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Why 2 vectors in $mathbbR^2$ can be span for the whole dimension, but not in higher dimensions
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Lemma: any two vectors in $mathbbR^2$ that are not scalar multiples of each other will span all of $mathbbR^2$.
why is true for this particular dimension but not necessarily for higher order dimensions?
linear-algebra vector-spaces
asked Jul 16 at 23:27
user6394019
30311
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up vote
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down vote
favorite
Lemma: any two vectors in $mathbbR^2$ that are not scalar multiples of each other will span all of $mathbbR^2$.
why is true for this particular dimension but not necessarily for higher order dimensions?
linear-algebra vector-spaces
asked Jul 16 at 23:27
user6394019
30311
4
That's exactly what the word "dimension" means. By definition. The dimension of a vector space is equal to the number of linearly independent vectors that spans all of the vector space.
– Lee Mosher
Jul 16 at 23:29
Now, perhaps you might want to know why $mathbb R^3$ has dimension $3$ and why $mathbb R^n$ has dimension $n$?
– Lee Mosher
Jul 16 at 23:30
1
It actually is true for higher dimensions, when appropriately generalised: any $n$ vectors (not two) from $mathbbR^n$ that are linearly independent (the $n$ vector version of not being scalar multiples of each other) will span all of $mathbbR^n$. As Lee Mosher points out, this can be generalised further to "finite-dimensional" vector spaces, but you need to understand what that means first. :-)
– Theo Bendit
Jul 16 at 23:49
Do you actually think two vectors can span all of $mathbb R^3?$
– zhw.
Jul 17 at 1:16
nahh you clearly misinterpreted the point. by the way it's clear to me that number of vetors increases respectively to order of dimension.... I was pointing out to the the vectors in any dimension of any order and why this claims(the claims in the title) is not necessarily true for other dimensions
– user6394019
Jul 17 at 5:50
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Lemma: any two vectors in $mathbbR^2$ that are not scalar multiples of each other will span all of $mathbbR^2$.
why is true for this particular dimension but not necessarily for higher order dimensions?
linear-algebra vector-spaces
asked Jul 16 at 23:27
user6394019
30311
Lemma: any two vectors in $mathbbR^2$ that are not scalar multiples of each other will span all of $mathbbR^2$.
why is true for this particular dimension but not necessarily for higher order dimensions?
linear-algebra vector-spaces
asked Jul 16 at 23:27
user6394019
30311
asked Jul 16 at 23:27
user6394019
30311
asked Jul 16 at 23:27
user6394019
30311
asked Jul 16 at 23:27
user6394019
30311
30311
4
That's exactly what the word "dimension" means. By definition. The dimension of a vector space is equal to the number of linearly independent vectors that spans all of the vector space.
– Lee Mosher
Jul 16 at 23:29
Now, perhaps you might want to know why $mathbb R^3$ has dimension $3$ and why $mathbb R^n$ has dimension $n$?
– Lee Mosher
Jul 16 at 23:30
1
It actually is true for higher dimensions, when appropriately generalised: any $n$ vectors (not two) from $mathbbR^n$ that are linearly independent (the $n$ vector version of not being scalar multiples of each other) will span all of $mathbbR^n$. As Lee Mosher points out, this can be generalised further to "finite-dimensional" vector spaces, but you need to understand what that means first. :-)
– Theo Bendit
Jul 16 at 23:49
Do you actually think two vectors can span all of $mathbb R^3?$
– zhw.
Jul 17 at 1:16
nahh you clearly misinterpreted the point. by the way it's clear to me that number of vetors increases respectively to order of dimension.... I was pointing out to the the vectors in any dimension of any order and why this claims(the claims in the title) is not necessarily true for other dimensions
– user6394019
Jul 17 at 5:50
add a comment |Â
4
That's exactly what the word "dimension" means. By definition. The dimension of a vector space is equal to the number of linearly independent vectors that spans all of the vector space.
– Lee Mosher
Jul 16 at 23:29
Now, perhaps you might want to know why $mathbb R^3$ has dimension $3$ and why $mathbb R^n$ has dimension $n$?
