Does there exist a continuous path between two sets of oriented basis for a vector space out of a collection...











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Let $V_1, V_2, dots, V_n$ be a collection of vector subspaces in $mathbb R^n$. For each $j=1, dots, n$, $dim(V_j) = m$ with $2 le m < n$. We also have the condition: for any collection of $lceil{frac n m}rceil$ vector spaces from ${V_1, dots, V_n}$, then $V_{k_1} + dots + V_{k_{lceil frac n m rceil}} = mathbb R^n$. Suppose we construct a basis $U = {u_1, dots, u_n}$ of $mathbb R^n$ in the manner: $u_j in V_j$ for each $j$. Now suppose we construct another basis $W = {w_1, dots, w_n}$ in the same manner, i.e., $w_j in V_j$ for each $j$. I am wondering whether $U$ is connected with $W$ in the sense: there is a path $gamma = gamma_1 times gamma_2 times dots times gamma_n$, where each $gamma_j: [0,1] to V_j$ is a continuous path connecting $v_j$ and $w_j$ in $V_j$ and for each $t$: $gamma(t)$ forms a basis for $mathbb R^n$. We assume the basis ${v_j}$ and ${w_j}$ have the same orientation.



The basis can be identified by $GL_n(mathbb R)_+$ or $GL_n(mathbb R)_-$ and we know they are connected. But is there a way to guarantee on the path, each column vector only varies in the corresponding subspace?





I asked a similar question here Constructing a continuous path between two sets of oriented basis for a vector space out of a collection of subspaces
. The conditions are stronger here.



An example of $V_1, dots, V_n$: suppose $n=5$, $m=2$. The construction I have in mind is: $V_i = text{span} ( (1, a_i, 0, 0, 0), (0, 0, 1, a_i, a_i^2))$. As long as $a_1 neq dots neq a_5 neq 0$, any three subspace would span $mathbb R^5$. For other cases, we can use similar idea.










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  • A necessary condition for the existence of this path is: $det([v_1,...,v_n])$ and $det([w_1,...,w_n])$ have the same sign. If we assume that, then the existence of this path is true when $n=3$.
    – André Porto
    Nov 6 at 18:29












  • Can you elaborate? I know $GL_n(mathbb R)_{pm}$ is connected. But I want a path such that each basis vector stays in the corresponding subspace.
    – user9527
    Nov 6 at 18:39










  • I just said that it is a necessary condition. May be not sufficient. In fact, I'm not convinced that there exists a set $V_1, ..., V_n$ satisfying your hypothesis. Even for $n=4$ and $m=2$. Do you have an example of sets satisfying these conditions? It's pretty simple when $n=3$. Can you show me an example for $n=4$ or greater?
    – André Porto
    Nov 8 at 13:54












  • @AndréPorto: I have in mind: for example $n=4, m=2$, we take each subspace to be $text{span}left(begin{pmatrix} 1 \ a_1\ 0 \ 0 end{pmatrix}, begin{pmatrix} 0 \ 0 \ 1 \ a_1 end{pmatrix}right)$ for some $a_1 in mathbb R^{times}$. If we take distinct numbers for each subspace and I am not mistaken, this should satisfy the assumption. For other dimensions, we can use the same idea.
    – user9527
    Nov 8 at 15:26








  • 1




    Just a comment: what you note as something like $span(V_1,ldots,V_k)$ (its meaning was properly made clear by you, btw) is simply what is known as the sum of the subspaces, that is$$V_1+cdots+V_k.$$
    – Alejandro Nasif Salum
    Nov 13 at 22:14

















up vote
12
down vote

favorite
2












Let $V_1, V_2, dots, V_n$ be a collection of vector subspaces in $mathbb R^n$. For each $j=1, dots, n$, $dim(V_j) = m$ with $2 le m < n$. We also have the condition: for any collection of $lceil{frac n m}rceil$ vector spaces from ${V_1, dots, V_n}$, then $V_{k_1} + dots + V_{k_{lceil frac n m rceil}} = mathbb R^n$. Suppose we construct a basis $U = {u_1, dots, u_n}$ of $mathbb R^n$ in the manner: $u_j in V_j$ for each $j$. Now suppose we construct another basis $W = {w_1, dots, w_n}$ in the same manner, i.e., $w_j in V_j$ for each $j$. I am wondering whether $U$ is connected with $W$ in the sense: there is a path $gamma = gamma_1 times gamma_2 times dots times gamma_n$, where each $gamma_j: [0,1] to V_j$ is a continuous path connecting $v_j$ and $w_j$ in $V_j$ and for each $t$: $gamma(t)$ forms a basis for $mathbb R^n$. We assume the basis ${v_j}$ and ${w_j}$ have the same orientation.



