Mixing
non-null vector space & basis
$begingroup$
Show that any non-null vector space has a basis.
What I am trying to do -Taking a spanning set which spans the vector space then if the spanning set is linearly independent then it form basis and we are done but if it is linearly dependent then we can find a vector of that set which can be written as a linear combination of other vectors we remove that vector now if the remaining set of vectors ar linearly independent then we are done otherwise repeat the same process till we reach a stage when only linearly independent set of vector remains left and thus it form's a basis. Don't know right or wrong.
linear-algebra
asked Jan 31 at 18:57
Daniel Daniel
114
$endgroup$
add a comment |
$begingroup$
Show that any non-null vector space has a basis.
What I am trying to do -Taking a spanning set which spans the vector space then if the spanning set is linearly independent then it form basis and we are done but if it is linearly dependent then we can find a vector of that set which can be written as a linear combination of other vectors we remove that vector now if the remaining set of vectors ar linearly independent then we are done otherwise repeat the same process till we reach a stage when only linearly independent set of vector remains left and thus it form's a basis. Don't know right or wrong.
linear-algebra
asked Jan 31 at 18:57
Daniel Daniel
114
$endgroup$
add a comment |
$begingroup$
Show that any non-null vector space has a basis.
What I am trying to do -Taking a spanning set which spans the vector space then if the spanning set is linearly independent then it form basis and we are done but if it is linearly dependent then we can find a vector of that set which can be written as a linear combination of other vectors we remove that vector now if the remaining set of vectors ar linearly independent then we are done otherwise repeat the same process till we reach a stage when only linearly independent set of vector remains left and thus it form's a basis. Don't know right or wrong.
linear-algebra
asked Jan 31 at 18:57
Daniel Daniel
114
$endgroup$
Show that any non-null vector space has a basis.
What I am trying to do -Taking a spanning set which spans the vector space then if the spanning set is linearly independent then it form basis and we are done but if it is linearly dependent then we can find a vector of that set which can be written as a linear combination of other vectors we remove that vector now if the remaining set of vectors ar linearly independent then we are done otherwise repeat the same process till we reach a stage when only linearly independent set of vector remains left and thus it form's a basis. Don't know right or wrong.
linear-algebra
linear-algebra
asked Jan 31 at 18:57
Daniel Daniel
114
asked Jan 31 at 18:57
Daniel Daniel
114
asked Jan 31 at 18:57
Daniel Daniel
114
asked Jan 31 at 18:57
Daniel Daniel
114
asked Jan 31 at 18:57
Daniel Daniel
114
114
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In your argument you are making a choice to remove elements and repeat the process. This requires axiom of choice for an infinite dimensional vector space.
Hint:
Take a non trivial vector space $V$. Since it is non trivial there is a non zero vector in it. Now make a set of linearly independent subsets of $V$. Show that this is partially ordered(poset) by inclusion and each poset has an upper bound(hint: think of nion). Apply Zorns lemma to show there is a maximal element.
answered Jan 31 at 19:08
Basanta R PahariBasanta R Pahari
1,761613
$endgroup$
$begingroup$
Is there are any other than what u have shown.As it is given where the use of lemme what u have stated is not allowed.
$endgroup$
– Daniel
Feb 1 at 2:04
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In your argument you are making a choice to remove elements and repeat the process. This requires axiom of choice for an infinite dimensional vector space.
Hint:
Take a non trivial vector space $V$. Since it is non trivial there is a non zero vector in it. Now make a set of linearly independent subsets of $V$. Show that this is partially ordered(poset) by inclusion and each poset has an upper bound(hint: think of nion). Apply Zorns lemma to show there is a maximal element.
answered Jan 31 at 19:08
Basanta R PahariBasanta R Pahari
1,761613
$endgroup$
$begingroup$
Is there are any other than what u have shown.As it is given where the use of lemme what u have stated is not allowed.
$endgroup$
– Daniel
Feb 1 at 2:04
add a comment |
$begingroup$
In your argument you are making a choice to remove elements and repeat the process. This requires axiom of choice for an infinite dimensional vector space.
Hint:
Take a non trivial vector space $V$. Since it is non trivial there is a non zero vector in it. Now make a set of linearly independent subsets of $V$. Show that this is partially ordered(poset) by inclusion and each poset has an upper bound(hint: think of nion). Apply Zorns lemma to show there is a maximal element.
answered Jan 31 at 19:08
Basanta R PahariBasanta R Pahari
1,761613
$endgroup$
$begingroup$
Is there are any other than what u have shown.As it is given where the use of lemme what u have stated is not allowed.
$endgroup$
– Daniel
Feb 1 at 2:04
add a comment |
$begingroup$
In your argument you are making a choice to remove elements and repeat the process. This requires axiom of choice for an infinite dimensional vector space.
Hint:
Take a non trivial vector space $V$. Since it is non trivial there is a non zero vector in it. Now make a set of linearly independent subsets of $V$. Show that this is partially ordered(poset) by inclusion and each poset has an upper bound(hint: think of nion). Apply Zorns lemma to show there is a maximal element.
answered Jan 31 at 19:08
Basanta R PahariBasanta R Pahari
1,761613
$endgroup$
In your argument you are making a choice to remove elements and repeat the process. This requires axiom of choice for an infinite dimensional vector space.
Hint:
Take a non trivial vector space $V$. Since it is non trivial there is a non zero vector in it. Now make a set of linearly independent subsets of $V$. Show that this is partially ordered(poset) by inclusion and each poset has an upper bound(hint: think of nion). Apply Zorns lemma to show there is a maximal element.
answered Jan 31 at 19:08
Basanta R PahariBasanta R Pahari
1,761613
answered Jan 31 at 19:08
Basanta R PahariBasanta R Pahari
1,761613
answered Jan 31 at 19:08
Basanta R PahariBasanta R Pahari
1,761613
answered Jan 31 at 19:08
Basanta R PahariBasanta R Pahari
1,761613
1,761613
$begingroup$
Is there are any other than what u have shown.As it is given where the use of lemme what u have stated is not allowed.
$endgroup$
– Daniel
Feb 1 at 2:04
add a comment |
$begingroup$
Is there are any other than what u have shown.As it is given where the use of lemme what u have stated is not allowed.
$endgroup$
– Daniel
Feb 1 at 2:04
$begingroup$
Is there are any other than what u have shown.As it is given where the use of lemme what u have stated is not allowed.
$endgroup$
– Daniel
Feb 1 at 2:04
$begingroup$
Is there are any other than what u have shown.As it is given where the use of lemme what u have stated is not allowed.
$endgroup$
– Daniel
Feb 1 at 2:04
add a comment |
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