Find dimension of even polynomials












1












$begingroup$


Let $V$ be a the vector space over $mathbb R$ of all polynomials with real coefficients. Let $W$ be the subset of all polynomials with only even powers in their expression.



So $p(X) in W$ means $p(x)=sum_{n=0}^k a_nx^{2n}$.



I showed that $W$ is a subspace of $V$, but I need to find out the dimension of $W$. Would I need to find a basis for $W$ first? How should I approach this?



Thank you.






linear-algebra vector-spaces







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    The dimension is infinite. The polynomials $1,x^2,x^4,x^6,dots$ are linearly independent.
    $endgroup$
    – André Nicolas
    Oct 15 '13 at 6:53










  • $begingroup$
    @AndréNicolas So would there be a basis?
    $endgroup$
    – Alti
    Oct 15 '13 at 7:05






  • 1




    $begingroup$
    Yes, the polynomials $1,x^2,x^4,x^6,dots$ are the simplest basis.
    $endgroup$
    – André Nicolas
    Oct 15 '13 at 7:09










  • $begingroup$
    @AndréNicolas I thought I would need a finite spanning subset in order to have a basis. Thank you
    $endgroup$
    – Alti
    Oct 15 '13 at 7:11






  • 1




    $begingroup$
    If $V$ was the space of all polynomials of degree say $le 5$, then a basis for your subspace would be ${1,x^2,x^4}$, the dimension would be $3$. But if $V$ is the infinite dimensional space of all polynomials, your $W$ is also infinite dimensional. Every vector space has a basis.
    $endgroup$
    – André Nicolas
    Oct 15 '13 at 7:16
















1












$begingroup$


Let $V$ be a the vector space over $mathbb R$ of all polynomials with real coefficients. Let $W$ be the subset of all polynomials with only even powers in their expression.



So $p(X) in W$ means $p(x)=sum_{n=0}^k a_nx^{2n}$.



I showed that $W$ is a subspace of $V$, but I need to find out the dimension of $W$. Would I need to find a basis for $W$ first? How should I approach this?



Thank you.






linear-algebra vector-spaces







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    The dimension is infinite. The polynomials $1,x^2,x^4,x^6,dots$ are linearly independent.
    $endgroup$
    – André Nicolas
    Oct 15 '13 at 6:53










  • $begingroup$
    @AndréNicolas So would there be a basis?
    $endgroup$
    – Alti
    Oct 15 '13 at 7:05






  • 1




    $begingroup$
    Yes, the polynomials $1,x^2,x^4,x^6,dots$ are the simplest basis.
    $endgroup$
    – André Nicolas
    Oct 15 '13 at 7:09










  • $begingroup$
    @AndréNicolas I thought I would need a finite spanning subset in order to have a basis. Thank you
    $endgroup$
    – Alti
    Oct 15 '13 at 7:11






  • 1




    $begingroup$
    If $V$ was the space of all polynomials of degree say $le 5$, then a basis for your subspace would be ${1,x^2,x^4}$, the dimension would be $3$. But if $V$ is the infinite dimensional space of all polynomials, your $W$ is also infinite dimensional. Every vector space has a basis.
    $endgroup$
    – André Nicolas
    Oct 15 '13 at 7:16














1












1








1





$begingroup$


Let $V$ be a the vector space over $mathbb R$ of all polynomials with real coefficients. Let $W$ be the subset of all polynomials with only even powers in their expression.



So $p(X) in W$ means $p(x)=sum_{n=0}^k a_nx^{2n}$.



I showed that $W$ is a subspace of $V$, but I need to find out the dimension of $W$. Would I need to find a basis for $W$ first? How should I approach this?



Thank you.






linear-algebra vector-spaces







share|cite|improve this question









$endgroup$




Let $V$ be a the vector space over $mathbb R$ of all polynomials with real coefficients. Let $W$ be the subset of all polynomials with only even powers in their expression.



So $p(X) in W$ means $p(x)=sum_{n=0}^k a_nx^{2n}$.



I showed that $W$ is a subspace of $V$, but I need to find out the dimension of $W$. Would I need to find a basis for $W$ first? How should I approach this?



Thank you.






