Agreement of a basis with a subspace
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I have difficulty understanding the following definition in E.B. Vinberg's algebra book on Chapter 5, Vector Spaces:
Definition 5.2. A basis of a space $V$ agrees with a subspace $U$ if $U$ is a linear span of some basis vectors (i.e., if it is one of the "coordinate subspaces" with respect to this basis).
Isn't it obvious that every basis of a vector space spans a subspace of that vector space? What distinguishes "agrees with a subspace" from "is spanned by the basis of a vector space"?
Thanks, your help is appreciated.
linear-algebra
edited Nov 22 at 13:33
Martin Sleziak
44.6k7115269
asked Nov 22 at 13:26
axis_y
141
add a comment |
up vote
2
down vote
favorite
I have difficulty understanding the following definition in E.B. Vinberg's algebra book on Chapter 5, Vector Spaces:
Definition 5.2. A basis of a space $V$ agrees with a subspace $U$ if $U$ is a linear span of some basis vectors (i.e., if it is one of the "coordinate subspaces" with respect to this basis).
Isn't it obvious that every basis of a vector space spans a subspace of that vector space? What distinguishes "agrees with a subspace" from "is spanned by the basis of a vector space"?
Thanks, your help is appreciated.
linear-algebra
edited Nov 22 at 13:33
Martin Sleziak
44.6k7115269
asked Nov 22 at 13:26
axis_y
141
In future, when you have a mathematical expression, please write the dollar sign both in front and behind it. Like so:$U$, not just$U. I edited the question for you this time.
– 5xum
Nov 22 at 13:28
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 27 at 10:50
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have difficulty understanding the following definition in E.B. Vinberg's algebra book on Chapter 5, Vector Spaces:
Definition 5.2. A basis of a space $V$ agrees with a subspace $U$ if $U$ is a linear span of some basis vectors (i.e., if it is one of the "coordinate subspaces" with respect to this basis).
Isn't it obvious that every basis of a vector space spans a subspace of that vector space? What distinguishes "agrees with a subspace" from "is spanned by the basis of a vector space"?
Thanks, your help is appreciated.
linear-algebra
edited Nov 22 at 13:33
Martin Sleziak
44.6k7115269
asked Nov 22 at 13:26
axis_y
141
I have difficulty understanding the following definition in E.B. Vinberg's algebra book on Chapter 5, Vector Spaces:
Definition 5.2. A basis of a space $V$ agrees with a subspace $U$ if $U$ is a linear span of some basis vectors (i.e., if it is one of the "coordinate subspaces" with respect to this basis).
Isn't it obvious that every basis of a vector space spans a subspace of that vector space? What distinguishes "agrees with a subspace" from "is spanned by the basis of a vector space"?
Thanks, your help is appreciated.
linear-algebra
linear-algebra
edited Nov 22 at 13:33
Martin Sleziak
44.6k7115269
asked Nov 22 at 13:26
axis_y
141
edited Nov 22 at 13:33
Martin Sleziak
44.6k7115269
asked Nov 22 at 13:26
axis_y
141
edited Nov 22 at 13:33
Martin Sleziak
44.6k7115269
edited Nov 22 at 13:33
Martin Sleziak
44.6k7115269
edited Nov 22 at 13:33
Martin Sleziak
44.6k7115269
44.6k7115269
asked Nov 22 at 13:26
axis_y
141
asked Nov 22 at 13:26
axis_y
141
asked Nov 22 at 13:26
axis_y
141
141
In future, when you have a mathematical expression, please write the dollar sign both in front and behind it. Like so:$U$, not just$U. I edited the question for you this time.
– 5xum
Nov 22 at 13:28
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 27 at 10:50
add a comment |
In future, when you have a mathematical expression, please write the dollar sign both in front and behind it. Like so:$U$, not just$U. I edited the question for you this time.
