Show that the vector space of polynomials R[x] is isomorphic to a proper subspace of itself
$begingroup$
Show that the vector space of polynomials R[x] is isomorphic to a proper
subspace of itself:
Vector Space Isomorphism exists when there exists a bijective (one-to-one and onto) linear mapping F:V $rightarrow$U.
the coefficient of the polynomials can be written as $(a_0,a_1,a_2...)$. But how to find the subspace?
What about instead to prove the dimension of the the two vector spaces is the same, which means isomorphic? But how to do it?
vector-space-isomorphism
asked May 11 '17 at 10:18
stedmoaoastedmoaoa
5311
$endgroup$
add a comment |
$begingroup$
Show that the vector space of polynomials R[x] is isomorphic to a proper
subspace of itself:
Vector Space Isomorphism exists when there exists a bijective (one-to-one and onto) linear mapping F:V $rightarrow$U.
the coefficient of the polynomials can be written as $(a_0,a_1,a_2...)$. But how to find the subspace?
What about instead to prove the dimension of the the two vector spaces is the same, which means isomorphic? But how to do it?
vector-space-isomorphism
asked May 11 '17 at 10:18
stedmoaoastedmoaoa
5311
$endgroup$
$begingroup$
Proper subspace means a subspace that is not $Bbb R[x]$.
$endgroup$
– Arnaud D.
May 11 '17 at 10:20
2
$begingroup$
Hint: consider $p(x) mapsto x p(x)$.
$endgroup$
– Kenny Wong
May 11 '17 at 10:28
1
$begingroup$
If it helps, you might want to forget the fact they're polynomials, and just think of them as sequences $(a_0, a_1, ldots)$ where each $a_i in R$. There is nothing polynomial-specific being used here.
$endgroup$
– Joppy
May 11 '17 at 13:26
add a comment |
$begingroup$
Show that the vector space of polynomials R[x] is isomorphic to a proper
subspace of itself:
Vector Space Isomorphism exists when there exists a bijective (one-to-one and onto) linear mapping F:V $rightarrow$U.
the coefficient of the polynomials can be written as $(a_0,a_1,a_2...)$. But how to find the subspace?
What about instead to prove the dimension of the the two vector spaces is the same, which means isomorphic? But how to do it?
vector-space-isomorphism
asked May 11 '17 at 10:18
stedmoaoastedmoaoa
5311
$endgroup$
Show that the vector space of polynomials R[x] is isomorphic to a proper
subspace of itself:
Vector Space Isomorphism exists when there exists a bijective (one-to-one and onto) linear mapping F:V $rightarrow$U.
the coefficient of the polynomials can be written as $(a_0,a_1,a_2...)$. But how to find the subspace?
What about instead to prove the dimension of the the two vector spaces is the same, which means isomorphic? But how to do it?
vector-space-isomorphism
vector-space-isomorphism
asked May 11 '17 at 10:18
stedmoaoastedmoaoa
5311
asked May 11 '17 at 10:18
stedmoaoastedmoaoa
5311
edited May 25 '17 at 11:35
stedmoaoa
asked May 11 '17 at 10:18
stedmoaoastedmoaoa
5311
asked May 11 '17 at 10:18
stedmoaoastedmoaoa
5311
asked May 11 '17 at 10:18
stedmoaoastedmoaoa
5311
5311
$begingroup$
Proper subspace means a subspace that is not $Bbb R[x]$.
$endgroup$
– Arnaud D.
May 11 '17 at 10:20
2
$begingroup$
Hint: consider $p(x) mapsto x p(x)$.
$endgroup$
– Kenny Wong
May 11 '17 at 10:28
1
$begingroup$
If it helps, you might want to forget the fact they're polynomials, and just think of them as sequences $(a_0, a_1, ldots)$ where each $a_i in R$. There is nothing polynomial-specific being used here.
$endgroup$
– Joppy
May 11 '17 at 13:26
add a comment |
$begingroup$
Proper subspace means a subspace that is not $Bbb R[x]$.
$endgroup$
– Arnaud D.
May 11 '17 at 10:20
2
$begingroup$
Hint: consider $p(x) mapsto x p(x)$.
$endgroup$
– Kenny Wong
May 11 '17 at 10:28
1
$begingroup$
If it helps, you might want to forget the fact they're polynomials, and just think of them as sequences $(a_0, a_1, ldots)$ where each $a_i in R$. There is nothing polynomial-specific being used here.
