Mixing
Given two subspaces $U,W$ of vector space $V$, how to show that $dim(U)+dim(W)=dim(U+W)+dim(Ucap W)$
$begingroup$
Let $U,W$ be subspaces of a vector space $V$. Show that
$$dim(U)+dim(W)=dim(U+W)+dim(Ucap W)$$
Hint: Show that the map given by $L:U×Wto V$ given by $L(u,w)=u-w$ is linear.
I can show that $L:U×Wto V$ given by $L(u,w)=u-w$ is a linear map. I also know that the dimension of $U×W$ is $dim(U)+dim(W)$. What do I do next? Any hints?
linear-algebra
edited Oct 1 '16 at 1:37
Parcly Taxel
42k1372101
asked Jan 29 '13 at 19:06
hasExamshasExams
94221529
$endgroup$
|
show 4 more comments
$begingroup$
Let $U,W$ be subspaces of a vector space $V$. Show that
$$dim(U)+dim(W)=dim(U+W)+dim(Ucap W)$$
Hint: Show that the map given by $L:U×Wto V$ given by $L(u,w)=u-w$ is linear.
I can show that $L:U×Wto V$ given by $L(u,w)=u-w$ is a linear map. I also know that the dimension of $U×W$ is $dim(U)+dim(W)$. What do I do next? Any hints?
linear-algebra
edited Oct 1 '16 at 1:37
Parcly Taxel
42k1372101
asked Jan 29 '13 at 19:06
hasExamshasExams
94221529
$endgroup$
1
$begingroup$
Show that $text{Dim}(U+W)= text{Dim}( U) + text{Dim} (W) -text{Dim} ( U cap W)$.
$endgroup$
– A.D
Jan 29 '13 at 19:12
$begingroup$
Isn't that same? I think Dimension of $u-w$(image of map), will be $text{Dim}(U+W)$, will the dimension of kernel of transformation will have $text{Dim}(Ucap W) $ ??
$endgroup$
– hasExams
Jan 29 '13 at 19:13
$begingroup$
You're on the right track. What are the nullspace and the range of $L$? Then conclude with the rank-nullity theorem: en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem
$endgroup$
– Julien
Jan 29 '13 at 19:13
$begingroup$
@julien am I right?? will they share the same null space?
$endgroup$
– hasExams
Jan 29 '13 at 19:14
1
$begingroup$
Share? There's only one map here. Its range is $L+W$. Its nullspace is isomorphic to $Lcap W$.
$endgroup$
– Julien
Jan 29 '13 at 19:16
|
show 4 more comments
$begingroup$
Let $U,W$ be subspaces of a vector space $V$. Show that
$$dim(U)+dim(W)=dim(U+W)+dim(Ucap W)$$
Hint: Show that the map given by $L:U×Wto V$ given by $L(u,w)=u-w$ is linear.
I can show that $L:U×Wto V$ given by $L(u,w)=u-w$ is a linear map. I also know that the dimension of $U×W$ is $dim(U)+dim(W)$. What do I do next? Any hints?
linear-algebra
edited Oct 1 '16 at 1:37
Parcly Taxel
42k1372101
asked Jan 29 '13 at 19:06
hasExamshasExams
94221529
$endgroup$
Let $U,W$ be subspaces of a vector space $V$. Show that
$$dim(U)+dim(W)=dim(U+W)+dim(Ucap W)$$
Hint: Show that the map given by $L:U×Wto V$ given by $L(u,w)=u-w$ is linear.
I can show that $L:U×Wto V$ given by $L(u,w)=u-w$ is a linear map. I also know that the dimension of $U×W$ is $dim(U)+dim(W)$. What do I do next? Any hints?
linear-algebra
linear-algebra
edited Oct 1 '16 at 1:37
Parcly Taxel
42k1372101
asked Jan 29 '13 at 19:06
hasExamshasExams
94221529
edited Oct 1 '16 at 1:37
Parcly Taxel
42k1372101
asked Jan 29 '13 at 19:06
hasExamshasExams
94221529
edited Oct 1 '16 at 1:37
Parcly Taxel
42k1372101
edited Oct 1 '16 at 1:37
Parcly Taxel
42k1372101
edited Oct 1 '16 at 1:37
Parcly Taxel
42k1372101
42k1372101
asked Jan 29 '13 at 19:06
hasExamshasExams
94221529
asked Jan 29 '13 at 19:06
hasExamshasExams
94221529
asked Jan 29 '13 at 19:06
hasExamshasExams
94221529
94221529
1
$begingroup$
Show that $text{Dim}(U+W)= text{Dim}( U) + text{Dim} (W) -text{Dim} ( U cap W)$.