– 5xum
Nov 22 at 13:28
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 27 at 10:50
In future, when you have a mathematical expression, please write the dollar sign both in front and behind it. Like so:
$U$, not just $U. I edited the question for you this time.– 5xum
Nov 22 at 13:28
In future, when you have a mathematical expression, please write the dollar sign both in front and behind it. Like so:
$U$, not just $U. I edited the question for you this time.– 5xum
Nov 22 at 13:28
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 27 at 10:50
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 27 at 10:50
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
When we are dealing with that definition, there are two objects given from the start: the subspace $U$ and the basis $B$. If, for instance $V=mathbb{R}^2$, $U={(x,x),|,xinmathbb{R}}$ and $B$ is the canonical basis, then $U$ does not agree with $B$.
edited Nov 22 at 13:33
Martin Sleziak
44.6k7115269
answered Nov 22 at 13:31
José Carlos Santos
146k22117217
I always find it comforting and a little amusing to see the same counterexample in another answer written in parallel to mine :).
– 5xum
Nov 22 at 13:34
Me too. But your answer was appeared first.
– José Carlos Santos
Nov 22 at 13:36
Sure, but they were clearly written in parallel. I mean, it's (arguably) the simplest possible counterexample, and the fact that we both went for it is further proof of that
– 5xum
Nov 22 at 13:40
I agree. It couldn't be simpler.
– José Carlos Santos
Nov 22 at 13:43
But, what about the subspace $z = 0$ with the basis ${(1,0,0), (0,1,0), (0,0,1)}$, do they agree?
– axis_y
Nov 22 at 14:06
|
show 3 more comments
up vote
0
down vote
It's easiest to see on an example. Let $V=mathbb R^2$.
The basis ${(1,0), (0,1)}$ agrees with the subspace $U={(x,0)|xinmathbb R}$.
The basis ${(1,0), (0,1)}$ does not agree with the subspace $U={(x,x)|xinmathbb R}$.
The answer to your question should now be more clear, but just in case:
Isn't it obvious that every basis of a vector space spans a subspace of that vector space?
Yes, this is obvious. If $B$ is a basis for $V$, then for every subset $Ssubseteq U$, $S$ spans some subspace of $V$. And on those subspaces, $U$ agrees with them. However, there can (and do) also exist subspaces of $V$ which are not spanned by any subset of $U$. We say that does not agree with those subspaces.
answered Nov 22 at 13:30
5xum
89.4k393161
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
When we are dealing with that definition, there are two objects given from the start: the subspace $U$ and the basis $B$. If, for instance $V=mathbb{R}^2$, $U={(x,x),|,xinmathbb{R}}$ and $B$ is the canonical basis, then $U$ does not agree with $B$.
edited Nov 22 at 13:33
Martin Sleziak
44.6k7115269
answered Nov 22 at 13:31
José Carlos Santos
146k22117217
I always find it comforting and a little amusing to see the same counterexample in another answer written in parallel to mine :).
– 5xum
Nov 22 at 13:34
Me too. But your answer was appeared first.
– José Carlos Santos
Nov 22 at 13:36
Sure, but they were clearly written in parallel. I mean, it's (arguably) the simplest possible counterexample, and the fact that we both went for it is further proof of that
– 5xum
Nov 22 at 13:40
I agree. It couldn't be simpler.
– José Carlos Santos
Nov 22 at 13:43
But, what about the subspace $z = 0$ with the basis ${(1,0,0), (0,1,0), (0,0,1)}$, do they agree?
– axis_y
Nov 22 at 14:06
|
show 3 more comments
up vote
1
down vote
When we are dealing with that definition, there are two objects given from the start: the subspace $U$ and the basis $B$. If, for instance $V=mathbb{R}^2$, $U={(x,x),|,xinmathbb{R}}$ and $B$ is the canonical basis, then $U$ does not agree with $B$.
edited Nov 22 at 13:33
Martin Sleziak
44.6k7115269
answered Nov 22 at 13:31
José Carlos Santos
146k22117217
I always find it comforting and a little amusing to see the same counterexample in another answer written in parallel to mine :).
– 5xum
Nov 22 at 13:34
Me too. But your answer was appeared first.
– José Carlos Santos
Nov 22 at 13:36
Sure, but they were clearly written in parallel. I mean, it's (arguably) the simplest possible counterexample, and the fact that we both went for it is further proof of that
– 5xum
Nov 22 at 13:40
I agree. It couldn't be simpler.
