Bases of linear operators on functions
0
$begingroup$
I can create many linear operators on functions by making linear combinations of derivatives:
$$
L = sum^infty_{n=0} alpha_nfrac{d^n}{dx^n}
$$
Because linear operators are closed under addition and scalar multiplication, have a zero element, ecetera, they seem like they should form a vector space. Clearly, $frac{d^n}{dx^n}$ forms a basis for at least a subspace of linear operators. What do I need to write to express a basis that spans all linear operators?
linear-algebra operator-theory
asked Dec 6 '18 at 15:23
Display NameDisplay Name
2438
$endgroup$
add a comment |
0
$begingroup$
I can create many linear operators on functions by making linear combinations of derivatives:
$$
L = sum^infty_{n=0} alpha_nfrac{d^n}{dx^n}
$$
Because linear operators are closed under addition and scalar multiplication, have a zero element, ecetera, they seem like they should form a vector space. Clearly, $frac{d^n}{dx^n}$ forms a basis for at least a subspace of linear operators. What do I need to write to express a basis that spans all linear operators?
linear-algebra operator-theory
asked Dec 6 '18 at 15:23
Display NameDisplay Name
2438
$endgroup$
2
$begingroup$
I think the question is too broad as well as unclear. Which vector space of functions are you referring to here? An infinite sum of differential operator may be undefined, unless all but finitely $alpha_n$ are zero. If your vector space is infinite dimensional, the space of linear operators on this vector space is uncountable, so I don't think you can "express" a basis of the space of linear operators.
$endgroup$
– user593746
Dec 6 '18 at 15:31
$begingroup$
Those remarks are definitely helpful, although I think I can have (and even express) an uncountable basis. One example is the Fourier kernel $e^{iomega t}$, which if you let the set be the function corresponding to every $omega in mathbb{R}$, forms a basis for a certain rigged Hilbert space. I think that the linear combination of operators will, in the end, probably be some kind of integral.
$endgroup$
– Display Name
Dec 6 '18 at 15:42
2
$begingroup$
I think you should explain what you mean by "basis" here. There are two meanings of the term "basis" used by functional analysts. First, the term "basis" could mean the usual definition of a basis of vector spaces, or sometimes called a Hamel basis. We only allow finite linear combinations of a Hamel basis. Second, the term "basis" could also mean a Schauder basis. In a separable infinite-dimensional Hilbert space, it has a countable Schauder basis, but not a countable Hamel basis. Your expression for $L$ makes me feel uncertain which definition of "basis" you are using.
$endgroup$
– user593746
Dec 6 '18 at 15:48
2
$begingroup$
And of course, it is possible to express a (Hamel) basis of an uncountable dimensional vector space, but you are not always so lucky. Try to express, for example, a basis of $Bbb{R}$ as a vector space over $Bbb{Q}$.
$endgroup$
– user593746
Dec 6 '18 at 15:52
add a comment |
0
0
0
$begingroup$
I can create many linear operators on functions by making linear combinations of derivatives:
$$
L = sum^infty_{n=0} alpha_nfrac{d^n}{dx^n}
$$
Because linear operators are closed under addition and scalar multiplication, have a zero element, ecetera, they seem like they should form a vector space. Clearly, $frac{d^n}{dx^n}$ forms a basis for at least a subspace of linear operators. What do I need to write to express a basis that spans all linear operators?
linear-algebra operator-theory
asked Dec 6 '18 at 15:23
Display NameDisplay Name
2438
$endgroup$
I can create many linear operators on functions by making linear combinations of derivatives:
$$
L = sum^infty_{n=0} alpha_nfrac{d^n}{dx^n}
$$
Because linear operators are closed under addition and scalar multiplication, have a zero element, ecetera, they seem like they should form a vector space. Clearly, $frac{d^n}{dx^n}$ forms a basis for at least a subspace of linear operators. What do I need to write to express a basis that spans all linear operators?
linear-algebra operator-theory
linear-algebra operator-theory
asked Dec 6 '18 at 15:23
Display NameDisplay Name
2438
asked Dec 6 '18 at 15:23
Display NameDisplay Name
2438
asked Dec 6 '18 at 15:23
Display NameDisplay Name
2438
asked Dec 6 '18 at 15:23
Display NameDisplay Name
2438
asked Dec 6 '18 at 15:23
Display NameDisplay Name
2438
2438
2
$begingroup$
I think the question is too broad as well as unclear. Which vector space of functions are you referring to here? An infinite sum of differential operator may be undefined, unless all but finitely $alpha_n$ are zero. If your vector space is infinite dimensional, the space of linear operators on this vector space is uncountable, so I don't think you can "express" a basis of the space of linear operators.
