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Can a set of vectors be linearly independent in one vector space, but be linearly dependent in another vector space?
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For example,
let S = x_1 in mathbb R$ be a subspace of $mathbb R^2$.
By definition, dim(S) = 1, and dim($mathbb R^2$) = 2.
Then the set (1, 1) only has one vector, so is it linearly independent in S, but is linearly dependent in $mathbb R^2$?
I know this doesn't make any sense, but we learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is in.
Since (1,1) is in both S and $mathbb R^2$, but the set (1, 1) only spans S, how come it is not only linearly independent in S and linearly dependent in $mathbb R^2$ (since (1,1) does not span $mathbb R^2$).
Sorry for this stupid question
linear-algebra
asked 1 hour ago
Tim Weah
484
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up vote
2
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For example,
let S = x_1 in mathbb R$ be a subspace of $mathbb R^2$.
By definition, dim(S) = 1, and dim($mathbb R^2$) = 2.
Then the set (1, 1) only has one vector, so is it linearly independent in S, but is linearly dependent in $mathbb R^2$?
I know this doesn't make any sense, but we learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is in.
Since (1,1) is in both S and $mathbb R^2$, but the set (1, 1) only spans S, how come it is not only linearly independent in S and linearly dependent in $mathbb R^2$ (since (1,1) does not span $mathbb R^2$).
Sorry for this stupid question
linear-algebra
asked 1 hour ago
Tim Weah
484
Vectors are linearly independent of other vectors in the same space so it doesn't make sense to discuss them in different spaces.
– CyclotomicField
1 hour ago
2
If the set $v_1,dots,v_n$ doesn't span $V$, then it need not be linearly dependent, unless $n=dim V$. In your case the set consists of one vector and $mathbbR^2$ has dimension $2$.
– egreg
1 hour ago
2
we learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is inYou must have misunderstood something. Two non-parallel vectors are linearly independent in $,Bbb R^3,$, for example, but they certainly don't span $,Bbb R^3,$.
– dxiv
1 hour ago
$ 1 $ is linearly independent in $mathbbR$ with the usual vector $mathbbR$-vector space structure; but if instead you use the vector space structure $x oplus y = x + y - 1$, $lambda odot x = lambda x - lambda + 1$ then $ 1 $ is linearly dependent with respect to this structure.
– Daniel Schepler
1 hour ago
On the other hand, if $W$ is a subspace of $V$ and $S subseteq W$, then $S$ is linearly independent as a subset of the vector space $W$ if and only if $S$ is linearly independent as a subset of the vector space $V$.
– Daniel Schepler
1 hour ago
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
For example,
let S = x_1 in mathbb R$ be a subspace of $mathbb R^2$.
By definition, dim(S) = 1, and dim($mathbb R^2$) = 2.
Then the set (1, 1) only has one vector, so is it linearly independent in S, but is linearly dependent in $mathbb R^2$?
I know this doesn't make any sense, but we learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is in.
Since (1,1) is in both S and $mathbb R^2$, but the set (1, 1) only spans S, how come it is not only linearly independent in S and linearly dependent in $mathbb R^2$ (since (1,1) does not span $mathbb R^2$).
Sorry for this stupid question
linear-algebra
asked 1 hour ago
Tim Weah
484
For example,
let S = x_1 in mathbb R$ be a subspace of $mathbb R^2$.
By definition, dim(S) = 1, and dim($mathbb R^2$) = 2.
Then the set (1, 1) only has one vector, so is it linearly independent in S, but is linearly dependent in $mathbb R^2$?
I know this doesn't make any sense, but we learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is in.
Since (1,1) is in both S and $mathbb R^2$, but the set (1, 1) only spans S, how come it is not only linearly independent in S and linearly dependent in $mathbb R^2$ (since (1,1) does not span $mathbb R^2$).
Sorry for this stupid question
linear-algebra
linear-algebra
asked 1 hour ago
Tim Weah
484
asked 1 hour ago
Tim Weah
484
asked 1 hour ago
Tim Weah
484
asked 1 hour ago
Tim Weah
484
asked 1 hour ago
Tim Weah
484
484
Vectors are linearly independent of other vectors in the same space so it doesn't make sense to discuss them in different spaces.
– CyclotomicField
1 hour ago
2
If the set $v_1,dots,v_n$ doesn't span $V$, then it need not be linearly dependent, unless $n=dim V$. In your case the set consists of one vector and $mathbbR^2$ has dimension $2$.
– egreg
1 hour ago
2
we learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is inYou must have misunderstood something. Two non-parallel vectors are linearly independent in $,Bbb R^3,$, for example, but they certainly don't span $,Bbb R^3,$.