– José Carlos Santos
Nov 22 at 13:43
But, what about the subspace $z = 0$ with the basis ${(1,0,0), (0,1,0), (0,0,1)}$, do they agree?
– axis_y
Nov 22 at 14:06
|
show 3 more comments
up vote
1
down vote
up vote
1
down vote
When we are dealing with that definition, there are two objects given from the start: the subspace $U$ and the basis $B$. If, for instance $V=mathbb{R}^2$, $U={(x,x),|,xinmathbb{R}}$ and $B$ is the canonical basis, then $U$ does not agree with $B$.
edited Nov 22 at 13:33
Martin Sleziak
44.6k7115269
answered Nov 22 at 13:31
José Carlos Santos
146k22117217
When we are dealing with that definition, there are two objects given from the start: the subspace $U$ and the basis $B$. If, for instance $V=mathbb{R}^2$, $U={(x,x),|,xinmathbb{R}}$ and $B$ is the canonical basis, then $U$ does not agree with $B$.
edited Nov 22 at 13:33
Martin Sleziak
44.6k7115269
answered Nov 22 at 13:31
José Carlos Santos
146k22117217
edited Nov 22 at 13:33
Martin Sleziak
44.6k7115269
edited Nov 22 at 13:33
Martin Sleziak
44.6k7115269
edited Nov 22 at 13:33
Martin Sleziak
44.6k7115269
44.6k7115269
answered Nov 22 at 13:31
José Carlos Santos
146k22117217
answered Nov 22 at 13:31
José Carlos Santos
146k22117217
answered Nov 22 at 13:31
José Carlos Santos
146k22117217
146k22117217
I always find it comforting and a little amusing to see the same counterexample in another answer written in parallel to mine :).
– 5xum
Nov 22 at 13:34
Me too. But your answer was appeared first.
– José Carlos Santos
Nov 22 at 13:36
Sure, but they were clearly written in parallel. I mean, it's (arguably) the simplest possible counterexample, and the fact that we both went for it is further proof of that
– 5xum
Nov 22 at 13:40
I agree. It couldn't be simpler.
– José Carlos Santos
Nov 22 at 13:43
But, what about the subspace $z = 0$ with the basis ${(1,0,0), (0,1,0), (0,0,1)}$, do they agree?
– axis_y
Nov 22 at 14:06
|
show 3 more comments
I always find it comforting and a little amusing to see the same counterexample in another answer written in parallel to mine :).
– 5xum
Nov 22 at 13:34
Me too. But your answer was appeared first.
– José Carlos Santos
Nov 22 at 13:36
Sure, but they were clearly written in parallel. I mean, it's (arguably) the simplest possible counterexample, and the fact that we both went for it is further proof of that
– 5xum
Nov 22 at 13:40
I agree. It couldn't be simpler.
– José Carlos Santos
Nov 22 at 13:43
But, what about the subspace $z = 0$ with the basis ${(1,0,0), (0,1,0), (0,0,1)}$, do they agree?
– axis_y
Nov 22 at 14:06
I always find it comforting and a little amusing to see the same counterexample in another answer written in parallel to mine :).
– 5xum
Nov 22 at 13:34
I always find it comforting and a little amusing to see the same counterexample in another answer written in parallel to mine :).
– 5xum
Nov 22 at 13:34
Me too. But your answer was appeared first.
– José Carlos Santos
Nov 22 at 13:36
Me too. But your answer was appeared first.
– José Carlos Santos
Nov 22 at 13:36
Sure, but they were clearly written in parallel. I mean, it's (arguably) the simplest possible counterexample, and the fact that we both went for it is further proof of that
– 5xum
Nov 22 at 13:40
Sure, but they were clearly written in parallel. I mean, it's (arguably) the simplest possible counterexample, and the fact that we both went for it is further proof of that
– 5xum
Nov 22 at 13:40
I agree. It couldn't be simpler.
– José Carlos Santos
Nov 22 at 13:43
I agree. It couldn't be simpler.
– José Carlos Santos
Nov 22 at 13:43
But, what about the subspace $z = 0$ with the basis ${(1,0,0), (0,1,0), (0,0,1)}$, do they agree?
