Mixing
Rank, dimension, basis
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I think I am a little bit confused with the terms in the title, so I hope you can correct me if I got it wrong...
$$
leftlbrace beginbmatrix x_1\0\0 endbmatrix : x_1 in mathbbR rightrbrace
$$ is a vector space. So far so good.
The dimension of the vector space is the number of vectors in the basis.
In class, I wrote down that a basis is
$$
B=(beginbmatrix1\0\0endbmatrix, beginbmatrix0\0\0endbmatrix, beginbmatrix0\0\0endbmatrix)
$$
But this does not seem logical to me now, as it should be the smallest number of independent vectors that span the vector space. And shouldn’t that be just
$
(beginbmatrix1\0\0endbmatrix)
$
So this is the place where I'm not sure: what is the dimension of the vector space? It is a line in $mathbb R^3$, and it would seem logical that the dimension of a line is $1$.
And what would the rank be? I totally confused myself.
Or is it like the dimension is $3$ and the rank is $1$. so the above solution for the basis would be correct...
Many thanks for your help
linear-algebra
edited 3 hours ago
user587192
7187
asked 4 hours ago
Lillys
236
add a comment |Â
up vote
2
down vote
favorite
I think I am a little bit confused with the terms in the title, so I hope you can correct me if I got it wrong...
$$
leftlbrace beginbmatrix x_1\0\0 endbmatrix : x_1 in mathbbR rightrbrace
$$ is a vector space. So far so good.
The dimension of the vector space is the number of vectors in the basis.
In class, I wrote down that a basis is
$$
B=(beginbmatrix1\0\0endbmatrix, beginbmatrix0\0\0endbmatrix, beginbmatrix0\0\0endbmatrix)
$$
But this does not seem logical to me now, as it should be the smallest number of independent vectors that span the vector space. And shouldn’t that be just
$
(beginbmatrix1\0\0endbmatrix)
$
So this is the place where I'm not sure: what is the dimension of the vector space? It is a line in $mathbb R^3$, and it would seem logical that the dimension of a line is $1$.
And what would the rank be? I totally confused myself.
Or is it like the dimension is $3$ and the rank is $1$. so the above solution for the basis would be correct...
Many thanks for your help
linear-algebra
edited 3 hours ago
user587192
7187
asked 4 hours ago
Lillys
236
A basis must be linearly independent, so it can't contain the zero vector. Your $B$ does (twice).
– Bungo
4 hours ago
Yes sorry i will add it
– Lillys
4 hours ago
"it should be the smallest number...shouldn't that just be $left(left[beginsmallmatrix1\0\0endsmallmatrixright]right)$" Yes, it should just be that. As mentioned, the zero vectors should not have been included. The dimension of the subspace spanned by your one vector is indeed one, despite the fact that the parent space is three-dimensional. When talking about dimension or percentage or whatever, you have to keep in mind "dimension of what", here we are talking about dimension of the subspace, not of the parent space.
– JMoravitz
4 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I think I am a little bit confused with the terms in the title, so I hope you can correct me if I got it wrong...
$$
leftlbrace beginbmatrix x_1\0\0 endbmatrix : x_1 in mathbbR rightrbrace
$$ is a vector space. So far so good.
The dimension of the vector space is the number of vectors in the basis.
In class, I wrote down that a basis is
$$
B=(beginbmatrix1\0\0endbmatrix, beginbmatrix0\0\0endbmatrix, beginbmatrix0\0\0endbmatrix)
$$
But this does not seem logical to me now, as it should be the smallest number of independent vectors that span the vector space. And shouldn’t that be just
$
(beginbmatrix1\0\0endbmatrix)
$
So this is the place where I'm not sure: what is the dimension of the vector space? It is a line in $mathbb R^3$, and it would seem logical that the dimension of a line is $1$.
And what would the rank be? I totally confused myself.
Or is it like the dimension is $3$ and the rank is $1$. so the above solution for the basis would be correct...