– Lee Mosher
Jul 16 at 23:30
1
It actually is true for higher dimensions, when appropriately generalised: any $n$ vectors (not two) from $mathbbR^n$ that are linearly independent (the $n$ vector version of not being scalar multiples of each other) will span all of $mathbbR^n$. As Lee Mosher points out, this can be generalised further to "finite-dimensional" vector spaces, but you need to understand what that means first. :-)
– Theo Bendit
Jul 16 at 23:49
Do you actually think two vectors can span all of $mathbb R^3?$
– zhw.
Jul 17 at 1:16
nahh you clearly misinterpreted the point. by the way it's clear to me that number of vetors increases respectively to order of dimension.... I was pointing out to the the vectors in any dimension of any order and why this claims(the claims in the title) is not necessarily true for other dimensions
– user6394019
Jul 17 at 5:50
4
4
That's exactly what the word "dimension" means. By definition. The dimension of a vector space is equal to the number of linearly independent vectors that spans all of the vector space.
– Lee Mosher
Jul 16 at 23:29
That's exactly what the word "dimension" means. By definition. The dimension of a vector space is equal to the number of linearly independent vectors that spans all of the vector space.
– Lee Mosher
Jul 16 at 23:29
Now, perhaps you might want to know why $mathbb R^3$ has dimension $3$ and why $mathbb R^n$ has dimension $n$?
– Lee Mosher
Jul 16 at 23:30
Now, perhaps you might want to know why $mathbb R^3$ has dimension $3$ and why $mathbb R^n$ has dimension $n$?
– Lee Mosher
Jul 16 at 23:30
1
1
It actually is true for higher dimensions, when appropriately generalised: any $n$ vectors (not two) from $mathbbR^n$ that are linearly independent (the $n$ vector version of not being scalar multiples of each other) will span all of $mathbbR^n$. As Lee Mosher points out, this can be generalised further to "finite-dimensional" vector spaces, but you need to understand what that means first. :-)
– Theo Bendit
Jul 16 at 23:49
It actually is true for higher dimensions, when appropriately generalised: any $n$ vectors (not two) from $mathbbR^n$ that are linearly independent (the $n$ vector version of not being scalar multiples of each other) will span all of $mathbbR^n$. As Lee Mosher points out, this can be generalised further to "finite-dimensional" vector spaces, but you need to understand what that means first. :-)
– Theo Bendit
Jul 16 at 23:49
Do you actually think two vectors can span all of $mathbb R^3?$
– zhw.
Jul 17 at 1:16
Do you actually think two vectors can span all of $mathbb R^3?$
– zhw.
Jul 17 at 1:16
nahh you clearly misinterpreted the point. by the way it's clear to me that number of vetors increases respectively to order of dimension.... I was pointing out to the the vectors in any dimension of any order and why this claims(the claims in the title) is not necessarily true for other dimensions
– user6394019
Jul 17 at 5:50
nahh you clearly misinterpreted the point. by the way it's clear to me that number of vetors increases respectively to order of dimension.... I was pointing out to the the vectors in any dimension of any order and why this claims(the claims in the title) is not necessarily true for other dimensions
– user6394019
Jul 17 at 5:50
add a comment |Â
2 Answers
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I am assuming you are referring to using n vectors for the $mathbbR^n$ dimension.
The requirement for a set of n vectors to span $mathbbR^n$ is that they are linearly independent- that is, no linear combination of all the vectors will equal the zero vector, unless all their coefficients are 0. Thinking graphically, this means that no combination of vectors will end up at the origin (in other words, if you graphically add the vectors, they can't reach where you started.)
For $mathbbR^2$, the only way for them to reach the origin (think on a 2-D plane) is if they are scalar multiples of each other. (Try working with it, you won't be able to reach the origin unless they are scalar multiples.) However, for three or more vectors, you can reach the origin by simply making a triangle, or square, or really any set of lines that ends where it starts. The vectors don't have to be scalar multiples of each other, they just have to form a triangle.
As an example, try $mathbbR^3$. Given the vectors x = (1,0,0), y = (0,1,0), z = (-1,-1,0), you can't span the dimension using x, y, and z. A quick look shows that any linear combination will yield the third coordinate as 0. Another quick look shows that the vectors are not scalar multiples of each other.
answered Jul 17 at 1:29
Ian Ng
1264
add a comment |Â
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Let $ngeq 3.$ Let $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n)$ belong to $Bbb R^n.$
Suppose that for every $z=(z_1,..,z_n)in Bbb R^n$ there exist $A,B in Bbb R$ such that $$(bullet)quad Ax_1+By_1=z_1 quad text and quad Ax_2+By_2=z_2.$$ (Note: Some $zin Bbb R^n$ would not belong to $Ax+By: A,Bin Bbb R$ if this were not so).