The basis can be identified by $GL_n(mathbb R)_+$ or $GL_n(mathbb R)_-$ and we know they are connected. But is there a way to guarantee on the path, each column vector only varies in the corresponding subspace?





I asked a similar question here Constructing a continuous path between two sets of oriented basis for a vector space out of a collection of subspaces
. The conditions are stronger here.



An example of $V_1, dots, V_n$: suppose $n=5$, $m=2$. The construction I have in mind is: $V_i = text{span} ( (1, a_i, 0, 0, 0), (0, 0, 1, a_i, a_i^2))$. As long as $a_1 neq dots neq a_5 neq 0$, any three subspace would span $mathbb R^5$. For other cases, we can use similar idea.










share|cite|improve this question
























  • A necessary condition for the existence of this path is: $det([v_1,...,v_n])$ and $det([w_1,...,w_n])$ have the same sign. If we assume that, then the existence of this path is true when $n=3$.
    – André Porto
    Nov 6 at 18:29












  • Can you elaborate? I know $GL_n(mathbb R)_{pm}$ is connected. But I want a path such that each basis vector stays in the corresponding subspace.
    – user9527
    Nov 6 at 18:39










  • I just said that it is a necessary condition. May be not sufficient. In fact, I'm not convinced that there exists a set $V_1, ..., V_n$ satisfying your hypothesis. Even for $n=4$ and $m=2$. Do you have an example of sets satisfying these conditions? It's pretty simple when $n=3$. Can you show me an example for $n=4$ or greater?
    – André Porto
    Nov 8 at 13:54












  • @AndréPorto: I have in mind: for example $n=4, m=2$, we take each subspace to be $text{span}left(begin{pmatrix} 1 \ a_1\ 0 \ 0 end{pmatrix}, begin{pmatrix} 0 \ 0 \ 1 \ a_1 end{pmatrix}right)$ for some $a_1 in mathbb R^{times}$. If we take distinct numbers for each subspace and I am not mistaken, this should satisfy the assumption. For other dimensions, we can use the same idea.
    – user9527
    Nov 8 at 15:26








  • 1




    Just a comment: what you note as something like $span(V_1,ldots,V_k)$ (its meaning was properly made clear by you, btw) is simply what is known as the sum of the subspaces, that is$$V_1+cdots+V_k.$$
    – Alejandro Nasif Salum
    Nov 13 at 22:14















up vote
12
down vote

favorite
2









up vote
12
down vote

favorite
2






2





Let $V_1, V_2, dots, V_n$ be a collection of vector subspaces in $mathbb R^n$. For each $j=1, dots, n$, $dim(V_j) = m$ with $2 le m < n$. We also have the condition: for any collection of $lceil{frac n m}rceil$ vector spaces from ${V_1, dots, V_n}$, then $V_{k_1} + dots + V_{k_{lceil frac n m rceil}} = mathbb R^n$. Suppose we construct a basis $U = {u_1, dots, u_n}$ of $mathbb R^n$ in the manner: $u_j in V_j$ for each $j$. Now suppose we construct another basis $W = {w_1, dots, w_n}$ in the same manner, i.e., $w_j in V_j$ for each $j$. I am wondering whether $U$ is connected with $W$ in the sense: there is a path $gamma = gamma_1 times gamma_2 times dots times gamma_n$, where each $gamma_j: [0,1] to V_j$ is a continuous path connecting $v_j$ and $w_j$ in $V_j$ and for each $t$: $gamma(t)$ forms a basis for $mathbb R^n$. We assume the basis ${v_j}$ and ${w_j}$ have the same orientation.



The basis can be identified by $GL_n(mathbb R)_+$ or $GL_n(mathbb R)_-$ and we know they are connected. But is there a way to guarantee on the path, each column vector only varies in the corresponding subspace?





I asked a similar question here Constructing a continuous path between two sets of oriented basis for a vector space out of a collection of subspaces
. The conditions are stronger here.



