Mixing
Intersection of subspaces in $mathbb{R}^4$
$begingroup$
If there are subspaces $U$ and $V$ in $mathbb{R}^4$, $4$-dim real vector space, and their dimensions are $2$ and $3$ respectively, is it impossible for their intersection to be of dimension $0$?
I think the geometrical interpretation is that the intersection of a line and $3$d space can not be a point, hence the answer is zero. Can you confirm this hypothesis and even generalize it to all dimensions?
Thank you.
vector-spaces dimension-theory
edited Jan 27 at 17:59
stressed out
6,5831939
asked Jan 27 at 17:55
Sunjin KimSunjin Kim
204
$endgroup$
add a comment |
$begingroup$
If there are subspaces $U$ and $V$ in $mathbb{R}^4$, $4$-dim real vector space, and their dimensions are $2$ and $3$ respectively, is it impossible for their intersection to be of dimension $0$?
I think the geometrical interpretation is that the intersection of a line and $3$d space can not be a point, hence the answer is zero. Can you confirm this hypothesis and even generalize it to all dimensions?
Thank you.
vector-spaces dimension-theory
edited Jan 27 at 17:59
stressed out
6,5831939
asked Jan 27 at 17:55
Sunjin KimSunjin Kim
204
$endgroup$
add a comment |
$begingroup$
If there are subspaces $U$ and $V$ in $mathbb{R}^4$, $4$-dim real vector space, and their dimensions are $2$ and $3$ respectively, is it impossible for their intersection to be of dimension $0$?
I think the geometrical interpretation is that the intersection of a line and $3$d space can not be a point, hence the answer is zero. Can you confirm this hypothesis and even generalize it to all dimensions?
Thank you.
vector-spaces dimension-theory
edited Jan 27 at 17:59
stressed out
6,5831939
asked Jan 27 at 17:55
Sunjin KimSunjin Kim
204
$endgroup$
If there are subspaces $U$ and $V$ in $mathbb{R}^4$, $4$-dim real vector space, and their dimensions are $2$ and $3$ respectively, is it impossible for their intersection to be of dimension $0$?
I think the geometrical interpretation is that the intersection of a line and $3$d space can not be a point, hence the answer is zero. Can you confirm this hypothesis and even generalize it to all dimensions?
Thank you.
vector-spaces dimension-theory
vector-spaces dimension-theory
edited Jan 27 at 17:59
stressed out
6,5831939
asked Jan 27 at 17:55
Sunjin KimSunjin Kim
204
edited Jan 27 at 17:59
stressed out
6,5831939
asked Jan 27 at 17:55
Sunjin KimSunjin Kim
204
edited Jan 27 at 17:59
stressed out
6,5831939
edited Jan 27 at 17:59
stressed out
6,5831939
edited Jan 27 at 17:59
stressed out
6,5831939
6,5831939
asked Jan 27 at 17:55
Sunjin KimSunjin Kim
204
asked Jan 27 at 17:55
Sunjin KimSunjin Kim
204
asked Jan 27 at 17:55
Sunjin KimSunjin Kim
204
204
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$dim (U+V)=dim (U)+dim (U)-dim (Ucap V).$
Since $dim(Bbb R^4)=4$ and $U+V$ is a subspace of $Bbb R^4$, $dim (Ucap V)=0$ will lead to a contradiction.
answered Jan 27 at 18:03
Thomas ShelbyThomas Shelby
4,4292726
$endgroup$
$begingroup$
The correct formula says $U+V$, not $Ucup V$. The later is not even a vector space in general.
$endgroup$
– stressed out
Jan 27 at 18:04
1
$begingroup$
@stressedout It was a typo. Is it fine now?
$endgroup$
– Thomas Shelby
Jan 27 at 18:05
1
$begingroup$
I noticed it was a typo, I just wanted you to fix it because it could've misled the OP. Yup. It's fine now. (+1)
$endgroup$
– stressed out
Jan 27 at 18:06
add a comment |
$begingroup$
Hint:
$$dim(Ucap V)+dim(U+V)=dim(U)+dim(V)$$
answered Jan 27 at 18:01
stressed outstressed out
6,5831939
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$dim (U+V)=dim (U)+dim (U)-dim (Ucap V).$
Since $dim(Bbb R^4)=4$ and $U+V$ is a subspace of $Bbb R^4$, $dim (Ucap V)=0$ will lead to a contradiction.
answered Jan 27 at 18:03
Thomas ShelbyThomas Shelby
4,4292726
$endgroup$
$begingroup$
The correct formula says $U+V$, not $Ucup V$. The later is not even a vector space in general.
$endgroup$
– stressed out
Jan 27 at 18:04
1
$begingroup$
@stressedout It was a typo. Is it fine now?
