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What is the significance of the fact that a set in a vector space is linearly dependent if it contains...
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A set $S$ in a vector space $V$ is linearly dependent if it contains finitely many linearly dependent vectors.
What is the significance of this definition? Obviously for finite sets, this is no different than the original definition. So what significance does it hold for infinite sets? I know it's not saying that infinite sets are always linearly independent, because that's not true (for example, the set of all real numbers, an infinite set, is linearly dependent).
Can you give me an example of a way in which this definition can be useful? Any help is appreciated!
linear-algebra vector-spaces
asked Jan 19 at 19:33
James RonaldJames Ronald
1297
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add a comment |
$begingroup$
A set $S$ in a vector space $V$ is linearly dependent if it contains finitely many linearly dependent vectors.
What is the significance of this definition? Obviously for finite sets, this is no different than the original definition. So what significance does it hold for infinite sets? I know it's not saying that infinite sets are always linearly independent, because that's not true (for example, the set of all real numbers, an infinite set, is linearly dependent).
Can you give me an example of a way in which this definition can be useful? Any help is appreciated!
linear-algebra vector-spaces
asked Jan 19 at 19:33
James RonaldJames Ronald
1297
$endgroup$
$begingroup$
An example where the result may be counterintuitive: over the vector space of continuous functions over $[0,1]$, the functions ${e^x,1,x,x^2,x^3,dots}$ are linearly independent. However, in a context where limits of sequences of functions are defined, we might say $$ e^x + sum_{k=0}^infty frac {-1}{n!} x^n = 0 $$ i.e. there is a non-trivial infinite linear combination such that $sum alpha_i v_i = 0$
$endgroup$
– Omnomnomnom
Jan 19 at 19:47
add a comment |
$begingroup$
A set $S$ in a vector space $V$ is linearly dependent if it contains finitely many linearly dependent vectors.
What is the significance of this definition? Obviously for finite sets, this is no different than the original definition. So what significance does it hold for infinite sets? I know it's not saying that infinite sets are always linearly independent, because that's not true (for example, the set of all real numbers, an infinite set, is linearly dependent).
Can you give me an example of a way in which this definition can be useful? Any help is appreciated!
linear-algebra vector-spaces
asked Jan 19 at 19:33
James RonaldJames Ronald
1297
$endgroup$
A set $S$ in a vector space $V$ is linearly dependent if it contains finitely many linearly dependent vectors.
What is the significance of this definition? Obviously for finite sets, this is no different than the original definition. So what significance does it hold for infinite sets? I know it's not saying that infinite sets are always linearly independent, because that's not true (for example, the set of all real numbers, an infinite set, is linearly dependent).
Can you give me an example of a way in which this definition can be useful? Any help is appreciated!
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Jan 19 at 19:33
James RonaldJames Ronald
1297
asked Jan 19 at 19:33
James RonaldJames Ronald
1297
asked Jan 19 at 19:33
James RonaldJames Ronald
1297
asked Jan 19 at 19:33
James RonaldJames Ronald
1297
asked Jan 19 at 19:33
James RonaldJames Ronald
1297
1297
$begingroup$
An example where the result may be counterintuitive: over the vector space of continuous functions over $[0,1]$, the functions ${e^x,1,x,x^2,x^3,dots}$ are linearly independent. However, in a context where limits of sequences of functions are defined, we might say $$ e^x + sum_{k=0}^infty frac {-1}{n!} x^n = 0 $$ i.e. there is a non-trivial infinite linear combination such that $sum alpha_i v_i = 0$
$endgroup$
– Omnomnomnom
Jan 19 at 19:47
add a comment |
$begingroup$
An example where the result may be counterintuitive: over the vector space of continuous functions over $[0,1]$, the functions ${e^x,1,x,x^2,x^3,dots}$ are linearly independent. However, in a context where limits of sequences of functions are defined, we might say $$ e^x + sum_{k=0}^infty frac {-1}{n!} x^n = 0 $$ i.e. there is a non-trivial infinite linear combination such that $sum alpha_i v_i = 0$
$endgroup$
– Omnomnomnom
Jan 19 at 19:47
$begingroup$
An example where the result may be counterintuitive: over the vector space of continuous functions over $[0,1]$, the functions ${e^x,1,x,x^2,x^3,dots}$ are linearly independent. However, in a context where limits of sequences of functions are defined, we might say $$ e^x + sum_{k=0}^infty frac {-1}{n!} x^n = 0 $$ i.e. there is a non-trivial infinite linear combination such that $sum alpha_i v_i = 0$
$endgroup$
– Omnomnomnom
Jan 19 at 19:47
$begingroup$
An example where the result may be counterintuitive: over the vector space of continuous functions over $[0,1]$, the functions ${e^x,1,x,x^2,x^3,dots}$ are linearly independent. However, in a context where limits of sequences of functions are defined, we might say $$ e^x + sum_{k=0}^infty frac {-1}{n!} x^n = 0 $$ i.e. there is a non-trivial infinite linear combination such that $sum alpha_i v_i = 0$
$endgroup$
– Omnomnomnom
Jan 19 at 19:47
add a comment |
2 Answers
2
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oldest
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I suppose that wherever you saw this, linear dependence is defined in two steps:
- It is defined first what is the meaning of saying that a finite subset $S$ of a vector space $V$ is linearly dependent.
