Is a vector space a subspace of itself?vector space and its subspaceProve: The set of all polynomials p with p(2) = p(3) is a vector spaceProving a vector space over itself have no subspacesVector Space vs SubspaceProving a subset is a subspace of a Vector SpaceAre all vector spaces also a subspace?Understand the definition of a vector subspaceProve that, with vector addition and scalar multiplication well-defined, $V/W$ becomes a vector space over $k$.Is the set of all exponential functions a subspace of the vector space of all continuous functions?I've seen two definitions of subspace; one involving vector spaces and one requiring linear combinations

Where else does the Shulchan Aruch quote an authority by name?

Patience, young "Padovan"

What is the offset in a seaplane's hull?

Need help identifying/translating a plaque in Tangier, Morocco

How can I add custom success page

Copycat chess is back

Is this food a bread or a loaf?

"My colleague's body is amazing"

map list to bin numbers

Can the Produce Flame cantrip be used to grapple, or as an unarmed strike, in the right circumstances?

What causes the sudden spool-up sound from an F-16 when enabling afterburner?

What are the advantages and disadvantages of running one shots compared to campaigns?

Is there a name of the flying bionic bird?

Are objects structures and/or vice versa?

What is it called when one voice type sings a 'solo'?

What is GPS' 19 year rollover and does it present a cybersecurity issue?

Why is the design of haulage companies so “special”?

Does bootstrapped regression allow for inference?

Are cabin dividers used to "hide" the flex of the airplane?

Where to refill my bottle in India?

Calculate Levenshtein distance between two strings in Python

Why do UK politicians seemingly ignore opinion polls on Brexit?

How is it possible for user's password to be changed after storage was encrypted? (on OS X, Android)

What is the meaning of "of trouble" in the following sentence?



Is a vector space a subspace of itself?


vector space and its subspaceProve: The set of all polynomials p with p(2) = p(3) is a vector spaceProving a vector space over itself have no subspacesVector Space vs SubspaceProving a subset is a subspace of a Vector SpaceAre all vector spaces also a subspace?Understand the definition of a vector subspaceProve that, with vector addition and scalar multiplication well-defined, $V/W$ becomes a vector space over $k$.Is the set of all exponential functions a subspace of the vector space of all continuous functions?I've seen two definitions of subspace; one involving vector spaces and one requiring linear combinations













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











$endgroup$







  • 1




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






  • 1




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







  • 1




    $begingroup$
    The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
    $endgroup$
    – Theo Bendit
    3 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
    1 hour ago















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











$endgroup$







  • 1




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






  • 1




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







  • 1




    $begingroup$
    The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
    $endgroup$
    – Theo Bendit
    3 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
    1 hour ago













1












1








1


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 30 mins ago









Eric Wofsey

192k14220352




192k14220352










asked 3 hours ago









mingming

4456




4456







  • 1




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






  • 1




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







  • 1




    $begingroup$
    The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
    $endgroup$
    – Theo Bendit
    3 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
    1 hour ago












  • 1




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






  • 1




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







  • 1




    $begingroup$
    The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
    $endgroup$
    – Theo Bendit
    3 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
    1 hour ago







1




1




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




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




1




1




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





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





1




1




$begingroup$
The term "proper" subspace is often used to denote a subspace space that is not the entire vector space.
$endgroup$
– Theo Bendit
3 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
3 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
1 hour 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
1 hour ago










2 Answers
2






active

oldest

votes


















7












$begingroup$

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









$endgroup$




















    2












    $begingroup$

    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









      7












      $begingroup$

      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









      $endgroup$

















        7












        $begingroup$

        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









        $endgroup$















          7












          7








          7





          $begingroup$

          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 3 hours ago









          José Carlos SantosJosé Carlos Santos

          173k23133241




          173k23133241





















              2












              $begingroup$

              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









              $endgroup$

















                2












                $begingroup$

                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









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  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 2 hours ago