Now if $(bullet)$ is solvable in $A,B$ for every $(z_1,z_2)$ then for every $(z_1,z_2)ne (0,0)$ there is a $ unique $ pair $ A,B$ that solves $(bullet).$
Consider $z=(z_1,...z_n)$ and $z'=(z'_1,...,z'_n)$ where $z_1=z_2=z'_1=z'_2=1$ but $z_3ne z'_3.$ Suppose $Ax+By=z$ and $A'x+B'y=z'.$ Then $A, B$ solves $(bullet)$ but $A', B'$ also solves $(bullet).$ So by the uniqueness we have $A=A'$ and $B=B' .$ But then $z=Ax+By=A'x+B'y=z',$ which is absurd , because $zne z'.$
So the presumption that $z$ and $z'$ both belong to the span of $x,y$ is untenable.
answered Jul 17 at 4:41
DanielWainfleet
31.7k31644
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I am assuming you are referring to using n vectors for the $mathbbR^n$ dimension.
The requirement for a set of n vectors to span $mathbbR^n$ is that they are linearly independent- that is, no linear combination of all the vectors will equal the zero vector, unless all their coefficients are 0. Thinking graphically, this means that no combination of vectors will end up at the origin (in other words, if you graphically add the vectors, they can't reach where you started.)
For $mathbbR^2$, the only way for them to reach the origin (think on a 2-D plane) is if they are scalar multiples of each other. (Try working with it, you won't be able to reach the origin unless they are scalar multiples.) However, for three or more vectors, you can reach the origin by simply making a triangle, or square, or really any set of lines that ends where it starts. The vectors don't have to be scalar multiples of each other, they just have to form a triangle.
As an example, try $mathbbR^3$. Given the vectors x = (1,0,0), y = (0,1,0), z = (-1,-1,0), you can't span the dimension using x, y, and z. A quick look shows that any linear combination will yield the third coordinate as 0. Another quick look shows that the vectors are not scalar multiples of each other.
answered Jul 17 at 1:29
Ian Ng
1264
add a comment |Â
up vote
1
down vote
I am assuming you are referring to using n vectors for the $mathbbR^n$ dimension.
The requirement for a set of n vectors to span $mathbbR^n$ is that they are linearly independent- that is, no linear combination of all the vectors will equal the zero vector, unless all their coefficients are 0. Thinking graphically, this means that no combination of vectors will end up at the origin (in other words, if you graphically add the vectors, they can't reach where you started.)
For $mathbbR^2$, the only way for them to reach the origin (think on a 2-D plane) is if they are scalar multiples of each other. (Try working with it, you won't be able to reach the origin unless they are scalar multiples.) However, for three or more vectors, you can reach the origin by simply making a triangle, or square, or really any set of lines that ends where it starts. The vectors don't have to be scalar multiples of each other, they just have to form a triangle.
As an example, try $mathbbR^3$. Given the vectors x = (1,0,0), y = (0,1,0), z = (-1,-1,0), you can't span the dimension using x, y, and z. A quick look shows that any linear combination will yield the third coordinate as 0. Another quick look shows that the vectors are not scalar multiples of each other.
answered Jul 17 at 1:29
Ian Ng
1264
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I am assuming you are referring to using n vectors for the $mathbbR^n$ dimension.
The requirement for a set of n vectors to span $mathbbR^n$ is that they are linearly independent- that is, no linear combination of all the vectors will equal the zero vector, unless all their coefficients are 0. Thinking graphically, this means that no combination of vectors will end up at the origin (in other words, if you graphically add the vectors, they can't reach where you started.)
For $mathbbR^2$, the only way for them to reach the origin (think on a 2-D plane) is if they are scalar multiples of each other. (Try working with it, you won't be able to reach the origin unless they are scalar multiples.) However, for three or more vectors, you can reach the origin by simply making a triangle, or square, or really any set of lines that ends where it starts. The vectors don't have to be scalar multiples of each other, they just have to form a triangle.