- Then, it is stated that an infinite subset $S$ of a vector space $V$ is linearly dependent if it contains a finite subset $S^star$ which is linearly dependent.
In any case, a set $S$ is linearly dependent if and only of there elements $v_1,ldots,v_nin S$ and scalars $lambda_1,ldots,lambda_n$, not all of which are $0$, such that $lambda_1 v_1+cdots+lambda_nv_n=0$. This applies in both cases (that is, both when $S$ is finite and when $S$ is infinite).
answered Jan 19 at 19:42
José Carlos SantosJosé Carlos Santos
164k22132235
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add a comment |
$begingroup$
Recall that a finite set $lbrace x_1,dots ,x_nrbracesubset V$ (where $V$ is a vector space over some field $k$) is called linearly dependent if we can find $alpha_1,dots ,alpha_n in k$ with at least one $alpha_i neq 0$ such that
$$
sum_{i=1}^n alpha_i x_i = 0.
$$
On the other hand, this definition makes no sense for an infinite set since the sum of an infinite set of vectors is not defined (in an arbitrary vector space).
By defining a linearly dependent infinite set to be a set containing finitely many dependent vectors, infinite linearly independent sets retain the useful properties of their finite counterparts. For instance, any linearly independent set is contained in a basis (assuming Zorn's lemma).
answered Jan 19 at 19:42
o.h.o.h.
4616
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2 Answers
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2 Answers
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$begingroup$
I suppose that wherever you saw this, linear dependence is defined in two steps:
- It is defined first what is the meaning of saying that a finite subset $S$ of a vector space $V$ is linearly dependent.
- Then, it is stated that an infinite subset $S$ of a vector space $V$ is linearly dependent if it contains a finite subset $S^star$ which is linearly dependent.
In any case, a set $S$ is linearly dependent if and only of there elements $v_1,ldots,v_nin S$ and scalars $lambda_1,ldots,lambda_n$, not all of which are $0$, such that $lambda_1 v_1+cdots+lambda_nv_n=0$. This applies in both cases (that is, both when $S$ is finite and when $S$ is infinite).
answered Jan 19 at 19:42
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
add a comment |
$begingroup$
I suppose that wherever you saw this, linear dependence is defined in two steps:
- It is defined first what is the meaning of saying that a finite subset $S$ of a vector space $V$ is linearly dependent.
- Then, it is stated that an infinite subset $S$ of a vector space $V$ is linearly dependent if it contains a finite subset $S^star$ which is linearly dependent.
In any case, a set $S$ is linearly dependent if and only of there elements $v_1,ldots,v_nin S$ and scalars $lambda_1,ldots,lambda_n$, not all of which are $0$, such that $lambda_1 v_1+cdots+lambda_nv_n=0$. This applies in both cases (that is, both when $S$ is finite and when $S$ is infinite).
answered Jan 19 at 19:42
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
add a comment |
$begingroup$
I suppose that wherever you saw this, linear dependence is defined in two steps:
- It is defined first what is the meaning of saying that a finite subset $S$ of a vector space $V$ is linearly dependent.
- Then, it is stated that an infinite subset $S$ of a vector space $V$ is linearly dependent if it contains a finite subset $S^star$ which is linearly dependent.
In any case, a set $S$ is linearly dependent if and only of there elements $v_1,ldots,v_nin S$ and scalars $lambda_1,ldots,lambda_n$, not all of which are $0$, such that $lambda_1 v_1+cdots+lambda_nv_n=0$. This applies in both cases (that is, both when $S$ is finite and when $S$ is infinite).
answered Jan 19 at 19:42
José Carlos SantosJosé Carlos Santos
164k22132235
$endgroup$
I suppose that wherever you saw this, linear dependence is defined in two steps:
- It is defined first what is the meaning of saying that a finite subset $S$ of a vector space $V$ is linearly dependent.
- Then, it is stated that an infinite subset $S$ of a vector space $V$ is linearly dependent if it contains a finite subset $S^star$ which is linearly dependent.
