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Extending Linear Functional over Polynomials
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As I've been reading over some papers on linear functionals, I've noted a large distinction is made between linear operators $T: mathbbC_n[z] to mathbbC[z]$ and linear operators $T: mathbbC[z] to mathbbC[z]$, where $mathbbC_n[z] subset mathbbC[z]$ is the set of complex polynomials of degree of most $n$.
As I've thought about this, it seems to me that if $T: mathbbC[z] to mathbbC[z]$ then surely we can view $T$ as a mapping from $mathbbC_n[z]$ to $mathbbC[z]$ by considering $T|_mathbbC_n[z]$, i.e. the restriction of $T$ to $mathbbC[z]$. However, I can't think of any concrete examples that show the converse is false, i.e. that a linear functional on $mathbbC_n[z]$ cannot be extended to a linear functional on $mathbbC[z]$.
The only examples I can think of do have an extension; for example, if we let $T(p(z)) = sqrtp^(n+1)(z)$ then surely $T$ is linear on $mathbbC_n[z]$ and is not linear on $mathbbC[z]$, but $T$ can be extended to $mathbbC[z]$ by defining $T(p(z)) = 1$. My question is thus as follows:
Does there exist a linear functional mapping $mathbbC_n[z]$ to $mathbbC[z]$ that does not extend to a linear functional mapping $mathbbC[z]$ to $mathbbC[z]$?
functional-analysis polynomials vector-spaces linear-transformations
asked Jul 23 at 9:23
Brevan Ellefsen
11.4k31449
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up vote
3
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As I've been reading over some papers on linear functionals, I've noted a large distinction is made between linear operators $T: mathbbC_n[z] to mathbbC[z]$ and linear operators $T: mathbbC[z] to mathbbC[z]$, where $mathbbC_n[z] subset mathbbC[z]$ is the set of complex polynomials of degree of most $n$.
As I've thought about this, it seems to me that if $T: mathbbC[z] to mathbbC[z]$ then surely we can view $T$ as a mapping from $mathbbC_n[z]$ to $mathbbC[z]$ by considering $T|_mathbbC_n[z]$, i.e. the restriction of $T$ to $mathbbC[z]$. However, I can't think of any concrete examples that show the converse is false, i.e. that a linear functional on $mathbbC_n[z]$ cannot be extended to a linear functional on $mathbbC[z]$.
The only examples I can think of do have an extension; for example, if we let $T(p(z)) = sqrtp^(n+1)(z)$ then surely $T$ is linear on $mathbbC_n[z]$ and is not linear on $mathbbC[z]$, but $T$ can be extended to $mathbbC[z]$ by defining $T(p(z)) = 1$. My question is thus as follows:
Does there exist a linear functional mapping $mathbbC_n[z]$ to $mathbbC[z]$ that does not extend to a linear functional mapping $mathbbC[z]$ to $mathbbC[z]$?
functional-analysis polynomials vector-spaces linear-transformations
asked Jul 23 at 9:23
Brevan Ellefsen
11.4k31449
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
As I've been reading over some papers on linear functionals, I've noted a large distinction is made between linear operators $T: mathbbC_n[z] to mathbbC[z]$ and linear operators $T: mathbbC[z] to mathbbC[z]$, where $mathbbC_n[z] subset mathbbC[z]$ is the set of complex polynomials of degree of most $n$.
As I've thought about this, it seems to me that if $T: mathbbC[z] to mathbbC[z]$ then surely we can view $T$ as a mapping from $mathbbC_n[z]$ to $mathbbC[z]$ by considering $T|_mathbbC_n[z]$, i.e. the restriction of $T$ to $mathbbC[z]$. However, I can't think of any concrete examples that show the converse is false, i.e. that a linear functional on $mathbbC_n[z]$ cannot be extended to a linear functional on $mathbbC[z]$.
The only examples I can think of do have an extension; for example, if we let $T(p(z)) = sqrtp^(n+1)(z)$ then surely $T$ is linear on $mathbbC_n[z]$ and is not linear on $mathbbC[z]$, but $T$ can be extended to $mathbbC[z]$ by defining $T(p(z)) = 1$. My question is thus as follows:
Does there exist a linear functional mapping $mathbbC_n[z]$ to $mathbbC[z]$ that does not extend to a linear functional mapping $mathbbC[z]$ to $mathbbC[z]$?
functional-analysis polynomials vector-spaces linear-transformations
asked Jul 23 at 9:23
Brevan Ellefsen
11.4k31449
As I've been reading over some papers on linear functionals, I've noted a large distinction is made between linear operators $T: mathbbC_n[z] to mathbbC[z]$ and linear operators $T: mathbbC[z] to mathbbC[z]$, where $mathbbC_n[z] subset mathbbC[z]$ is the set of complex polynomials of degree of most $n$.
