Prove a set of functions is a sub space











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Let $mathbb{R}^mathbb{R} $ be a vector-space of functions $f:mathbb{R}→mathbb{R}$.



The set of functions $f:mathbb{R}→mathbb{R}$ so that $f(-1) + f(1) = 0$ are an example of subspace?



I know I have to prove 0 is in the set, the sum of two vectors (functions) is in the set and the scalar product is in the set.
I have 9 different groups to check whether they are sub spaces or not, but really don't know the method to proceed with the question, so I would like this as an example.



Thanks!






linear-algebra functions







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  • If you add two functions who both satisfy your property, check they they too satisfy the property. If you multiply a function which satisfies your property by a real number, check that the result too satisfies the property.
    – JMoravitz
    Nov 22 at 20:28















up vote
0
down vote

favorite












Let $mathbb{R}^mathbb{R} $ be a vector-space of functions $f:mathbb{R}→mathbb{R}$.



The set of functions $f:mathbb{R}→mathbb{R}$ so that $f(-1) + f(1) = 0$ are an example of subspace?



I know I have to prove 0 is in the set, the sum of two vectors (functions) is in the set and the scalar product is in the set.
I have 9 different groups to check whether they are sub spaces or not, but really don't know the method to proceed with the question, so I would like this as an example.



Thanks!






linear-algebra functions







share|cite|improve this question






















  • If you add two functions who both satisfy your property, check they they too satisfy the property. If you multiply a function which satisfies your property by a real number, check that the result too satisfies the property.
    – JMoravitz
    Nov 22 at 20:28













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $mathbb{R}^mathbb{R} $ be a vector-space of functions $f:mathbb{R}→mathbb{R}$.



The set of functions $f:mathbb{R}→mathbb{R}$ so that $f(-1) + f(1) = 0$ are an example of subspace?



I know I have to prove 0 is in the set, the sum of two vectors (functions) is in the set and the scalar product is in the set.
I have 9 different groups to check whether they are sub spaces or not, but really don't know the method to proceed with the question, so I would like this as an example.



Thanks!






linear-algebra functions







share|cite|improve this question













Let $mathbb{R}^mathbb{R} $ be a vector-space of functions $f:mathbb{R}→mathbb{R}$.



The set of functions $f:mathbb{R}→mathbb{R}$ so that $f(-1) + f(1) = 0$ are an example of subspace?



I know I have to prove 0 is in the set, the sum of two vectors (functions) is in the set and the scalar product is in the set.
I have 9 different groups to check whether they are sub spaces or not, but really don't know the method to proceed with the question, so I would like this as an example.



Thanks!






linear-algebra functions




linear-algebra functions






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asked Nov 22 at 20:21









Tegernako

596




596












  • If you add two functions who both satisfy your property, check they they too satisfy the property. If you multiply a function which satisfies your property by a real number, check that the result too satisfies the property.
    – JMoravitz
    Nov 22 at 20:28


















  • If you add two functions who both satisfy your property, check they they too satisfy the property. If you multiply a function which satisfies your property by a real number, check that the result too satisfies the property.
    – JMoravitz
    Nov 22 at 20:28
















If you add two functions who both satisfy your property, check they they too satisfy the property. If you multiply a function which satisfies your property by a real number, check that the result too satisfies the property.
– JMoravitz
Nov 22 at 20:28




If you add two functions who both satisfy your property, check they they too satisfy the property. If you multiply a function which satisfies your property by a real number, check that the result too satisfies the property.
– JMoravitz
Nov 22 at 20:28










2 Answers
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Yes that is. We define $S$ as following $$S={f|f(1)+f(-1)=0}$$simply $0in S$. Now if $f$ and $g$ are in $S$ we have $$f(1)+f(-1)=0\g(1)+g(-1)=0$$therefore$$(af+bg)(1)+(af+bg)(-1)=a(f(1)+f(-1))+b(g(1)+g(-1))=0$$for any $a,binBbb R$ therefore $S$ is a subspace






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  • I see, this clarified it for me. Then by the following logic, if we have the group f(0) + f(1) = 1, it is not a subspace because: (f+g)(0) + (f+g)(1) = (f(0) + f(1)) + (g(0) + g(1)) = 2 correct?
    – Tegernako
    Nov 23 at 6:28












  • That's right. Furthermore $0notin S$ which is another argument ....
    – Mostafa Ayaz
    Nov 23 at 7:26


















up vote
0
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$f(x)=0$ for all $x$ implies $fin S :={f|f(1)+f(-1)=0} $ and if $g,hin S$ and $a,binmathbb{R}$ then $ag(1)+bh(1)+ag(-1)+bh(-1)=a(g(1)+g(-1))+b(h(1)+h(-1))=0$. So, $ag+bhin S$. So, $S$ is a sub-space.






