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Finding a basis of an infinite dimensional vector space with a given vector
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Finding a basis of an infinite dimensional vector space with a given vector
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$begingroup$
If $K$ is a field and $V=K^n$ a finite dimensional $K$-vector space with basis $A={e_1,dots,e_n}$, then, given a vector $vin V$ we can find a new basis $E$ of $V$ such that $vin E$; this is done as follows:
First we form the set ${v}cup A$ and observe that it has $n+1$ elements, therefore it spans all of $V$. Now we start the following process: We set $E_0={v}$. If $e_1notintext{span}(E_0)$, then we set $E_1={v}cup{e_1}$. Otherwise, $E_1=E_0$. Then, if $e_2notintext{span}(E_1)$, we set $E_2=E_1cup{e_2}$; otherwise, $E_2=E_1$. We go on until we eliminate every element of $A$; the set $E_n$ is our basis $E$; obviously it is linearly independent and note that it is impossible to "rule out" two elements of $A$, since if $e,e'$ were ruled out, that would mean that there are not-all-zero scalars $lambda_i,mu_i$ such that $e=v+sumlambda_ie_i$, $e'=v+summu_ie_i$, then $e=e'+sum(lambda_i-mu_i)e_i$, which is impossible, therefore $E$ has $n$ elements.
My question is, can we do the same for infinite dimensional vector spaces? That is, given a vector $v$ on an infinite dimensional $K$-vector space $V$, can we construct (feels like it is too much to ask)/ can we prove the existence of a basis $E$ such that $vin E$?
P.S: I live in a world where the axiom of choice is true, so $V$ does have a basis.
linear-algebra set-theory axiom-of-choice
edited 1 hour ago
Robert Shore
1,32815
asked 2 hours ago
JustDroppedInJustDroppedIn
2,054420
$endgroup$
add a comment |
$begingroup$
If $K$ is a field and $V=K^n$ a finite dimensional $K$-vector space with basis $A={e_1,dots,e_n}$, then, given a vector $vin V$ we can find a new basis $E$ of $V$ such that $vin E$; this is done as follows:
First we form the set ${v}cup A$ and observe that it has $n+1$ elements, therefore it spans all of $V$. Now we start the following process: We set $E_0={v}$. If $e_1notintext{span}(E_0)$, then we set $E_1={v}cup{e_1}$. Otherwise, $E_1=E_0$. Then, if $e_2notintext{span}(E_1)$, we set $E_2=E_1cup{e_2}$; otherwise, $E_2=E_1$. We go on until we eliminate every element of $A$; the set $E_n$ is our basis $E$; obviously it is linearly independent and note that it is impossible to "rule out" two elements of $A$, since if $e,e'$ were ruled out, that would mean that there are not-all-zero scalars $lambda_i,mu_i$ such that $e=v+sumlambda_ie_i$, $e'=v+summu_ie_i$, then $e=e'+sum(lambda_i-mu_i)e_i$, which is impossible, therefore $E$ has $n$ elements.
My question is, can we do the same for infinite dimensional vector spaces? That is, given a vector $v$ on an infinite dimensional $K$-vector space $V$, can we construct (feels like it is too much to ask)/ can we prove the existence of a basis $E$ such that $vin E$?
P.S: I live in a world where the axiom of choice is true, so $V$ does have a basis.
linear-algebra set-theory axiom-of-choice
edited 1 hour ago
Robert Shore
1,32815
asked 2 hours ago
JustDroppedInJustDroppedIn
2,054420
$endgroup$
add a comment |
$begingroup$
If $K$ is a field and $V=K^n$ a finite dimensional $K$-vector space with basis $A={e_1,dots,e_n}$, then, given a vector $vin V$ we can find a new basis $E$ of $V$ such that $vin E$; this is done as follows:
First we form the set ${v}cup A$ and observe that it has $n+1$ elements, therefore it spans all of $V$. Now we start the following process: We set $E_0={v}$. If $e_1notintext{span}(E_0)$, then we set $E_1={v}cup{e_1}$. Otherwise, $E_1=E_0$. Then, if $e_2notintext{span}(E_1)$, we set $E_2=E_1cup{e_2}$; otherwise, $E_2=E_1$. We go on until we eliminate every element of $A$; the set $E_n$ is our basis $E$; obviously it is linearly independent and note that it is impossible to "rule out" two elements of $A$, since if $e,e'$ were ruled out, that would mean that there are not-all-zero scalars $lambda_i,mu_i$ such that $e=v+sumlambda_ie_i$, $e'=v+summu_ie_i$, then $e=e'+sum(lambda_i-mu_i)e_i$, which is impossible, therefore $E$ has $n$ elements.
