspace. Please explain what V1 - V10 means. If you are working from a particular text, a citation to that text would be helpful, too.
    $endgroup$
    – Xander Henderson
    9 hours ago






  • 3




    $begingroup$
    @CompuChip That's not a usage of "proper set" that I'm accustomed to, but I can certainly imagine it being used by certain fields of maths. I would instead say "proper, non-trivial" to mean not the whole set/space, and not the empty set/$0$ subspace.
    $endgroup$
    – Theo Bendit
    3 hours ago








1




1




$begingroup$
How do you define a subspace of a vector space?
$endgroup$
– Brian
10 hours ago




$begingroup$
How do you define a subspace of a vector space?
$endgroup$
– Brian
10 hours ago




3




3




$begingroup$
Is a set a subset of itself?? What’s V1-V10?
$endgroup$
– J. W. Tanner
10 hours ago






$begingroup$
Is a set a subset of itself?? What’s V1-V10?
$endgroup$
– J. W. Tanner
10 hours ago






2




2




$begingroup$
The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
$endgroup$
– Theo Bendit
10 hours ago




$begingroup$
The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
$endgroup$
– Theo Bendit
10 hours ago




4




4




$begingroup$
As other commenters have noted, your question lacks context. Please edit your question to include more context, lest your question be closed. Please give a definition of a subspace. Please explain what V1 - V10 means. If you are working from a particular text, a citation to that text would be helpful, too.
$endgroup$
– Xander Henderson
9 hours ago




$begingroup$
As other commenters have noted, your question lacks context. Please edit your question to include more context, lest your question be closed. Please give a definition of a subspace. Please explain what V1 - V10 means. If you are working from a particular text, a citation to that text would be helpful, too.
$endgroup$
– Xander Henderson
9 hours ago




3




3




$begingroup$
@CompuChip That's not a usage of "proper set" that I'm accustomed to, but I can certainly imagine it being used by certain fields of maths. I would instead say "proper, non-trivial" to mean not the whole set/space, and not the empty set/$0$ subspace.
$endgroup$
– Theo Bendit
3 hours ago




$begingroup$
@CompuChip That's not a usage of "proper set" that I'm accustomed to, but I can certainly imagine it being used by certain fields of maths. I would instead say "proper, non-trivial" to mean not the whole set/space, and not the empty set/$0$ subspace.
$endgroup$
– Theo Bendit
3 hours ago










2 Answers
2






active

oldest

votes


















10












$begingroup$

Yes, every vector space is a vector subspace of itself, since it is a non-empty subset of itself which is closed with respect to addition and with respect to product by scalars.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    I'm guessing that V1 - V10 are the axioms for proving vector spaces.



    To prove something is a vector space, independent of any other vector spaces you know of, you are required to prove all of the axioms in the definition. Not all operations that call themselves $+$ are worthy addition operations; just because you denote it $+$ does not mean it is (for example) associative, or has an additive identity.



    There is a lot to prove, because there's a lot to gain. Vector spaces have a simply enormous amount of structure, and that structure gives us a really rich theory and powerful tools. If you have an object that you wish to understand better, and you can show it is a vector space (or at least, related to a vector space), then you'll instantly have some serious mathematical firepower at your fingertips.



    Subspaces give us a shortcut to proving a vector space. If you have a subset of a known vector space, then you can prove just $3$ properties, rather than $10$. We can skip a lot of the steps because somebody has already done them previously when showing the larger vector space is indeed a vector space. You don't need to show, for example, $v + w = w + v$ for all $v, w$ in your subset, because we already know this is true for all vectors in the larger vector space.



    I'm writing this, not as a direct answer to your question (which Jose Carlos Santos has answered already), but because confusion like this often stems from some sloppiness on the above point. I've seen many students (and, lamentably, several instructors) fail to grasp that showing the subspace conditions on a set that is not clearly a subset of a known vector space does not prove a vector space. The shortcut works because somebody has already established most of the axioms beforehand, but if this is not true, then the argument is a fallacy.



    You can absolutely apply the subspace conditions on the whole of a vector space provided you've proven it's a vector space already with axioms V1 - V10.






    share|cite|improve this answer









    $endgroup$














      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%2f3180447%2fis-a-vector-space-a-subspace-of-itself%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      10












      $begingroup$

      Yes, every vector space is a vector subspace of itself, since it is a non-empty subset of itself which is closed with respect to addition and with respect to product by scalars.






      share|cite|improve this answer









      $endgroup$


















        10












        $begingroup$

        Yes, every vector space is a vector subspace of itself, since it is a non-empty subset of itself which is closed with respect to addition and with respect to product by scalars.






        share|cite|improve this answer









        $endgroup$
















          10












          10








          10





          $begingroup$

          Yes, every vector space is a vector subspace of itself, since it is a non-empty subset of itself which is closed with respect to addition and with respect to product by scalars.






          share|cite|improve this answer









          $endgroup$



          Yes, every vector space is a vector subspace of itself, since it is a non-empty subset of itself which is closed with respect to addition and with respect to product by scalars.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 10 hours ago









          José Carlos SantosJosé Carlos Santos

          173k23133242




          173k23133242























              4












              $begingroup$

              I'm guessing that V1 - V10 are the axioms for proving vector spaces.



