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How to write this vector in terms of Hamel basis? [closed]
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Let $(c,c,c,...)$ be a vector in $l_infty$, where $c ne 0$. How to write this vector in terms of Hamel basis?
linear-algebra functional-analysis normed-spaces
asked Jul 29 at 5:34
Infinity
513213
closed as off-topic by Shailesh, Jyrki Lahtonen, Taroccoesbrocco, John Ma, mechanodroid Jul 29 at 12:35
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Taroccoesbrocco, mechanodroid
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Let $(c,c,c,...)$ be a vector in $l_infty$, where $c ne 0$. How to write this vector in terms of Hamel basis?
linear-algebra functional-analysis normed-spaces
asked Jul 29 at 5:34
Infinity
513213
closed as off-topic by Shailesh, Jyrki Lahtonen, Taroccoesbrocco, John Ma, mechanodroid Jul 29 at 12:35
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Taroccoesbrocco, mechanodroid
2
Depends which basis. For instance, there is a basis that has $(c,c,c,ldots)$ as an element, so it's pretty easy in that one.
– spaceisdarkgreen
Jul 29 at 5:44
2
As far as I know, nobody has actually "seen/described explicitly" such a Hamel basis. Ah, the miracle of Zorn's Lemma.
– max_zorn
Jul 29 at 6:02
1
I sincerely doubt that you would be expected to answer such a question. For one, you wouldn't be asked to write $(c,c,c)$ in terms of a basis of $BbbR^3$ without being given the basis first. And, nobody has ever exhibited a Hamel basis of $ell_infty$. They only exist by virtue of the Axiom of Choice, you know.
– Jyrki Lahtonen
Jul 29 at 6:03
Possible duplicate of What is the basis of the vector space $l^infty$?
– Jyrki Lahtonen
Jul 29 at 6:09
@max_zorn ... if you do say so yourself. ;-)
– Theo Bendit
Jul 29 at 6:15
add a comment |Â
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Let $(c,c,c,...)$ be a vector in $l_infty$, where $c ne 0$. How to write this vector in terms of Hamel basis?
linear-algebra functional-analysis normed-spaces
asked Jul 29 at 5:34
Infinity
513213
Let $(c,c,c,...)$ be a vector in $l_infty$, where $c ne 0$. How to write this vector in terms of Hamel basis?
linear-algebra functional-analysis normed-spaces
asked Jul 29 at 5:34
Infinity
513213
asked Jul 29 at 5:34
Infinity
513213
asked Jul 29 at 5:34
Infinity
513213
asked Jul 29 at 5:34
Infinity
513213
513213
closed as off-topic by Shailesh, Jyrki Lahtonen, Taroccoesbrocco, John Ma, mechanodroid Jul 29 at 12:35
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Taroccoesbrocco, mechanodroid
closed as off-topic by Shailesh, Jyrki Lahtonen, Taroccoesbrocco, John Ma, mechanodroid Jul 29 at 12:35
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Taroccoesbrocco, mechanodroid
2
Depends which basis. For instance, there is a basis that has $(c,c,c,ldots)$ as an element, so it's pretty easy in that one.
– spaceisdarkgreen
Jul 29 at 5:44
2
As far as I know, nobody has actually "seen/described explicitly" such a Hamel basis. Ah, the miracle of Zorn's Lemma.
– max_zorn
Jul 29 at 6:02
1
I sincerely doubt that you would be expected to answer such a question. For one, you wouldn't be asked to write $(c,c,c)$ in terms of a basis of $BbbR^3$ without being given the basis first. And, nobody has ever exhibited a Hamel basis of $ell_infty$. They only exist by virtue of the Axiom of Choice, you know.
– Jyrki Lahtonen
Jul 29 at 6:03
Possible duplicate of What is the basis of the vector space $l^infty$?
– Jyrki Lahtonen
Jul 29 at 6:09
@max_zorn ... if you do say so yourself. ;-)
– Theo Bendit
Jul 29 at 6:15
add a comment |Â
2
Depends which basis. For instance, there is a basis that has $(c,c,c,ldots)$ as an element, so it's pretty easy in that one.
– spaceisdarkgreen
Jul 29 at 5:44
2
As far as I know, nobody has actually "seen/described explicitly" such a Hamel basis. Ah, the miracle of Zorn's Lemma.
– max_zorn
Jul 29 at 6:02
1
I sincerely doubt that you would be expected to answer such a question. For one, you wouldn't be asked to write $(c,c,c)$ in terms of a basis of $BbbR^3$ without being given the basis first. And, nobody has ever exhibited a Hamel basis of $ell_infty$. They only exist by virtue of the Axiom of Choice, you know.
– Jyrki Lahtonen
Jul 29 at 6:03
Possible duplicate of What is the basis of the vector space $l^infty$?
