Proof that every finite number of subspaces with the same dimension of $mathbb R^n$ have a common complement
$begingroup$
The question is as follows:
Let there be a number of $m$ subspaces of $mathbb R^n$ that all have the same dimension.
Prove that all of these subspaces have a common complement.
I'm a little bit stuck on this question.
The cases where the dimension of the subspaces is equal to $n$ or the union of the subspaces is a subspace itself is trivial.
But apart from that I'm not sure on how to move on.
Any suggestions?
Sorry for bad English and formatting; it's my first question and not in my first language.
linear-algebra vector-spaces
edited Jan 10 at 13:17
Paul Frost
11.4k3934
asked Jan 10 at 3:06
GauntGaunt
61
$endgroup$
add a comment |
$begingroup$
The question is as follows:
Let there be a number of $m$ subspaces of $mathbb R^n$ that all have the same dimension.
Prove that all of these subspaces have a common complement.
I'm a little bit stuck on this question.
The cases where the dimension of the subspaces is equal to $n$ or the union of the subspaces is a subspace itself is trivial.
But apart from that I'm not sure on how to move on.
Any suggestions?
Sorry for bad English and formatting; it's my first question and not in my first language.
linear-algebra vector-spaces
edited Jan 10 at 13:17
Paul Frost
11.4k3934
asked Jan 10 at 3:06
GauntGaunt
61
$endgroup$
$begingroup$
@copper They do have a common complement, in the direct sum sense. Of course it's not the orthogonal complement.
$endgroup$
– Matt Samuel
Jan 10 at 4:05
$begingroup$
@MattSamuel; Thanks, I was stuck in my Hilbertian mindset...
$endgroup$
– copper.hat
Jan 10 at 4:12
$begingroup$
I guess you need them to be proper subspaces right
$endgroup$
– M. Van
Jan 10 at 13:12
add a comment |
$begingroup$
The question is as follows:
Let there be a number of $m$ subspaces of $mathbb R^n$ that all have the same dimension.
Prove that all of these subspaces have a common complement.
I'm a little bit stuck on this question.
The cases where the dimension of the subspaces is equal to $n$ or the union of the subspaces is a subspace itself is trivial.
But apart from that I'm not sure on how to move on.
Any suggestions?
Sorry for bad English and formatting; it's my first question and not in my first language.
linear-algebra vector-spaces
edited Jan 10 at 13:17
Paul Frost
11.4k3934
asked Jan 10 at 3:06
GauntGaunt
61
$endgroup$
The question is as follows:
Let there be a number of $m$ subspaces of $mathbb R^n$ that all have the same dimension.
Prove that all of these subspaces have a common complement.
I'm a little bit stuck on this question.
The cases where the dimension of the subspaces is equal to $n$ or the union of the subspaces is a subspace itself is trivial.
But apart from that I'm not sure on how to move on.
Any suggestions?
Sorry for bad English and formatting; it's my first question and not in my first language.
linear-algebra vector-spaces
linear-algebra vector-spaces
edited Jan 10 at 13:17
Paul Frost
11.4k3934
asked Jan 10 at 3:06
GauntGaunt
61
edited Jan 10 at 13:17
Paul Frost
11.4k3934
asked Jan 10 at 3:06
GauntGaunt
61
edited Jan 10 at 13:17
Paul Frost
11.4k3934
edited Jan 10 at 13:17
Paul Frost
11.4k3934
edited Jan 10 at 13:17
Paul Frost
11.4k3934
11.4k3934
asked Jan 10 at 3:06
GauntGaunt
61
asked Jan 10 at 3:06
GauntGaunt
61
asked Jan 10 at 3:06
GauntGaunt
61
61
$begingroup$
@copper They do have a common complement, in the direct sum sense. Of course it's not the orthogonal complement.
$endgroup$
– Matt Samuel
Jan 10 at 4:05
$begingroup$
@MattSamuel; Thanks, I was stuck in my Hilbertian mindset...
$endgroup$
– copper.hat
Jan 10 at 4:12
$begingroup$
I guess you need them to be proper subspaces right
$endgroup$
– M. Van
Jan 10 at 13:12
add a comment |
$begingroup$
@copper They do have a common complement, in the direct sum sense. Of course it's not the orthogonal complement.
$endgroup$
– Matt Samuel
Jan 10 at 4:05
$begingroup$
@MattSamuel; Thanks, I was stuck in my Hilbertian mindset...
$endgroup$
– copper.hat
Jan 10 at 4:12
$begingroup$
I guess you need them to be proper subspaces right
$endgroup$
– M. Van
Jan 10 at 13:12
$begingroup$
@copper They do have a common complement, in the direct sum sense. Of course it's not the orthogonal complement.
$endgroup$
– Matt Samuel
Jan 10 at 4:05
$begingroup$
@copper They do have a common complement, in the direct sum sense. Of course it's not the orthogonal complement.
$endgroup$
– Matt Samuel
Jan 10 at 4:05
$begingroup$
@MattSamuel; Thanks, I was stuck in my Hilbertian mindset...