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










asked Oct 15 '13 at 6:52









AltiAlti

1,16832245




1,16832245








  • 2




    $begingroup$
    The dimension is infinite. The polynomials $1,x^2,x^4,x^6,dots$ are linearly independent.
    $endgroup$
    – André Nicolas
    Oct 15 '13 at 6:53










  • $begingroup$
    @AndréNicolas So would there be a basis?
    $endgroup$
    – Alti
    Oct 15 '13 at 7:05






  • 1




    $begingroup$
    Yes, the polynomials $1,x^2,x^4,x^6,dots$ are the simplest basis.
    $endgroup$
    – André Nicolas
    Oct 15 '13 at 7:09










  • $begingroup$
    @AndréNicolas I thought I would need a finite spanning subset in order to have a basis. Thank you
    $endgroup$
    – Alti
    Oct 15 '13 at 7:11






  • 1




    $begingroup$
    If $V$ was the space of all polynomials of degree say $le 5$, then a basis for your subspace would be ${1,x^2,x^4}$, the dimension would be $3$. But if $V$ is the infinite dimensional space of all polynomials, your $W$ is also infinite dimensional. Every vector space has a basis.
    $endgroup$
    – André Nicolas
    Oct 15 '13 at 7:16














  • 2




    $begingroup$
    The dimension is infinite. The polynomials $1,x^2,x^4,x^6,dots$ are linearly independent.
    $endgroup$
    – André Nicolas
    Oct 15 '13 at 6:53










  • $begingroup$
    @AndréNicolas So would there be a basis?
    $endgroup$
    – Alti
    Oct 15 '13 at 7:05






  • 1




    $begingroup$
    Yes, the polynomials $1,x^2,x^4,x^6,dots$ are the simplest basis.
    $endgroup$
    – André Nicolas
    Oct 15 '13 at 7:09










  • $begingroup$
    @AndréNicolas I thought I would need a finite spanning subset in order to have a basis. Thank you
    $endgroup$
    – Alti
    Oct 15 '13 at 7:11






  • 1




    $begingroup$
    If $V$ was the space of all polynomials of degree say $le 5$, then a basis for your subspace would be ${1,x^2,x^4}$, the dimension would be $3$. But if $V$ is the infinite dimensional space of all polynomials, your $W$ is also infinite dimensional. Every vector space has a basis.
    $endgroup$
    – André Nicolas
    Oct 15 '13 at 7:16








2




2




$begingroup$
The dimension is infinite. The polynomials $1,x^2,x^4,x^6,dots$ are linearly independent.
$endgroup$
– André Nicolas
Oct 15 '13 at 6:53




$begingroup$
The dimension is infinite. The polynomials $1,x^2,x^4,x^6,dots$ are linearly independent.
$endgroup$
– André Nicolas
Oct 15 '13 at 6:53












$begingroup$
@AndréNicolas So would there be a basis?
$endgroup$
– Alti
Oct 15 '13 at 7:05




$begingroup$
@AndréNicolas So would there be a basis?
$endgroup$
– Alti
Oct 15 '13 at 7:05




1




1




$begingroup$
Yes, the polynomials $1,x^2,x^4,x^6,dots$ are the simplest basis.
$endgroup$
– André Nicolas
Oct 15 '13 at 7:09




$begingroup$
Yes, the polynomials $1,x^2,x^4,x^6,dots$ are the simplest basis.
$endgroup$
– André Nicolas
Oct 15 '13 at 7:09












$begingroup$
@AndréNicolas I thought I would need a finite spanning subset in order to have a basis. Thank you
$endgroup$
– Alti
Oct 15 '13 at 7:11




$begingroup$
@AndréNicolas I thought I would need a finite spanning subset in order to have a basis. Thank you
$endgroup$
– Alti
Oct 15 '13 at 7:11




1




1




$begingroup$
If $V$ was the space of all polynomials of degree say $le 5$, then a basis for your subspace would be ${1,x^2,x^4}$, the dimension would be $3$. But if $V$ is the infinite dimensional space of all polynomials, your $W$ is also infinite dimensional. Every vector space has a basis.
$endgroup$
– André Nicolas
Oct 15 '13 at 7:16




$begingroup$
If $V$ was the space of all polynomials of degree say $le 5$, then a basis for your subspace would be ${1,x^2,x^4}$, the dimension would be $3$. But if $V$ is the infinite dimensional space of all polynomials, your $W$ is also infinite dimensional. Every vector space has a basis.
$endgroup$
– André Nicolas
Oct 15 '13 at 7:16










2 Answers
2






active

oldest

votes


















3












$begingroup$

Your space $W$ is infinite dimensional. It is easy to see that every polynomial in $W$ can be expressed as a (finite) linear combination of the polynomials $1, x^2, x^4, x^6, x^8,dots$.