– dxiv
1 hour ago
$ 1 $ is linearly independent in $mathbbR$ with the usual vector $mathbbR$-vector space structure; but if instead you use the vector space structure $x oplus y = x + y - 1$, $lambda odot x = lambda x - lambda + 1$ then $ 1 $ is linearly dependent with respect to this structure.
– Daniel Schepler
1 hour ago
On the other hand, if $W$ is a subspace of $V$ and $S subseteq W$, then $S$ is linearly independent as a subset of the vector space $W$ if and only if $S$ is linearly independent as a subset of the vector space $V$.
– Daniel Schepler
1 hour ago
 |Â
show 1 more comment
Vectors are linearly independent of other vectors in the same space so it doesn't make sense to discuss them in different spaces.
– CyclotomicField
1 hour ago
2
If the set $v_1,dots,v_n$ doesn't span $V$, then it need not be linearly dependent, unless $n=dim V$. In your case the set consists of one vector and $mathbbR^2$ has dimension $2$.
– egreg
1 hour ago
2
we learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is inYou must have misunderstood something. Two non-parallel vectors are linearly independent in $,Bbb R^3,$, for example, but they certainly don't span $,Bbb R^3,$.
– dxiv
1 hour ago
$ 1 $ is linearly independent in $mathbbR$ with the usual vector $mathbbR$-vector space structure; but if instead you use the vector space structure $x oplus y = x + y - 1$, $lambda odot x = lambda x - lambda + 1$ then $ 1 $ is linearly dependent with respect to this structure.
– Daniel Schepler
1 hour ago
On the other hand, if $W$ is a subspace of $V$ and $S subseteq W$, then $S$ is linearly independent as a subset of the vector space $W$ if and only if $S$ is linearly independent as a subset of the vector space $V$.
– Daniel Schepler
1 hour ago
Vectors are linearly independent of other vectors in the same space so it doesn't make sense to discuss them in different spaces.
– CyclotomicField
1 hour ago
Vectors are linearly independent of other vectors in the same space so it doesn't make sense to discuss them in different spaces.
– CyclotomicField
1 hour ago
2
2
If the set $v_1,dots,v_n$ doesn't span $V$, then it need not be linearly dependent, unless $n=dim V$. In your case the set consists of one vector and $mathbbR^2$ has dimension $2$.
– egreg
1 hour ago
If the set $v_1,dots,v_n$ doesn't span $V$, then it need not be linearly dependent, unless $n=dim V$. In your case the set consists of one vector and $mathbbR^2$ has dimension $2$.
– egreg
1 hour ago
2
2
we learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is in You must have misunderstood something. Two non-parallel vectors are linearly independent in $,Bbb R^3,$, for example, but they certainly don't span $,Bbb R^3,$.– dxiv
1 hour ago
we learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is in You must have misunderstood something. Two non-parallel vectors are linearly independent in $,Bbb R^3,$, for example, but they certainly don't span $,Bbb R^3,$.– dxiv
1 hour ago
$ 1 $ is linearly independent in $mathbbR$ with the usual vector $mathbbR$-vector space structure; but if instead you use the vector space structure $x oplus y = x + y - 1$, $lambda odot x = lambda x - lambda + 1$ then $ 1 $ is linearly dependent with respect to this structure.
– Daniel Schepler
1 hour ago
$ 1 $ is linearly independent in $mathbbR$ with the usual vector $mathbbR$-vector space structure; but if instead you use the vector space structure $x oplus y = x + y - 1$, $lambda odot x = lambda x - lambda + 1$ then $ 1 $ is linearly dependent with respect to this structure.
– Daniel Schepler
1 hour ago
On the other hand, if $W$ is a subspace of $V$ and $S subseteq W$, then $S$ is linearly independent as a subset of the vector space $W$ if and only if $S$ is linearly independent as a subset of the vector space $V$.
– Daniel Schepler
1 hour ago
On the other hand, if $W$ is a subspace of $V$ and $S subseteq W$, then $S$ is linearly independent as a subset of the vector space $W$ if and only if $S$ is linearly independent as a subset of the vector space $V$.
– Daniel Schepler
1 hour ago
 |Â
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5 Answers
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Suppose $v_1,v_2,dots,v_n$ is a basis of $V$. Then
$$
v_2,dots,v_n
$$
doesn't span $V$, but it is linearly independent.
A set of vectors may fail to span $V$, but it can still be linearly independent.
The only case when you can infer linear dependence from the fact that a set fails to span $V$ is when the set has the same number of elements as the dimension of $V$.