As an example, try $mathbbR^3$. Given the vectors x = (1,0,0), y = (0,1,0), z = (-1,-1,0), you can't span the dimension using x, y, and z. A quick look shows that any linear combination will yield the third coordinate as 0. Another quick look shows that the vectors are not scalar multiples of each other.
answered Jul 17 at 1:29
Ian Ng
1264
I am assuming you are referring to using n vectors for the $mathbbR^n$ dimension.
The requirement for a set of n vectors to span $mathbbR^n$ is that they are linearly independent- that is, no linear combination of all the vectors will equal the zero vector, unless all their coefficients are 0. Thinking graphically, this means that no combination of vectors will end up at the origin (in other words, if you graphically add the vectors, they can't reach where you started.)
For $mathbbR^2$, the only way for them to reach the origin (think on a 2-D plane) is if they are scalar multiples of each other. (Try working with it, you won't be able to reach the origin unless they are scalar multiples.) However, for three or more vectors, you can reach the origin by simply making a triangle, or square, or really any set of lines that ends where it starts. The vectors don't have to be scalar multiples of each other, they just have to form a triangle.
As an example, try $mathbbR^3$. Given the vectors x = (1,0,0), y = (0,1,0), z = (-1,-1,0), you can't span the dimension using x, y, and z. A quick look shows that any linear combination will yield the third coordinate as 0. Another quick look shows that the vectors are not scalar multiples of each other.
answered Jul 17 at 1:29
Ian Ng
1264
answered Jul 17 at 1:29
Ian Ng
1264
answered Jul 17 at 1:29
Ian Ng
1264
answered Jul 17 at 1:29
Ian Ng
1264
1264
add a comment |Â
add a comment |Â
up vote
0
down vote
Let $ngeq 3.$ Let $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n)$ belong to $Bbb R^n.$
Suppose that for every $z=(z_1,..,z_n)in Bbb R^n$ there exist $A,B in Bbb R$ such that $$(bullet)quad Ax_1+By_1=z_1 quad text and quad Ax_2+By_2=z_2.$$ (Note: Some $zin Bbb R^n$ would not belong to $Ax+By: A,Bin Bbb R$ if this were not so).
Now if $(bullet)$ is solvable in $A,B$ for every $(z_1,z_2)$ then for every $(z_1,z_2)ne (0,0)$ there is a $ unique $ pair $ A,B$ that solves $(bullet).$
Consider $z=(z_1,...z_n)$ and $z'=(z'_1,...,z'_n)$ where $z_1=z_2=z'_1=z'_2=1$ but $z_3ne z'_3.$ Suppose $Ax+By=z$ and $A'x+B'y=z'.$ Then $A, B$ solves $(bullet)$ but $A', B'$ also solves $(bullet).$ So by the uniqueness we have $A=A'$ and $B=B' .$ But then $z=Ax+By=A'x+B'y=z',$ which is absurd , because $zne z'.$
So the presumption that $z$ and $z'$ both belong to the span of $x,y$ is untenable.
answered Jul 17 at 4:41
DanielWainfleet
31.7k31644
add a comment |Â
up vote
0
down vote
Let $ngeq 3.$ Let $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n)$ belong to $Bbb R^n.$
Suppose that for every $z=(z_1,..,z_n)in Bbb R^n$ there exist $A,B in Bbb R$ such that $$(bullet)quad Ax_1+By_1=z_1 quad text and quad Ax_2+By_2=z_2.$$ (Note: Some $zin Bbb R^n$ would not belong to $Ax+By: A,Bin Bbb R$ if this were not so).
Now if $(bullet)$ is solvable in $A,B$ for every $(z_1,z_2)$ then for every $(z_1,z_2)ne (0,0)$ there is a $ unique $ pair $ A,B$ that solves $(bullet).$
Consider $z=(z_1,...z_n)$ and $z'=(z'_1,...,z'_n)$ where $z_1=z_2=z'_1=z'_2=1$ but $z_3ne z'_3.$ Suppose $Ax+By=z$ and $A'x+B'y=z'.$ Then $A, B$ solves $(bullet)$ but $A', B'$ also solves $(bullet).$ So by the uniqueness we have $A=A'$ and $B=B' .$ But then $z=Ax+By=A'x+B'y=z',$ which is absurd , because $zne z'.$
So the presumption that $z$ and $z'$ both belong to the span of $x,y$ is untenable.
answered Jul 17 at 4:41
DanielWainfleet
31.7k31644
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $ngeq 3.$ Let $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n)$ belong to $Bbb R^n.$
Suppose that for every $z=(z_1,..,z_n)in Bbb R^n$ there exist $A,B in Bbb R$ such that $$(bullet)quad Ax_1+By_1=z_1 quad text and quad Ax_2+By_2=z_2.$$ (Note: Some $zin Bbb R^n$ would not belong to $Ax+By: A,Bin Bbb R$ if this were not so).
