Let $V=mathbb{R}^3$ and $W={ (x,y,z): x+y+z=0}$ Describe $V/W$ geometrically and contrsuct an explicit...
$begingroup$
For an isomorphism I let $phi:W^perpto V/W$ be defined as $phi(x)=[x]$
To show $phi$ is injective suppose $xneq y$ for $x,y in W^perp$.
Since $W={ (x,y,z): x+y+z=0}$, $(1,1,1)$ is the vector normal to the plane $W$.
And so $x=a(1,1,1)$ and $y=b(1,1,1)$ where $aneq b$.
Then $x-y=(a-b,a-b,a-b)$ and $a-b+a-b+a-b=3(a-b)neq 0$
so $x-y notin W$ and $[x]neq [y]$. Thus $phi$ is injective.
I'm not sure how to show it's onto.
To show $phi$ is onto, let $[y] in V/W$
then maybe I have that $y=w+v$ for some vector $win W$ and $vin V$. So that $phi (w+v)=[w+v]=[y]$?
I think describing V/W geometrically would mean showing $V/W=[span(1,1,1)]$
linear-algebra quotient-spaces
asked Dec 9 '18 at 21:38
AColoredReptileAColoredReptile
23728
$endgroup$
add a comment |
$begingroup$
For an isomorphism I let $phi:W^perpto V/W$ be defined as $phi(x)=[x]$
To show $phi$ is injective suppose $xneq y$ for $x,y in W^perp$.
Since $W={ (x,y,z): x+y+z=0}$, $(1,1,1)$ is the vector normal to the plane $W$.
And so $x=a(1,1,1)$ and $y=b(1,1,1)$ where $aneq b$.
Then $x-y=(a-b,a-b,a-b)$ and $a-b+a-b+a-b=3(a-b)neq 0$
so $x-y notin W$ and $[x]neq [y]$. Thus $phi$ is injective.
I'm not sure how to show it's onto.
To show $phi$ is onto, let $[y] in V/W$
then maybe I have that $y=w+v$ for some vector $win W$ and $vin V$. So that $phi (w+v)=[w+v]=[y]$?
I think describing V/W geometrically would mean showing $V/W=[span(1,1,1)]$
linear-algebra quotient-spaces
asked Dec 9 '18 at 21:38
AColoredReptileAColoredReptile
23728
$endgroup$
$begingroup$
You get surjectivity "for free" due to rank nullity and I'm pretty sure describing $V/W$ geometrically is as you said.
$endgroup$
– Dionel Jaime
Dec 9 '18 at 22:04
$begingroup$
Why do I get onto for free? From this $dim(V)=dim(V/W)+dim(W)$? I guess I don't understand.
$endgroup$
– AColoredReptile
Dec 9 '18 at 22:09
add a comment |
$begingroup$
For an isomorphism I let $phi:W^perpto V/W$ be defined as $phi(x)=[x]$
To show $phi$ is injective suppose $xneq y$ for $x,y in W^perp$.
Since $W={ (x,y,z): x+y+z=0}$, $(1,1,1)$ is the vector normal to the plane $W$.
And so $x=a(1,1,1)$ and $y=b(1,1,1)$ where $aneq b$.
Then $x-y=(a-b,a-b,a-b)$ and $a-b+a-b+a-b=3(a-b)neq 0$
so $x-y notin W$ and $[x]neq [y]$. Thus $phi$ is injective.
I'm not sure how to show it's onto.
To show $phi$ is onto, let $[y] in V/W$
then maybe I have that $y=w+v$ for some vector $win W$ and $vin V$. So that $phi (w+v)=[w+v]=[y]$?
I think describing V/W geometrically would mean showing $V/W=[span(1,1,1)]$
linear-algebra quotient-spaces
asked Dec 9 '18 at 21:38
AColoredReptileAColoredReptile
23728
$endgroup$
For an isomorphism I let $phi:W^perpto V/W$ be defined as $phi(x)=[x]$
To show $phi$ is injective suppose $xneq y$ for $x,y in W^perp$.
Since $W={ (x,y,z): x+y+z=0}$, $(1,1,1)$ is the vector normal to the plane $W$.
And so $x=a(1,1,1)$ and $y=b(1,1,1)$ where $aneq b$.
Then $x-y=(a-b,a-b,a-b)$ and $a-b+a-b+a-b=3(a-b)neq 0$
so $x-y notin W$ and $[x]neq [y]$. Thus $phi$ is injective.