In your case this condition is not satisfied, because $(1,1)$ consists of one element, but the dimension of $mathbbR^2$ is two.
answered 1 hour ago
egreg
167k1281190
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You said you “learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is in.†This isn’t correct, unless “the vector space that it is in†means “the smallest vector space that it is in,†which would not be a typical reader’s understanding. But even with that understanding, the statement is not useful, because the fact is not special to linearly independent sets. Every set of vectors spans the smallest vector space that contains them: a set of vectors spans its span. (That’s practically the definition of span.)
I suspect what you were supposed to learn in class was that “A set of vectors in a vector space $V$ can only be a basis for $V$ if the set spans $V$.†(In addition, it must be a linearly independent set of vectors.)
In particular, the set $(1,1)$ is a linearly independent set, whether it is considered as a set of vectors in $mathbb R^2$ or as a set of vectors in what you call $S$. The fact that $(1,1)$ does not span $mathbb R^2$ does not tell you anything about the linear independence of the set. (And by the way, any set containing only one nonzero vector is a linearly independent set of vectors.)
answered 1 hour ago
Steve Kass
10.5k11428
Slight correction in first paragraph: A set is not linearly independent if it spans the smallest space that it's contained in...
– Chris Custer
46 mins ago
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0
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As egreg explained, your example is wrong.
Suppose $V$ and $W$ are both subspaces of a vector space $X$.
Then any set of vectors $v_1, ldots, v_k$ in the intersection $V cap W$ that is linearly dependent as a subset of vector space $V$ is also linearly dependent as a subset of $W$. This is because both linear dependence statements are equivalent to the existence of scalars $c_1, ldots, c_k$, not all $0$, such that $c_1 v_1 + ldots + c_k v_k = 0$.
answered 1 hour ago
Robert Israel
308k22201444
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Firstly, the definition
... that a set of vectors can only be linearly independent if it spans the vector space that it is in.
is not correct.
A set $S$ of vectors is linearly dependent if there is some $v in S$ that can be written as a linear combination of other vectors from $S$. It is linearly independent otherwise.
More formally, a set of vectors $S = v_1,ldots,v_k$ is linearly independent if
$$sum_i=1^kalpha_iv_i = 0 quad Rightarrow quad alpha_1= cdots = alpha_k=0$$
That is, the only way we can make zero from vectors in $S$ is by multiplying every vector by zero.
It is easy to see that if we remove vectors from linearly independent $S$, it stays linearly independent. Conversely, if we add a vector to $S$, it may become dependent. The maximum cardinality (number of elements) of a linearly independent set of vectors in a $n$-dimensional vectors space $V$ is $n$ and this set is then called a basis for $V$. It has the property that it spans entire $V$, that is - every $v in V$ can be represented as a linear combination of vectors from basis.
Note that the cardinality of a set of vectors tells us nothing precise about its linear (in)dependence.
example 1: $(1,1),(2,2)$ is linearly dependent in $mathbbR^2$ because $2 cdot (1,1) = (2,2)$.
example 2: $(1,1)$ is linearly independent in $mathbbR^2$ but not a basis for $mathbbR^2$ because $|S| = 1 neq 2 = dim mathbbR^2$..
example 3: $(1,1),(2,3)$ is linearly independent in $mathbbR^2$ and also a basis for $mathbbR^2$ because $|S| = 2 = dim mathbbR^2$.
example 4: $(1,1),(2,3),v$ must be linearly dependent in $mathbbR^2$, for any $v in mathbbR^2$.
example 5: $(1,1)$ is linearly independent in your space $S := (x_1,x_1)$ and a basis for it.
answered 1 hour ago
Sandro LovniÄÂki
20615
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I think it's safe to say that if a set of vectors is linearly independent in one space, then it will be in any space that contains it.
This seems clear for any superspace, $Wsupset V$...
Also, when a different space contains our space... For instance, the spaces $V=x$-axis, $W=xy$-plane and $U=xz$-plane. We simply get that $(1,0)$ is linearly independent in (actually) all three spaces.
The reason is that the defining condition, namely that only the trivial linear combination is zero, is independent of the surrounding space...
One could say, for instance, that a set is linearly independent $iff$ it is a basis for the space it spans. Again, notice there is no mention of any (larger) surrounding space.
As pointed out by @egreg, if there are $n$ vectors, and if $operatornamedim V=n$, then it is true (that the set is linearly independent iff it spans $V$).
