Vector space over a field












2














Lets say i have a vector space V over a field $Bbb F$, does $Bbb F$ only determine the scalar multiplication i can do, or does it also determine what value my vector coefficients (maybe not the right word) can take?



An example would be V over $Bbb R$, example of vector in V would be x = (x1,x2,x3), does x1, x2, x3 have to be a member of $Bbb R$ too? x1, x2, x3 is what im calling as coeffcients.






linear-algebra







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  • 1




    Can you be more precise about what you mean by "my vector coefficients"? Or, do you have a specific example of a vector field where this confusion arises?
    – Omnomnomnom
    2 hours ago










  • If what you mean by "vector coefficients" are the entries in an $n$-tuple thought of as a vector, then yes, the entries come from the field.
    – Ethan Bolker
    2 hours ago






  • 2




    Not every vector space consists of $n$-tuples, but even limiting ourselves to such vector spaces, the entries of such an $n$-tuple only have to come from an extension of the field $mathbf{F}$. For example, the set $mathbf{C}^2$ consisting of ordered pairs of complex numbers is a vector space over $mathbf{R}$.
    – Keenan Kidwell
    2 hours ago
















2














Lets say i have a vector space V over a field $Bbb F$, does $Bbb F$ only determine the scalar multiplication i can do, or does it also determine what value my vector coefficients (maybe not the right word) can take?



An example would be V over $Bbb R$, example of vector in V would be x = (x1,x2,x3), does x1, x2, x3 have to be a member of $Bbb R$ too? x1, x2, x3 is what im calling as coeffcients.






linear-algebra







share|cite|improve this question









New contributor




5Six is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    Can you be more precise about what you mean by "my vector coefficients"? Or, do you have a specific example of a vector field where this confusion arises?
    – Omnomnomnom
    2 hours ago










  • If what you mean by "vector coefficients" are the entries in an $n$-tuple thought of as a vector, then yes, the entries come from the field.
    – Ethan Bolker
    2 hours ago






  • 2




    Not every vector space consists of $n$-tuples, but even limiting ourselves to such vector spaces, the entries of such an $n$-tuple only have to come from an extension of the field $mathbf{F}$. For example, the set $mathbf{C}^2$ consisting of ordered pairs of complex numbers is a vector space over $mathbf{R}$.
    – Keenan Kidwell
    2 hours ago














2












2








2







Lets say i have a vector space V over a field $Bbb F$, does $Bbb F$ only determine the scalar multiplication i can do, or does it also determine what value my vector coefficients (maybe not the right word) can take?



An example would be V over $Bbb R$, example of vector in V would be x = (x1,x2,x3), does x1, x2, x3 have to be a member of $Bbb R$ too? x1, x2, x3 is what im calling as coeffcients.






linear-algebra







share|cite|improve this question









New contributor




5Six is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Lets say i have a vector space V over a field $Bbb F$, does $Bbb F$ only determine the scalar multiplication i can do, or does it also determine what value my vector coefficients (maybe not the right word) can take?



An example would be V over $Bbb R$, example of vector in V would be x = (x1,x2,x3), does x1, x2, x3 have to be a member of $Bbb R$ too? x1, x2, x3 is what im calling as coeffcients.






linear-algebra




linear-algebra






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5Six is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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Check out our Code of Conduct.









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edited 2 hours ago





















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asked 2 hours ago









5Six

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5Six is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.








  • 1




    Can you be more precise about what you mean by "my vector coefficients"? Or, do you have a specific example of a vector field where this confusion arises?
    – Omnomnomnom
    2 hours ago










  • If what you mean by "vector coefficients" are the entries in an $n$-tuple thought of as a vector, then yes, the entries come from the field.
    – Ethan Bolker
    2 hours ago






  • 2




    Not every vector space consists of $n$-tuples, but even limiting ourselves to such vector spaces, the entries of such an $n$-tuple only have to come from an extension of the field $mathbf{F}$. For example, the set $mathbf{C}^2$ consisting of ordered pairs of complex numbers is a vector space over $mathbf{R}$.
    – Keenan Kidwell
    2 hours ago














  • 1




    Can you be more precise about what you mean by "my vector coefficients"? Or, do you have a specific example of a vector field where this confusion arises?
    – Omnomnomnom
    2 hours ago