– axis_y
Nov 22 at 14:06
But, what about the subspace $z = 0$ with the basis ${(1,0,0), (0,1,0), (0,0,1)}$, do they agree?
– axis_y
Nov 22 at 14:06
|
show 3 more comments
up vote
0
down vote
It's easiest to see on an example. Let $V=mathbb R^2$.
The basis ${(1,0), (0,1)}$ agrees with the subspace $U={(x,0)|xinmathbb R}$.
The basis ${(1,0), (0,1)}$ does not agree with the subspace $U={(x,x)|xinmathbb R}$.
The answer to your question should now be more clear, but just in case:
Isn't it obvious that every basis of a vector space spans a subspace of that vector space?
Yes, this is obvious. If $B$ is a basis for $V$, then for every subset $Ssubseteq U$, $S$ spans some subspace of $V$. And on those subspaces, $U$ agrees with them. However, there can (and do) also exist subspaces of $V$ which are not spanned by any subset of $U$. We say that does not agree with those subspaces.
answered Nov 22 at 13:30
5xum
89.4k393161
add a comment |
up vote
0
down vote
It's easiest to see on an example. Let $V=mathbb R^2$.
The basis ${(1,0), (0,1)}$ agrees with the subspace $U={(x,0)|xinmathbb R}$.
The basis ${(1,0), (0,1)}$ does not agree with the subspace $U={(x,x)|xinmathbb R}$.
The answer to your question should now be more clear, but just in case:
Isn't it obvious that every basis of a vector space spans a subspace of that vector space?
Yes, this is obvious. If $B$ is a basis for $V$, then for every subset $Ssubseteq U$, $S$ spans some subspace of $V$. And on those subspaces, $U$ agrees with them. However, there can (and do) also exist subspaces of $V$ which are not spanned by any subset of $U$. We say that does not agree with those subspaces.
answered Nov 22 at 13:30
5xum
89.4k393161
add a comment |
up vote
0
down vote
up vote
0
down vote
It's easiest to see on an example. Let $V=mathbb R^2$.
The basis ${(1,0), (0,1)}$ agrees with the subspace $U={(x,0)|xinmathbb R}$.
The basis ${(1,0), (0,1)}$ does not agree with the subspace $U={(x,x)|xinmathbb R}$.
The answer to your question should now be more clear, but just in case:
Isn't it obvious that every basis of a vector space spans a subspace of that vector space?
Yes, this is obvious. If $B$ is a basis for $V$, then for every subset $Ssubseteq U$, $S$ spans some subspace of $V$. And on those subspaces, $U$ agrees with them. However, there can (and do) also exist subspaces of $V$ which are not spanned by any subset of $U$. We say that does not agree with those subspaces.
answered Nov 22 at 13:30
5xum
89.4k393161
It's easiest to see on an example. Let $V=mathbb R^2$.
The basis ${(1,0), (0,1)}$ agrees with the subspace $U={(x,0)|xinmathbb R}$.
The basis ${(1,0), (0,1)}$ does not agree with the subspace $U={(x,x)|xinmathbb R}$.
The answer to your question should now be more clear, but just in case:
Isn't it obvious that every basis of a vector space spans a subspace of that vector space?
Yes, this is obvious. If $B$ is a basis for $V$, then for every subset $Ssubseteq U$, $S$ spans some subspace of $V$. And on those subspaces, $U$ agrees with them. However, there can (and do) also exist subspaces of $V$ which are not spanned by any subset of $U$. We say that does not agree with those subspaces.
answered Nov 22 at 13:30
5xum
89.4k393161
answered Nov 22 at 13:30
5xum
89.4k393161
answered Nov 22 at 13:30
5xum
89.4k393161
answered Nov 22 at 13:30
5xum
89.4k393161
89.4k393161
add a comment |
add a comment |
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In future, when you have a mathematical expression, please write the dollar sign both in front and behind it. Like so:
$U$, not just$U. I edited the question for you this time.– 5xum
Nov 22 at 13:28
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
– 5xum
Nov 27 at 10:50