$endgroup$
– copper.hat
Jan 10 at 4:12
$begingroup$
@MattSamuel; Thanks, I was stuck in my Hilbertian mindset...
$endgroup$
– copper.hat
Jan 10 at 4:12
$begingroup$
I guess you need them to be proper subspaces right
$endgroup$
– M. Van
Jan 10 at 13:12
$begingroup$
I guess you need them to be proper subspaces right
$endgroup$
– M. Van
Jan 10 at 13:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is a proof by induction:
Let $d$ be the dimension of the subspaces.
Since $mathbb{R}^n = mathbb{R}^n + {0}$ we see that the cases $d=0,d=n$ are straightforward.
Suppose $0<d<n$ and $S_1,...,S_d$ are the subspaces. Pick a point $x_{d+1}$ that is not in any of the subspaces. Then $S_k+operatorname{sp}{x_{d+1}}$ are a collection of distinct subspaces of dimension $d+1$.
If $d+1=n$ then $operatorname{sp} {x_{d+1}}$ is a complement, otherwise repeat the process with the $d+1$ dimensional subspaces.
This will result in a collection of points $x_{d+1},...,x_n$ such that
$mathbb{R}^n = S_k + operatorname{sp}{x_{d+1},...,x_n}$.
answered Jan 10 at 4:37
copper.hatcopper.hat
127k559160
$endgroup$
$begingroup$
Why is it possible to choose some $x_{d+1} $ outside all $S_i$?
$endgroup$
– Jens Schwaiger
Jan 10 at 16:57
$begingroup$
@JensSchwaiger: The cheap answer is that subspaces have measure zero, so their union has measure zero. Let me see i I can come up with a more constructive answer.
$endgroup$
– copper.hat
Jan 10 at 17:14
$begingroup$
See mathcounterexamples.net/… for example.
$endgroup$
– Jens Schwaiger
Jan 11 at 5:13
$begingroup$
@JensSchwaiger: Thanks!
$endgroup$
– copper.hat
Jan 11 at 5:54
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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$begingroup$
Here is a proof by induction:
Let $d$ be the dimension of the subspaces.
Since $mathbb{R}^n = mathbb{R}^n + {0}$ we see that the cases $d=0,d=n$ are straightforward.
Suppose $0<d<n$ and $S_1,...,S_d$ are the subspaces. Pick a point $x_{d+1}$ that is not in any of the subspaces. Then $S_k+operatorname{sp}{x_{d+1}}$ are a collection of distinct subspaces of dimension $d+1$.
If $d+1=n$ then $operatorname{sp} {x_{d+1}}$ is a complement, otherwise repeat the process with the $d+1$ dimensional subspaces.
This will result in a collection of points $x_{d+1},...,x_n$ such that
$mathbb{R}^n = S_k + operatorname{sp}{x_{d+1},...,x_n}$.
answered Jan 10 at 4:37
copper.hatcopper.hat
127k559160
$endgroup$
$begingroup$
Why is it possible to choose some $x_{d+1} $ outside all $S_i$?
$endgroup$
– Jens Schwaiger
Jan 10 at 16:57
$begingroup$
@JensSchwaiger: The cheap answer is that subspaces have measure zero, so their union has measure zero. Let me see i I can come up with a more constructive answer.
$endgroup$
– copper.hat
Jan 10 at 17:14
$begingroup$
See mathcounterexamples.net/… for example.
$endgroup$
– Jens Schwaiger
Jan 11 at 5:13
$begingroup$
@JensSchwaiger: Thanks!
$endgroup$
– copper.hat
Jan 11 at 5:54
add a comment |
$begingroup$
Here is a proof by induction:
Let $d$ be the dimension of the subspaces.
Since $mathbb{R}^n = mathbb{R}^n + {0}$ we see that the cases $d=0,d=n$ are straightforward.
Suppose $0<d<n$ and $S_1,...,S_d$ are the subspaces. Pick a point $x_{d+1}$ that is not in any of the subspaces. Then $S_k+operatorname{sp}{x_{d+1}}$ are a collection of distinct subspaces of dimension $d+1$.
If $d+1=n$ then $operatorname{sp} {x_{d+1}}$ is a complement, otherwise repeat the process with the $d+1$ dimensional subspaces.
This will result in a collection of points $x_{d+1},...,x_n$ such that
$mathbb{R}^n = S_k + operatorname{sp}{x_{d+1},...,x_n}$.
answered Jan 10 at 4:37
copper.hatcopper.hat
127k559160
$endgroup$
$begingroup$
Why is it possible to choose some $x_{d+1} $ outside all $S_i$?
$endgroup$
– Jens Schwaiger
Jan 10 at 16:57
$begingroup$
@JensSchwaiger: The cheap answer is that subspaces have measure zero, so their union has measure zero. Let me see i I can come up with a more constructive answer.
$endgroup$
– copper.hat
Jan 10 at 17:14
$begingroup$
See mathcounterexamples.net/… for example.
$endgroup$
– Jens Schwaiger
Jan 11 at 5:13
$begingroup$
@JensSchwaiger: Thanks!
$endgroup$
– copper.hat
Jan 11 at 5:54
add a comment |
$begingroup$
Here is a proof by induction:
Let $d$ be the dimension of the subspaces.