$endgroup$
– A.D
Jan 29 '13 at 19:12
$begingroup$
Isn't that same? I think Dimension of $u-w$(image of map), will be $text{Dim}(U+W)$, will the dimension of kernel of transformation will have $text{Dim}(Ucap W) $ ??
$endgroup$
– hasExams
Jan 29 '13 at 19:13
$begingroup$
You're on the right track. What are the nullspace and the range of $L$? Then conclude with the rank-nullity theorem: en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem
$endgroup$
– Julien
Jan 29 '13 at 19:13
$begingroup$
@julien am I right?? will they share the same null space?
$endgroup$
– hasExams
Jan 29 '13 at 19:14
1
$begingroup$
Share? There's only one map here. Its range is $L+W$. Its nullspace is isomorphic to $Lcap W$.
$endgroup$
– Julien
Jan 29 '13 at 19:16
|
show 4 more comments
1
$begingroup$
Show that $text{Dim}(U+W)= text{Dim}( U) + text{Dim} (W) -text{Dim} ( U cap W)$.
$endgroup$
– A.D
Jan 29 '13 at 19:12
$begingroup$
Isn't that same? I think Dimension of $u-w$(image of map), will be $text{Dim}(U+W)$, will the dimension of kernel of transformation will have $text{Dim}(Ucap W) $ ??
$endgroup$
– hasExams
Jan 29 '13 at 19:13
$begingroup$
You're on the right track. What are the nullspace and the range of $L$? Then conclude with the rank-nullity theorem: en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem
$endgroup$
– Julien
Jan 29 '13 at 19:13
$begingroup$
@julien am I right?? will they share the same null space?
$endgroup$
– hasExams
Jan 29 '13 at 19:14
1
$begingroup$
Share? There's only one map here. Its range is $L+W$. Its nullspace is isomorphic to $Lcap W$.
$endgroup$
– Julien
Jan 29 '13 at 19:16
1
1
$begingroup$
Show that $text{Dim}(U+W)= text{Dim}( U) + text{Dim} (W) -text{Dim} ( U cap W)$.
$endgroup$
– A.D
Jan 29 '13 at 19:12
$begingroup$
Show that $text{Dim}(U+W)= text{Dim}( U) + text{Dim} (W) -text{Dim} ( U cap W)$.
$endgroup$
– A.D
Jan 29 '13 at 19:12
$begingroup$
Isn't that same? I think Dimension of $u-w$(image of map), will be $text{Dim}(U+W)$, will the dimension of kernel of transformation will have $text{Dim}(Ucap W) $ ??
$endgroup$
– hasExams
Jan 29 '13 at 19:13
$begingroup$
Isn't that same? I think Dimension of $u-w$(image of map), will be $text{Dim}(U+W)$, will the dimension of kernel of transformation will have $text{Dim}(Ucap W) $ ??
$endgroup$
– hasExams
Jan 29 '13 at 19:13
$begingroup$
You're on the right track. What are the nullspace and the range of $L$? Then conclude with the rank-nullity theorem: en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem
$endgroup$
– Julien
Jan 29 '13 at 19:13
$begingroup$
You're on the right track. What are the nullspace and the range of $L$? Then conclude with the rank-nullity theorem: en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem
$endgroup$
– Julien
Jan 29 '13 at 19:13
$begingroup$
@julien am I right?? will they share the same null space?
$endgroup$
– hasExams
Jan 29 '13 at 19:14
$begingroup$
@julien am I right?? will they share the same null space?
$endgroup$
– hasExams
Jan 29 '13 at 19:14
1
1
$begingroup$
Share? There's only one map here. Its range is $L+W$. Its nullspace is isomorphic to $Lcap W$.
$endgroup$
– Julien
Jan 29 '13 at 19:16
$begingroup$
Share? There's only one map here. Its range is $L+W$. Its nullspace is isomorphic to $Lcap W$.
$endgroup$
– Julien
Jan 29 '13 at 19:16
|
show 4 more comments
3 Answers
3
active
oldest
votes
$begingroup$
The range of the map $L$ is clearly $U+W$.
Now the nullspace is:
$$
mbox{Ker} ;L={(u,w);;; u=win Ucap W}
$$
so it is isomorphic to $Ucap W$ via the map $vlongmapsto (v,v)$.
