Proof that every finite number of subspaces with the same dimension of $mathbb R^n$ have a common complement












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    @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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    @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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 13:17









Paul Frost

11.4k3934




11.4k3934










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


















  • $begingroup$
    @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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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
$mathbb{R}^n = S_k + operatorname{sp}{x_{d+1},...,x_n}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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
$mathbb{R}^n = S_k + operatorname{sp}{x_{d+1},...,x_n}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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
$mathbb{R}^n = S_k + operatorname{sp}{x_{d+1},...,x_n}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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
$mathbb{R}^n = S_k + operatorname{sp}{x_{d+1},...,x_n}$.






share|cite|improve this answer









$endgroup$



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
$mathbb{R}^n = S_k + operatorname{sp}{x_{d+1},...,x_n}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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