Many thanks for your help
linear-algebra
edited 3 hours ago
user587192
7187
asked 4 hours ago
Lillys
236
I think I am a little bit confused with the terms in the title, so I hope you can correct me if I got it wrong...
$$
leftlbrace beginbmatrix x_1\0\0 endbmatrix : x_1 in mathbbR rightrbrace
$$ is a vector space. So far so good.
The dimension of the vector space is the number of vectors in the basis.
In class, I wrote down that a basis is
$$
B=(beginbmatrix1\0\0endbmatrix, beginbmatrix0\0\0endbmatrix, beginbmatrix0\0\0endbmatrix)
$$
But this does not seem logical to me now, as it should be the smallest number of independent vectors that span the vector space. And shouldn’t that be just
$
(beginbmatrix1\0\0endbmatrix)
$
So this is the place where I'm not sure: what is the dimension of the vector space? It is a line in $mathbb R^3$, and it would seem logical that the dimension of a line is $1$.
And what would the rank be? I totally confused myself.
Or is it like the dimension is $3$ and the rank is $1$. so the above solution for the basis would be correct...
Many thanks for your help
linear-algebra
linear-algebra
edited 3 hours ago
user587192
7187
asked 4 hours ago
Lillys
236
edited 3 hours ago
user587192
7187
asked 4 hours ago
Lillys
236
edited 3 hours ago
user587192
7187
edited 3 hours ago
user587192
7187
edited 3 hours ago
user587192
7187
7187
asked 4 hours ago
Lillys
236
asked 4 hours ago
Lillys
236
asked 4 hours ago
Lillys
236
236
A basis must be linearly independent, so it can't contain the zero vector. Your $B$ does (twice).
– Bungo
4 hours ago
Yes sorry i will add it
– Lillys
4 hours ago
"it should be the smallest number...shouldn't that just be $left(left[beginsmallmatrix1\0\0endsmallmatrixright]right)$" Yes, it should just be that. As mentioned, the zero vectors should not have been included. The dimension of the subspace spanned by your one vector is indeed one, despite the fact that the parent space is three-dimensional. When talking about dimension or percentage or whatever, you have to keep in mind "dimension of what", here we are talking about dimension of the subspace, not of the parent space.
– JMoravitz
4 hours ago
add a comment |Â
A basis must be linearly independent, so it can't contain the zero vector. Your $B$ does (twice).
– Bungo
4 hours ago
Yes sorry i will add it
– Lillys
4 hours ago
"it should be the smallest number...shouldn't that just be $left(left[beginsmallmatrix1\0\0endsmallmatrixright]right)$" Yes, it should just be that. As mentioned, the zero vectors should not have been included. The dimension of the subspace spanned by your one vector is indeed one, despite the fact that the parent space is three-dimensional. When talking about dimension or percentage or whatever, you have to keep in mind "dimension of what", here we are talking about dimension of the subspace, not of the parent space.
– JMoravitz
4 hours ago
A basis must be linearly independent, so it can't contain the zero vector. Your $B$ does (twice).
– Bungo
4 hours ago
A basis must be linearly independent, so it can't contain the zero vector. Your $B$ does (twice).
– Bungo
4 hours ago
Yes sorry i will add it
– Lillys
4 hours ago
Yes sorry i will add it
– Lillys
4 hours ago
"it should be the smallest number...shouldn't that just be $left(left[beginsmallmatrix1\0\0endsmallmatrixright]right)$" Yes, it should just be that. As mentioned, the zero vectors should not have been included. The dimension of the subspace spanned by your one vector is indeed one, despite the fact that the parent space is three-dimensional. When talking about dimension or percentage or whatever, you have to keep in mind "dimension of what", here we are talking about dimension of the subspace, not of the parent space.
– JMoravitz
4 hours ago
"it should be the smallest number...shouldn't that just be $left(left[beginsmallmatrix1\0\0endsmallmatrixright]right)$" Yes, it should just be that. As mentioned, the zero vectors should not have been included. The dimension of the subspace spanned by your one vector is indeed one, despite the fact that the parent space is three-dimensional. When talking about dimension or percentage or whatever, you have to keep in mind "dimension of what", here we are talking about dimension of the subspace, not of the parent space.