  • If what you mean by "vector coefficients" are the entries in an $n$-tuple thought of as a vector, then yes, the entries come from the field.
    – Ethan Bolker
    2 hours ago






  • 2




    Not every vector space consists of $n$-tuples, but even limiting ourselves to such vector spaces, the entries of such an $n$-tuple only have to come from an extension of the field $mathbf{F}$. For example, the set $mathbf{C}^2$ consisting of ordered pairs of complex numbers is a vector space over $mathbf{R}$.
    – Keenan Kidwell
    2 hours ago








1




1




Can you be more precise about what you mean by "my vector coefficients"? Or, do you have a specific example of a vector field where this confusion arises?
– Omnomnomnom
2 hours ago




Can you be more precise about what you mean by "my vector coefficients"? Or, do you have a specific example of a vector field where this confusion arises?
– Omnomnomnom
2 hours ago












If what you mean by "vector coefficients" are the entries in an $n$-tuple thought of as a vector, then yes, the entries come from the field.
– Ethan Bolker
2 hours ago




If what you mean by "vector coefficients" are the entries in an $n$-tuple thought of as a vector, then yes, the entries come from the field.
– Ethan Bolker
2 hours ago




2




2




Not every vector space consists of $n$-tuples, but even limiting ourselves to such vector spaces, the entries of such an $n$-tuple only have to come from an extension of the field $mathbf{F}$. For example, the set $mathbf{C}^2$ consisting of ordered pairs of complex numbers is a vector space over $mathbf{R}$.
– Keenan Kidwell
2 hours ago




Not every vector space consists of $n$-tuples, but even limiting ourselves to such vector spaces, the entries of such an $n$-tuple only have to come from an extension of the field $mathbf{F}$. For example, the set $mathbf{C}^2$ consisting of ordered pairs of complex numbers is a vector space over $mathbf{R}$.
– Keenan Kidwell
2 hours ago










2 Answers
2






active

oldest

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3














Often when we write
$pmatrix{1 \ 2 \3}$ we mean that this represents some coordinates and actually must be interpreted as: $$1 cdot v_1 + 2 cdot v_2 +3 cdot v_3$$



Where the $v_i$ are the basis vectors in which you represent this coordinate vector. Notice that the coefficients are definitely elements of my field $mathbb{F}$. This has nice properties, for instance, once you use some coordinate vectors for a given system you can show linear independence of that system by showing linear independence for the coordinate vectors.



Also see






share|cite|improve this answer





















  • So in my example x1, x2, x3 has to be members of R? if i am understanding correctly
    – 5Six
    2 hours ago





















1














If I am interpreting your question correctly, the answer is that yes $mathbb{F}$ determines determines both.



But I need to make sure I understand what you mean by "vector coefficients". I assume you are referring to those ocurring in linear combinations of vectors. So if $w$ and $v$ are vectors then $aw+bv$ is a linear combination of $w$ and $v$ with coefficients $a$ and $b$. Indeed, in general the "coefficients" here are assumed to be elements of your field $mathbb{F}$. Nobody uses the term "vector coefficients". Only the term coefficient is used, as these coefficients are understood to belong to your field of scalars.






share|cite|improve this answer





















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    2 Answers
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    2 Answers
    2






    active

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    active

    oldest

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    active

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    3














    Often when we write
    $pmatrix{1 \ 2 \3}$ we mean that this represents some coordinates and actually must be interpreted as: $$1 cdot v_1 + 2 cdot v_2 +3 cdot v_3$$



    Where the $v_i$ are the basis vectors in which you represent this coordinate vector. Notice that the coefficients are definitely elements of my field $mathbb{F}$. This has nice properties, for instance, once you use some coordinate vectors for a given system you can show linear independence of that system by showing linear independence for the coordinate vectors.



    Also see






    share|cite|improve this answer





















    • So in my example x1, x2, x3 has to be members of R? if i am understanding correctly
      – 5Six
      2 hours ago


















    3














    Often when we write
    $pmatrix{1 \ 2 \3}$ we mean that this represents some coordinates and actually must be interpreted as: $$1 cdot v_1 + 2 cdot v_2 +3 cdot v_3$$



    Where the $v_i$ are the basis vectors in which you represent this coordinate vector. Notice that the coefficients are definitely elements of my field $mathbb{F}$. This has nice properties, for instance, once you use some coordinate vectors for a given system you can show linear independence of that system by showing linear independence for the coordinate vectors.