By the rank-nullity theorem applied to $L$, we find:
$$
mbox{rank};L+mbox{null};L=mbox{dim};(Utimes W)
$$
which yields the desired formula, which is sometimes called Grassmann formula..
answered Jan 29 '13 at 19:26
JulienJulien
38.8k358130
$endgroup$
add a comment |
$begingroup$
Hint: assume $dim ( U cap W)=k$ and let $B_{ ( U cap W)}={[x_1,...x_k]}$ then expand this base for W and U let these base are $$B_U={[x_1,...x_k,y_1,..,y_n]}$$ $$B_w={[x_1,...x_k,z_1,..,z_m]}$$ then prove C={$x_1,...x_k$,$y_1$,..,$y_n$,$z_1$,..,$z_m$} is base for W+U you can show C generate W+U and C is independent.
edited Jan 29 '13 at 19:28
guest196883
5,0171241
answered Jan 29 '13 at 19:16
M.HM.H
7,29711654
$endgroup$
$begingroup$
i thought about it too, but this is a bit awkward (since i was working on linear transformation than vector spaces). thanks anyway!!
$endgroup$
– hasExams
Jan 29 '13 at 19:23
$begingroup$
@testuser How exactly do you claim that this is awkward? This is the most straightforward way to solve it.
$endgroup$
– guest196883
Jan 29 '13 at 19:24
$begingroup$
@DoctorBatmanGod context :P
$endgroup$
– hasExams
Jan 29 '13 at 19:25
add a comment |
$begingroup$
Does the following suffice to prove it?
If we proceed to show the symmetric differences:
$dim(U times W) = [ dim(U setminus W) + dim(U cap W) ] + [ dim(W setminus U) + dim(U cap W) ] $,
And we have:
$dim(U oplus W) = dim(U setminus W) + dim(W setminus U) + dim(U cap W)$
Thus,
$dim(U times W) = dim(U oplus W) + dim(U cap W)$
answered Jan 16 at 4:33
freehumoristfreehumorist
351214
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
$begingroup$
The range of the map $L$ is clearly $U+W$.
Now the nullspace is:
$$
mbox{Ker} ;L={(u,w);;; u=win Ucap W}
$$
so it is isomorphic to $Ucap W$ via the map $vlongmapsto (v,v)$.
By the rank-nullity theorem applied to $L$, we find:
$$
mbox{rank};L+mbox{null};L=mbox{dim};(Utimes W)
$$
which yields the desired formula, which is sometimes called Grassmann formula..
answered Jan 29 '13 at 19:26
JulienJulien
38.8k358130
$endgroup$
add a comment |
$begingroup$
The range of the map $L$ is clearly $U+W$.
Now the nullspace is:
$$
mbox{Ker} ;L={(u,w);;; u=win Ucap W}
$$
so it is isomorphic to $Ucap W$ via the map $vlongmapsto (v,v)$.
By the rank-nullity theorem applied to $L$, we find:
$$
mbox{rank};L+mbox{null};L=mbox{dim};(Utimes W)
$$
which yields the desired formula, which is sometimes called Grassmann formula..
answered Jan 29 '13 at 19:26
JulienJulien
38.8k358130
$endgroup$
add a comment |
$begingroup$
The range of the map $L$ is clearly $U+W$.
Now the nullspace is:
$$
mbox{Ker} ;L={(u,w);;; u=win Ucap W}
$$
so it is isomorphic to $Ucap W$ via the map $vlongmapsto (v,v)$.
By the rank-nullity theorem applied to $L$, we find:
$$
mbox{rank};L+mbox{null};L=mbox{dim};(Utimes W)
$$
which yields the desired formula, which is sometimes called Grassmann formula..
answered Jan 29 '13 at 19:26
JulienJulien
38.8k358130
$endgroup$
The range of the map $L$ is clearly $U+W$.
Now the nullspace is:
$$
mbox{Ker} ;L={(u,w);;; u=win Ucap W}
$$
so it is isomorphic to $Ucap W$ via the map $vlongmapsto (v,v)$.
By the rank-nullity theorem applied to $L$, we find:
$$
mbox{rank};L+mbox{null};L=mbox{dim};(Utimes W)
$$
which yields the desired formula, which is sometimes called Grassmann formula..
answered Jan 29 '13 at 19:26
JulienJulien
38.8k358130
answered Jan 29 '13 at 19:26
JulienJulien
38.8k358130
answered Jan 29 '13 at 19:26
JulienJulien
38.8k358130
answered Jan 29 '13 at 19:26
JulienJulien
38.8k358130
38.8k358130
add a comment |
add a comment |
$begingroup$
Hint: assume $dim ( U cap W)=k$ and let $B_{ ( U cap W)}={[x_1,...x_k]}$ then expand this base for W and U let these base are $$B_U={[x_1,...x_k,y_1,..,y_n]}$$ $$B_w={[x_1,...x_k,z_1,..,z_m]}$$ then prove C={$x_1,...x_k$,$y_1$,..,$y_n$,$z_1$,..,$z_m$} is base for W+U you can show C generate W+U and C is independent.
edited Jan 29 '13 at 19:28
guest196883
5,0171241
answered Jan 29 '13 at 19:16
M.HM.H
7,29711654
$endgroup$
$begingroup$
i thought about it too, but this is a bit awkward (since i was working on linear transformation than vector spaces). thanks anyway!!