An example of $V_1, dots, V_n$: suppose $n=5$, $m=2$. The construction I have in mind is: $V_i = text{span} ( (1, a_i, 0, 0, 0), (0, 0, 1, a_i, a_i^2))$. As long as $a_1 neq dots neq a_5 neq 0$, any three subspace would span $mathbb R^5$. For other cases, we can use similar idea.










share|cite|improve this question















Let $V_1, V_2, dots, V_n$ be a collection of vector subspaces in $mathbb R^n$. For each $j=1, dots, n$, $dim(V_j) = m$ with $2 le m < n$. We also have the condition: for any collection of $lceil{frac n m}rceil$ vector spaces from ${V_1, dots, V_n}$, then $V_{k_1} + dots + V_{k_{lceil frac n m rceil}} = mathbb R^n$. Suppose we construct a basis $U = {u_1, dots, u_n}$ of $mathbb R^n$ in the manner: $u_j in V_j$ for each $j$. Now suppose we construct another basis $W = {w_1, dots, w_n}$ in the same manner, i.e., $w_j in V_j$ for each $j$. I am wondering whether $U$ is connected with $W$ in the sense: there is a path $gamma = gamma_1 times gamma_2 times dots times gamma_n$, where each $gamma_j: [0,1] to V_j$ is a continuous path connecting $v_j$ and $w_j$ in $V_j$ and for each $t$: $gamma(t)$ forms a basis for $mathbb R^n$. We assume the basis ${v_j}$ and ${w_j}$ have the same orientation.



The basis can be identified by $GL_n(mathbb R)_+$ or $GL_n(mathbb R)_-$ and we know they are connected. But is there a way to guarantee on the path, each column vector only varies in the corresponding subspace?





I asked a similar question here Constructing a continuous path between two sets of oriented basis for a vector space out of a collection of subspaces
. The conditions are stronger here.



An example of $V_1, dots, V_n$: suppose $n=5$, $m=2$. The construction I have in mind is: $V_i = text{span} ( (1, a_i, 0, 0, 0), (0, 0, 1, a_i, a_i^2))$. As long as $a_1 neq dots neq a_5 neq 0$, any three subspace would span $mathbb R^5$. For other cases, we can use similar idea.







linear-algebra general-topology path-connected






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share|cite|improve this question













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edited Nov 14 at 23:47

























asked Nov 5 at 23:59









user9527

1,1541628




1,1541628












  • A necessary condition for the existence of this path is: $det([v_1,...,v_n])$ and $det([w_1,...,w_n])$ have the same sign. If we assume that, then the existence of this path is true when $n=3$.
    – André Porto
    Nov 6 at 18:29












  • Can you elaborate? I know $GL_n(mathbb R)_{pm}$ is connected. But I want a path such that each basis vector stays in the corresponding subspace.
    – user9527
    Nov 6 at 18:39










  • I just said that it is a necessary condition. May be not sufficient. In fact, I'm not convinced that there exists a set $V_1, ..., V_n$ satisfying your hypothesis. Even for $n=4$ and $m=2$. Do you have an example of sets satisfying these conditions? It's pretty simple when $n=3$. Can you show me an example for $n=4$ or greater?
    – André Porto
    Nov 8 at 13:54












  • @AndréPorto: I have in mind: for example $n=4, m=2$, we take each subspace to be $text{span}left(begin{pmatrix} 1 \ a_1\ 0 \ 0 end{pmatrix}, begin{pmatrix} 0 \ 0 \ 1 \ a_1 end{pmatrix}right)$ for some $a_1 in mathbb R^{times}$. If we take distinct numbers for each subspace and I am not mistaken, this should satisfy the assumption. For other dimensions, we can use the same idea.
    – user9527
    Nov 8 at 15:26








  • 1




    Just a comment: what you note as something like $span(V_1,ldots,V_k)$ (its meaning was properly made clear by you, btw) is simply what is known as the sum of the subspaces, that is$$V_1+cdots+V_k.$$
    – Alejandro Nasif Salum
    Nov 13 at 22:14




















  • A necessary condition for the existence of this path is: $det([v_1,...,v_n])$ and $det([w_1,...,w_n])$ have the same sign. If we assume that, then the existence of this path is true when $n=3$.
    – André Porto
    Nov 6 at 18:29












  • Can you elaborate? I know $GL_n(mathbb R)_{pm}$ is connected. But I want a path such that each basis vector stays in the corresponding subspace.
    – user9527
    Nov 6 at 18:39