My question is, can we do the same for infinite dimensional vector spaces? That is, given a vector $v$ on an infinite dimensional $K$-vector space $V$, can we construct (feels like it is too much to ask)/ can we prove the existence of a basis $E$ such that $vin E$?
P.S: I live in a world where the axiom of choice is true, so $V$ does have a basis.
linear-algebra set-theory axiom-of-choice
edited 1 hour ago
Robert Shore
1,32815
asked 2 hours ago
JustDroppedInJustDroppedIn
2,054420
$endgroup$
If $K$ is a field and $V=K^n$ a finite dimensional $K$-vector space with basis $A={e_1,dots,e_n}$, then, given a vector $vin V$ we can find a new basis $E$ of $V$ such that $vin E$; this is done as follows:
First we form the set ${v}cup A$ and observe that it has $n+1$ elements, therefore it spans all of $V$. Now we start the following process: We set $E_0={v}$. If $e_1notintext{span}(E_0)$, then we set $E_1={v}cup{e_1}$. Otherwise, $E_1=E_0$. Then, if $e_2notintext{span}(E_1)$, we set $E_2=E_1cup{e_2}$; otherwise, $E_2=E_1$. We go on until we eliminate every element of $A$; the set $E_n$ is our basis $E$; obviously it is linearly independent and note that it is impossible to "rule out" two elements of $A$, since if $e,e'$ were ruled out, that would mean that there are not-all-zero scalars $lambda_i,mu_i$ such that $e=v+sumlambda_ie_i$, $e'=v+summu_ie_i$, then $e=e'+sum(lambda_i-mu_i)e_i$, which is impossible, therefore $E$ has $n$ elements.
My question is, can we do the same for infinite dimensional vector spaces? That is, given a vector $v$ on an infinite dimensional $K$-vector space $V$, can we construct (feels like it is too much to ask)/ can we prove the existence of a basis $E$ such that $vin E$?
P.S: I live in a world where the axiom of choice is true, so $V$ does have a basis.
linear-algebra set-theory axiom-of-choice
linear-algebra set-theory axiom-of-choice
edited 1 hour ago
Robert Shore
1,32815
asked 2 hours ago
JustDroppedInJustDroppedIn
2,054420
edited 1 hour ago
Robert Shore
1,32815
asked 2 hours ago
JustDroppedInJustDroppedIn
2,054420
edited 1 hour ago
Robert Shore
1,32815
edited 1 hour ago
Robert Shore
1,32815
edited 1 hour ago
Robert Shore
1,32815
1,32815
asked 2 hours ago
JustDroppedInJustDroppedIn
2,054420
asked 2 hours ago
JustDroppedInJustDroppedIn
2,054420
asked 2 hours ago
JustDroppedInJustDroppedIn
2,054420
2,054420
add a comment |
add a comment |
2 Answers
2
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$begingroup$
Yes (as long as $v neq 0$). You prove that a basis exists via Zorn's Lemma, proving that a maximal element (which must exist by Zorn's Lemma) in the collection of linearly independent subsets of $V$ (partially ordered by inclusion) is a basis. This proof shows that any linearly independent set can be extended to a basis -- just take a maximal element that extends the linearly independent set. But the proof necessarily uses the Axiom of Choice so it won't be constructive.
answered 1 hour ago
Robert ShoreRobert Shore
1,32815
$endgroup$
add a comment |
$begingroup$
Let $A$ be a basis of $V$ (using AC) and $v in V$. Then $v$ is a finite linear combination of elements of $A$. wlog assume that combination is
$$
v = a + k_2a_2 + cdots + k_na_n .
$$
Then $B = A/{a} cup {v}$ is a basis: it spans, and any finite subset is linearly independent.