              To prove something is a vector space, independent of any other vector spaces you know of, you are required to prove all of the axioms in the definition. Not all operations that call themselves $+$ are worthy addition operations; just because you denote it $+$ does not mean it is (for example) associative, or has an additive identity.



              There is a lot to prove, because there's a lot to gain. Vector spaces have a simply enormous amount of structure, and that structure gives us a really rich theory and powerful tools. If you have an object that you wish to understand better, and you can show it is a vector space (or at least, related to a vector space), then you'll instantly have some serious mathematical firepower at your fingertips.



              Subspaces give us a shortcut to proving a vector space. If you have a subset of a known vector space, then you can prove just $3$ properties, rather than $10$. We can skip a lot of the steps because somebody has already done them previously when showing the larger vector space is indeed a vector space. You don't need to show, for example, $v + w = w + v$ for all $v, w$ in your subset, because we already know this is true for all vectors in the larger vector space.



              I'm writing this, not as a direct answer to your question (which Jose Carlos Santos has answered already), but because confusion like this often stems from some sloppiness on the above point. I've seen many students (and, lamentably, several instructors) fail to grasp that showing the subspace conditions on a set that is not clearly a subset of a known vector space does not prove a vector space. The shortcut works because somebody has already established most of the axioms beforehand, but if this is not true, then the argument is a fallacy.



              You can absolutely apply the subspace conditions on the whole of a vector space provided you've proven it's a vector space already with axioms V1 - V10.






              share|cite|improve this answer









              $endgroup$


















                4












                $begingroup$

                I'm guessing that V1 - V10 are the axioms for proving vector spaces.



                To prove something is a vector space, independent of any other vector spaces you know of, you are required to prove all of the axioms in the definition. Not all operations that call themselves $+$ are worthy addition operations; just because you denote it $+$ does not mean it is (for example) associative, or has an additive identity.



                There is a lot to prove, because there's a lot to gain. Vector spaces have a simply enormous amount of structure, and that structure gives us a really rich theory and powerful tools. If you have an object that you wish to understand better, and you can show it is a vector space (or at least, related to a vector space), then you'll instantly have some serious mathematical firepower at your fingertips.



                Subspaces give us a shortcut to proving a vector space. If you have a subset of a known vector space, then you can prove just $3$ properties, rather than $10$. We can skip a lot of the steps because somebody has already done them previously when showing the larger vector space is indeed a vector space. You don't need to show, for example, $v + w = w + v$ for all $v, w$ in your subset, because we already know this is true for all vectors in the larger vector space.



                I'm writing this, not as a direct answer to your question (which Jose Carlos Santos has answered already), but because confusion like this often stems from some sloppiness on the above point. I've seen many students (and, lamentably, several instructors) fail to grasp that showing the subspace conditions on a set that is not clearly a subset of a known vector space does not prove a vector space. The shortcut works because somebody has already established most of the axioms beforehand, but if this is not true, then the argument is a fallacy.



                You can absolutely apply the subspace conditions on the whole of a vector space provided you've proven it's a vector space already with axioms V1 - V10.






                share|cite|improve this answer









                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  I'm guessing that V1 - V10 are the axioms for proving vector spaces.



                  To prove something is a vector space, independent of any other vector spaces you know of, you are required to prove all of the axioms in the definition. Not all operations that call themselves $+$ are worthy addition operations; just because you denote it $+$ does not mean it is (for example) associative, or has an additive identity.



                  There is a lot to prove, because there's a lot to gain. Vector spaces have a simply enormous amount of structure, and that structure gives us a really rich theory and powerful tools. If you have an object that you wish to understand better, and you can show it is a vector space (or at least, related to a vector space), then you'll instantly have some serious mathematical firepower at your fingertips.



                  Subspaces give us a shortcut to proving a vector space. If you have a subset of a known vector space, then you can prove just $3$ properties, rather than $10$. We can skip a lot of the steps because somebody has already done them previously when showing the larger vector space is indeed a vector space. You don't need to show, for example, $v + w = w + v$ for all $v, w$ in your subset, because we already know this is true for all vectors in the larger vector space.



                  I'm writing this, not as a direct answer to your question (which Jose Carlos Santos has answered already), but because confusion like this often stems from some sloppiness on the above point. I've seen many students (and, lamentably, several instructors) fail to grasp that showing the subspace conditions on a set that is not clearly a subset of a known vector space does not prove a vector space. The shortcut works because somebody has already established most of the axioms beforehand, but if this is not true, then the argument is a fallacy.