Since $mathbb{R}^n = mathbb{R}^n + {0}$ we see that the cases $d=0,d=n$ are straightforward.
Suppose $0<d<n$ and $S_1,...,S_d$ are the subspaces. Pick a point $x_{d+1}$ that is not in any of the subspaces. Then $S_k+operatorname{sp}{x_{d+1}}$ are a collection of distinct subspaces of dimension $d+1$.
If $d+1=n$ then $operatorname{sp} {x_{d+1}}$ is a complement, otherwise repeat the process with the $d+1$ dimensional subspaces.
This will result in a collection of points $x_{d+1},...,x_n$ such that
$mathbb{R}^n = S_k + operatorname{sp}{x_{d+1},...,x_n}$.
answered Jan 10 at 4:37
copper.hatcopper.hat
127k559160
$endgroup$
Here is a proof by induction:
Let $d$ be the dimension of the subspaces.
Since $mathbb{R}^n = mathbb{R}^n + {0}$ we see that the cases $d=0,d=n$ are straightforward.
Suppose $0<d<n$ and $S_1,...,S_d$ are the subspaces. Pick a point $x_{d+1}$ that is not in any of the subspaces. Then $S_k+operatorname{sp}{x_{d+1}}$ are a collection of distinct subspaces of dimension $d+1$.
If $d+1=n$ then $operatorname{sp} {x_{d+1}}$ is a complement, otherwise repeat the process with the $d+1$ dimensional subspaces.
This will result in a collection of points $x_{d+1},...,x_n$ such that
$mathbb{R}^n = S_k + operatorname{sp}{x_{d+1},...,x_n}$.
answered Jan 10 at 4:37
copper.hatcopper.hat
127k559160
answered Jan 10 at 4:37
copper.hatcopper.hat
127k559160
answered Jan 10 at 4:37
copper.hatcopper.hat
127k559160
answered Jan 10 at 4:37
copper.hatcopper.hat
127k559160
127k559160
$begingroup$
Why is it possible to choose some $x_{d+1} $ outside all $S_i$?
$endgroup$
– Jens Schwaiger
Jan 10 at 16:57
$begingroup$
@JensSchwaiger: The cheap answer is that subspaces have measure zero, so their union has measure zero. Let me see i I can come up with a more constructive answer.
$endgroup$
– copper.hat
Jan 10 at 17:14
$begingroup$
See mathcounterexamples.net/… for example.
$endgroup$
– Jens Schwaiger
Jan 11 at 5:13
$begingroup$
@JensSchwaiger: Thanks!
$endgroup$
– copper.hat
Jan 11 at 5:54
add a comment |
$begingroup$
Why is it possible to choose some $x_{d+1} $ outside all $S_i$?
$endgroup$
– Jens Schwaiger
Jan 10 at 16:57
$begingroup$
@JensSchwaiger: The cheap answer is that subspaces have measure zero, so their union has measure zero. Let me see i I can come up with a more constructive answer.
$endgroup$
– copper.hat
Jan 10 at 17:14
$begingroup$
See mathcounterexamples.net/… for example.
$endgroup$
– Jens Schwaiger
Jan 11 at 5:13
$begingroup$
@JensSchwaiger: Thanks!
$endgroup$
– copper.hat
Jan 11 at 5:54
$begingroup$
Why is it possible to choose some $x_{d+1} $ outside all $S_i$?
$endgroup$
– Jens Schwaiger
Jan 10 at 16:57
$begingroup$
Why is it possible to choose some $x_{d+1} $ outside all $S_i$?
$endgroup$
– Jens Schwaiger
Jan 10 at 16:57
$begingroup$
@JensSchwaiger: The cheap answer is that subspaces have measure zero, so their union has measure zero. Let me see i I can come up with a more constructive answer.
$endgroup$
– copper.hat
Jan 10 at 17:14
$begingroup$
@JensSchwaiger: The cheap answer is that subspaces have measure zero, so their union has measure zero. Let me see i I can come up with a more constructive answer.
$endgroup$
– copper.hat
Jan 10 at 17:14
$begingroup$
See mathcounterexamples.net/… for example.
$endgroup$
– Jens Schwaiger
Jan 11 at 5:13
$begingroup$
See mathcounterexamples.net/… for example.
$endgroup$
– Jens Schwaiger
Jan 11 at 5:13
$begingroup$
@JensSchwaiger: Thanks!
$endgroup$
– copper.hat
Jan 11 at 5:54
$begingroup$
@JensSchwaiger: Thanks!
$endgroup$
– copper.hat
Jan 11 at 5:54
add a comment |
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$begingroup$
@copper They do have a common complement, in the direct sum sense. Of course it's not the orthogonal complement.
$endgroup$
– Matt Samuel
Jan 10 at 4:05
$begingroup$
@MattSamuel; Thanks, I was stuck in my Hilbertian mindset...
$endgroup$
– copper.hat
Jan 10 at 4:12
$begingroup$
I guess you need them to be proper subspaces right
$endgroup$
– M. Van
Jan 10 at 13:12