$endgroup$
– hasExams
Jan 29 '13 at 19:23
$begingroup$
@testuser How exactly do you claim that this is awkward? This is the most straightforward way to solve it.
$endgroup$
– guest196883
Jan 29 '13 at 19:24
$begingroup$
@DoctorBatmanGod context :P
$endgroup$
– hasExams
Jan 29 '13 at 19:25
add a comment |
$begingroup$
Hint: assume $dim ( U cap W)=k$ and let $B_{ ( U cap W)}={[x_1,...x_k]}$ then expand this base for W and U let these base are $$B_U={[x_1,...x_k,y_1,..,y_n]}$$ $$B_w={[x_1,...x_k,z_1,..,z_m]}$$ then prove C={$x_1,...x_k$,$y_1$,..,$y_n$,$z_1$,..,$z_m$} is base for W+U you can show C generate W+U and C is independent.
edited Jan 29 '13 at 19:28
guest196883
5,0171241
answered Jan 29 '13 at 19:16
M.HM.H
7,29711654
$endgroup$
$begingroup$
i thought about it too, but this is a bit awkward (since i was working on linear transformation than vector spaces). thanks anyway!!
$endgroup$
– hasExams
Jan 29 '13 at 19:23
$begingroup$
@testuser How exactly do you claim that this is awkward? This is the most straightforward way to solve it.
$endgroup$
– guest196883
Jan 29 '13 at 19:24
$begingroup$
@DoctorBatmanGod context :P
$endgroup$
– hasExams
Jan 29 '13 at 19:25
add a comment |
$begingroup$
Hint: assume $dim ( U cap W)=k$ and let $B_{ ( U cap W)}={[x_1,...x_k]}$ then expand this base for W and U let these base are $$B_U={[x_1,...x_k,y_1,..,y_n]}$$ $$B_w={[x_1,...x_k,z_1,..,z_m]}$$ then prove C={$x_1,...x_k$,$y_1$,..,$y_n$,$z_1$,..,$z_m$} is base for W+U you can show C generate W+U and C is independent.
edited Jan 29 '13 at 19:28
guest196883
5,0171241
answered Jan 29 '13 at 19:16
M.HM.H
7,29711654
$endgroup$
Hint: assume $dim ( U cap W)=k$ and let $B_{ ( U cap W)}={[x_1,...x_k]}$ then expand this base for W and U let these base are $$B_U={[x_1,...x_k,y_1,..,y_n]}$$ $$B_w={[x_1,...x_k,z_1,..,z_m]}$$ then prove C={$x_1,...x_k$,$y_1$,..,$y_n$,$z_1$,..,$z_m$} is base for W+U you can show C generate W+U and C is independent.
edited Jan 29 '13 at 19:28
guest196883
5,0171241
answered Jan 29 '13 at 19:16
M.HM.H
7,29711654
edited Jan 29 '13 at 19:28
guest196883
5,0171241
edited Jan 29 '13 at 19:28
guest196883
5,0171241
edited Jan 29 '13 at 19:28
guest196883
5,0171241
5,0171241
answered Jan 29 '13 at 19:16
M.HM.H
7,29711654
answered Jan 29 '13 at 19:16
M.HM.H
7,29711654
answered Jan 29 '13 at 19:16
M.HM.H
7,29711654
7,29711654
$begingroup$
i thought about it too, but this is a bit awkward (since i was working on linear transformation than vector spaces). thanks anyway!!
$endgroup$
– hasExams
Jan 29 '13 at 19:23
$begingroup$
@testuser How exactly do you claim that this is awkward? This is the most straightforward way to solve it.
$endgroup$
– guest196883
Jan 29 '13 at 19:24
$begingroup$
@DoctorBatmanGod context :P
$endgroup$
– hasExams
Jan 29 '13 at 19:25
add a comment |
$begingroup$
i thought about it too, but this is a bit awkward (since i was working on linear transformation than vector spaces). thanks anyway!!