I'm not sure how to show it's onto.
To show $phi$ is onto, let $[y] in V/W$
then maybe I have that $y=w+v$ for some vector $win W$ and $vin V$. So that $phi (w+v)=[w+v]=[y]$?
I think describing V/W geometrically would mean showing $V/W=[span(1,1,1)]$
linear-algebra quotient-spaces
linear-algebra quotient-spaces
asked Dec 9 '18 at 21:38
AColoredReptileAColoredReptile
23728
asked Dec 9 '18 at 21:38
AColoredReptileAColoredReptile
23728
asked Dec 9 '18 at 21:38
AColoredReptileAColoredReptile
23728
asked Dec 9 '18 at 21:38
AColoredReptileAColoredReptile
23728
asked Dec 9 '18 at 21:38
AColoredReptileAColoredReptile
23728
23728
$begingroup$
You get surjectivity "for free" due to rank nullity and I'm pretty sure describing $V/W$ geometrically is as you said.
$endgroup$
– Dionel Jaime
Dec 9 '18 at 22:04
$begingroup$
Why do I get onto for free? From this $dim(V)=dim(V/W)+dim(W)$? I guess I don't understand.
$endgroup$
– AColoredReptile
Dec 9 '18 at 22:09
add a comment |
$begingroup$
You get surjectivity "for free" due to rank nullity and I'm pretty sure describing $V/W$ geometrically is as you said.
$endgroup$
– Dionel Jaime
Dec 9 '18 at 22:04
$begingroup$
Why do I get onto for free? From this $dim(V)=dim(V/W)+dim(W)$? I guess I don't understand.
$endgroup$
– AColoredReptile
Dec 9 '18 at 22:09
$begingroup$
You get surjectivity "for free" due to rank nullity and I'm pretty sure describing $V/W$ geometrically is as you said.
$endgroup$
– Dionel Jaime
Dec 9 '18 at 22:04
$begingroup$
You get surjectivity "for free" due to rank nullity and I'm pretty sure describing $V/W$ geometrically is as you said.
$endgroup$
– Dionel Jaime
Dec 9 '18 at 22:04
$begingroup$
Why do I get onto for free? From this $dim(V)=dim(V/W)+dim(W)$? I guess I don't understand.
$endgroup$
– AColoredReptile
Dec 9 '18 at 22:09
$begingroup$
Why do I get onto for free? From this $dim(V)=dim(V/W)+dim(W)$? I guess I don't understand.
$endgroup$
– AColoredReptile
Dec 9 '18 at 22:09
add a comment |
3 Answers
3
active
oldest
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$begingroup$
$W^perp$ is one dimensional since it's isomorphic to a one dimensional space, namely, $span{(1,1,1)}$.
Since $dim(W) = 2$, we have that $dim(V/W) = 1$
You have an injective map, $phi$, between two one dimensional spaces, $W^perp $ and $V/W$
Rank nullity says the the dimension of the image of $phi$ is one and hence is all of $V/W$
answered Dec 9 '18 at 22:14
Dionel JaimeDionel Jaime
1,748514
$endgroup$
add a comment |
$begingroup$
As Jaime pointed out, surjectivity is free via rank-nullity. If you want a more elementary proof, here it is: Let $[x]inmathbb{R}^3/W$. If $yin W$ then $[x+y]=[x]$. Hence, we need only show that $W^perp+W=mathbb{R}^3$. Indeed, suppose $(a,b,c)inmathbb{R}^3$. Set $d:=frac{a+b+c}{3}$, and observe that $(d,d,d)+(a-d,b-d,c-d)=(a,b,c)$ with $a-d+b-d+c-d=0$.
As for the geometry, well, there is no natural geometric description of $mathbb{R}^3/W$, as it does not sit in any natural geometric space. However since $mathbb{R}^3/W$ is isomorphic to $W^perp$, and $W^perp$ is just $mathbb{R}(1,1,1)$, that is as good as a geometric description.
answered Dec 9 '18 at 22:11
Ben WBen W
2,283615
$endgroup$
add a comment |
$begingroup$
Geometrically, you can think of W as the plane through the origin with normal vector (1,1,1). You can think of the quotient space as translates of this plane along the normal vector. So you have this infinite family of parallel planes indexed by how far it's translated from W, which will be a scalar multiple of (1,1,1). Thats why it's 1-dimensional.
answered Dec 9 '18 at 22:25
Joel PereiraJoel Pereira
75919
$endgroup$
add a comment |
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3 Answers
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3 Answers
3
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$begingroup$
$W^perp$ is one dimensional since it's isomorphic to a one dimensional space, namely, $span{(1,1,1)}$.
