Intersection of subspaces












1












$begingroup$


I am working on an example of vector spaces. I have the following question:



Let ${V_1,V_2,ldots,V_t}$ be a family of $n$-dimensional subspaces and the dimension intersection of any $n$ subspace is at least one. Is it true that



$dim bigcap^{i=t}_{i=1}V_igeq1$?



I have calculated $dim bigcap^{i=t}_{i=1}V_i$ for several spaces and all of spaces satisfied in $dim bigcap^{i=t}_{i=1}V_igeq1$.



Can anybody take counterexample?






linear-algebra vector-spaces







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$endgroup$












  • $begingroup$
    Each $V_i$ is $n$-dimensional. But you are taking exactly $n$ to take intersections?
    $endgroup$
    – Sigur
    Jun 30 '12 at 21:03










  • $begingroup$
    @Sigur yes exactly
    $endgroup$
    – Babak Miraftab
    Jun 30 '12 at 21:06
















1












$begingroup$


I am working on an example of vector spaces. I have the following question:



Let ${V_1,V_2,ldots,V_t}$ be a family of $n$-dimensional subspaces and the dimension intersection of any $n$ subspace is at least one. Is it true that



$dim bigcap^{i=t}_{i=1}V_igeq1$?



I have calculated $dim bigcap^{i=t}_{i=1}V_i$ for several spaces and all of spaces satisfied in $dim bigcap^{i=t}_{i=1}V_igeq1$.



Can anybody take counterexample?






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Each $V_i$ is $n$-dimensional. But you are taking exactly $n$ to take intersections?
    $endgroup$
    – Sigur
    Jun 30 '12 at 21:03










  • $begingroup$
    @Sigur yes exactly
    $endgroup$
    – Babak Miraftab
    Jun 30 '12 at 21:06














1












1








1





$begingroup$


I am working on an example of vector spaces. I have the following question:



Let ${V_1,V_2,ldots,V_t}$ be a family of $n$-dimensional subspaces and the dimension intersection of any $n$ subspace is at least one. Is it true that



$dim bigcap^{i=t}_{i=1}V_igeq1$?



I have calculated $dim bigcap^{i=t}_{i=1}V_i$ for several spaces and all of spaces satisfied in $dim bigcap^{i=t}_{i=1}V_igeq1$.



Can anybody take counterexample?






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$




I am working on an example of vector spaces. I have the following question:



Let ${V_1,V_2,ldots,V_t}$ be a family of $n$-dimensional subspaces and the dimension intersection of any $n$ subspace is at least one. Is it true that



$dim bigcap^{i=t}_{i=1}V_igeq1$?



I have calculated $dim bigcap^{i=t}_{i=1}V_i$ for several spaces and all of spaces satisfied in $dim bigcap^{i=t}_{i=1}V_igeq1$.



Can anybody take counterexample?






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 Dec 26 '18 at 12:08







Babak Miraftab

















asked Jun 30 '12 at 20:57









Babak MiraftabBabak Miraftab

5,38212148




5,38212148












  • $begingroup$
    Each $V_i$ is $n$-dimensional. But you are taking exactly $n$ to take intersections?
    $endgroup$
    – Sigur
    Jun 30 '12 at 21:03










  • $begingroup$
    @Sigur yes exactly
    $endgroup$
    – Babak Miraftab
    Jun 30 '12 at 21:06


















  • $begingroup$
    Each $V_i$ is $n$-dimensional. But you are taking exactly $n$ to take intersections?
    $endgroup$
    – Sigur
    Jun 30 '12 at 21:03










  • $begingroup$
    @Sigur yes exactly
    $endgroup$
    – Babak Miraftab
    Jun 30 '12 at 21:06
















$begingroup$
Each $V_i$ is $n$-dimensional. But you are taking exactly $n$ to take intersections?
$endgroup$
– Sigur
Jun 30 '12 at 21:03




$begingroup$
Each $V_i$ is $n$-dimensional. But you are taking exactly $n$ to take intersections?
$endgroup$
– Sigur
Jun 30 '12 at 21:03












$begingroup$
@Sigur yes exactly
$endgroup$
– Babak Miraftab
Jun 30 '12 at 21:06




$begingroup$
@Sigur yes exactly
$endgroup$
– Babak Miraftab
Jun 30 '12 at 21:06










1 Answer
1






active

oldest

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9












$begingroup$

Consider the vector space $mathbb{R}^3$. Let $t = 3$ and $n = 2$. Consider $V_1$ to be the $xy$ plane. $V_2$ to be the $xz$ plane. $V_3$ to be the $yz$ plane. All of these are $2$ dimensional subspaces. The intersection of any pair is one dimensional line. However the intersection of all three is just the origin which is not $1$-dimensional.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok thanks,it is very helpful
    $endgroup$
    – Babak Miraftab
    Jun 30 '12 at 21:07











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1 Answer
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active

oldest

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active

oldest

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active

oldest

votes









9












$begingroup$

Consider the vector space $mathbb{R}^3$. Let $t = 3$ and $n = 2$. Consider $V_1$ to be the $xy$ plane. $V_2$ to be the $xz$ plane. $V_3$ to be the $yz$ plane. All of these are $2$ dimensional subspaces. The intersection of any pair is one dimensional line. However the intersection of all three is just the origin which is not $1$-dimensional.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok thanks,it is very helpful
    $endgroup$
    – Babak Miraftab
    Jun 30 '12 at 21:07
















9












$begingroup$

Consider the vector space $mathbb{R}^3$. Let $t = 3$ and $n = 2$. Consider $V_1$ to be the $xy$ plane. $V_2$ to be the $xz$ plane. $V_3$ to be the $yz$ plane. All of these are $2$ dimensional subspaces. The intersection of any pair is one dimensional line. However the intersection of all three is just the origin which is not $1$-dimensional.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok thanks,it is very helpful
    $endgroup$
    – Babak Miraftab
    Jun 30 '12 at 21:07














9












9








9





$begingroup$

Consider the vector space $mathbb{R}^3$. Let $t = 3$ and $n = 2$. Consider $V_1$ to be the $xy$ plane. $V_2$ to be the $xz$ plane. $V_3$ to be the $yz$ plane. All of these are $2$ dimensional subspaces. The intersection of any pair is one dimensional line. However the intersection of all three is just the origin which is not $1$-dimensional.






share|cite|improve this answer









$endgroup$



Consider the vector space $mathbb{R}^3$. Let $t = 3$ and $n = 2$. Consider $V_1$ to be the $xy$ plane. $V_2$ to be the $xz$ plane. $V_3$ to be the $yz$ plane. All of these are $2$ dimensional subspaces. The intersection of any pair is one dimensional line. However the intersection of all three is just the origin which is not $1$-dimensional.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jun 30 '12 at 21:03









WilliamWilliam

17.3k22256




17.3k22256












  • $begingroup$
    ok thanks,it is very helpful
    $endgroup$
    – Babak Miraftab
    Jun 30 '12 at 21:07


















  • $begingroup$
    ok thanks,it is very helpful
    $endgroup$
    – Babak Miraftab
    Jun 30 '12 at 21:07
















$begingroup$
ok thanks,it is very helpful
$endgroup$
– Babak Miraftab
Jun 30 '12 at 21:07




$begingroup$
ok thanks,it is very helpful
$endgroup$
– Babak Miraftab
Jun 30 '12 at 21:07


















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