Finding a basis of an infinite dimensional vector space with a given vector
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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 $vcup A$ 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 $e_1notintextspan(E_0)$, then we set $E_1=vcupe_1$. Otherwise, $E_1=E_0$. Then, if $e_2notintextspan(E_1)$, we set $E_2=E_1cupe_2$; 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 Feb 20 at 3:28
Robert Shore
3,030222
asked Feb 20 at 3:04
JustDroppedInJustDroppedIn
2,163420
$endgroup$
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 $vcup A$ 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 $e_1notintextspan(E_0)$, then we set $E_1=vcupe_1$. Otherwise, $E_1=E_0$. Then, if $e_2notintextspan(E_1)$, we set $E_2=E_1cupe_2$; 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 Feb 20 at 3:28
Robert Shore
3,030222
asked Feb 20 at 3:04
JustDroppedInJustDroppedIn
2,163420
$endgroup$
$begingroup$
BTW. A property $P(x)$, which any set $x$ may or may not have, is called a property of finite character when $P(x)iff forall$ finite $ysubset x,(P(y))$ holds for every $x$. For example, being a linearly independent subset of $V$ is a property of finite character. The Teichmuller-Tukey Lemma: If $P$ is of finite character and $P(x)$ for some $xin X$ then there is a $subset$-maximal $Ysubset X$ such that $xin Y$ and $P(Y).$ In the axiom system ZF, this "lemma" is equivalent to the axiom of choice.
$endgroup$
– DanielWainfleet
Feb 28 at 10:41
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 $vcup A$ 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 $e_1notintextspan(E_0)$, then we set $E_1=vcupe_1$. Otherwise, $E_1=E_0$. Then, if $e_2notintextspan(E_1)$, we set $E_2=E_1cupe_2$; 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 Feb 20 at 3:28
Robert Shore
3,030222
asked Feb 20 at 3:04
JustDroppedInJustDroppedIn
2,163420
$endgroup$
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 $vcup A$ 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 $e_1notintextspan(E_0)$, then we set $E_1=vcupe_1$. Otherwise, $E_1=E_0$. Then, if $e_2notintextspan(E_1)$, we set $E_2=E_1cupe_2$; 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 Feb 20 at 3:28
Robert Shore
3,030222
asked Feb 20 at 3:04
JustDroppedInJustDroppedIn
2,163420
edited Feb 20 at 3:28
Robert Shore
3,030222
asked Feb 20 at 3:04
JustDroppedInJustDroppedIn
2,163420
edited Feb 20 at 3:28
Robert Shore
3,030222
edited Feb 20 at 3:28
Robert Shore
3,030222
edited Feb 20 at 3:28
Robert Shore
3,030222
3,030222
asked Feb 20 at 3:04
JustDroppedInJustDroppedIn
2,163420
asked Feb 20 at 3:04
JustDroppedInJustDroppedIn
2,163420
asked Feb 20 at 3:04
JustDroppedInJustDroppedIn
2,163420
2,163420
$begingroup$
BTW. A property $P(x)$, which any set $x$ may or may not have, is called a property of finite character when $P(x)iff forall$ finite $ysubset x,(P(y))$ holds for every $x$. For example, being a linearly independent subset of $V$ is a property of finite character. The Teichmuller-Tukey Lemma: If $P$ is of finite character and $P(x)$ for some $xin X$ then there is a $subset$-maximal $Ysubset X$ such that $xin Y$ and $P(Y).$ In the axiom system ZF, this "lemma" is equivalent to the axiom of choice.
$endgroup$
– DanielWainfleet
Feb 28 at 10:41
add a comment |
$begingroup$
BTW. A property $P(x)$, which any set $x$ may or may not have, is called a property of finite character when $P(x)iff forall$ finite $ysubset x,(P(y))$ holds for every $x$. For example, being a linearly independent subset of $V$ is a property of finite character. The Teichmuller-Tukey Lemma: If $P$ is of finite character and $P(x)$ for some $xin X$ then there is a $subset$-maximal $Ysubset X$ such that $xin Y$ and $P(Y).$ In the axiom system ZF, this "lemma" is equivalent to the axiom of choice.