Since $dim(W) = 2$, we have that $dim(V/W) = 1$
You have an injective map, $phi$, between two one dimensional spaces, $W^perp $ and $V/W$
Rank nullity says the the dimension of the image of $phi$ is one and hence is all of $V/W$
answered Dec 9 '18 at 22:14
Dionel JaimeDionel Jaime
1,748514
$endgroup$
add a comment |
$begingroup$
$W^perp$ is one dimensional since it's isomorphic to a one dimensional space, namely, $span{(1,1,1)}$.
Since $dim(W) = 2$, we have that $dim(V/W) = 1$
You have an injective map, $phi$, between two one dimensional spaces, $W^perp $ and $V/W$
Rank nullity says the the dimension of the image of $phi$ is one and hence is all of $V/W$
answered Dec 9 '18 at 22:14
Dionel JaimeDionel Jaime
1,748514
$endgroup$
add a comment |
$begingroup$
$W^perp$ is one dimensional since it's isomorphic to a one dimensional space, namely, $span{(1,1,1)}$.
Since $dim(W) = 2$, we have that $dim(V/W) = 1$
You have an injective map, $phi$, between two one dimensional spaces, $W^perp $ and $V/W$
Rank nullity says the the dimension of the image of $phi$ is one and hence is all of $V/W$
answered Dec 9 '18 at 22:14
Dionel JaimeDionel Jaime
1,748514
$endgroup$
$W^perp$ is one dimensional since it's isomorphic to a one dimensional space, namely, $span{(1,1,1)}$.
Since $dim(W) = 2$, we have that $dim(V/W) = 1$
You have an injective map, $phi$, between two one dimensional spaces, $W^perp $ and $V/W$
Rank nullity says the the dimension of the image of $phi$ is one and hence is all of $V/W$
answered Dec 9 '18 at 22:14
Dionel JaimeDionel Jaime
1,748514
answered Dec 9 '18 at 22:14
Dionel JaimeDionel Jaime
1,748514
answered Dec 9 '18 at 22:14
Dionel JaimeDionel Jaime
1,748514
answered Dec 9 '18 at 22:14
Dionel JaimeDionel Jaime
1,748514
1,748514
add a comment |
add a comment |
$begingroup$
As Jaime pointed out, surjectivity is free via rank-nullity. If you want a more elementary proof, here it is: Let $[x]inmathbb{R}^3/W$. If $yin W$ then $[x+y]=[x]$. Hence, we need only show that $W^perp+W=mathbb{R}^3$. Indeed, suppose $(a,b,c)inmathbb{R}^3$. Set $d:=frac{a+b+c}{3}$, and observe that $(d,d,d)+(a-d,b-d,c-d)=(a,b,c)$ with $a-d+b-d+c-d=0$.
As for the geometry, well, there is no natural geometric description of $mathbb{R}^3/W$, as it does not sit in any natural geometric space. However since $mathbb{R}^3/W$ is isomorphic to $W^perp$, and $W^perp$ is just $mathbb{R}(1,1,1)$, that is as good as a geometric description.
answered Dec 9 '18 at 22:11
Ben WBen W
2,283615
$endgroup$
add a comment |
$begingroup$
As Jaime pointed out, surjectivity is free via rank-nullity. If you want a more elementary proof, here it is: Let $[x]inmathbb{R}^3/W$. If $yin W$ then $[x+y]=[x]$. Hence, we need only show that $W^perp+W=mathbb{R}^3$. Indeed, suppose $(a,b,c)inmathbb{R}^3$. Set $d:=frac{a+b+c}{3}$, and observe that $(d,d,d)+(a-d,b-d,c-d)=(a,b,c)$ with $a-d+b-d+c-d=0$.