– JMoravitz
4 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
The differences:
A basis is a subset of the vector space with special properties: it has to span the vector space, and it has to be linearly independent.
The initial set of three elements you gave fails to be linearly independent, but it does span the space you specified. In that case you just call it a generating set.
The dimension of a finite dimensional vector space is a cardinal number: it is the cardinality of a basis (any basis!)
The rank of a linear transformation is the dimension of its image. That is, if you have a linear transformation $f:Vto W$, the rank of $f$ is $dim(f(V))$. This is the most common usage of the word "rank" in regular linear algebra. I can also imagine some authors unfortunately using "rank" as a synonym for dimension, but hopefully that is not very common.
So the three things refer to each other (dimension is the size of a basis, rank is the dimension of the image) but you can see they are different things.
So there where im not sure bc what is the dimension of the vectorspace, it is a line in R^3, and it would seem logical that the dimension of a line is 1.
The vector space you mentioned does indeed have dimension $1$. It is a subspace of a vector space of dimension $3$ ($mathbb R^3$), but it does not have dimension $3$ itself. Its bases only have $1$ element, but every basis of $mathbb R^3$ has three elements. The only relationship between the dimension of a vector space $V$ and a subspace $Wsubseteq V$ is that $dim(W)leq dim(V)$, which your example demonstrates.
answered 4 hours ago
rschwieb
103k1299237
I also have a problem where i was asked to determine the rank auf a set of vectors. Am i correct to say it is the number of linear independent vectors? Many thanks for your answer
– Lillys
3 hours ago
1
@Lillys I have never heard of that, but I would interpret it as the dimension of the span of those vectors, as you say.
– rschwieb
3 hours ago
add a comment |Â
up vote
2
down vote
You are correct in finding just one vector in your basis.
The vectors in a basis should be linearly independent and $0$ vectors are not allowed in your basis for the same reason.
Usually the rank is defined if you have a linear transformation or a matrix.
I did not see a matrix or a linear transformation in your statement to find the rank of it.
answered 4 hours ago
Mohammad Riazi-Kermani
38.3k41957
add a comment |Â
up vote
2
down vote
You don't want to think of
$$beginbmatrix x_1\0\0 endbmatrix$$
being a vector space. A vector space is a set of vectors (that satisfy certain conditions), so you want
$$leftlbrace beginbmatrix x_1\0\0 endbmatrix : x_1 in mathbbR rightrbrace$$
Second, a basis is a collection of linearly independent vectors that span the vector space. So you should not have any zero vectors in your basis.
Then your basis for this vector space will contain a single vector, and so is one-dimensional.
As for rank, you don't typically talk about the rank of a vector space; the rank of a matrix is the dimension of the column space of that matrix.
answered 4 hours ago
Morgan Rodgers
9,50121338
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The differences:
A basis is a subset of the vector space with special properties: it has to span the vector space, and it has to be linearly independent.
The initial set of three elements you gave fails to be linearly independent, but it does span the space you specified. In that case you just call it a generating set.
The dimension of a finite dimensional vector space is a cardinal number: it is the cardinality of a basis (any basis!)
The rank of a linear transformation is the dimension of its image. That is, if you have a linear transformation $f:Vto W$, the rank of $f$ is $dim(f(V))$. This is the most common usage of the word "rank" in regular linear algebra. I can also imagine some authors unfortunately using "rank" as a synonym for dimension, but hopefully that is not very common.
So the three things refer to each other (dimension is the size of a basis, rank is the dimension of the image) but you can see they are different things.
So there where im not sure bc what is the dimension of the vectorspace, it is a line in R^3, and it would seem logical that the dimension of a line is 1.