$endgroup$
– Joppy
May 11 '17 at 13:26
$begingroup$
Proper subspace means a subspace that is not $Bbb R[x]$.
$endgroup$
– Arnaud D.
May 11 '17 at 10:20
$begingroup$
Proper subspace means a subspace that is not $Bbb R[x]$.
$endgroup$
– Arnaud D.
May 11 '17 at 10:20
2
2
$begingroup$
Hint: consider $p(x) mapsto x p(x)$.
$endgroup$
– Kenny Wong
May 11 '17 at 10:28
$begingroup$
Hint: consider $p(x) mapsto x p(x)$.
$endgroup$
– Kenny Wong
May 11 '17 at 10:28
1
1
$begingroup$
If it helps, you might want to forget the fact they're polynomials, and just think of them as sequences $(a_0, a_1, ldots)$ where each $a_i in R$. There is nothing polynomial-specific being used here.
$endgroup$
– Joppy
May 11 '17 at 13:26
$begingroup$
If it helps, you might want to forget the fact they're polynomials, and just think of them as sequences $(a_0, a_1, ldots)$ where each $a_i in R$. There is nothing polynomial-specific being used here.
$endgroup$
– Joppy
May 11 '17 at 13:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Consider $p(x) mapsto p(x^2)$.
This acts on a sequence of coefficients (which is all a polynomial is) by inserting zeros between them:
$$(a_0,a_1,a_2,dots,a_n,0,0,0,dots) mapsto (a_0,0,a_1,0,a_2,dots,a_{n-1},0,a_n,0,0,0,dots)$$
and so you can recover one from the other.
answered May 11 '17 at 10:53
lhflhf
168k11172404
$endgroup$
$begingroup$
Can you show a bit more hint I am very new to this topic. Thank you!
$endgroup$
– stedmoaoa
May 11 '17 at 12:13
$begingroup$
So you mean for the polynomials, just inserts 0 in between and it constructs a subspace that it is isomorphic to?
$endgroup$
– stedmoaoa
May 25 '17 at 10:18
$begingroup$
I got the idea but how to show that it is one-to-one and onto?
$endgroup$
– stedmoaoa
May 25 '17 at 10:20
$begingroup$
@stedmoaoa, just use the function definition I gave.
$endgroup$
– lhf
May 25 '17 at 11:02
add a comment |
$begingroup$
First, note that you can't do this with finite dimensional vector spaces. You have to come up with a linear transformation which is one-to-one but not onto. Any linear transformation is determined by what happens to the vectors in any basis. Use the standard basis for $R[x]$, that is, $x^n$ where $nge 0$. You have to map each $x^n$ to a polynomial such that the mapping is one-to-one. One way to do this is to map $x^n$ to a polynomial of degree $n+1$. You can check this ensures that only the zero polynomial maps to zero. You can check that it is not onto because, for example, $x^0$ is not the image of any polynomial. These two facts depend on looking at the leading term of polynomials and seeing that no nonzero polynomial is a linear combination of polynomials of lesser degrees.
answered May 11 '17 at 20:33
SomosSomos
14.9k11337
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Consider $p(x) mapsto p(x^2)$.
This acts on a sequence of coefficients (which is all a polynomial is) by inserting zeros between them:
$$(a_0,a_1,a_2,dots,a_n,0,0,0,dots) mapsto (a_0,0,a_1,0,a_2,dots,a_{n-1},0,a_n,0,0,0,dots)$$
and so you can recover one from the other.
answered May 11 '17 at 10:53
lhflhf
168k11172404
$endgroup$
$begingroup$
Can you show a bit more hint I am very new to this topic. Thank you!
$endgroup$
– stedmoaoa
May 11 '17 at 12:13
$begingroup$
So you mean for the polynomials, just inserts 0 in between and it constructs a subspace that it is isomorphic to?
$endgroup$
– stedmoaoa
May 25 '17 at 10:18
$begingroup$
I got the idea but how to show that it is one-to-one and onto?
$endgroup$
– stedmoaoa
May 25 '17 at 10:20
$begingroup$
@stedmoaoa, just use the function definition I gave.
$endgroup$
– lhf
May 25 '17 at 11:02
add a comment |
$begingroup$
Hint: Consider $p(x) mapsto p(x^2)$.