$endgroup$
– user593746
Dec 6 '18 at 15:31
$begingroup$
Those remarks are definitely helpful, although I think I can have (and even express) an uncountable basis. One example is the Fourier kernel $e^{iomega t}$, which if you let the set be the function corresponding to every $omega in mathbb{R}$, forms a basis for a certain rigged Hilbert space. I think that the linear combination of operators will, in the end, probably be some kind of integral.
$endgroup$
– Display Name
Dec 6 '18 at 15:42
2
$begingroup$
I think you should explain what you mean by "basis" here. There are two meanings of the term "basis" used by functional analysts. First, the term "basis" could mean the usual definition of a basis of vector spaces, or sometimes called a Hamel basis. We only allow finite linear combinations of a Hamel basis. Second, the term "basis" could also mean a Schauder basis. In a separable infinite-dimensional Hilbert space, it has a countable Schauder basis, but not a countable Hamel basis. Your expression for $L$ makes me feel uncertain which definition of "basis" you are using.
$endgroup$
– user593746
Dec 6 '18 at 15:48
2
$begingroup$
And of course, it is possible to express a (Hamel) basis of an uncountable dimensional vector space, but you are not always so lucky. Try to express, for example, a basis of $Bbb{R}$ as a vector space over $Bbb{Q}$.
$endgroup$
– user593746
Dec 6 '18 at 15:52
add a comment |
2
$begingroup$
I think the question is too broad as well as unclear. Which vector space of functions are you referring to here? An infinite sum of differential operator may be undefined, unless all but finitely $alpha_n$ are zero. If your vector space is infinite dimensional, the space of linear operators on this vector space is uncountable, so I don't think you can "express" a basis of the space of linear operators.
$endgroup$
– user593746
Dec 6 '18 at 15:31
$begingroup$
Those remarks are definitely helpful, although I think I can have (and even express) an uncountable basis. One example is the Fourier kernel $e^{iomega t}$, which if you let the set be the function corresponding to every $omega in mathbb{R}$, forms a basis for a certain rigged Hilbert space. I think that the linear combination of operators will, in the end, probably be some kind of integral.
$endgroup$
– Display Name
Dec 6 '18 at 15:42
2
$begingroup$
I think you should explain what you mean by "basis" here. There are two meanings of the term "basis" used by functional analysts. First, the term "basis" could mean the usual definition of a basis of vector spaces, or sometimes called a Hamel basis. We only allow finite linear combinations of a Hamel basis. Second, the term "basis" could also mean a Schauder basis. In a separable infinite-dimensional Hilbert space, it has a countable Schauder basis, but not a countable Hamel basis. Your expression for $L$ makes me feel uncertain which definition of "basis" you are using.
$endgroup$
– user593746
Dec 6 '18 at 15:48
2
$begingroup$
And of course, it is possible to express a (Hamel) basis of an uncountable dimensional vector space, but you are not always so lucky. Try to express, for example, a basis of $Bbb{R}$ as a vector space over $Bbb{Q}$.
$endgroup$
– user593746
Dec 6 '18 at 15:52
2
2
$begingroup$
I think the question is too broad as well as unclear. Which vector space of functions are you referring to here? An infinite sum of differential operator may be undefined, unless all but finitely $alpha_n$ are zero. If your vector space is infinite dimensional, the space of linear operators on this vector space is uncountable, so I don't think you can "express" a basis of the space of linear operators.
$endgroup$
– user593746
Dec 6 '18 at 15:31
$begingroup$
I think the question is too broad as well as unclear. Which vector space of functions are you referring to here? An infinite sum of differential operator may be undefined, unless all but finitely $alpha_n$ are zero. If your vector space is infinite dimensional, the space of linear operators on this vector space is uncountable, so I don't think you can "express" a basis of the space of linear operators.