$endgroup$
– hasExams
Jan 29 '13 at 19:23
$begingroup$
@testuser How exactly do you claim that this is awkward? This is the most straightforward way to solve it.
$endgroup$
– guest196883
Jan 29 '13 at 19:24
$begingroup$
@DoctorBatmanGod context :P
$endgroup$
– hasExams
Jan 29 '13 at 19:25
$begingroup$
i thought about it too, but this is a bit awkward (since i was working on linear transformation than vector spaces). thanks anyway!!
$endgroup$
– hasExams
Jan 29 '13 at 19:23
$begingroup$
i thought about it too, but this is a bit awkward (since i was working on linear transformation than vector spaces). thanks anyway!!
$endgroup$
– hasExams
Jan 29 '13 at 19:23
$begingroup$
@testuser How exactly do you claim that this is awkward? This is the most straightforward way to solve it.
$endgroup$
– guest196883
Jan 29 '13 at 19:24
$begingroup$
@testuser How exactly do you claim that this is awkward? This is the most straightforward way to solve it.
$endgroup$
– guest196883
Jan 29 '13 at 19:24
$begingroup$
@DoctorBatmanGod context :P
$endgroup$
– hasExams
Jan 29 '13 at 19:25
$begingroup$
@DoctorBatmanGod context :P
$endgroup$
– hasExams
Jan 29 '13 at 19:25
add a comment |
$begingroup$
Does the following suffice to prove it?
If we proceed to show the symmetric differences:
$dim(U times W) = [ dim(U setminus W) + dim(U cap W) ] + [ dim(W setminus U) + dim(U cap W) ] $,
And we have:
$dim(U oplus W) = dim(U setminus W) + dim(W setminus U) + dim(U cap W)$
Thus,
$dim(U times W) = dim(U oplus W) + dim(U cap W)$
answered Jan 16 at 4:33
freehumoristfreehumorist
351214
$endgroup$
add a comment |
$begingroup$
Does the following suffice to prove it?
If we proceed to show the symmetric differences:
$dim(U times W) = [ dim(U setminus W) + dim(U cap W) ] + [ dim(W setminus U) + dim(U cap W) ] $,
And we have:
$dim(U oplus W) = dim(U setminus W) + dim(W setminus U) + dim(U cap W)$
Thus,
$dim(U times W) = dim(U oplus W) + dim(U cap W)$
answered Jan 16 at 4:33
freehumoristfreehumorist
351214
$endgroup$
add a comment |
$begingroup$
Does the following suffice to prove it?
If we proceed to show the symmetric differences:
$dim(U times W) = [ dim(U setminus W) + dim(U cap W) ] + [ dim(W setminus U) + dim(U cap W) ] $,
And we have:
$dim(U oplus W) = dim(U setminus W) + dim(W setminus U) + dim(U cap W)$
Thus,
$dim(U times W) = dim(U oplus W) + dim(U cap W)$
answered Jan 16 at 4:33
freehumoristfreehumorist
351214
$endgroup$
Does the following suffice to prove it?
If we proceed to show the symmetric differences:
$dim(U times W) = [ dim(U setminus W) + dim(U cap W) ] + [ dim(W setminus U) + dim(U cap W) ] $,
And we have:
$dim(U oplus W) = dim(U setminus W) + dim(W setminus U) + dim(U cap W)$
Thus,
$dim(U times W) = dim(U oplus W) + dim(U cap W)$
answered Jan 16 at 4:33
freehumoristfreehumorist
351214
answered Jan 16 at 4:33
freehumoristfreehumorist
351214
answered Jan 16 at 4:33
freehumoristfreehumorist
351214
answered Jan 16 at 4:33
freehumoristfreehumorist
351214
351214
add a comment |
add a comment |
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1
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Show that $text{Dim}(U+W)= text{Dim}( U) + text{Dim} (W) -text{Dim} ( U cap W)$.
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– A.D
Jan 29 '13 at 19:12
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Isn't that same? I think Dimension of $u-w$(image of map), will be $text{Dim}(U+W)$, will the dimension of kernel of transformation will have $text{Dim}(Ucap W) $ ??
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– hasExams
Jan 29 '13 at 19:13
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You're on the right track. What are the nullspace and the range of $L$? Then conclude with the rank-nullity theorem: en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem
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– Julien
Jan 29 '13 at 19:13
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@julien am I right?? will they share the same null space?
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– hasExams
Jan 29 '13 at 19:14
1
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Share? There's only one map here. Its range is $L+W$. Its nullspace is isomorphic to $Lcap W$.
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– Julien
Jan 29 '13 at 19:16