In light of @Daniel Schepler's comments, we should note that changing the base field, or the vector space "structure", we can get some different results. This shows your question may be a little better than you originally thought (though it was an error that led you to it)...
answered 57 mins ago
Chris Custer
6,5992622
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5 Answers
5
active
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5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Suppose $v_1,v_2,dots,v_n$ is a basis of $V$. Then
$$
v_2,dots,v_n
$$
doesn't span $V$, but it is linearly independent.
A set of vectors may fail to span $V$, but it can still be linearly independent.
The only case when you can infer linear dependence from the fact that a set fails to span $V$ is when the set has the same number of elements as the dimension of $V$.
In your case this condition is not satisfied, because $(1,1)$ consists of one element, but the dimension of $mathbbR^2$ is two.
answered 1 hour ago
egreg
167k1281190
add a comment |Â
up vote
2
down vote
Suppose $v_1,v_2,dots,v_n$ is a basis of $V$. Then
$$
v_2,dots,v_n
$$
doesn't span $V$, but it is linearly independent.
A set of vectors may fail to span $V$, but it can still be linearly independent.
The only case when you can infer linear dependence from the fact that a set fails to span $V$ is when the set has the same number of elements as the dimension of $V$.
In your case this condition is not satisfied, because $(1,1)$ consists of one element, but the dimension of $mathbbR^2$ is two.
answered 1 hour ago
egreg
167k1281190
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Suppose $v_1,v_2,dots,v_n$ is a basis of $V$. Then
$$
v_2,dots,v_n
$$
doesn't span $V$, but it is linearly independent.
A set of vectors may fail to span $V$, but it can still be linearly independent.
The only case when you can infer linear dependence from the fact that a set fails to span $V$ is when the set has the same number of elements as the dimension of $V$.
In your case this condition is not satisfied, because $(1,1)$ consists of one element, but the dimension of $mathbbR^2$ is two.
answered 1 hour ago
egreg
167k1281190
Suppose $v_1,v_2,dots,v_n$ is a basis of $V$. Then
$$
v_2,dots,v_n
$$
doesn't span $V$, but it is linearly independent.
A set of vectors may fail to span $V$, but it can still be linearly independent.
The only case when you can infer linear dependence from the fact that a set fails to span $V$ is when the set has the same number of elements as the dimension of $V$.
In your case this condition is not satisfied, because $(1,1)$ consists of one element, but the dimension of $mathbbR^2$ is two.
answered 1 hour ago
egreg
167k1281190
answered 1 hour ago
egreg
167k1281190
answered 1 hour ago
egreg
167k1281190
answered 1 hour ago
egreg
167k1281190
167k1281190
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add a comment |Â
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2
down vote
You said you “learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is in.†This isn’t correct, unless “the vector space that it is in†means “the smallest vector space that it is in,†which would not be a typical reader’s understanding. But even with that understanding, the statement is not useful, because the fact is not special to linearly independent sets. Every set of vectors spans the smallest vector space that contains them: a set of vectors spans its span. (That’s practically the definition of span.)
I suspect what you were supposed to learn in class was that “A set of vectors in a vector space $V$ can only be a basis for $V$ if the set spans $V$.†(In addition, it must be a linearly independent set of vectors.)
In particular, the set $(1,1)$ is a linearly independent set, whether it is considered as a set of vectors in $mathbb R^2$ or as a set of vectors in what you call $S$. The fact that $(1,1)$ does not span $mathbb R^2$ does not tell you anything about the linear independence of the set. (And by the way, any set containing only one nonzero vector is a linearly independent set of vectors.)
answered 1 hour ago
Steve Kass
10.5k11428
Slight correction in first paragraph: A set is not linearly independent if it spans the smallest space that it's contained in...
– Chris Custer
46 mins ago
add a comment |Â
up vote
2
down vote
You said you “learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is in.†This isn’t correct, unless “the vector space that it is in†means “the smallest vector space that it is in,†which would not be a typical reader’s understanding. But even with that understanding, the statement is not useful, because the fact is not special to linearly independent sets. Every set of vectors spans the smallest vector space that contains them: a set of vectors spans its span. (That’s practically the definition of span.)
I suspect what you were supposed to learn in class was that “A set of vectors in a vector space $V$ can only be a basis for $V$ if the set spans $V$.†(In addition, it must be a linearly independent set of vectors.)
In particular, the set $(1,1)$ is a linearly independent set, whether it is considered as a set of vectors in $mathbb R^2$ or as a set of vectors in what you call $S$. The fact that $(1,1)$ does not span $mathbb R^2$ does not tell you anything about the linear independence of the set. (And by the way, any set containing only one nonzero vector is a linearly independent set of vectors.)
answered 1 hour ago
Steve Kass
10.5k11428
Slight correction in first paragraph: A set is not linearly independent if it spans the smallest space that it's contained in...