In any case, a set $S$ is linearly dependent if and only of there elements $v_1,ldots,v_nin S$ and scalars $lambda_1,ldots,lambda_n$, not all of which are $0$, such that $lambda_1 v_1+cdots+lambda_nv_n=0$. This applies in both cases (that is, both when $S$ is finite and when $S$ is infinite).
answered Jan 19 at 19:42
José Carlos SantosJosé Carlos Santos
164k22132235
edited Jan 20 at 14:28
answered Jan 19 at 19:42
José Carlos SantosJosé Carlos Santos
164k22132235
answered Jan 19 at 19:42
José Carlos SantosJosé Carlos Santos
164k22132235
answered Jan 19 at 19:42
José Carlos SantosJosé Carlos Santos
164k22132235
164k22132235
add a comment |
add a comment |
$begingroup$
Recall that a finite set $lbrace x_1,dots ,x_nrbracesubset V$ (where $V$ is a vector space over some field $k$) is called linearly dependent if we can find $alpha_1,dots ,alpha_n in k$ with at least one $alpha_i neq 0$ such that
$$
sum_{i=1}^n alpha_i x_i = 0.
$$
On the other hand, this definition makes no sense for an infinite set since the sum of an infinite set of vectors is not defined (in an arbitrary vector space).
By defining a linearly dependent infinite set to be a set containing finitely many dependent vectors, infinite linearly independent sets retain the useful properties of their finite counterparts. For instance, any linearly independent set is contained in a basis (assuming Zorn's lemma).
answered Jan 19 at 19:42
o.h.o.h.
4616
$endgroup$
add a comment |
$begingroup$
Recall that a finite set $lbrace x_1,dots ,x_nrbracesubset V$ (where $V$ is a vector space over some field $k$) is called linearly dependent if we can find $alpha_1,dots ,alpha_n in k$ with at least one $alpha_i neq 0$ such that
$$
sum_{i=1}^n alpha_i x_i = 0.
$$
On the other hand, this definition makes no sense for an infinite set since the sum of an infinite set of vectors is not defined (in an arbitrary vector space).
By defining a linearly dependent infinite set to be a set containing finitely many dependent vectors, infinite linearly independent sets retain the useful properties of their finite counterparts. For instance, any linearly independent set is contained in a basis (assuming Zorn's lemma).
answered Jan 19 at 19:42
o.h.o.h.
4616
$endgroup$
add a comment |
$begingroup$
Recall that a finite set $lbrace x_1,dots ,x_nrbracesubset V$ (where $V$ is a vector space over some field $k$) is called linearly dependent if we can find $alpha_1,dots ,alpha_n in k$ with at least one $alpha_i neq 0$ such that
$$
sum_{i=1}^n alpha_i x_i = 0.
$$
On the other hand, this definition makes no sense for an infinite set since the sum of an infinite set of vectors is not defined (in an arbitrary vector space).
By defining a linearly dependent infinite set to be a set containing finitely many dependent vectors, infinite linearly independent sets retain the useful properties of their finite counterparts. For instance, any linearly independent set is contained in a basis (assuming Zorn's lemma).
answered Jan 19 at 19:42
o.h.o.h.
4616
$endgroup$
Recall that a finite set $lbrace x_1,dots ,x_nrbracesubset V$ (where $V$ is a vector space over some field $k$) is called linearly dependent if we can find $alpha_1,dots ,alpha_n in k$ with at least one $alpha_i neq 0$ such that
$$
sum_{i=1}^n alpha_i x_i = 0.
$$
On the other hand, this definition makes no sense for an infinite set since the sum of an infinite set of vectors is not defined (in an arbitrary vector space).
By defining a linearly dependent infinite set to be a set containing finitely many dependent vectors, infinite linearly independent sets retain the useful properties of their finite counterparts. For instance, any linearly independent set is contained in a basis (assuming Zorn's lemma).
answered Jan 19 at 19:42
o.h.o.h.
4616
edited Jan 19 at 19:52
answered Jan 19 at 19:42
o.h.o.h.
4616
answered Jan 19 at 19:42
o.h.o.h.
4616
answered Jan 19 at 19:42
o.h.o.h.
4616
4616
add a comment |
add a comment |
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$begingroup$
An example where the result may be counterintuitive: over the vector space of continuous functions over $[0,1]$, the functions ${e^x,1,x,x^2,x^3,dots}$ are linearly independent. However, in a context where limits of sequences of functions are defined, we might say $$ e^x + sum_{k=0}^infty frac {-1}{n!} x^n = 0 $$ i.e. there is a non-trivial infinite linear combination such that $sum alpha_i v_i = 0$
$endgroup$
– Omnomnomnom
Jan 19 at 19:47