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







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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















up vote
2
down vote

favorite
1












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







share|cite|improve this question
























  • 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













up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





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







share|cite|improve this question















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






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edited Nov 22 at 13:33









Martin Sleziak

44.6k7115269




44.6k7115269










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


















  • 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










2 Answers
2






active

oldest

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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 $B$.






share|cite|improve this answer























  • 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




















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.






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    2 Answers
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    2 Answers
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    up vote
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    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 $B$.






    share|cite|improve this answer























    • 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

















    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 $B$.






    share|cite|improve this answer























    • 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















    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 $B$.






    share|cite|improve this answer














    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 $B$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 22 at 13:33









    Martin Sleziak

    44.6k7115269




    44.6k7115269










    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




















    • 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












    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.






    share|cite|improve this answer

























      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.






      share|cite|improve this answer























        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 13:30









        5xum

        89.4k393161




        89.4k393161






























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