$endgroup$
– user593746
Dec 6 '18 at 15:31
$begingroup$
Those remarks are definitely helpful, although I think I can have (and even express) an uncountable basis. One example is the Fourier kernel $e^{iomega t}$, which if you let the set be the function corresponding to every $omega in mathbb{R}$, forms a basis for a certain rigged Hilbert space. I think that the linear combination of operators will, in the end, probably be some kind of integral.
$endgroup$
– Display Name
Dec 6 '18 at 15:42
$begingroup$
Those remarks are definitely helpful, although I think I can have (and even express) an uncountable basis. One example is the Fourier kernel $e^{iomega t}$, which if you let the set be the function corresponding to every $omega in mathbb{R}$, forms a basis for a certain rigged Hilbert space. I think that the linear combination of operators will, in the end, probably be some kind of integral.
$endgroup$
– Display Name
Dec 6 '18 at 15:42
2
2
$begingroup$
I think you should explain what you mean by "basis" here. There are two meanings of the term "basis" used by functional analysts. First, the term "basis" could mean the usual definition of a basis of vector spaces, or sometimes called a Hamel basis. We only allow finite linear combinations of a Hamel basis. Second, the term "basis" could also mean a Schauder basis. In a separable infinite-dimensional Hilbert space, it has a countable Schauder basis, but not a countable Hamel basis. Your expression for $L$ makes me feel uncertain which definition of "basis" you are using.
$endgroup$
– user593746
Dec 6 '18 at 15:48
$begingroup$
I think you should explain what you mean by "basis" here. There are two meanings of the term "basis" used by functional analysts. First, the term "basis" could mean the usual definition of a basis of vector spaces, or sometimes called a Hamel basis. We only allow finite linear combinations of a Hamel basis. Second, the term "basis" could also mean a Schauder basis. In a separable infinite-dimensional Hilbert space, it has a countable Schauder basis, but not a countable Hamel basis. Your expression for $L$ makes me feel uncertain which definition of "basis" you are using.
$endgroup$
– user593746
Dec 6 '18 at 15:48
2
2
$begingroup$
And of course, it is possible to express a (Hamel) basis of an uncountable dimensional vector space, but you are not always so lucky. Try to express, for example, a basis of $Bbb{R}$ as a vector space over $Bbb{Q}$.
$endgroup$
– user593746
Dec 6 '18 at 15:52
$begingroup$
And of course, it is possible to express a (Hamel) basis of an uncountable dimensional vector space, but you are not always so lucky. Try to express, for example, a basis of $Bbb{R}$ as a vector space over $Bbb{Q}$.
$endgroup$
– user593746
Dec 6 '18 at 15:52
add a comment |
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2
$begingroup$
I think the question is too broad as well as unclear. Which vector space of functions are you referring to here? An infinite sum of differential operator may be undefined, unless all but finitely $alpha_n$ are zero. If your vector space is infinite dimensional, the space of linear operators on this vector space is uncountable, so I don't think you can "express" a basis of the space of linear operators.
$endgroup$
– user593746
Dec 6 '18 at 15:31
$begingroup$
Those remarks are definitely helpful, although I think I can have (and even express) an uncountable basis. One example is the Fourier kernel $e^{iomega t}$, which if you let the set be the function corresponding to every $omega in mathbb{R}$, forms a basis for a certain rigged Hilbert space. I think that the linear combination of operators will, in the end, probably be some kind of integral.
$endgroup$
– Display Name
Dec 6 '18 at 15:42
2
$begingroup$
I think you should explain what you mean by "basis" here. There are two meanings of the term "basis" used by functional analysts. First, the term "basis" could mean the usual definition of a basis of vector spaces, or sometimes called a Hamel basis. We only allow finite linear combinations of a Hamel basis. Second, the term "basis" could also mean a Schauder basis. In a separable infinite-dimensional Hilbert space, it has a countable Schauder basis, but not a countable Hamel basis. Your expression for $L$ makes me feel uncertain which definition of "basis" you are using.
$endgroup$
– user593746
Dec 6 '18 at 15:48
2
$begingroup$
And of course, it is possible to express a (Hamel) basis of an uncountable dimensional vector space, but you are not always so lucky. Try to express, for example, a basis of $Bbb{R}$ as a vector space over $Bbb{Q}$.
$endgroup$
– user593746
Dec 6 '18 at 15:52