The vector space you mentioned does indeed have dimension $1$. It is a subspace of a vector space of dimension $3$ ($mathbb R^3$), but it does not have dimension $3$ itself. Its bases only have $1$ element, but every basis of $mathbb R^3$ has three elements. The only relationship between the dimension of a vector space $V$ and a subspace $Wsubseteq V$ is that $dim(W)leq dim(V)$, which your example demonstrates.
answered 4 hours ago
rschwieb
103k1299237
I also have a problem where i was asked to determine the rank auf a set of vectors. Am i correct to say it is the number of linear independent vectors? Many thanks for your answer
– Lillys
3 hours ago
1
@Lillys I have never heard of that, but I would interpret it as the dimension of the span of those vectors, as you say.
– rschwieb
3 hours ago
add a comment |Â
up vote
3
down vote
The differences:
A basis is a subset of the vector space with special properties: it has to span the vector space, and it has to be linearly independent.
The initial set of three elements you gave fails to be linearly independent, but it does span the space you specified. In that case you just call it a generating set.
The dimension of a finite dimensional vector space is a cardinal number: it is the cardinality of a basis (any basis!)
The rank of a linear transformation is the dimension of its image. That is, if you have a linear transformation $f:Vto W$, the rank of $f$ is $dim(f(V))$. This is the most common usage of the word "rank" in regular linear algebra. I can also imagine some authors unfortunately using "rank" as a synonym for dimension, but hopefully that is not very common.
So the three things refer to each other (dimension is the size of a basis, rank is the dimension of the image) but you can see they are different things.
So there where im not sure bc what is the dimension of the vectorspace, it is a line in R^3, and it would seem logical that the dimension of a line is 1.
The vector space you mentioned does indeed have dimension $1$. It is a subspace of a vector space of dimension $3$ ($mathbb R^3$), but it does not have dimension $3$ itself. Its bases only have $1$ element, but every basis of $mathbb R^3$ has three elements. The only relationship between the dimension of a vector space $V$ and a subspace $Wsubseteq V$ is that $dim(W)leq dim(V)$, which your example demonstrates.
answered 4 hours ago
rschwieb
103k1299237
I also have a problem where i was asked to determine the rank auf a set of vectors. Am i correct to say it is the number of linear independent vectors? Many thanks for your answer
– Lillys
3 hours ago
1
@Lillys I have never heard of that, but I would interpret it as the dimension of the span of those vectors, as you say.
– rschwieb
3 hours ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The differences:
A basis is a subset of the vector space with special properties: it has to span the vector space, and it has to be linearly independent.
The initial set of three elements you gave fails to be linearly independent, but it does span the space you specified. In that case you just call it a generating set.
The dimension of a finite dimensional vector space is a cardinal number: it is the cardinality of a basis (any basis!)
The rank of a linear transformation is the dimension of its image. That is, if you have a linear transformation $f:Vto W$, the rank of $f$ is $dim(f(V))$. This is the most common usage of the word "rank" in regular linear algebra. I can also imagine some authors unfortunately using "rank" as a synonym for dimension, but hopefully that is not very common.
So the three things refer to each other (dimension is the size of a basis, rank is the dimension of the image) but you can see they are different things.
So there where im not sure bc what is the dimension of the vectorspace, it is a line in R^3, and it would seem logical that the dimension of a line is 1.
The vector space you mentioned does indeed have dimension $1$. It is a subspace of a vector space of dimension $3$ ($mathbb R^3$), but it does not have dimension $3$ itself. Its bases only have $1$ element, but every basis of $mathbb R^3$ has three elements. The only relationship between the dimension of a vector space $V$ and a subspace $Wsubseteq V$ is that $dim(W)leq dim(V)$, which your example demonstrates.
answered 4 hours ago
rschwieb
103k1299237
The differences:
A basis is a subset of the vector space with special properties: it has to span the vector space, and it has to be linearly independent.
The initial set of three elements you gave fails to be linearly independent, but it does span the space you specified. In that case you just call it a generating set.
The dimension of a finite dimensional vector space is a cardinal number: it is the cardinality of a basis (any basis!)
