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Is a vector space a subspace?


vector space and its subspaceProve: The set of all polynomials p with p(2) = p(3) is a vector spaceVector Space vs SubspaceProving a subset is a subspace of a Vector SpaceAre all vector spaces also a subspace?Understand the definition of a vector subspaceProving 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?Linear Algebra - Zero subspace vs empty subspace













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 confuse 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











$endgroup$











  • $begingroup$
    How do you define a subspace of a vector space?
    $endgroup$
    – Brian
    1 hour ago










  • $begingroup$
    Is a set a subset of itself?? What’s V1-V10?
    $endgroup$
    – J. W. Tanner
    1 hour 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
    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 confuse 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











$endgroup$











  • $begingroup$
    How do you define a subspace of a vector space?
    $endgroup$
    – Brian
    1 hour ago










  • $begingroup$
    Is a set a subset of itself?? What’s V1-V10?
    $endgroup$
    – J. W. Tanner
    1 hour 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
    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 confuse 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 confuse 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 1 hour ago









José Carlos Santos

173k23133241




173k23133241










asked 1 hour ago









mingming

4456




4456











  • $begingroup$
    How do you define a subspace of a vector space?
    $endgroup$
    – Brian
    1 hour ago










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
















  • $begingroup$
    How do you define a subspace of a vector space?
    $endgroup$
    – Brian
    1 hour ago










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















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




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












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





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




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










2 Answers
2






active

oldest

votes


















6












$begingroup$

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









$endgroup$




















    1












    $begingroup$

    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









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      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









      $endgroup$

















        6












        $begingroup$

        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









        $endgroup$















          6












          6








          6





          $begingroup$

          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 1 hour ago









          José Carlos SantosJosé Carlos Santos

          173k23133241




          173k23133241





















              1












              $begingroup$

              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









              $endgroup$

















                1












                $begingroup$

                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









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  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 37 mins ago









                  Theo BenditTheo Bendit

                  20.7k12354




                  20.7k12354



























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