    Also see






    share|cite|improve this answer





















    • So in my example x1, x2, x3 has to be members of R? if i am understanding correctly
      – 5Six
      2 hours ago
















    3












    3








    3






    Often when we write
    $pmatrix{1 \ 2 \3}$ we mean that this represents some coordinates and actually must be interpreted as: $$1 cdot v_1 + 2 cdot v_2 +3 cdot v_3$$



    Where the $v_i$ are the basis vectors in which you represent this coordinate vector. Notice that the coefficients are definitely elements of my field $mathbb{F}$. This has nice properties, for instance, once you use some coordinate vectors for a given system you can show linear independence of that system by showing linear independence for the coordinate vectors.



    Also see






    share|cite|improve this answer












    Often when we write
    $pmatrix{1 \ 2 \3}$ we mean that this represents some coordinates and actually must be interpreted as: $$1 cdot v_1 + 2 cdot v_2 +3 cdot v_3$$



    Where the $v_i$ are the basis vectors in which you represent this coordinate vector. Notice that the coefficients are definitely elements of my field $mathbb{F}$. This has nice properties, for instance, once you use some coordinate vectors for a given system you can show linear independence of that system by showing linear independence for the coordinate vectors.



    Also see







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 2 hours ago









    Wesley Strik

    1,536422




    1,536422












    • So in my example x1, x2, x3 has to be members of R? if i am understanding correctly
      – 5Six
      2 hours ago




















    • So in my example x1, x2, x3 has to be members of R? if i am understanding correctly
      – 5Six
      2 hours ago


















    So in my example x1, x2, x3 has to be members of R? if i am understanding correctly
    – 5Six
    2 hours ago






    So in my example x1, x2, x3 has to be members of R? if i am understanding correctly
    – 5Six
    2 hours ago













    1














    If I am interpreting your question correctly, the answer is that yes $mathbb{F}$ determines determines both.



    But I need to make sure I understand what you mean by "vector coefficients". I assume you are referring to those ocurring in linear combinations of vectors. So if $w$ and $v$ are vectors then $aw+bv$ is a linear combination of $w$ and $v$ with coefficients $a$ and $b$. Indeed, in general the "coefficients" here are assumed to be elements of your field $mathbb{F}$. Nobody uses the term "vector coefficients". Only the term coefficient is used, as these coefficients are understood to belong to your field of scalars.






    share|cite|improve this answer


























      1














      If I am interpreting your question correctly, the answer is that yes $mathbb{F}$ determines determines both.



      But I need to make sure I understand what you mean by "vector coefficients". I assume you are referring to those ocurring in linear combinations of vectors. So if $w$ and $v$ are vectors then $aw+bv$ is a linear combination of $w$ and $v$ with coefficients $a$ and $b$. Indeed, in general the "coefficients" here are assumed to be elements of your field $mathbb{F}$. Nobody uses the term "vector coefficients". Only the term coefficient is used, as these coefficients are understood to belong to your field of scalars.






      share|cite|improve this answer
























        1












        1








        1






        If I am interpreting your question correctly, the answer is that yes $mathbb{F}$ determines determines both.



        But I need to make sure I understand what you mean by "vector coefficients". I assume you are referring to those ocurring in linear combinations of vectors. So if $w$ and $v$ are vectors then $aw+bv$ is a linear combination of $w$ and $v$ with coefficients $a$ and $b$. Indeed, in general the "coefficients" here are assumed to be elements of your field $mathbb{F}$. Nobody uses the term "vector coefficients". Only the term coefficient is used, as these coefficients are understood to belong to your field of scalars.






        share|cite|improve this answer












        If I am interpreting your question correctly, the answer is that yes $mathbb{F}$ determines determines both.



        But I need to make sure I understand what you mean by "vector coefficients". I assume you are referring to those ocurring in linear combinations of vectors. So if $w$ and $v$ are vectors then $aw+bv$ is a linear combination of $w$ and $v$ with coefficients $a$ and $b$. Indeed, in general the "coefficients" here are assumed to be elements of your field $mathbb{F}$. Nobody uses the term "vector coefficients". Only the term coefficient is used, as these coefficients are understood to belong to your field of scalars.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        clmundergrad

        1765




        1765






















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