The rank of a linear transformation is the dimension of its image. That is, if you have a linear transformation $f:Vto W$, the rank of $f$ is $dim(f(V))$. This is the most common usage of the word "rank" in regular linear algebra. I can also imagine some authors unfortunately using "rank" as a synonym for dimension, but hopefully that is not very common.
So the three things refer to each other (dimension is the size of a basis, rank is the dimension of the image) but you can see they are different things.
So there where im not sure bc what is the dimension of the vectorspace, it is a line in R^3, and it would seem logical that the dimension of a line is 1.
The vector space you mentioned does indeed have dimension $1$. It is a subspace of a vector space of dimension $3$ ($mathbb R^3$), but it does not have dimension $3$ itself. Its bases only have $1$ element, but every basis of $mathbb R^3$ has three elements. The only relationship between the dimension of a vector space $V$ and a subspace $Wsubseteq V$ is that $dim(W)leq dim(V)$, which your example demonstrates.
answered 4 hours ago
rschwieb
103k1299237
edited 3 hours ago
answered 4 hours ago
rschwieb
103k1299237
answered 4 hours ago
rschwieb
103k1299237
answered 4 hours ago
rschwieb
103k1299237
103k1299237
I also have a problem where i was asked to determine the rank auf a set of vectors. Am i correct to say it is the number of linear independent vectors? Many thanks for your answer
– Lillys
3 hours ago
1
@Lillys I have never heard of that, but I would interpret it as the dimension of the span of those vectors, as you say.
– rschwieb
3 hours ago
add a comment |Â
I also have a problem where i was asked to determine the rank auf a set of vectors. Am i correct to say it is the number of linear independent vectors? Many thanks for your answer
– Lillys
3 hours ago
1
@Lillys I have never heard of that, but I would interpret it as the dimension of the span of those vectors, as you say.
– rschwieb
3 hours ago
I also have a problem where i was asked to determine the rank auf a set of vectors. Am i correct to say it is the number of linear independent vectors? Many thanks for your answer
– Lillys
3 hours ago
I also have a problem where i was asked to determine the rank auf a set of vectors. Am i correct to say it is the number of linear independent vectors? Many thanks for your answer
– Lillys
3 hours ago
1
1
@Lillys I have never heard of that, but I would interpret it as the dimension of the span of those vectors, as you say.
– rschwieb
3 hours ago
@Lillys I have never heard of that, but I would interpret it as the dimension of the span of those vectors, as you say.
– rschwieb
3 hours ago
add a comment |Â
up vote
2
down vote
You are correct in finding just one vector in your basis.
The vectors in a basis should be linearly independent and $0$ vectors are not allowed in your basis for the same reason.
Usually the rank is defined if you have a linear transformation or a matrix.
I did not see a matrix or a linear transformation in your statement to find the rank of it.
answered 4 hours ago
Mohammad Riazi-Kermani
38.3k41957
add a comment |Â
up vote
2
down vote
You are correct in finding just one vector in your basis.
The vectors in a basis should be linearly independent and $0$ vectors are not allowed in your basis for the same reason.
Usually the rank is defined if you have a linear transformation or a matrix.
I did not see a matrix or a linear transformation in your statement to find the rank of it.
answered 4 hours ago
Mohammad Riazi-Kermani
38.3k41957
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You are correct in finding just one vector in your basis.
The vectors in a basis should be linearly independent and $0$ vectors are not allowed in your basis for the same reason.
Usually the rank is defined if you have a linear transformation or a matrix.
I did not see a matrix or a linear transformation in your statement to find the rank of it.
answered 4 hours ago
Mohammad Riazi-Kermani
38.3k41957
You are correct in finding just one vector in your basis.
The vectors in a basis should be linearly independent and $0$ vectors are not allowed in your basis for the same reason.
Usually the rank is defined if you have a linear transformation or a matrix.