                  Theo BenditTheo Bendit

                  20.7k12354




                  20.7k12354



























                      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

                      Are there any AGPL-style licences that require source code modifications to be public? Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Force derivative works to be publicAre there any GPL like licenses for Apple App Store?Do you violate the GPL if you provide source code that cannot be compiled?GPL - is it distribution to use libraries in an appliance loaned to customers?Distributing App for free which uses GPL'ed codeModifications of server software under GPL, with web/CLI interfaceDoes using an AGPLv3-licensed library prevent me from dual-licensing my own source code?Can I publish only select code under GPLv3 from a private project?Is there published precedent regarding the scope of covered work that uses AGPL software?If MIT licensed code links to GPL licensed code what should be the license of the resulting binary program?If I use a public API endpoint that has its source code licensed under AGPL in my app, do I need to disclose my source?

                      Image
                      Flight departed from the gate 5 min before scheduled departure time. Refund options How to show a density matrix is in a pure/mixed state? Can we cancel the equality mark here? How to ask rejected full-time candidates to apply to teach individual courses? Are there any AGPL-style licences that require source code modifications to be public? Why did Israel vote against lifting the American embargo on Cuba? What should one know about term logic before studying propositional and predicate logic? Why is Rajasthan pro BJP in the LS elections but not in the state elections? New Order #6: Easter Egg Question on Gÿongy' lemma proof How to resize main filesystem Is it OK if I do not take the receipt in Germany? My mentor says to set image to Fine instead of RAW — how is this different from JPG? Can I feed enough spin up electron to a black hole to affect it's angular momentum? How do I find my Spellcasting Ability for my D&D character? Is my guitar...
                      Read more

                      2013 GY136 Descoberta | Órbita | Referências Menu de navegação«List Of Centaurs and Scattered-Disk Objects»«List of Known Trans-Neptunian Objects»

                      Image
                      Objetos do disco dispersoObjetos transnetunianos objeto transnetunianodisco dispersoSistema Solarcorpo celesteressonância orbitalplanetaplanetamagnitude absolutadiâmetro9 de abril2013Outer Solar System Origins Surveyórbitaexcentricidadesemieixo maiorUAperiélioSolafélio 2013 GY136 Data da descoberta 9 de abril de 2013 Descoberto por Outer Solar System Origins Survey Categoria Transnetuniano Objeto do disco disperso Elementos orbitais Semieixo maior 55,961 UA Perélio 32,595 UA Afélio 79,327 UA Excentricidade 0,418 Inclinação 10,9° Características físicas Dimensões 133 km Magnitude absoluta 7,6 ver 2013 GY 136 , também escrito como 2013 GY136 , é um objeto transnetuniano que está localizado no disco disperso, uma região do Sistema Solar. Este corpo celeste está em uma ressonância orbital de 2:5 com o planeta planeta. Ele possui uma magnitude absoluta de 7,6 [ 1 ] e tem um diâmetro estimad...
                      Read more

                      Metrô de Los Teques Índice Linhas | Estações | Ver também | Referências Ligações externas | Menu de navegação«INSTITUCIÓN»«Mapa de rutas»originalMetrô de Los TequesC.A. Metro Los Teques |Alcaldía de Guaicaipuro – Sitio OficialGobernacion de Mirandaeeeeeee

                      Image
                      Metrô de Los TequesFundações na Venezuela em 2006 metropolitanoLos TequesGuaicaipuroMirandaGoverno da VenezuelaLinha 1Linha 2estaçõesLinha 1Linha 2IndependenciaSan AntonioLinha 3GuaicaipuroLibertadorEstação Las AdjuntasMetrô de Caracasalgarismocorestaçõesnomecorestaçõesestações Metrô de Los Teques Uma das composições que operam no sistema. Informações Proprietário Governo da Venezuela Local [[ Los Teques, Miranda]] País variant = Tipo de transporte Metropolitano Número de linhas 2 Número de estações 5 Website [1] Funcionamento Início de funcionamento 3 de novembro de 2006 (12 anos) [ 1 ] Operadora(s) C. A. Metro Los Teques Dados técnicos Extensão do sistema 11,2 km Bitola Bitola padrão (1435 mm) Velocidade máxima 80 km/h ...
                      Read more