As I've thought about this, it seems to me that if $T: mathbbC[z] to mathbbC[z]$ then surely we can view $T$ as a mapping from $mathbbC_n[z]$ to $mathbbC[z]$ by considering $T|_mathbbC_n[z]$, i.e. the restriction of $T$ to $mathbbC[z]$. However, I can't think of any concrete examples that show the converse is false, i.e. that a linear functional on $mathbbC_n[z]$ cannot be extended to a linear functional on $mathbbC[z]$.
The only examples I can think of do have an extension; for example, if we let $T(p(z)) = sqrtp^(n+1)(z)$ then surely $T$ is linear on $mathbbC_n[z]$ and is not linear on $mathbbC[z]$, but $T$ can be extended to $mathbbC[z]$ by defining $T(p(z)) = 1$. My question is thus as follows:
Does there exist a linear functional mapping $mathbbC_n[z]$ to $mathbbC[z]$ that does not extend to a linear functional mapping $mathbbC[z]$ to $mathbbC[z]$?
functional-analysis polynomials vector-spaces linear-transformations
asked Jul 23 at 9:23
Brevan Ellefsen
11.4k31449
edited Jul 23 at 10:19
asked Jul 23 at 9:23
Brevan Ellefsen
11.4k31449
asked Jul 23 at 9:23
Brevan Ellefsen
11.4k31449
asked Jul 23 at 9:23
Brevan Ellefsen
11.4k31449
11.4k31449
add a comment |Â
add a comment |Â
1 Answer
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No, there is not such a linear functional. Because if $alphacolonmathbbC_n[z]longrightarrowmathbbC[z]$ is a linear functional, you can define $AcolonmathbbC[z]longrightarrowmathbbC[z]$ by$$A(a_0+a_1z+cdots+a_mz^m)=begincasesalpha(a_0+a_1z+cdots+a_mz^m)&text if mgeqslant n\alpha(a_0+a_1z+cdots+a_nz^n)&text otherwise.endcases$$Then $A$ is linear and $A|_mathbbC_n[z]=alpha$.
More generally, if $V$ and $W$ are vector spaces and $U$ is a vector subspace of $V$, then every linear map $fcolon Ulongrightarrow W$ can be extended to a linear map $Fcolon Vlongrightarrow W$.
answered Jul 23 at 9:30
José Carlos Santos
113k1698176
Forgive my confusion, but if $alpha(p(z)) = expleft(p^(n)(z)right)$, then how are you proposing extending $alpha$?
– Brevan Ellefsen
Jul 23 at 9:46
Since $alpha(0)=1neq0$, $alpha$ is not linear.
– José Carlos Santos
Jul 23 at 9:48
You are absolutely correct. Forgive my naiveness in this - I haven't need to touch linear algebra in a long time - it just so happens a bit of complex analysis I'm studying now relies on it. So if we were to, say, consider $alpha(p(z)) = sqrtp^(n+1)(z)$ (which should just be the zero map when restricted to $n$th degree polynomials if I'm not mistaken), then how do I extend it? I seem to be missing something here, because if $m > n$ I would naively think that $alpha(a_0 + cdots a_m z^m)$ would return $sqrtfracd^n+1dz^n+1(a_0 + cdots a_m z^m)$ which doesnt seem linear
– Brevan Ellefsen
Jul 23 at 10:00
Since $alphabigl(p(z)bigr)=0$, you can extend it to the null function from $mathbbC[z]$ into itself.
– José Carlos Santos
Jul 23 at 10:08
Ok, I think that makes sense. So when defining $A$ as we do in your post, I shouldn't be thinking about the extension using the same definition of $alpha$ as I started with? e.g. even though $alpha(p(z)) = sqrtp^(n+1)(z) = 0$ on $mathbbC_n[z]$, in extending $alpha$ I need to choose the appropriate definition $alpha = 0$? That seems to be like an equivalence relation over the mappings equal on $mathbbC_n[z]$, but then how do we know which definition of $alpha$ will work?
– Brevan Ellefsen
Jul 23 at 10:14
 |Â
show 5 more comments
1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
No, there is not such a linear functional. Because if $alphacolonmathbbC_n[z]longrightarrowmathbbC[z]$ is a linear functional, you can define $AcolonmathbbC[z]longrightarrowmathbbC[z]$ by$$A(a_0+a_1z+cdots+a_mz^m)=begincasesalpha(a_0+a_1z+cdots+a_mz^m)&text if mgeqslant n\alpha(a_0+a_1z+cdots+a_nz^n)&text otherwise.endcases$$Then $A$ is linear and $A|_mathbbC_n[z]=alpha$.