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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    up vote
    0
    down vote



    accepted










    Yes that is. We define $S$ as following $$S={f|f(1)+f(-1)=0}$$simply $0in S$. Now if $f$ and $g$ are in $S$ we have $$f(1)+f(-1)=0\g(1)+g(-1)=0$$therefore$$(af+bg)(1)+(af+bg)(-1)=a(f(1)+f(-1))+b(g(1)+g(-1))=0$$for any $a,binBbb R$ therefore $S$ is a subspace






    share|cite|improve this answer





















    • I see, this clarified it for me. Then by the following logic, if we have the group f(0) + f(1) = 1, it is not a subspace because: (f+g)(0) + (f+g)(1) = (f(0) + f(1)) + (g(0) + g(1)) = 2 correct?
      – Tegernako
      Nov 23 at 6:28












    • That's right. Furthermore $0notin S$ which is another argument ....
      – Mostafa Ayaz
      Nov 23 at 7:26















    up vote
    0
    down vote



    accepted










    Yes that is. We define $S$ as following $$S={f|f(1)+f(-1)=0}$$simply $0in S$. Now if $f$ and $g$ are in $S$ we have $$f(1)+f(-1)=0\g(1)+g(-1)=0$$therefore$$(af+bg)(1)+(af+bg)(-1)=a(f(1)+f(-1))+b(g(1)+g(-1))=0$$for any $a,binBbb R$ therefore $S$ is a subspace






    share|cite|improve this answer





















    • I see, this clarified it for me. Then by the following logic, if we have the group f(0) + f(1) = 1, it is not a subspace because: (f+g)(0) + (f+g)(1) = (f(0) + f(1)) + (g(0) + g(1)) = 2 correct?
      – Tegernako
      Nov 23 at 6:28












    • That's right. Furthermore $0notin S$ which is another argument ....
      – Mostafa Ayaz
      Nov 23 at 7:26













    up vote
    0
    down vote



    accepted







    up vote
    0
    down vote



    accepted






    Yes that is. We define $S$ as following $$S={f|f(1)+f(-1)=0}$$simply $0in S$. Now if $f$ and $g$ are in $S$ we have $$f(1)+f(-1)=0\g(1)+g(-1)=0$$therefore$$(af+bg)(1)+(af+bg)(-1)=a(f(1)+f(-1))+b(g(1)+g(-1))=0$$for any $a,binBbb R$ therefore $S$ is a subspace






    share|cite|improve this answer












    Yes that is. We define $S$ as following $$S={f|f(1)+f(-1)=0}$$simply $0in S$. Now if $f$ and $g$ are in $S$ we have $$f(1)+f(-1)=0\g(1)+g(-1)=0$$therefore$$(af+bg)(1)+(af+bg)(-1)=a(f(1)+f(-1))+b(g(1)+g(-1))=0$$for any $a,binBbb R$ therefore $S$ is a subspace







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 22 at 21:29









    Mostafa Ayaz

    13.6k3836




    13.6k3836












    • I see, this clarified it for me. Then by the following logic, if we have the group f(0) + f(1) = 1, it is not a subspace because: (f+g)(0) + (f+g)(1) = (f(0) + f(1)) + (g(0) + g(1)) = 2 correct?
      – Tegernako
      Nov 23 at 6:28












    • That's right. Furthermore $0notin S$ which is another argument ....
      – Mostafa Ayaz
      Nov 23 at 7:26


















    • I see, this clarified it for me. Then by the following logic, if we have the group f(0) + f(1) = 1, it is not a subspace because: (f+g)(0) + (f+g)(1) = (f(0) + f(1)) + (g(0) + g(1)) = 2 correct?
      – Tegernako
      Nov 23 at 6:28












    • That's right. Furthermore $0notin S$ which is another argument ....
      – Mostafa Ayaz
      Nov 23 at 7:26
















    I see, this clarified it for me. Then by the following logic, if we have the group f(0) + f(1) = 1, it is not a subspace because: (f+g)(0) + (f+g)(1) = (f(0) + f(1)) + (g(0) + g(1)) = 2 correct?
    – Tegernako
    Nov 23 at 6:28






    I see, this clarified it for me. Then by the following logic, if we have the group f(0) + f(1) = 1, it is not a subspace because: (f+g)(0) + (f+g)(1) = (f(0) + f(1)) + (g(0) + g(1)) = 2 correct?
    – Tegernako
    Nov 23 at 6:28














    That's right. Furthermore $0notin S$ which is another argument ....
    – Mostafa Ayaz
    Nov 23 at 7:26




    That's right. Furthermore $0notin S$ which is another argument ....
    – Mostafa Ayaz
    Nov 23 at 7:26










    up vote
    0
    down vote













    $f(x)=0$ for all $x$ implies $fin S :={f|f(1)+f(-1)=0} $ and if $g,hin S$ and $a,binmathbb{R}$ then $ag(1)+bh(1)+ag(-1)+bh(-1)=a(g(1)+g(-1))+b(h(1)+h(-1))=0$. So, $ag+bhin S$. So, $S$ is a sub-space.






    share|cite|improve this answer

























      up vote
      0
      down vote













      $f(x)=0$ for all $x$ implies $fin S :={f|f(1)+f(-1)=0} $ and if $g,hin S$ and $a,binmathbb{R}$ then $ag(1)+bh(1)+ag(-1)+bh(-1)=a(g(1)+g(-1))+b(h(1)+h(-1))=0$. So, $ag+bhin S$. So, $S$ is a sub-space.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        $f(x)=0$ for all $x$ implies $fin S :={f|f(1)+f(-1)=0} $ and if $g,hin S$ and $a,binmathbb{R}$ then $ag(1)+bh(1)+ag(-1)+bh(-1)=a(g(1)+g(-1))+b(h(1)+h(-1))=0$. So, $ag+bhin S$. So, $S$ is a sub-space.






        share|cite|improve this answer












        $f(x)=0$ for all $x$ implies $fin S :={f|f(1)+f(-1)=0} $ and if $g,hin S$ and $a,binmathbb{R}$ then $ag(1)+bh(1)+ag(-1)+bh(-1)=a(g(1)+g(-1))+b(h(1)+h(-1))=0$. So, $ag+bhin S$. So, $S$ is a sub-space.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 20:27









        John_Wick

        1,199111




        1,199111






























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