– Chris Custer
46 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You said you “learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is in.†This isn’t correct, unless “the vector space that it is in†means “the smallest vector space that it is in,†which would not be a typical reader’s understanding. But even with that understanding, the statement is not useful, because the fact is not special to linearly independent sets. Every set of vectors spans the smallest vector space that contains them: a set of vectors spans its span. (That’s practically the definition of span.)
I suspect what you were supposed to learn in class was that “A set of vectors in a vector space $V$ can only be a basis for $V$ if the set spans $V$.†(In addition, it must be a linearly independent set of vectors.)
In particular, the set $(1,1)$ is a linearly independent set, whether it is considered as a set of vectors in $mathbb R^2$ or as a set of vectors in what you call $S$. The fact that $(1,1)$ does not span $mathbb R^2$ does not tell you anything about the linear independence of the set. (And by the way, any set containing only one nonzero vector is a linearly independent set of vectors.)
answered 1 hour ago
Steve Kass
10.5k11428
You said you “learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is in.†This isn’t correct, unless “the vector space that it is in†means “the smallest vector space that it is in,†which would not be a typical reader’s understanding. But even with that understanding, the statement is not useful, because the fact is not special to linearly independent sets. Every set of vectors spans the smallest vector space that contains them: a set of vectors spans its span. (That’s practically the definition of span.)
I suspect what you were supposed to learn in class was that “A set of vectors in a vector space $V$ can only be a basis for $V$ if the set spans $V$.†(In addition, it must be a linearly independent set of vectors.)
In particular, the set $(1,1)$ is a linearly independent set, whether it is considered as a set of vectors in $mathbb R^2$ or as a set of vectors in what you call $S$. The fact that $(1,1)$ does not span $mathbb R^2$ does not tell you anything about the linear independence of the set. (And by the way, any set containing only one nonzero vector is a linearly independent set of vectors.)
answered 1 hour ago
Steve Kass
10.5k11428
answered 1 hour ago
Steve Kass
10.5k11428
answered 1 hour ago
Steve Kass
10.5k11428
answered 1 hour ago
Steve Kass
10.5k11428
10.5k11428
Slight correction in first paragraph: A set is not linearly independent if it spans the smallest space that it's contained in...
– Chris Custer
46 mins ago
add a comment |Â
Slight correction in first paragraph: A set is not linearly independent if it spans the smallest space that it's contained in...
– Chris Custer
46 mins ago
Slight correction in first paragraph: A set is not linearly independent if it spans the smallest space that it's contained in...
– Chris Custer
46 mins ago
Slight correction in first paragraph: A set is not linearly independent if it spans the smallest space that it's contained in...
– Chris Custer
46 mins ago
add a comment |Â
up vote
0
down vote
As egreg explained, your example is wrong.
Suppose $V$ and $W$ are both subspaces of a vector space $X$.
Then any set of vectors $v_1, ldots, v_k$ in the intersection $V cap W$ that is linearly dependent as a subset of vector space $V$ is also linearly dependent as a subset of $W$. This is because both linear dependence statements are equivalent to the existence of scalars $c_1, ldots, c_k$, not all $0$, such that $c_1 v_1 + ldots + c_k v_k = 0$.
answered 1 hour ago
Robert Israel
308k22201444
add a comment |Â
up vote
0
down vote
As egreg explained, your example is wrong.
Suppose $V$ and $W$ are both subspaces of a vector space $X$.
Then any set of vectors $v_1, ldots, v_k$ in the intersection $V cap W$ that is linearly dependent as a subset of vector space $V$ is also linearly dependent as a subset of $W$. This is because both linear dependence statements are equivalent to the existence of scalars $c_1, ldots, c_k$, not all $0$, such that $c_1 v_1 + ldots + c_k v_k = 0$.
answered 1 hour ago
Robert Israel
308k22201444
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As egreg explained, your example is wrong.
Suppose $V$ and $W$ are both subspaces of a vector space $X$.
Then any set of vectors $v_1, ldots, v_k$ in the intersection $V cap W$ that is linearly dependent as a subset of vector space $V$ is also linearly dependent as a subset of $W$. This is because both linear dependence statements are equivalent to the existence of scalars $c_1, ldots, c_k$, not all $0$, such that $c_1 v_1 + ldots + c_k v_k = 0$.
answered 1 hour ago
Robert Israel
308k22201444
As egreg explained, your example is wrong.
Suppose $V$ and $W$ are both subspaces of a vector space $X$.