Now if $(bullet)$ is solvable in $A,B$ for every $(z_1,z_2)$ then for every $(z_1,z_2)ne (0,0)$ there is a $ unique $ pair $ A,B$ that solves $(bullet).$
Consider $z=(z_1,...z_n)$ and $z'=(z'_1,...,z'_n)$ where $z_1=z_2=z'_1=z'_2=1$ but $z_3ne z'_3.$ Suppose $Ax+By=z$ and $A'x+B'y=z'.$ Then $A, B$ solves $(bullet)$ but $A', B'$ also solves $(bullet).$ So by the uniqueness we have $A=A'$ and $B=B' .$ But then $z=Ax+By=A'x+B'y=z',$ which is absurd , because $zne z'.$
So the presumption that $z$ and $z'$ both belong to the span of $x,y$ is untenable.
answered Jul 17 at 4:41
DanielWainfleet
31.7k31644
Let $ngeq 3.$ Let $x=(x_1,...,x_n)$ and $y=(y_1,...,y_n)$ belong to $Bbb R^n.$
Suppose that for every $z=(z_1,..,z_n)in Bbb R^n$ there exist $A,B in Bbb R$ such that $$(bullet)quad Ax_1+By_1=z_1 quad text and quad Ax_2+By_2=z_2.$$ (Note: Some $zin Bbb R^n$ would not belong to $Ax+By: A,Bin Bbb R$ if this were not so).
Now if $(bullet)$ is solvable in $A,B$ for every $(z_1,z_2)$ then for every $(z_1,z_2)ne (0,0)$ there is a $ unique $ pair $ A,B$ that solves $(bullet).$
Consider $z=(z_1,...z_n)$ and $z'=(z'_1,...,z'_n)$ where $z_1=z_2=z'_1=z'_2=1$ but $z_3ne z'_3.$ Suppose $Ax+By=z$ and $A'x+B'y=z'.$ Then $A, B$ solves $(bullet)$ but $A', B'$ also solves $(bullet).$ So by the uniqueness we have $A=A'$ and $B=B' .$ But then $z=Ax+By=A'x+B'y=z',$ which is absurd , because $zne z'.$
So the presumption that $z$ and $z'$ both belong to the span of $x,y$ is untenable.
answered Jul 17 at 4:41
DanielWainfleet
31.7k31644
answered Jul 17 at 4:41
DanielWainfleet
31.7k31644
answered Jul 17 at 4:41
DanielWainfleet
31.7k31644
answered Jul 17 at 4:41
DanielWainfleet
31.7k31644
31.7k31644
add a comment |Â
add a comment |Â
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4
That's exactly what the word "dimension" means. By definition. The dimension of a vector space is equal to the number of linearly independent vectors that spans all of the vector space.
– Lee Mosher
Jul 16 at 23:29
Now, perhaps you might want to know why $mathbb R^3$ has dimension $3$ and why $mathbb R^n$ has dimension $n$?
– Lee Mosher
Jul 16 at 23:30
1
It actually is true for higher dimensions, when appropriately generalised: any $n$ vectors (not two) from $mathbbR^n$ that are linearly independent (the $n$ vector version of not being scalar multiples of each other) will span all of $mathbbR^n$. As Lee Mosher points out, this can be generalised further to "finite-dimensional" vector spaces, but you need to understand what that means first. :-)
– Theo Bendit
Jul 16 at 23:49
Do you actually think two vectors can span all of $mathbb R^3?$
– zhw.
Jul 17 at 1:16
nahh you clearly misinterpreted the point. by the way it's clear to me that number of vetors increases respectively to order of dimension.... I was pointing out to the the vectors in any dimension of any order and why this claims(the claims in the title) is not necessarily true for other dimensions
– user6394019
Jul 17 at 5:50