I did not see a matrix or a linear transformation in your statement to find the rank of it.
answered 4 hours ago
Mohammad Riazi-Kermani
38.3k41957
answered 4 hours ago
Mohammad Riazi-Kermani
38.3k41957
answered 4 hours ago
Mohammad Riazi-Kermani
38.3k41957
answered 4 hours ago
Mohammad Riazi-Kermani
38.3k41957
38.3k41957
add a comment |Â
add a comment |Â
up vote
2
down vote
You don't want to think of
$$beginbmatrix x_1\0\0 endbmatrix$$
being a vector space. A vector space is a set of vectors (that satisfy certain conditions), so you want
$$leftlbrace beginbmatrix x_1\0\0 endbmatrix : x_1 in mathbbR rightrbrace$$
Second, a basis is a collection of linearly independent vectors that span the vector space. So you should not have any zero vectors in your basis.
Then your basis for this vector space will contain a single vector, and so is one-dimensional.
As for rank, you don't typically talk about the rank of a vector space; the rank of a matrix is the dimension of the column space of that matrix.
answered 4 hours ago
Morgan Rodgers
9,50121338
add a comment |Â
up vote
2
down vote
You don't want to think of
$$beginbmatrix x_1\0\0 endbmatrix$$
being a vector space. A vector space is a set of vectors (that satisfy certain conditions), so you want
$$leftlbrace beginbmatrix x_1\0\0 endbmatrix : x_1 in mathbbR rightrbrace$$
Second, a basis is a collection of linearly independent vectors that span the vector space. So you should not have any zero vectors in your basis.
Then your basis for this vector space will contain a single vector, and so is one-dimensional.
As for rank, you don't typically talk about the rank of a vector space; the rank of a matrix is the dimension of the column space of that matrix.
answered 4 hours ago
Morgan Rodgers
9,50121338
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You don't want to think of
$$beginbmatrix x_1\0\0 endbmatrix$$
being a vector space. A vector space is a set of vectors (that satisfy certain conditions), so you want
$$leftlbrace beginbmatrix x_1\0\0 endbmatrix : x_1 in mathbbR rightrbrace$$
Second, a basis is a collection of linearly independent vectors that span the vector space. So you should not have any zero vectors in your basis.
Then your basis for this vector space will contain a single vector, and so is one-dimensional.
As for rank, you don't typically talk about the rank of a vector space; the rank of a matrix is the dimension of the column space of that matrix.
answered 4 hours ago
Morgan Rodgers
9,50121338
You don't want to think of
$$beginbmatrix x_1\0\0 endbmatrix$$
being a vector space. A vector space is a set of vectors (that satisfy certain conditions), so you want
$$leftlbrace beginbmatrix x_1\0\0 endbmatrix : x_1 in mathbbR rightrbrace$$
Second, a basis is a collection of linearly independent vectors that span the vector space. So you should not have any zero vectors in your basis.
Then your basis for this vector space will contain a single vector, and so is one-dimensional.
As for rank, you don't typically talk about the rank of a vector space; the rank of a matrix is the dimension of the column space of that matrix.
answered 4 hours ago
Morgan Rodgers
9,50121338
answered 4 hours ago
Morgan Rodgers
9,50121338
answered 4 hours ago
Morgan Rodgers
9,50121338
answered 4 hours ago
Morgan Rodgers
9,50121338
9,50121338
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2988818%2frank-dimension-basis%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
A basis must be linearly independent, so it can't contain the zero vector. Your $B$ does (twice).
– Bungo
4 hours ago
Yes sorry i will add it
– Lillys
4 hours ago
"it should be the smallest number...shouldn't that just be $left(left[beginsmallmatrix1\0\0endsmallmatrixright]right)$" Yes, it should just be that. As mentioned, the zero vectors should not have been included. The dimension of the subspace spanned by your one vector is indeed one, despite the fact that the parent space is three-dimensional. When talking about dimension or percentage or whatever, you have to keep in mind "dimension of what", here we are talking about dimension of the subspace, not of the parent space.
– JMoravitz
4 hours ago