answered 1 hour ago
Ethan BolkerEthan Bolker
43.5k551116
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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votes
$begingroup$
Yes (as long as $v neq 0$). You prove that a basis exists via Zorn's Lemma, proving that a maximal element (which must exist by Zorn's Lemma) in the collection of linearly independent subsets of $V$ (partially ordered by inclusion) is a basis. This proof shows that any linearly independent set can be extended to a basis -- just take a maximal element that extends the linearly independent set. But the proof necessarily uses the Axiom of Choice so it won't be constructive.
answered 1 hour ago
Robert ShoreRobert Shore
1,32815
$endgroup$
add a comment |
$begingroup$
Yes (as long as $v neq 0$). You prove that a basis exists via Zorn's Lemma, proving that a maximal element (which must exist by Zorn's Lemma) in the collection of linearly independent subsets of $V$ (partially ordered by inclusion) is a basis. This proof shows that any linearly independent set can be extended to a basis -- just take a maximal element that extends the linearly independent set. But the proof necessarily uses the Axiom of Choice so it won't be constructive.
answered 1 hour ago
Robert ShoreRobert Shore
1,32815
$endgroup$
add a comment |
$begingroup$
Yes (as long as $v neq 0$). You prove that a basis exists via Zorn's Lemma, proving that a maximal element (which must exist by Zorn's Lemma) in the collection of linearly independent subsets of $V$ (partially ordered by inclusion) is a basis. This proof shows that any linearly independent set can be extended to a basis -- just take a maximal element that extends the linearly independent set. But the proof necessarily uses the Axiom of Choice so it won't be constructive.
answered 1 hour ago
Robert ShoreRobert Shore
1,32815
$endgroup$
Yes (as long as $v neq 0$). You prove that a basis exists via Zorn's Lemma, proving that a maximal element (which must exist by Zorn's Lemma) in the collection of linearly independent subsets of $V$ (partially ordered by inclusion) is a basis. This proof shows that any linearly independent set can be extended to a basis -- just take a maximal element that extends the linearly independent set. But the proof necessarily uses the Axiom of Choice so it won't be constructive.
answered 1 hour ago
Robert ShoreRobert Shore
1,32815
edited 1 hour ago
answered 1 hour ago
Robert ShoreRobert Shore
1,32815
answered 1 hour ago
Robert ShoreRobert Shore
1,32815
answered 1 hour ago
Robert ShoreRobert Shore
1,32815
1,32815
add a comment |
add a comment |
$begingroup$
Let $A$ be a basis of $V$ (using AC) and $v in V$. Then $v$ is a finite linear combination of elements of $A$. wlog assume that combination is
$$
v = a + k_2a_2 + cdots + k_na_n .
$$
Then $B = A/{a} cup {v}$ is a basis: it spans, and any finite subset is linearly independent.
answered 1 hour ago
Ethan BolkerEthan Bolker
43.5k551116
$endgroup$
add a comment |
$begingroup$
Let $A$ be a basis of $V$ (using AC) and $v in V$. Then $v$ is a finite linear combination of elements of $A$. wlog assume that combination is
$$
v = a + k_2a_2 + cdots + k_na_n .
$$
Then $B = A/{a} cup {v}$ is a basis: it spans, and any finite subset is linearly independent.
answered 1 hour ago
Ethan BolkerEthan Bolker
43.5k551116
$endgroup$
add a comment |
$begingroup$
Let $A$ be a basis of $V$ (using AC) and $v in V$. Then $v$ is a finite linear combination of elements of $A$. wlog assume that combination is
$$
v = a + k_2a_2 + cdots + k_na_n .
$$
Then $B = A/{a} cup {v}$ is a basis: it spans, and any finite subset is linearly independent.
answered 1 hour ago
Ethan BolkerEthan Bolker
43.5k551116
$endgroup$
Let $A$ be a basis of $V$ (using AC) and $v in V$. Then $v$ is a finite linear combination of elements of $A$. wlog assume that combination is
$$
v = a + k_2a_2 + cdots + k_na_n .
$$
Then $B = A/{a} cup {v}$ is a basis: it spans, and any finite subset is linearly independent.
answered 1 hour ago
Ethan BolkerEthan Bolker
43.5k551116
answered 1 hour ago
Ethan BolkerEthan Bolker
43.5k551116
answered 1 hour ago
Ethan BolkerEthan Bolker
43.5k551116
answered 1 hour ago
Ethan BolkerEthan Bolker
43.5k551116
43.5k551116
add a comment |
add a comment |
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