For completeness, one should show that these polynomials form a linearly independent set, and are therefore a basis for $W$. This follows immediately from the definition of equality for polynomials. For an alternative argument, suppose that $P(x)=a_0+a_1x^2+a_2x^4+cdots +a_nx^{2n}$ is identically $0$. We show that all the $a_i$ are $0$.



If the $a_i$ are not all $0$, we may assume without loss of generality that $a_nne 0$. From
$$0equiv a_0+a_1x^2+a_2x^4+cdots +a_nx^{2n},$$
by differentiating $2n$ times we get
$$0equiv (a_n)(2n)!,$$
which is false.






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    The best way I could think to do it (I at first got a little tripped up equating two sides and wanting to cancel like terms to keep linear independence, but that's not the case here since we want even functions i.e. $f(-x)= f(x)$, so it damn sure better be the same on both sides).



    Looks like no one else here made the same mistake I did, but now for coming up with the dimension you can say if $n=$even $Rightarrow dim(W)= n/2+1$ (plus one for your constant term). If $n=$odd $Rightarrow dim(W)=(n+1)/2$ (again the $+1$ for the constant term).






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Welcome to MSE. Your answer adds nothing new to the already existing answers.
      $endgroup$
      – José Carlos Santos
      Dec 22 '18 at 19:07











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Your space $W$ is infinite dimensional. It is easy to see that every polynomial in $W$ can be expressed as a (finite) linear combination of the polynomials $1, x^2, x^4, x^6, x^8,dots$.



    For completeness, one should show that these polynomials form a linearly independent set, and are therefore a basis for $W$. This follows immediately from the definition of equality for polynomials. For an alternative argument, suppose that $P(x)=a_0+a_1x^2+a_2x^4+cdots +a_nx^{2n}$ is identically $0$. We show that all the $a_i$ are $0$.



    If the $a_i$ are not all $0$, we may assume without loss of generality that $a_nne 0$. From
    $$0equiv a_0+a_1x^2+a_2x^4+cdots +a_nx^{2n},$$
    by differentiating $2n$ times we get
    $$0equiv (a_n)(2n)!,$$
    which is false.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Your space $W$ is infinite dimensional. It is easy to see that every polynomial in $W$ can be expressed as a (finite) linear combination of the polynomials $1, x^2, x^4, x^6, x^8,dots$.



      For completeness, one should show that these polynomials form a linearly independent set, and are therefore a basis for $W$. This follows immediately from the definition of equality for polynomials. For an alternative argument, suppose that $P(x)=a_0+a_1x^2+a_2x^4+cdots +a_nx^{2n}$ is identically $0$. We show that all the $a_i$ are $0$.



      If the $a_i$ are not all $0$, we may assume without loss of generality that $a_nne 0$. From
      $$0equiv a_0+a_1x^2+a_2x^4+cdots +a_nx^{2n},$$
      by differentiating $2n$ times we get
      $$0equiv (a_n)(2n)!,$$
      which is false.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Your space $W$ is infinite dimensional. It is easy to see that every polynomial in $W$ can be expressed as a (finite) linear combination of the polynomials $1, x^2, x^4, x^6, x^8,dots$.



        For completeness, one should show that these polynomials form a linearly independent set, and are therefore a basis for $W$. This follows immediately from the definition of equality for polynomials. For an alternative argument, suppose that $P(x)=a_0+a_1x^2+a_2x^4+cdots +a_nx^{2n}$ is identically $0$. We show that all the $a_i$ are $0$.



        If the $a_i$ are not all $0$, we may assume without loss of generality that $a_nne 0$. From
        $$0equiv a_0+a_1x^2+a_2x^4+cdots +a_nx^{2n},$$
        by differentiating $2n$ times we get
        $$0equiv (a_n)(2n)!,$$
        which is false.






        share|cite|improve this answer









        $endgroup$



        Your space $W$ is infinite dimensional. It is easy to see that every polynomial in $W$ can be expressed as a (finite) linear combination of the polynomials $1, x^2, x^4, x^6, x^8,dots$.