– Jyrki Lahtonen
Jul 29 at 6:09
@max_zorn ... if you do say so yourself. ;-)
– Theo Bendit
Jul 29 at 6:15
2
2
Depends which basis. For instance, there is a basis that has $(c,c,c,ldots)$ as an element, so it's pretty easy in that one.
– spaceisdarkgreen
Jul 29 at 5:44
Depends which basis. For instance, there is a basis that has $(c,c,c,ldots)$ as an element, so it's pretty easy in that one.
– spaceisdarkgreen
Jul 29 at 5:44
2
2
As far as I know, nobody has actually "seen/described explicitly" such a Hamel basis. Ah, the miracle of Zorn's Lemma.
– max_zorn
Jul 29 at 6:02
As far as I know, nobody has actually "seen/described explicitly" such a Hamel basis. Ah, the miracle of Zorn's Lemma.
– max_zorn
Jul 29 at 6:02
1
1
I sincerely doubt that you would be expected to answer such a question. For one, you wouldn't be asked to write $(c,c,c)$ in terms of a basis of $BbbR^3$ without being given the basis first. And, nobody has ever exhibited a Hamel basis of $ell_infty$. They only exist by virtue of the Axiom of Choice, you know.
– Jyrki Lahtonen
Jul 29 at 6:03
I sincerely doubt that you would be expected to answer such a question. For one, you wouldn't be asked to write $(c,c,c)$ in terms of a basis of $BbbR^3$ without being given the basis first. And, nobody has ever exhibited a Hamel basis of $ell_infty$. They only exist by virtue of the Axiom of Choice, you know.
– Jyrki Lahtonen
Jul 29 at 6:03
Possible duplicate of What is the basis of the vector space $l^infty$?
– Jyrki Lahtonen
Jul 29 at 6:09
Possible duplicate of What is the basis of the vector space $l^infty$?
– Jyrki Lahtonen
Jul 29 at 6:09
@max_zorn ... if you do say so yourself. ;-)
– Theo Bendit
Jul 29 at 6:15
@max_zorn ... if you do say so yourself. ;-)
– Theo Bendit
Jul 29 at 6:15
add a comment |Â
1 Answer
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It's not clear exactly which misconceptions you have here, so I'll just say a few things and hopefully it will clarify.
The set $(1,0,0,ldots),(0,1,0,0,ldots), ldots$ is not a Hamel basis for $l_infty$ as is evidenced by the fact that $(c,c,ldots)$ cannot be written as a finite linear combination of those vectors.
There is no such thing as the Hamel basis. There are generally many of them. There is also not generally a canonical basis, so saying "the" doesn't even work implicitly.
It is generally impossible to explicitly exhibit a basis (and $l_infty$ is not an exception), so there's no way to answer your question in any kind of generality. However, from the usual general proof that a basis exists, it's reasonably clear that, for some given element $v$ of your vector space $V,$ that there is a basis that contains $v$ as an element. More generally, for any linearly independent subset $Lsubseteq V,$ there is a basis $B$ with $Lsubseteq B.$ (The usual argument applies Zorn's lemma to the set of all linearly independent subsets of $V,$ and we can just modify it to instead consider the set of all linearly independent subsets of $V$ that contain $L$ as a subset.) So there are certainly bases in which your vector $(c,c,ldots)$ has a simple expression.
answered Jul 29 at 6:14
spaceisdarkgreen
27.2k21546
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
It's not clear exactly which misconceptions you have here, so I'll just say a few things and hopefully it will clarify.
The set $(1,0,0,ldots),(0,1,0,0,ldots), ldots$ is not a Hamel basis for $l_infty$ as is evidenced by the fact that $(c,c,ldots)$ cannot be written as a finite linear combination of those vectors.
There is no such thing as the Hamel basis. There are generally many of them. There is also not generally a canonical basis, so saying "the" doesn't even work implicitly.
It is generally impossible to explicitly exhibit a basis (and $l_infty$ is not an exception), so there's no way to answer your question in any kind of generality. However, from the usual general proof that a basis exists, it's reasonably clear that, for some given element $v$ of your vector space $V,$ that there is a basis that contains $v$ as an element. More generally, for any linearly independent subset $Lsubseteq V,$ there is a basis $B$ with $Lsubseteq B.$ (The usual argument applies Zorn's lemma to the set of all linearly independent subsets of $V,$ and we can just modify it to instead consider the set of all linearly independent subsets of $V$ that contain $L$ as a subset.) So there are certainly bases in which your vector $(c,c,ldots)$ has a simple expression.
answered Jul 29 at 6:14
spaceisdarkgreen
27.2k21546
add a comment |Â
up vote
2
down vote
It's not clear exactly which misconceptions you have here, so I'll just say a few things and hopefully it will clarify.