More generally, if $V$ and $W$ are vector spaces and $U$ is a vector subspace of $V$, then every linear map $fcolon Ulongrightarrow W$ can be extended to a linear map $Fcolon Vlongrightarrow W$.
answered Jul 23 at 9:30
José Carlos Santos
113k1698176
Forgive my confusion, but if $alpha(p(z)) = expleft(p^(n)(z)right)$, then how are you proposing extending $alpha$?
– Brevan Ellefsen
Jul 23 at 9:46
Since $alpha(0)=1neq0$, $alpha$ is not linear.
– José Carlos Santos
Jul 23 at 9:48
You are absolutely correct. Forgive my naiveness in this - I haven't need to touch linear algebra in a long time - it just so happens a bit of complex analysis I'm studying now relies on it. So if we were to, say, consider $alpha(p(z)) = sqrtp^(n+1)(z)$ (which should just be the zero map when restricted to $n$th degree polynomials if I'm not mistaken), then how do I extend it? I seem to be missing something here, because if $m > n$ I would naively think that $alpha(a_0 + cdots a_m z^m)$ would return $sqrtfracd^n+1dz^n+1(a_0 + cdots a_m z^m)$ which doesnt seem linear
– Brevan Ellefsen
Jul 23 at 10:00
Since $alphabigl(p(z)bigr)=0$, you can extend it to the null function from $mathbbC[z]$ into itself.
– José Carlos Santos
Jul 23 at 10:08
Ok, I think that makes sense. So when defining $A$ as we do in your post, I shouldn't be thinking about the extension using the same definition of $alpha$ as I started with? e.g. even though $alpha(p(z)) = sqrtp^(n+1)(z) = 0$ on $mathbbC_n[z]$, in extending $alpha$ I need to choose the appropriate definition $alpha = 0$? That seems to be like an equivalence relation over the mappings equal on $mathbbC_n[z]$, but then how do we know which definition of $alpha$ will work?
– Brevan Ellefsen
Jul 23 at 10:14
 |Â
show 5 more comments
up vote
2
down vote
No, there is not such a linear functional. Because if $alphacolonmathbbC_n[z]longrightarrowmathbbC[z]$ is a linear functional, you can define $AcolonmathbbC[z]longrightarrowmathbbC[z]$ by$$A(a_0+a_1z+cdots+a_mz^m)=begincasesalpha(a_0+a_1z+cdots+a_mz^m)&text if mgeqslant n\alpha(a_0+a_1z+cdots+a_nz^n)&text otherwise.endcases$$Then $A$ is linear and $A|_mathbbC_n[z]=alpha$.
More generally, if $V$ and $W$ are vector spaces and $U$ is a vector subspace of $V$, then every linear map $fcolon Ulongrightarrow W$ can be extended to a linear map $Fcolon Vlongrightarrow W$.
answered Jul 23 at 9:30
José Carlos Santos
113k1698176
Forgive my confusion, but if $alpha(p(z)) = expleft(p^(n)(z)right)$, then how are you proposing extending $alpha$?
– Brevan Ellefsen
Jul 23 at 9:46
Since $alpha(0)=1neq0$, $alpha$ is not linear.
– José Carlos Santos
Jul 23 at 9:48
You are absolutely correct. Forgive my naiveness in this - I haven't need to touch linear algebra in a long time - it just so happens a bit of complex analysis I'm studying now relies on it. So if we were to, say, consider $alpha(p(z)) = sqrtp^(n+1)(z)$ (which should just be the zero map when restricted to $n$th degree polynomials if I'm not mistaken), then how do I extend it? I seem to be missing something here, because if $m > n$ I would naively think that $alpha(a_0 + cdots a_m z^m)$ would return $sqrtfracd^n+1dz^n+1(a_0 + cdots a_m z^m)$ which doesnt seem linear
– Brevan Ellefsen
Jul 23 at 10:00
Since $alphabigl(p(z)bigr)=0$, you can extend it to the null function from $mathbbC[z]$ into itself.
– José Carlos Santos
Jul 23 at 10:08
Ok, I think that makes sense. So when defining $A$ as we do in your post, I shouldn't be thinking about the extension using the same definition of $alpha$ as I started with? e.g. even though $alpha(p(z)) = sqrtp^(n+1)(z) = 0$ on $mathbbC_n[z]$, in extending $alpha$ I need to choose the appropriate definition $alpha = 0$? That seems to be like an equivalence relation over the mappings equal on $mathbbC_n[z]$, but then how do we know which definition of $alpha$ will work?