This acts on a sequence of coefficients (which is all a polynomial is) by inserting zeros between them:
$$(a_0,a_1,a_2,dots,a_n,0,0,0,dots) mapsto (a_0,0,a_1,0,a_2,dots,a_{n-1},0,a_n,0,0,0,dots)$$
and so you can recover one from the other.
answered May 11 '17 at 10:53
lhflhf
168k11172404
$endgroup$
$begingroup$
Can you show a bit more hint I am very new to this topic. Thank you!
$endgroup$
– stedmoaoa
May 11 '17 at 12:13
$begingroup$
So you mean for the polynomials, just inserts 0 in between and it constructs a subspace that it is isomorphic to?
$endgroup$
– stedmoaoa
May 25 '17 at 10:18
$begingroup$
I got the idea but how to show that it is one-to-one and onto?
$endgroup$
– stedmoaoa
May 25 '17 at 10:20
$begingroup$
@stedmoaoa, just use the function definition I gave.
$endgroup$
– lhf
May 25 '17 at 11:02
add a comment |
$begingroup$
Hint: Consider $p(x) mapsto p(x^2)$.
This acts on a sequence of coefficients (which is all a polynomial is) by inserting zeros between them:
$$(a_0,a_1,a_2,dots,a_n,0,0,0,dots) mapsto (a_0,0,a_1,0,a_2,dots,a_{n-1},0,a_n,0,0,0,dots)$$
and so you can recover one from the other.
answered May 11 '17 at 10:53
lhflhf
168k11172404
$endgroup$
Hint: Consider $p(x) mapsto p(x^2)$.
This acts on a sequence of coefficients (which is all a polynomial is) by inserting zeros between them:
$$(a_0,a_1,a_2,dots,a_n,0,0,0,dots) mapsto (a_0,0,a_1,0,a_2,dots,a_{n-1},0,a_n,0,0,0,dots)$$
and so you can recover one from the other.
answered May 11 '17 at 10:53
lhflhf
168k11172404
edited May 11 '17 at 14:48
answered May 11 '17 at 10:53
lhflhf
168k11172404
answered May 11 '17 at 10:53
lhflhf
168k11172404
answered May 11 '17 at 10:53
lhflhf
168k11172404
168k11172404
$begingroup$
Can you show a bit more hint I am very new to this topic. Thank you!
$endgroup$
– stedmoaoa
May 11 '17 at 12:13
$begingroup$
So you mean for the polynomials, just inserts 0 in between and it constructs a subspace that it is isomorphic to?
$endgroup$
– stedmoaoa
May 25 '17 at 10:18
$begingroup$
I got the idea but how to show that it is one-to-one and onto?
$endgroup$
– stedmoaoa
May 25 '17 at 10:20
$begingroup$
@stedmoaoa, just use the function definition I gave.
$endgroup$
– lhf
May 25 '17 at 11:02
add a comment |
$begingroup$
Can you show a bit more hint I am very new to this topic. Thank you!
$endgroup$
– stedmoaoa
May 11 '17 at 12:13
$begingroup$
So you mean for the polynomials, just inserts 0 in between and it constructs a subspace that it is isomorphic to?
$endgroup$
– stedmoaoa
May 25 '17 at 10:18
$begingroup$
I got the idea but how to show that it is one-to-one and onto?
$endgroup$
– stedmoaoa
May 25 '17 at 10:20
$begingroup$
@stedmoaoa, just use the function definition I gave.
$endgroup$
– lhf
May 25 '17 at 11:02
$begingroup$
Can you show a bit more hint I am very new to this topic. Thank you!
$endgroup$
– stedmoaoa
May 11 '17 at 12:13
$begingroup$
Can you show a bit more hint I am very new to this topic. Thank you!
$endgroup$
– stedmoaoa
May 11 '17 at 12:13
$begingroup$
So you mean for the polynomials, just inserts 0 in between and it constructs a subspace that it is isomorphic to?
$endgroup$
– stedmoaoa
May 25 '17 at 10:18
$begingroup$
So you mean for the polynomials, just inserts 0 in between and it constructs a subspace that it is isomorphic to?
$endgroup$
– stedmoaoa
May 25 '17 at 10:18
$begingroup$
I got the idea but how to show that it is one-to-one and onto?
$endgroup$
– stedmoaoa
May 25 '17 at 10:20
$begingroup$
I got the idea but how to show that it is one-to-one and onto?