  • I just said that it is a necessary condition. May be not sufficient. In fact, I'm not convinced that there exists a set $V_1, ..., V_n$ satisfying your hypothesis. Even for $n=4$ and $m=2$. Do you have an example of sets satisfying these conditions? It's pretty simple when $n=3$. Can you show me an example for $n=4$ or greater?
    – André Porto
    Nov 8 at 13:54












  • @AndréPorto: I have in mind: for example $n=4, m=2$, we take each subspace to be $text{span}left(begin{pmatrix} 1 \ a_1\ 0 \ 0 end{pmatrix}, begin{pmatrix} 0 \ 0 \ 1 \ a_1 end{pmatrix}right)$ for some $a_1 in mathbb R^{times}$. If we take distinct numbers for each subspace and I am not mistaken, this should satisfy the assumption. For other dimensions, we can use the same idea.
    – user9527
    Nov 8 at 15:26








  • 1




    Just a comment: what you note as something like $span(V_1,ldots,V_k)$ (its meaning was properly made clear by you, btw) is simply what is known as the sum of the subspaces, that is$$V_1+cdots+V_k.$$
    – Alejandro Nasif Salum
    Nov 13 at 22:14


















A necessary condition for the existence of this path is: $det([v_1,...,v_n])$ and $det([w_1,...,w_n])$ have the same sign. If we assume that, then the existence of this path is true when $n=3$.
– André Porto
Nov 6 at 18:29






A necessary condition for the existence of this path is: $det([v_1,...,v_n])$ and $det([w_1,...,w_n])$ have the same sign. If we assume that, then the existence of this path is true when $n=3$.
– André Porto
Nov 6 at 18:29














Can you elaborate? I know $GL_n(mathbb R)_{pm}$ is connected. But I want a path such that each basis vector stays in the corresponding subspace.
– user9527
Nov 6 at 18:39




Can you elaborate? I know $GL_n(mathbb R)_{pm}$ is connected. But I want a path such that each basis vector stays in the corresponding subspace.
– user9527
Nov 6 at 18:39












I just said that it is a necessary condition. May be not sufficient. In fact, I'm not convinced that there exists a set $V_1, ..., V_n$ satisfying your hypothesis. Even for $n=4$ and $m=2$. Do you have an example of sets satisfying these conditions? It's pretty simple when $n=3$. Can you show me an example for $n=4$ or greater?
– André Porto
Nov 8 at 13:54






I just said that it is a necessary condition. May be not sufficient. In fact, I'm not convinced that there exists a set $V_1, ..., V_n$ satisfying your hypothesis. Even for $n=4$ and $m=2$. Do you have an example of sets satisfying these conditions? It's pretty simple when $n=3$. Can you show me an example for $n=4$ or greater?
– André Porto
Nov 8 at 13:54














@AndréPorto: I have in mind: for example $n=4, m=2$, we take each subspace to be $text{span}left(begin{pmatrix} 1 \ a_1\ 0 \ 0 end{pmatrix}, begin{pmatrix} 0 \ 0 \ 1 \ a_1 end{pmatrix}right)$ for some $a_1 in mathbb R^{times}$. If we take distinct numbers for each subspace and I am not mistaken, this should satisfy the assumption. For other dimensions, we can use the same idea.
– user9527
Nov 8 at 15:26






@AndréPorto: I have in mind: for example $n=4, m=2$, we take each subspace to be $text{span}left(begin{pmatrix} 1 \ a_1\ 0 \ 0 end{pmatrix}, begin{pmatrix} 0 \ 0 \ 1 \ a_1 end{pmatrix}right)$ for some $a_1 in mathbb R^{times}$. If we take distinct numbers for each subspace and I am not mistaken, this should satisfy the assumption. For other dimensions, we can use the same idea.
– user9527
Nov 8 at 15:26






1




1




Just a comment: what you note as something like $span(V_1,ldots,V_k)$ (its meaning was properly made clear by you, btw) is simply what is known as the sum of the subspaces, that is$$V_1+cdots+V_k.$$
– Alejandro Nasif Salum
Nov 13 at 22:14






Just a comment: what you note as something like $span(V_1,ldots,V_k)$ (its meaning was properly made clear by you, btw) is simply what is known as the sum of the subspaces, that is$$V_1+cdots+V_k.$$
– Alejandro Nasif Salum
Nov 13 at 22:14

















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