Vector space over a field
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
asked 2 hours ago
5Six
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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
asked 2 hours ago
5Six
184
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
1 hour 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
1 hour ago
add a comment |
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
asked 2 hours ago
5Six
184
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
asked 2 hours ago
5Six
184
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.
asked 2 hours ago
5Six
184
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.
edited 1 hour ago
asked 2 hours ago
5Six
184
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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.
asked 2 hours ago
5Six
184
asked 2 hours ago
5Six
184
184
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.
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.
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
1 hour 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
1 hour ago
add a comment |
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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
add a comment |
2 Answers
2
active
oldest
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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
answered 1 hour ago
Wesley Strik
1,536422
So in my example x1, x2, x3 has to be members of R? if i am understanding correctly
– 5Six
1 hour ago
add a comment |
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.
answered 1 hour ago
clmundergrad
1765
add a comment |
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2 Answers
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oldest
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2 Answers
2
active
oldest
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votes
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
answered 1 hour ago
Wesley Strik
1,536422
So in my example x1, x2, x3 has to be members of R? if i am understanding correctly
– 5Six
1 hour ago
add a comment |
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
answered 1 hour ago
Wesley Strik
1,536422
So in my example x1, x2, x3 has to be members of R? if i am understanding correctly
– 5Six
1 hour ago
add a comment |
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
answered 1 hour ago
Wesley Strik
1,536422
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
answered 1 hour ago
Wesley Strik
1,536422
answered 1 hour ago
Wesley Strik
1,536422
answered 1 hour ago
Wesley Strik
1,536422
answered 1 hour 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
1 hour ago
add a comment |
So in my example x1, x2, x3 has to be members of R? if i am understanding correctly
– 5Six
1 hour ago
So in my example x1, x2, x3 has to be members of R? if i am understanding correctly
– 5Six
1 hour ago
So in my example x1, x2, x3 has to be members of R? if i am understanding correctly
– 5Six
1 hour ago
add a comment |
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.
answered 1 hour ago
clmundergrad
1765
add a comment |
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.
answered 1 hour ago
clmundergrad
1765
add a comment |
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.
answered 1 hour ago
clmundergrad
1765
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.
answered 1 hour ago
clmundergrad
1765
answered 1 hour ago
clmundergrad
1765
answered 1 hour ago
clmundergrad
1765
answered 1 hour ago
clmundergrad
1765
1765
add a comment |
add a comment |
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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
1 hour 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
1 hour ago