– Brevan Ellefsen
Jul 23 at 10:14
 |Â
show 5 more comments
up vote
2
down vote
up vote
2
down vote
No, there is not such a linear functional. Because if $alphacolonmathbbC_n[z]longrightarrowmathbbC[z]$ is a linear functional, you can define $AcolonmathbbC[z]longrightarrowmathbbC[z]$ by$$A(a_0+a_1z+cdots+a_mz^m)=begincasesalpha(a_0+a_1z+cdots+a_mz^m)&text if mgeqslant n\alpha(a_0+a_1z+cdots+a_nz^n)&text otherwise.endcases$$Then $A$ is linear and $A|_mathbbC_n[z]=alpha$.
More generally, if $V$ and $W$ are vector spaces and $U$ is a vector subspace of $V$, then every linear map $fcolon Ulongrightarrow W$ can be extended to a linear map $Fcolon Vlongrightarrow W$.
answered Jul 23 at 9:30
José Carlos Santos
113k1698176
No, there is not such a linear functional. Because if $alphacolonmathbbC_n[z]longrightarrowmathbbC[z]$ is a linear functional, you can define $AcolonmathbbC[z]longrightarrowmathbbC[z]$ by$$A(a_0+a_1z+cdots+a_mz^m)=begincasesalpha(a_0+a_1z+cdots+a_mz^m)&text if mgeqslant n\alpha(a_0+a_1z+cdots+a_nz^n)&text otherwise.endcases$$Then $A$ is linear and $A|_mathbbC_n[z]=alpha$.
More generally, if $V$ and $W$ are vector spaces and $U$ is a vector subspace of $V$, then every linear map $fcolon Ulongrightarrow W$ can be extended to a linear map $Fcolon Vlongrightarrow W$.
answered Jul 23 at 9:30
José Carlos Santos
113k1698176
answered Jul 23 at 9:30
José Carlos Santos
113k1698176
answered Jul 23 at 9:30
José Carlos Santos
113k1698176
answered Jul 23 at 9:30
José Carlos Santos
113k1698176
113k1698176
Forgive my confusion, but if $alpha(p(z)) = expleft(p^(n)(z)right)$, then how are you proposing extending $alpha$?
– Brevan Ellefsen
Jul 23 at 9:46
Since $alpha(0)=1neq0$, $alpha$ is not linear.
– José Carlos Santos
Jul 23 at 9:48
You are absolutely correct. Forgive my naiveness in this - I haven't need to touch linear algebra in a long time - it just so happens a bit of complex analysis I'm studying now relies on it. So if we were to, say, consider $alpha(p(z)) = sqrtp^(n+1)(z)$ (which should just be the zero map when restricted to $n$th degree polynomials if I'm not mistaken), then how do I extend it? I seem to be missing something here, because if $m > n$ I would naively think that $alpha(a_0 + cdots a_m z^m)$ would return $sqrtfracd^n+1dz^n+1(a_0 + cdots a_m z^m)$ which doesnt seem linear
– Brevan Ellefsen
Jul 23 at 10:00
Since $alphabigl(p(z)bigr)=0$, you can extend it to the null function from $mathbbC[z]$ into itself.
– José Carlos Santos
Jul 23 at 10:08
Ok, I think that makes sense. So when defining $A$ as we do in your post, I shouldn't be thinking about the extension using the same definition of $alpha$ as I started with? e.g. even though $alpha(p(z)) = sqrtp^(n+1)(z) = 0$ on $mathbbC_n[z]$, in extending $alpha$ I need to choose the appropriate definition $alpha = 0$? That seems to be like an equivalence relation over the mappings equal on $mathbbC_n[z]$, but then how do we know which definition of $alpha$ will work?
– Brevan Ellefsen
Jul 23 at 10:14
 |Â
show 5 more comments
Forgive my confusion, but if $alpha(p(z)) = expleft(p^(n)(z)right)$, then how are you proposing extending $alpha$?
– Brevan Ellefsen
Jul 23 at 9:46
Since $alpha(0)=1neq0$, $alpha$ is not linear.