Then any set of vectors $v_1, ldots, v_k$ in the intersection $V cap W$ that is linearly dependent as a subset of vector space $V$ is also linearly dependent as a subset of $W$. This is because both linear dependence statements are equivalent to the existence of scalars $c_1, ldots, c_k$, not all $0$, such that $c_1 v_1 + ldots + c_k v_k = 0$.
answered 1 hour ago
Robert Israel
308k22201444
answered 1 hour ago
Robert Israel
308k22201444
answered 1 hour ago
Robert Israel
308k22201444
answered 1 hour ago
Robert Israel
308k22201444
308k22201444
add a comment |Â
add a comment |Â
up vote
0
down vote
Firstly, the definition
... that a set of vectors can only be linearly independent if it spans the vector space that it is in.
is not correct.
A set $S$ of vectors is linearly dependent if there is some $v in S$ that can be written as a linear combination of other vectors from $S$. It is linearly independent otherwise.
More formally, a set of vectors $S = v_1,ldots,v_k$ is linearly independent if
$$sum_i=1^kalpha_iv_i = 0 quad Rightarrow quad alpha_1= cdots = alpha_k=0$$
That is, the only way we can make zero from vectors in $S$ is by multiplying every vector by zero.
It is easy to see that if we remove vectors from linearly independent $S$, it stays linearly independent. Conversely, if we add a vector to $S$, it may become dependent. The maximum cardinality (number of elements) of a linearly independent set of vectors in a $n$-dimensional vectors space $V$ is $n$ and this set is then called a basis for $V$. It has the property that it spans entire $V$, that is - every $v in V$ can be represented as a linear combination of vectors from basis.
Note that the cardinality of a set of vectors tells us nothing precise about its linear (in)dependence.
example 1: $(1,1),(2,2)$ is linearly dependent in $mathbbR^2$ because $2 cdot (1,1) = (2,2)$.
example 2: $(1,1)$ is linearly independent in $mathbbR^2$ but not a basis for $mathbbR^2$ because $|S| = 1 neq 2 = dim mathbbR^2$..
example 3: $(1,1),(2,3)$ is linearly independent in $mathbbR^2$ and also a basis for $mathbbR^2$ because $|S| = 2 = dim mathbbR^2$.
example 4: $(1,1),(2,3),v$ must be linearly dependent in $mathbbR^2$, for any $v in mathbbR^2$.
example 5: $(1,1)$ is linearly independent in your space $S := (x_1,x_1)$ and a basis for it.
answered 1 hour ago
Sandro LovniÄÂki
20615
add a comment |Â
up vote
0
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Firstly, the definition
... that a set of vectors can only be linearly independent if it spans the vector space that it is in.
is not correct.
A set $S$ of vectors is linearly dependent if there is some $v in S$ that can be written as a linear combination of other vectors from $S$. It is linearly independent otherwise.
More formally, a set of vectors $S = v_1,ldots,v_k$ is linearly independent if
$$sum_i=1^kalpha_iv_i = 0 quad Rightarrow quad alpha_1= cdots = alpha_k=0$$
That is, the only way we can make zero from vectors in $S$ is by multiplying every vector by zero.
It is easy to see that if we remove vectors from linearly independent $S$, it stays linearly independent. Conversely, if we add a vector to $S$, it may become dependent. The maximum cardinality (number of elements) of a linearly independent set of vectors in a $n$-dimensional vectors space $V$ is $n$ and this set is then called a basis for $V$. It has the property that it spans entire $V$, that is - every $v in V$ can be represented as a linear combination of vectors from basis.
Note that the cardinality of a set of vectors tells us nothing precise about its linear (in)dependence.
example 1: $(1,1),(2,2)$ is linearly dependent in $mathbbR^2$ because $2 cdot (1,1) = (2,2)$.
example 2: $(1,1)$ is linearly independent in $mathbbR^2$ but not a basis for $mathbbR^2$ because $|S| = 1 neq 2 = dim mathbbR^2$..
example 3: $(1,1),(2,3)$ is linearly independent in $mathbbR^2$ and also a basis for $mathbbR^2$ because $|S| = 2 = dim mathbbR^2$.
example 4: $(1,1),(2,3),v$ must be linearly dependent in $mathbbR^2$, for any $v in mathbbR^2$.
example 5: $(1,1)$ is linearly independent in your space $S := (x_1,x_1)$ and a basis for it.
answered 1 hour ago
Sandro LovniÄÂki
20615
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Firstly, the definition
... that a set of vectors can only be linearly independent if it spans the vector space that it is in.
is not correct.
A set $S$ of vectors is linearly dependent if there is some $v in S$ that can be written as a linear combination of other vectors from $S$. It is linearly independent otherwise.