$endgroup$
– DanielWainfleet
Feb 28 at 10:41
$begingroup$
BTW. A property $P(x)$, which any set $x$ may or may not have, is called a property of finite character when $P(x)iff forall$ finite $ysubset x,(P(y))$ holds for every $x$. For example, being a linearly independent subset of $V$ is a property of finite character. The Teichmuller-Tukey Lemma: If $P$ is of finite character and $P(x)$ for some $xin X$ then there is a $subset$-maximal $Ysubset X$ such that $xin Y$ and $P(Y).$ In the axiom system ZF, this "lemma" is equivalent to the axiom of choice.
$endgroup$
– DanielWainfleet
Feb 28 at 10:41
$begingroup$
BTW. A property $P(x)$, which any set $x$ may or may not have, is called a property of finite character when $P(x)iff forall$ finite $ysubset x,(P(y))$ holds for every $x$. For example, being a linearly independent subset of $V$ is a property of finite character. The Teichmuller-Tukey Lemma: If $P$ is of finite character and $P(x)$ for some $xin X$ then there is a $subset$-maximal $Ysubset X$ such that $xin Y$ and $P(Y).$ In the axiom system ZF, this "lemma" is equivalent to the axiom of choice.
$endgroup$
– DanielWainfleet
Feb 28 at 10:41
add a comment |
2 Answers
2
active
oldest
votes
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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 Feb 20 at 3:11
Robert ShoreRobert Shore
3,030222
$endgroup$
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 Feb 20 at 3:31
Ethan BolkerEthan Bolker
45k553120
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
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oldest
votes
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oldest
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 Feb 20 at 3:11
Robert ShoreRobert Shore
3,030222
$endgroup$
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 Feb 20 at 3:11
Robert ShoreRobert Shore
3,030222
$endgroup$
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 Feb 20 at 3:11
Robert ShoreRobert Shore
3,030222
$endgroup$
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 Feb 20 at 3:11
Robert ShoreRobert Shore
3,030222
edited Feb 20 at 3:36
answered Feb 20 at 3:11
Robert ShoreRobert Shore
3,030222
answered Feb 20 at 3:11
Robert ShoreRobert Shore
3,030222
answered Feb 20 at 3:11
Robert ShoreRobert Shore
3,030222
3,030222
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 Feb 20 at 3:31
Ethan BolkerEthan Bolker
45k553120
$endgroup$
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 Feb 20 at 3:31
Ethan BolkerEthan Bolker
45k553120
$endgroup$
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 Feb 20 at 3:31
Ethan BolkerEthan Bolker
45k553120
$endgroup$
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 Feb 20 at 3:31
Ethan BolkerEthan Bolker
45k553120
answered Feb 20 at 3:31
Ethan BolkerEthan Bolker
45k553120
answered Feb 20 at 3:31
Ethan BolkerEthan Bolker
45k553120
answered Feb 20 at 3:31
Ethan BolkerEthan Bolker
45k553120
45k553120
add a comment |
add a comment |
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$begingroup$
BTW. A property $P(x)$, which any set $x$ may or may not have, is called a property of finite character when $P(x)iff forall$ finite $ysubset x,(P(y))$ holds for every $x$. For example, being a linearly independent subset of $V$ is a property of finite character. The Teichmuller-Tukey Lemma: If $P$ is of finite character and $P(x)$ for some $xin X$ then there is a $subset$-maximal $Ysubset X$ such that $xin Y$ and $P(Y).$ In the axiom system ZF, this "lemma" is equivalent to the axiom of choice.
$endgroup$
– DanielWainfleet
Feb 28 at 10:41