Find dimension of even polynomials
$begingroup$
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
asked Oct 15 '13 at 6:52
AltiAlti
1,16832245
$endgroup$
|
show 1 more comment
$begingroup$
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
asked Oct 15 '13 at 6:52
AltiAlti
1,16832245
$endgroup$
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
|
show 1 more comment
$begingroup$
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
asked Oct 15 '13 at 6:52
AltiAlti
1,16832245
$endgroup$
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
asked Oct 15 '13 at 6:52
AltiAlti
1,16832245
asked Oct 15 '13 at 6:52
AltiAlti
1,16832245
asked Oct 15 '13 at 6:52
AltiAlti
1,16832245
asked Oct 15 '13 at 6:52
AltiAlti
1,16832245
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
|
show 1 more comment
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
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Oct 15 '13 at 8:04
André NicolasAndré Nicolas
452k36425810
$endgroup$
add a comment |
$begingroup$
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).
edited Dec 22 '18 at 20:14
dantopa
6,46942243
answered Dec 22 '18 at 18:46
ARichardsonARichardson
12
$endgroup$
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Dec 22 '18 at 19:07
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
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oldest
votes
$begingroup$
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.
answered Oct 15 '13 at 8:04
André NicolasAndré Nicolas
452k36425810
$endgroup$
add a comment |
$begingroup$
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.
answered Oct 15 '13 at 8:04
André NicolasAndré Nicolas
452k36425810
$endgroup$
add a comment |
$begingroup$
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.
answered Oct 15 '13 at 8:04
André NicolasAndré Nicolas
452k36425810
$endgroup$
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.
answered Oct 15 '13 at 8:04
André NicolasAndré Nicolas
452k36425810
answered Oct 15 '13 at 8:04
André NicolasAndré Nicolas
452k36425810
answered Oct 15 '13 at 8:04
André NicolasAndré Nicolas
452k36425810
answered Oct 15 '13 at 8:04
André NicolasAndré Nicolas
452k36425810
452k36425810
add a comment |
add a comment |
$begingroup$
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).
edited Dec 22 '18 at 20:14
dantopa
6,46942243
answered Dec 22 '18 at 18:46
ARichardsonARichardson
12
$endgroup$
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Dec 22 '18 at 19:07
add a comment |
$begingroup$
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).
edited Dec 22 '18 at 20:14
dantopa
6,46942243
answered Dec 22 '18 at 18:46
ARichardsonARichardson
12
$endgroup$
$begingroup$
Welcome to MSE. Your answer adds nothing new to the already existing answers.
$endgroup$
– José Carlos Santos
Dec 22 '18 at 19:07
add a comment |
$begingroup$
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).
edited Dec 22 '18 at 20:14
dantopa
6,46942243
answered Dec 22 '18 at 18:46
ARichardsonARichardson
12
$endgroup$
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).
edited Dec 22 '18 at 20:14
dantopa
6,46942243
answered Dec 22 '18 at 18:46
ARichardsonARichardson
12
edited Dec 22 '18 at 20:14
dantopa
6,46942243
edited Dec 22 '18 at 20:14
dantopa
6,46942243
edited Dec 22 '18 at 20:14
dantopa
6,46942243
6,46942243
answered Dec 22 '18 at 18:46
ARichardsonARichardson
12
answered Dec 22 '18 at 18:46
ARichardsonARichardson
12
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
add a comment |
$begingroup$
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
add a comment |
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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