More formally, a set of vectors $S = v_1,ldots,v_k$ is linearly independent if
$$sum_i=1^kalpha_iv_i = 0 quad Rightarrow quad alpha_1= cdots = alpha_k=0$$
That is, the only way we can make zero from vectors in $S$ is by multiplying every vector by zero.
It is easy to see that if we remove vectors from linearly independent $S$, it stays linearly independent. Conversely, if we add a vector to $S$, it may become dependent. The maximum cardinality (number of elements) of a linearly independent set of vectors in a $n$-dimensional vectors space $V$ is $n$ and this set is then called a basis for $V$. It has the property that it spans entire $V$, that is - every $v in V$ can be represented as a linear combination of vectors from basis.
Note that the cardinality of a set of vectors tells us nothing precise about its linear (in)dependence.
example 1: $(1,1),(2,2)$ is linearly dependent in $mathbbR^2$ because $2 cdot (1,1) = (2,2)$.
example 2: $(1,1)$ is linearly independent in $mathbbR^2$ but not a basis for $mathbbR^2$ because $|S| = 1 neq 2 = dim mathbbR^2$..
example 3: $(1,1),(2,3)$ is linearly independent in $mathbbR^2$ and also a basis for $mathbbR^2$ because $|S| = 2 = dim mathbbR^2$.
example 4: $(1,1),(2,3),v$ must be linearly dependent in $mathbbR^2$, for any $v in mathbbR^2$.
example 5: $(1,1)$ is linearly independent in your space $S := (x_1,x_1)$ and a basis for it.
answered 1 hour ago
Sandro LovniÄÂki
20615
Firstly, the definition
... that a set of vectors can only be linearly independent if it spans the vector space that it is in.
is not correct.
A set $S$ of vectors is linearly dependent if there is some $v in S$ that can be written as a linear combination of other vectors from $S$. It is linearly independent otherwise.
More formally, a set of vectors $S = v_1,ldots,v_k$ is linearly independent if
$$sum_i=1^kalpha_iv_i = 0 quad Rightarrow quad alpha_1= cdots = alpha_k=0$$
That is, the only way we can make zero from vectors in $S$ is by multiplying every vector by zero.
It is easy to see that if we remove vectors from linearly independent $S$, it stays linearly independent. Conversely, if we add a vector to $S$, it may become dependent. The maximum cardinality (number of elements) of a linearly independent set of vectors in a $n$-dimensional vectors space $V$ is $n$ and this set is then called a basis for $V$. It has the property that it spans entire $V$, that is - every $v in V$ can be represented as a linear combination of vectors from basis.
Note that the cardinality of a set of vectors tells us nothing precise about its linear (in)dependence.
example 1: $(1,1),(2,2)$ is linearly dependent in $mathbbR^2$ because $2 cdot (1,1) = (2,2)$.
example 2: $(1,1)$ is linearly independent in $mathbbR^2$ but not a basis for $mathbbR^2$ because $|S| = 1 neq 2 = dim mathbbR^2$..
example 3: $(1,1),(2,3)$ is linearly independent in $mathbbR^2$ and also a basis for $mathbbR^2$ because $|S| = 2 = dim mathbbR^2$.
example 4: $(1,1),(2,3),v$ must be linearly dependent in $mathbbR^2$, for any $v in mathbbR^2$.
example 5: $(1,1)$ is linearly independent in your space $S := (x_1,x_1)$ and a basis for it.
answered 1 hour ago
Sandro LovniÄÂki
20615
edited 1 hour ago
answered 1 hour ago
Sandro LovniÄÂki
20615
answered 1 hour ago
Sandro LovniÄÂki
20615
answered 1 hour ago
Sandro LovniÄÂki
20615
20615
add a comment |Â
add a comment |Â
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0
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I think it's safe to say that if a set of vectors is linearly independent in one space, then it will be in any space that contains it.
This seems clear for any superspace, $Wsupset V$...
Also, when a different space contains our space... For instance, the spaces $V=x$-axis, $W=xy$-plane and $U=xz$-plane. We simply get that $(1,0)$ is linearly independent in (actually) all three spaces.
The reason is that the defining condition, namely that only the trivial linear combination is zero, is independent of the surrounding space...
One could say, for instance, that a set is linearly independent $iff$ it is a basis for the space it spans. Again, notice there is no mention of any (larger) surrounding space.
As pointed out by @egreg, if there are $n$ vectors, and if $operatornamedim V=n$, then it is true (that the set is linearly independent iff it spans $V$).