$endgroup$
– stedmoaoa
May 25 '17 at 10:20
$begingroup$
@stedmoaoa, just use the function definition I gave.
$endgroup$
– lhf
May 25 '17 at 11:02
$begingroup$
@stedmoaoa, just use the function definition I gave.
$endgroup$
– lhf
May 25 '17 at 11:02
add a comment |
$begingroup$
First, note that you can't do this with finite dimensional vector spaces. You have to come up with a linear transformation which is one-to-one but not onto. Any linear transformation is determined by what happens to the vectors in any basis. Use the standard basis for $R[x]$, that is, $x^n$ where $nge 0$. You have to map each $x^n$ to a polynomial such that the mapping is one-to-one. One way to do this is to map $x^n$ to a polynomial of degree $n+1$. You can check this ensures that only the zero polynomial maps to zero. You can check that it is not onto because, for example, $x^0$ is not the image of any polynomial. These two facts depend on looking at the leading term of polynomials and seeing that no nonzero polynomial is a linear combination of polynomials of lesser degrees.
answered May 11 '17 at 20:33
SomosSomos
14.9k11337
$endgroup$
add a comment |
$begingroup$
First, note that you can't do this with finite dimensional vector spaces. You have to come up with a linear transformation which is one-to-one but not onto. Any linear transformation is determined by what happens to the vectors in any basis. Use the standard basis for $R[x]$, that is, $x^n$ where $nge 0$. You have to map each $x^n$ to a polynomial such that the mapping is one-to-one. One way to do this is to map $x^n$ to a polynomial of degree $n+1$. You can check this ensures that only the zero polynomial maps to zero. You can check that it is not onto because, for example, $x^0$ is not the image of any polynomial. These two facts depend on looking at the leading term of polynomials and seeing that no nonzero polynomial is a linear combination of polynomials of lesser degrees.
answered May 11 '17 at 20:33
SomosSomos
14.9k11337
$endgroup$
add a comment |
$begingroup$
First, note that you can't do this with finite dimensional vector spaces. You have to come up with a linear transformation which is one-to-one but not onto. Any linear transformation is determined by what happens to the vectors in any basis. Use the standard basis for $R[x]$, that is, $x^n$ where $nge 0$. You have to map each $x^n$ to a polynomial such that the mapping is one-to-one. One way to do this is to map $x^n$ to a polynomial of degree $n+1$. You can check this ensures that only the zero polynomial maps to zero. You can check that it is not onto because, for example, $x^0$ is not the image of any polynomial. These two facts depend on looking at the leading term of polynomials and seeing that no nonzero polynomial is a linear combination of polynomials of lesser degrees.
answered May 11 '17 at 20:33
SomosSomos
14.9k11337
$endgroup$
First, note that you can't do this with finite dimensional vector spaces. You have to come up with a linear transformation which is one-to-one but not onto. Any linear transformation is determined by what happens to the vectors in any basis. Use the standard basis for $R[x]$, that is, $x^n$ where $nge 0$. You have to map each $x^n$ to a polynomial such that the mapping is one-to-one. One way to do this is to map $x^n$ to a polynomial of degree $n+1$. You can check this ensures that only the zero polynomial maps to zero. You can check that it is not onto because, for example, $x^0$ is not the image of any polynomial. These two facts depend on looking at the leading term of polynomials and seeing that no nonzero polynomial is a linear combination of polynomials of lesser degrees.
answered May 11 '17 at 20:33
SomosSomos
14.9k11337
answered May 11 '17 at 20:33
SomosSomos
14.9k11337
answered May 11 '17 at 20:33
SomosSomos
14.9k11337
answered May 11 '17 at 20:33
SomosSomos
14.9k11337
14.9k11337
add a comment |
add a comment |
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$begingroup$
Proper subspace means a subspace that is not $Bbb R[x]$.
$endgroup$
– Arnaud D.
May 11 '17 at 10:20
2
$begingroup$
Hint: consider $p(x) mapsto x p(x)$.
$endgroup$
– Kenny Wong
May 11 '17 at 10:28
1
$begingroup$
If it helps, you might want to forget the fact they're polynomials, and just think of them as sequences $(a_0, a_1, ldots)$ where each $a_i in R$. There is nothing polynomial-specific being used here.
$endgroup$
– Joppy
May 11 '17 at 13:26