– José Carlos Santos
Jul 23 at 9:48
You are absolutely correct. Forgive my naiveness in this - I haven't need to touch linear algebra in a long time - it just so happens a bit of complex analysis I'm studying now relies on it. So if we were to, say, consider $alpha(p(z)) = sqrtp^(n+1)(z)$ (which should just be the zero map when restricted to $n$th degree polynomials if I'm not mistaken), then how do I extend it? I seem to be missing something here, because if $m > n$ I would naively think that $alpha(a_0 + cdots a_m z^m)$ would return $sqrtfracd^n+1dz^n+1(a_0 + cdots a_m z^m)$ which doesnt seem linear
– Brevan Ellefsen
Jul 23 at 10:00
Since $alphabigl(p(z)bigr)=0$, you can extend it to the null function from $mathbbC[z]$ into itself.
– José Carlos Santos
Jul 23 at 10:08
Ok, I think that makes sense. So when defining $A$ as we do in your post, I shouldn't be thinking about the extension using the same definition of $alpha$ as I started with? e.g. even though $alpha(p(z)) = sqrtp^(n+1)(z) = 0$ on $mathbbC_n[z]$, in extending $alpha$ I need to choose the appropriate definition $alpha = 0$? That seems to be like an equivalence relation over the mappings equal on $mathbbC_n[z]$, but then how do we know which definition of $alpha$ will work?
– Brevan Ellefsen
Jul 23 at 10:14
Forgive my confusion, but if $alpha(p(z)) = expleft(p^(n)(z)right)$, then how are you proposing extending $alpha$?
– Brevan Ellefsen
Jul 23 at 9:46
Forgive my confusion, but if $alpha(p(z)) = expleft(p^(n)(z)right)$, then how are you proposing extending $alpha$?
– Brevan Ellefsen
Jul 23 at 9:46
Since $alpha(0)=1neq0$, $alpha$ is not linear.
– José Carlos Santos
Jul 23 at 9:48
Since $alpha(0)=1neq0$, $alpha$ is not linear.
– José Carlos Santos
Jul 23 at 9:48
You are absolutely correct. Forgive my naiveness in this - I haven't need to touch linear algebra in a long time - it just so happens a bit of complex analysis I'm studying now relies on it. So if we were to, say, consider $alpha(p(z)) = sqrtp^(n+1)(z)$ (which should just be the zero map when restricted to $n$th degree polynomials if I'm not mistaken), then how do I extend it? I seem to be missing something here, because if $m > n$ I would naively think that $alpha(a_0 + cdots a_m z^m)$ would return $sqrtfracd^n+1dz^n+1(a_0 + cdots a_m z^m)$ which doesnt seem linear
– Brevan Ellefsen
Jul 23 at 10:00
You are absolutely correct. Forgive my naiveness in this - I haven't need to touch linear algebra in a long time - it just so happens a bit of complex analysis I'm studying now relies on it. So if we were to, say, consider $alpha(p(z)) = sqrtp^(n+1)(z)$ (which should just be the zero map when restricted to $n$th degree polynomials if I'm not mistaken), then how do I extend it? I seem to be missing something here, because if $m > n$ I would naively think that $alpha(a_0 + cdots a_m z^m)$ would return $sqrtfracd^n+1dz^n+1(a_0 + cdots a_m z^m)$ which doesnt seem linear
– Brevan Ellefsen
Jul 23 at 10:00
Since $alphabigl(p(z)bigr)=0$, you can extend it to the null function from $mathbbC[z]$ into itself.
– José Carlos Santos
Jul 23 at 10:08
Since $alphabigl(p(z)bigr)=0$, you can extend it to the null function from $mathbbC[z]$ into itself.
– José Carlos Santos
Jul 23 at 10:08
Ok, I think that makes sense. So when defining $A$ as we do in your post, I shouldn't be thinking about the extension using the same definition of $alpha$ as I started with? e.g. even though $alpha(p(z)) = sqrtp^(n+1)(z) = 0$ on $mathbbC_n[z]$, in extending $alpha$ I need to choose the appropriate definition $alpha = 0$? That seems to be like an equivalence relation over the mappings equal on $mathbbC_n[z]$, but then how do we know which definition of $alpha$ will work?
– Brevan Ellefsen
Jul 23 at 10:14
Ok, I think that makes sense. So when defining $A$ as we do in your post, I shouldn't be thinking about the extension using the same definition of $alpha$ as I started with? e.g. even though $alpha(p(z)) = sqrtp^(n+1)(z) = 0$ on $mathbbC_n[z]$, in extending $alpha$ I need to choose the appropriate definition $alpha = 0$? That seems to be like an equivalence relation over the mappings equal on $mathbbC_n[z]$, but then how do we know which definition of $alpha$ will work?
– Brevan Ellefsen
Jul 23 at 10:14
 |Â
show 5 more comments
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