In light of @Daniel Schepler's comments, we should note that changing the base field, or the vector space "structure", we can get some different results. This shows your question may be a little better than you originally thought (though it was an error that led you to it)...
answered 57 mins ago
Chris Custer
6,5992622
add a comment |Â
up vote
0
down vote
I think it's safe to say that if a set of vectors is linearly independent in one space, then it will be in any space that contains it.
This seems clear for any superspace, $Wsupset V$...
Also, when a different space contains our space... For instance, the spaces $V=x$-axis, $W=xy$-plane and $U=xz$-plane. We simply get that $(1,0)$ is linearly independent in (actually) all three spaces.
The reason is that the defining condition, namely that only the trivial linear combination is zero, is independent of the surrounding space...
One could say, for instance, that a set is linearly independent $iff$ it is a basis for the space it spans. Again, notice there is no mention of any (larger) surrounding space.
As pointed out by @egreg, if there are $n$ vectors, and if $operatornamedim V=n$, then it is true (that the set is linearly independent iff it spans $V$).
In light of @Daniel Schepler's comments, we should note that changing the base field, or the vector space "structure", we can get some different results. This shows your question may be a little better than you originally thought (though it was an error that led you to it)...
answered 57 mins ago
Chris Custer
6,5992622
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think it's safe to say that if a set of vectors is linearly independent in one space, then it will be in any space that contains it.
This seems clear for any superspace, $Wsupset V$...
Also, when a different space contains our space... For instance, the spaces $V=x$-axis, $W=xy$-plane and $U=xz$-plane. We simply get that $(1,0)$ is linearly independent in (actually) all three spaces.
The reason is that the defining condition, namely that only the trivial linear combination is zero, is independent of the surrounding space...
One could say, for instance, that a set is linearly independent $iff$ it is a basis for the space it spans. Again, notice there is no mention of any (larger) surrounding space.
As pointed out by @egreg, if there are $n$ vectors, and if $operatornamedim V=n$, then it is true (that the set is linearly independent iff it spans $V$).
In light of @Daniel Schepler's comments, we should note that changing the base field, or the vector space "structure", we can get some different results. This shows your question may be a little better than you originally thought (though it was an error that led you to it)...
answered 57 mins ago
Chris Custer
6,5992622
I think it's safe to say that if a set of vectors is linearly independent in one space, then it will be in any space that contains it.
This seems clear for any superspace, $Wsupset V$...
Also, when a different space contains our space... For instance, the spaces $V=x$-axis, $W=xy$-plane and $U=xz$-plane. We simply get that $(1,0)$ is linearly independent in (actually) all three spaces.
The reason is that the defining condition, namely that only the trivial linear combination is zero, is independent of the surrounding space...
One could say, for instance, that a set is linearly independent $iff$ it is a basis for the space it spans. Again, notice there is no mention of any (larger) surrounding space.
As pointed out by @egreg, if there are $n$ vectors, and if $operatornamedim V=n$, then it is true (that the set is linearly independent iff it spans $V$).
In light of @Daniel Schepler's comments, we should note that changing the base field, or the vector space "structure", we can get some different results. This shows your question may be a little better than you originally thought (though it was an error that led you to it)...
answered 57 mins ago
Chris Custer
6,5992622
edited 11 mins ago
answered 57 mins ago
Chris Custer
6,5992622
answered 57 mins ago
Chris Custer
6,5992622
answered 57 mins ago
Chris Custer
6,5992622
6,5992622
add a comment |Â
add a comment |Â
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Vectors are linearly independent of other vectors in the same space so it doesn't make sense to discuss them in different spaces.
– CyclotomicField
1 hour ago
2
If the set $v_1,dots,v_n$ doesn't span $V$, then it need not be linearly dependent, unless $n=dim V$. In your case the set consists of one vector and $mathbbR^2$ has dimension $2$.
– egreg
1 hour ago
2
we learned in class that a set of vectors can only be linearly independent if it spans the vector space that it is inYou must have misunderstood something. Two non-parallel vectors are linearly independent in $,Bbb R^3,$, for example, but they certainly don't span $,Bbb R^3,$.– dxiv
1 hour ago
$ 1 $ is linearly independent in $mathbbR$ with the usual vector $mathbbR$-vector space structure; but if instead you use the vector space structure $x oplus y = x + y - 1$, $lambda odot x = lambda x - lambda + 1$ then $ 1 $ is linearly dependent with respect to this structure.
– Daniel Schepler
1 hour ago
On the other hand, if $W$ is a subspace of $V$ and $S subseteq W$, then $S$ is linearly independent as a subset of the vector space $W$ if and only if $S$ is linearly independent as a subset of the vector space $V$.
– Daniel Schepler
1 hour ago