$endgroup$
– Thomas Shelby
Jan 27 at 18:05
1
$begingroup$
I noticed it was a typo, I just wanted you to fix it because it could've misled the OP. Yup. It's fine now. (+1)
$endgroup$
– stressed out
Jan 27 at 18:06
add a comment |
$begingroup$
$dim (U+V)=dim (U)+dim (U)-dim (Ucap V).$
Since $dim(Bbb R^4)=4$ and $U+V$ is a subspace of $Bbb R^4$, $dim (Ucap V)=0$ will lead to a contradiction.
answered Jan 27 at 18:03
Thomas ShelbyThomas Shelby
4,4292726
$endgroup$
$begingroup$
The correct formula says $U+V$, not $Ucup V$. The later is not even a vector space in general.
$endgroup$
– stressed out
Jan 27 at 18:04
1
$begingroup$
@stressedout It was a typo. Is it fine now?
$endgroup$
– Thomas Shelby
Jan 27 at 18:05
1
$begingroup$
I noticed it was a typo, I just wanted you to fix it because it could've misled the OP. Yup. It's fine now. (+1)
$endgroup$
– stressed out
Jan 27 at 18:06
add a comment |
$begingroup$
$dim (U+V)=dim (U)+dim (U)-dim (Ucap V).$
Since $dim(Bbb R^4)=4$ and $U+V$ is a subspace of $Bbb R^4$, $dim (Ucap V)=0$ will lead to a contradiction.
answered Jan 27 at 18:03
Thomas ShelbyThomas Shelby
4,4292726
$endgroup$
$dim (U+V)=dim (U)+dim (U)-dim (Ucap V).$
Since $dim(Bbb R^4)=4$ and $U+V$ is a subspace of $Bbb R^4$, $dim (Ucap V)=0$ will lead to a contradiction.
answered Jan 27 at 18:03
Thomas ShelbyThomas Shelby
4,4292726
answered Jan 27 at 18:03
Thomas ShelbyThomas Shelby
4,4292726
answered Jan 27 at 18:03
Thomas ShelbyThomas Shelby
4,4292726
answered Jan 27 at 18:03
Thomas ShelbyThomas Shelby
4,4292726
4,4292726
$begingroup$
The correct formula says $U+V$, not $Ucup V$. The later is not even a vector space in general.
$endgroup$
– stressed out
Jan 27 at 18:04
1
$begingroup$
@stressedout It was a typo. Is it fine now?
$endgroup$
– Thomas Shelby
Jan 27 at 18:05
1
$begingroup$
I noticed it was a typo, I just wanted you to fix it because it could've misled the OP. Yup. It's fine now. (+1)
$endgroup$
– stressed out
Jan 27 at 18:06
add a comment |
$begingroup$
The correct formula says $U+V$, not $Ucup V$. The later is not even a vector space in general.
$endgroup$
– stressed out
Jan 27 at 18:04
1
$begingroup$
@stressedout It was a typo. Is it fine now?
$endgroup$
– Thomas Shelby
Jan 27 at 18:05
1
$begingroup$
I noticed it was a typo, I just wanted you to fix it because it could've misled the OP. Yup. It's fine now. (+1)
$endgroup$
– stressed out
Jan 27 at 18:06
$begingroup$
The correct formula says $U+V$, not $Ucup V$. The later is not even a vector space in general.
$endgroup$
– stressed out
Jan 27 at 18:04
$begingroup$
The correct formula says $U+V$, not $Ucup V$. The later is not even a vector space in general.
$endgroup$
– stressed out
Jan 27 at 18:04
1
1
$begingroup$
@stressedout It was a typo. Is it fine now?
$endgroup$
– Thomas Shelby
Jan 27 at 18:05
$begingroup$
@stressedout It was a typo. Is it fine now?
$endgroup$
– Thomas Shelby
Jan 27 at 18:05
1
1
$begingroup$
I noticed it was a typo, I just wanted you to fix it because it could've misled the OP. Yup. It's fine now. (+1)
$endgroup$
– stressed out
Jan 27 at 18:06
$begingroup$
I noticed it was a typo, I just wanted you to fix it because it could've misled the OP. Yup. It's fine now. (+1)
$endgroup$
– stressed out
Jan 27 at 18:06
add a comment |
$begingroup$
Hint:
$$dim(Ucap V)+dim(U+V)=dim(U)+dim(V)$$
answered Jan 27 at 18:01
stressed outstressed out
6,5831939
$endgroup$
add a comment |
$begingroup$
Hint:
$$dim(Ucap V)+dim(U+V)=dim(U)+dim(V)$$
answered Jan 27 at 18:01
stressed outstressed out
6,5831939
$endgroup$
add a comment |
$begingroup$
Hint:
$$dim(Ucap V)+dim(U+V)=dim(U)+dim(V)$$
answered Jan 27 at 18:01
stressed outstressed out
6,5831939
$endgroup$
Hint:
$$dim(Ucap V)+dim(U+V)=dim(U)+dim(V)$$
answered Jan 27 at 18:01
stressed outstressed out
6,5831939
answered Jan 27 at 18:01
stressed outstressed out
6,5831939
answered Jan 27 at 18:01
stressed outstressed out
6,5831939
answered Jan 27 at 18:01
stressed outstressed out
6,5831939
6,5831939
add a comment |
add a comment |
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