The set $(1,0,0,ldots),(0,1,0,0,ldots), ldots$ is not a Hamel basis for $l_infty$ as is evidenced by the fact that $(c,c,ldots)$ cannot be written as a finite linear combination of those vectors.
There is no such thing as the Hamel basis. There are generally many of them. There is also not generally a canonical basis, so saying "the" doesn't even work implicitly.
It is generally impossible to explicitly exhibit a basis (and $l_infty$ is not an exception), so there's no way to answer your question in any kind of generality. However, from the usual general proof that a basis exists, it's reasonably clear that, for some given element $v$ of your vector space $V,$ that there is a basis that contains $v$ as an element. More generally, for any linearly independent subset $Lsubseteq V,$ there is a basis $B$ with $Lsubseteq B.$ (The usual argument applies Zorn's lemma to the set of all linearly independent subsets of $V,$ and we can just modify it to instead consider the set of all linearly independent subsets of $V$ that contain $L$ as a subset.) So there are certainly bases in which your vector $(c,c,ldots)$ has a simple expression.
answered Jul 29 at 6:14
spaceisdarkgreen
27.2k21546
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It's not clear exactly which misconceptions you have here, so I'll just say a few things and hopefully it will clarify.
The set $(1,0,0,ldots),(0,1,0,0,ldots), ldots$ is not a Hamel basis for $l_infty$ as is evidenced by the fact that $(c,c,ldots)$ cannot be written as a finite linear combination of those vectors.
There is no such thing as the Hamel basis. There are generally many of them. There is also not generally a canonical basis, so saying "the" doesn't even work implicitly.
It is generally impossible to explicitly exhibit a basis (and $l_infty$ is not an exception), so there's no way to answer your question in any kind of generality. However, from the usual general proof that a basis exists, it's reasonably clear that, for some given element $v$ of your vector space $V,$ that there is a basis that contains $v$ as an element. More generally, for any linearly independent subset $Lsubseteq V,$ there is a basis $B$ with $Lsubseteq B.$ (The usual argument applies Zorn's lemma to the set of all linearly independent subsets of $V,$ and we can just modify it to instead consider the set of all linearly independent subsets of $V$ that contain $L$ as a subset.) So there are certainly bases in which your vector $(c,c,ldots)$ has a simple expression.
answered Jul 29 at 6:14
spaceisdarkgreen
27.2k21546
It's not clear exactly which misconceptions you have here, so I'll just say a few things and hopefully it will clarify.
The set $(1,0,0,ldots),(0,1,0,0,ldots), ldots$ is not a Hamel basis for $l_infty$ as is evidenced by the fact that $(c,c,ldots)$ cannot be written as a finite linear combination of those vectors.
There is no such thing as the Hamel basis. There are generally many of them. There is also not generally a canonical basis, so saying "the" doesn't even work implicitly.
It is generally impossible to explicitly exhibit a basis (and $l_infty$ is not an exception), so there's no way to answer your question in any kind of generality. However, from the usual general proof that a basis exists, it's reasonably clear that, for some given element $v$ of your vector space $V,$ that there is a basis that contains $v$ as an element. More generally, for any linearly independent subset $Lsubseteq V,$ there is a basis $B$ with $Lsubseteq B.$ (The usual argument applies Zorn's lemma to the set of all linearly independent subsets of $V,$ and we can just modify it to instead consider the set of all linearly independent subsets of $V$ that contain $L$ as a subset.) So there are certainly bases in which your vector $(c,c,ldots)$ has a simple expression.
answered Jul 29 at 6:14
spaceisdarkgreen
27.2k21546
edited Jul 29 at 6:22
answered Jul 29 at 6:14
spaceisdarkgreen
27.2k21546
answered Jul 29 at 6:14
spaceisdarkgreen
27.2k21546
answered Jul 29 at 6:14
spaceisdarkgreen
27.2k21546
27.2k21546
add a comment |Â
add a comment |Â
2
Depends which basis. For instance, there is a basis that has $(c,c,c,ldots)$ as an element, so it's pretty easy in that one.
– spaceisdarkgreen
Jul 29 at 5:44
2
As far as I know, nobody has actually "seen/described explicitly" such a Hamel basis. Ah, the miracle of Zorn's Lemma.
– max_zorn
Jul 29 at 6:02
1
I sincerely doubt that you would be expected to answer such a question. For one, you wouldn't be asked to write $(c,c,c)$ in terms of a basis of $BbbR^3$ without being given the basis first. And, nobody has ever exhibited a Hamel basis of $ell_infty$. They only exist by virtue of the Axiom of Choice, you know.
– Jyrki Lahtonen
Jul 29 at 6:03
Possible duplicate of What is the basis of the vector space $l^infty$?
– Jyrki Lahtonen
Jul 29 at 6:09
@max_zorn ... if you do say so yourself. ;-)
– Theo Bendit
Jul 29 at 6:15