                  You can absolutely apply the subspace conditions on the whole of a vector space provided you've proven it's a vector space already with axioms V1 - V10.






                  share|cite|improve this answer









                  $endgroup$



                  I'm guessing that V1 - V10 are the axioms for proving vector spaces.



                  To prove something is a vector space, independent of any other vector spaces you know of, you are required to prove all of the axioms in the definition. Not all operations that call themselves $+$ are worthy addition operations; just because you denote it $+$ does not mean it is (for example) associative, or has an additive identity.



                  There is a lot to prove, because there's a lot to gain. Vector spaces have a simply enormous amount of structure, and that structure gives us a really rich theory and powerful tools. If you have an object that you wish to understand better, and you can show it is a vector space (or at least, related to a vector space), then you'll instantly have some serious mathematical firepower at your fingertips.



                  Subspaces give us a shortcut to proving a vector space. If you have a subset of a known vector space, then you can prove just $3$ properties, rather than $10$. We can skip a lot of the steps because somebody has already done them previously when showing the larger vector space is indeed a vector space. You don't need to show, for example, $v + w = w + v$ for all $v, w$ in your subset, because we already know this is true for all vectors in the larger vector space.



                  I'm writing this, not as a direct answer to your question (which Jose Carlos Santos has answered already), but because confusion like this often stems from some sloppiness on the above point. I've seen many students (and, lamentably, several instructors) fail to grasp that showing the subspace conditions on a set that is not clearly a subset of a known vector space does not prove a vector space. The shortcut works because somebody has already established most of the axioms beforehand, but if this is not true, then the argument is a fallacy.



                  You can absolutely apply the subspace conditions on the whole of a vector space provided you've proven it's a vector space already with axioms V1 - V10.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 10 hours ago









                  Theo BenditTheo Bendit

                  20.8k12354




                  20.8k12354






























                      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%2f3180447%2fis-a-vector-space-a-subspace-of-itself%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

                      Lallio

                      Image
                      Coordenadas: 45° 40' N 9° 38' E Lallio      Comuna    Lallio Localização de Lallio na Itália Coordenadas 45° 40' N 9° 38' E Região Lombardia Província Bérgamo Área - Total 2 km² Altitude 216 m População  - Total 3 826     • Densidade 1 913 hab./km² Outros dados Comunas limítrofes Bergamo, Dalmine, Stezzano, Treviolo Código ISTAT 016123 Código cadastral E422 Código postal 24040 Prefixo telefônico 035 Website www.comune.lallio.bg.it Lallio é uma comuna italiana da região da Lombardia, província de Bérgamo, com cerca de 3.826 habitantes. Estende-se por uma área de 2 km², tendo uma densidade populacional de 1913 hab/km². Faz fronteira com Bergamo, Dalmine, Stezzano, Treviolo. [ 1 ] [ 2 ] [ 3 ] Demografia | Variação demográfica do município entre 1861 e 2011 [ 3 ] Fonte : Istituto Nazionale di Statistica (ISTAT) - Elabor...
                      Read more

                      Unable to find Lightning Node

                      Image
                      2 I have successfully installed and synced my lightning node (Raspiblitz). I see my public ip on port 8333 has 8 conns. However, when I check my node on 1ml.com my node is not found. I did forward port 8333 and 9735 on my router. Should I connect to a peer or create a channel before my node is visible for the outside? lightning-network lightning-network-daemon share | improve this question asked 1 hour ago Bart DV Bart DV 31 3 add a comment  |  ...
                      Read more

                      Futebolista

                      Image
                      Ronaldo primeiro jogador a ser eleito por três vezes o melhor jogador do mundo pela FIFA. Ronaldinho Gaúcho atuando pelo Barcelona onde foi duas vezes eleito o melhor jogador do mundo pela FIFA. Kaká atuando pelo Milan onde foi eleito em 2007 o melhor jogador do mundo pela FIFA. Lothar Matthäus primeiro jogador a ser eleito o melhor jogador do mundo pela FIFA em 1991 pela Inter de Milão. Neymar durante partida contra a Escócia, em 27 de março de 2011. Crianças praticando futebol. Robinho, futebolista brasileiro, em 15 de novembro de 2006. Atletas do Chelsea, da Inglaterra. A venda de camisas com nome e número dos jogadores é uma grande fonte de arrecadação. Jürgen Klinsmann atuou como jogador e foi treinador da Alemanha. Estádio de futebol, principal local onde laboram os jogadores. O jogador de futebol , ou Futebolista , é um atleta profissional de futebol, e cuja prática do desporto coletivo é fundamental. Estes trabalhadores sã...
                      Read more