As for the geometry, well, there is no natural geometric description of $mathbb{R}^3/W$, as it does not sit in any natural geometric space. However since $mathbb{R}^3/W$ is isomorphic to $W^perp$, and $W^perp$ is just $mathbb{R}(1,1,1)$, that is as good as a geometric description.
answered Dec 9 '18 at 22:11
Ben WBen W
2,283615
$endgroup$
add a comment |
$begingroup$
As Jaime pointed out, surjectivity is free via rank-nullity. If you want a more elementary proof, here it is: Let $[x]inmathbb{R}^3/W$. If $yin W$ then $[x+y]=[x]$. Hence, we need only show that $W^perp+W=mathbb{R}^3$. Indeed, suppose $(a,b,c)inmathbb{R}^3$. Set $d:=frac{a+b+c}{3}$, and observe that $(d,d,d)+(a-d,b-d,c-d)=(a,b,c)$ with $a-d+b-d+c-d=0$.
As for the geometry, well, there is no natural geometric description of $mathbb{R}^3/W$, as it does not sit in any natural geometric space. However since $mathbb{R}^3/W$ is isomorphic to $W^perp$, and $W^perp$ is just $mathbb{R}(1,1,1)$, that is as good as a geometric description.
answered Dec 9 '18 at 22:11
Ben WBen W
2,283615
$endgroup$
As Jaime pointed out, surjectivity is free via rank-nullity. If you want a more elementary proof, here it is: Let $[x]inmathbb{R}^3/W$. If $yin W$ then $[x+y]=[x]$. Hence, we need only show that $W^perp+W=mathbb{R}^3$. Indeed, suppose $(a,b,c)inmathbb{R}^3$. Set $d:=frac{a+b+c}{3}$, and observe that $(d,d,d)+(a-d,b-d,c-d)=(a,b,c)$ with $a-d+b-d+c-d=0$.
As for the geometry, well, there is no natural geometric description of $mathbb{R}^3/W$, as it does not sit in any natural geometric space. However since $mathbb{R}^3/W$ is isomorphic to $W^perp$, and $W^perp$ is just $mathbb{R}(1,1,1)$, that is as good as a geometric description.
answered Dec 9 '18 at 22:11
Ben WBen W
2,283615
answered Dec 9 '18 at 22:11
Ben WBen W
2,283615
answered Dec 9 '18 at 22:11
Ben WBen W
2,283615
answered Dec 9 '18 at 22:11
Ben WBen W
2,283615
2,283615
add a comment |
add a comment |
$begingroup$
Geometrically, you can think of W as the plane through the origin with normal vector (1,1,1). You can think of the quotient space as translates of this plane along the normal vector. So you have this infinite family of parallel planes indexed by how far it's translated from W, which will be a scalar multiple of (1,1,1). Thats why it's 1-dimensional.
answered Dec 9 '18 at 22:25
Joel PereiraJoel Pereira
75919
$endgroup$
add a comment |
$begingroup$
Geometrically, you can think of W as the plane through the origin with normal vector (1,1,1). You can think of the quotient space as translates of this plane along the normal vector. So you have this infinite family of parallel planes indexed by how far it's translated from W, which will be a scalar multiple of (1,1,1). Thats why it's 1-dimensional.
answered Dec 9 '18 at 22:25
Joel PereiraJoel Pereira
75919
$endgroup$
add a comment |
$begingroup$
Geometrically, you can think of W as the plane through the origin with normal vector (1,1,1). You can think of the quotient space as translates of this plane along the normal vector. So you have this infinite family of parallel planes indexed by how far it's translated from W, which will be a scalar multiple of (1,1,1). Thats why it's 1-dimensional.
answered Dec 9 '18 at 22:25
Joel PereiraJoel Pereira
75919
$endgroup$
Geometrically, you can think of W as the plane through the origin with normal vector (1,1,1). You can think of the quotient space as translates of this plane along the normal vector. So you have this infinite family of parallel planes indexed by how far it's translated from W, which will be a scalar multiple of (1,1,1). Thats why it's 1-dimensional.
answered Dec 9 '18 at 22:25
Joel PereiraJoel Pereira
75919
answered Dec 9 '18 at 22:25
Joel PereiraJoel Pereira
75919
answered Dec 9 '18 at 22:25
Joel PereiraJoel Pereira
75919
answered Dec 9 '18 at 22:25
Joel PereiraJoel Pereira
75919
75919
add a comment |
add a comment |
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$begingroup$
You get surjectivity "for free" due to rank nullity and I'm pretty sure describing $V/W$ geometrically is as you said.
$endgroup$
– Dionel Jaime
Dec 9 '18 at 22:04
$begingroup$
Why do I get onto for free? From this $dim(V)=dim(V/W)+dim(W)$? I guess I don't understand.
$endgroup$
– AColoredReptile
Dec 9 '18 at 22:09