        For completeness, one should show that these polynomials form a linearly independent set, and are therefore a basis for $W$. This follows immediately from the definition of equality for polynomials. For an alternative argument, suppose that $P(x)=a_0+a_1x^2+a_2x^4+cdots +a_nx^{2n}$ is identically $0$. We show that all the $a_i$ are $0$.



        If the $a_i$ are not all $0$, we may assume without loss of generality that $a_nne 0$. From
        $$0equiv a_0+a_1x^2+a_2x^4+cdots +a_nx^{2n},$$
        by differentiating $2n$ times we get
        $$0equiv (a_n)(2n)!,$$
        which is false.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Oct 15 '13 at 8:04









        André NicolasAndré Nicolas

        452k36425810




        452k36425810























            -1












            $begingroup$

            The best way I could think to do it (I at first got a little tripped up equating two sides and wanting to cancel like terms to keep linear independence, but that's not the case here since we want even functions i.e. $f(-x)= f(x)$, so it damn sure better be the same on both sides).



            Looks like no one else here made the same mistake I did, but now for coming up with the dimension you can say if $n=$even $Rightarrow dim(W)= n/2+1$ (plus one for your constant term). If $n=$odd $Rightarrow dim(W)=(n+1)/2$ (again the $+1$ for the constant term).






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Welcome to MSE. Your answer adds nothing new to the already existing answers.
              $endgroup$
              – José Carlos Santos
              Dec 22 '18 at 19:07
















            -1












            $begingroup$

            The best way I could think to do it (I at first got a little tripped up equating two sides and wanting to cancel like terms to keep linear independence, but that's not the case here since we want even functions i.e. $f(-x)= f(x)$, so it damn sure better be the same on both sides).



            Looks like no one else here made the same mistake I did, but now for coming up with the dimension you can say if $n=$even $Rightarrow dim(W)= n/2+1$ (plus one for your constant term). If $n=$odd $Rightarrow dim(W)=(n+1)/2$ (again the $+1$ for the constant term).






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Welcome to MSE. Your answer adds nothing new to the already existing answers.
              $endgroup$
              – José Carlos Santos
              Dec 22 '18 at 19:07














            -1












            -1








            -1





            $begingroup$

            The best way I could think to do it (I at first got a little tripped up equating two sides and wanting to cancel like terms to keep linear independence, but that's not the case here since we want even functions i.e. $f(-x)= f(x)$, so it damn sure better be the same on both sides).



            Looks like no one else here made the same mistake I did, but now for coming up with the dimension you can say if $n=$even $Rightarrow dim(W)= n/2+1$ (plus one for your constant term). If $n=$odd $Rightarrow dim(W)=(n+1)/2$ (again the $+1$ for the constant term).






            share|cite|improve this answer











            $endgroup$



            The best way I could think to do it (I at first got a little tripped up equating two sides and wanting to cancel like terms to keep linear independence, but that's not the case here since we want even functions i.e. $f(-x)= f(x)$, so it damn sure better be the same on both sides).



            Looks like no one else here made the same mistake I did, but now for coming up with the dimension you can say if $n=$even $Rightarrow dim(W)= n/2+1$ (plus one for your constant term). If $n=$odd $Rightarrow dim(W)=(n+1)/2$ (again the $+1$ for the constant term).







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 22 '18 at 20:14









            dantopa

            6,46942243




            6,46942243










            answered Dec 22 '18 at 18:46









            ARichardsonARichardson

            12




            12












            • $begingroup$
              Welcome to MSE. Your answer adds nothing new to the already existing answers.
              $endgroup$
              – José Carlos Santos
              Dec 22 '18 at 19:07


















            • $begingroup$
              Welcome to MSE. Your answer adds nothing new to the already existing answers.
              $endgroup$
              – José Carlos Santos
              Dec 22 '18 at 19:07
















            $begingroup$
            Welcome to MSE. Your answer adds nothing new to the already existing answers.
            $endgroup$
            – José Carlos Santos
            Dec 22 '18 at 19:07




            $begingroup$
            Welcome to MSE. Your answer adds nothing new to the already existing answers.
            $endgroup$
            – José Carlos Santos
            Dec 22 '18 at 19:07


















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