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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 $m < n$. Suppose we can 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 $\mathbb R^n$.

Intuitively, it seems the path can be chosen within each subspace. But I failed to formally state this: I was thinking to continuously choose a path $\gamma_j: [0,1] \to V_j$ such that $\gamma_j(0) = v_j$ and $\gamma_j(1) = w_j$ but get lost on whether or not we can guarantee the linearly independence in the process. I am not sure whether orientation would be relevant, but if so let us assume the basis $\{v_j\}$ and $\{w_j\}$ have the same orientation.


Edit: As pointed out by Paul Frost, if $m=1$, this is not possible. But I would love to see a general case for $m \ge 2$.

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    $\begingroup$ How do you mean this 'whether $U$ is connected with $W$? $\endgroup$ – Berci Oct 18 '18 at 23:39
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    $\begingroup$ @Berci: I edited the question to make it more clear. Thanks. $\endgroup$ – user1101010 Oct 19 '18 at 2:09
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    $\begingroup$ The set of all bases of $\mathbb{R}^n$ can be identified with the group $GL_n(\mathbb{R})$. It has two (path) components corresponding to the two possible orientations. Hence any two bases having the same orientation can be connected by a path $\gamma$ as in your question, but we do not necessarily have control by subspaces $V_j$ as you desire. $\endgroup$ – Paul Frost Oct 19 '18 at 10:12
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It is not even possible for $m = n-1$.

Let $E = \{ e_1,\dots,e_n \}$ be the standard basis of $\mathbb{R}^n$. Let $V$ be the the subspace generated by $\{ e_1,\dots,e_{n-1} \}$ (i.e. $V = \mathbb{R}^{n-1} \times \{ 0\}$) and $V_j = V$ for $j =1,\dots,n-1$. Let $V_n$ be the subspace generated by $\{ e_1,\dots,e_{n-2},e_n \}$. Finally let $U = E$ and $W = \{ e_1,\dots,-e_{n-1},-e_n \}$. These are two bases of $\mathbb{R}^n$ which have the same orientation.

Assume there exist paths $\gamma_j$ as desired. Then $\gamma_n$ is a path in $V_n \subset \mathbb{R}^n$ such that $\gamma_n(0) = e_n$, $\gamma_n(1) = -e_n$. If $p_n : \mathbb{R}^n \to \mathbb{R}$ denotes the projection $p_n(x_1,\dots,x_n) = x_n$, we see that $p_n\gamma_n(0) = 1$ and $p_n\gamma_n(1) = -1$. Since $p_n\gamma_n$ is continuous, there exists $t \in I$ such that $p_n\gamma_n(t) = 0$ which means $\gamma_n(t) \in V$. Hence $\gamma(t) \in V$. This shows that $\gamma(t)$ cannot be a basis of $\mathbb{R}^n$ which is a contradiction.

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In general it is not true. Let $U =\{u_1,\dots,u_n \}$ be a basis of $\mathbb{R}^n$ and $V_j$ be the one-dimensional subspace generated by $u_j$. But now for any collection $\epsilon_j \in \{ 0,1 \}$ the $w_j = (-1)^{\epsilon_j}u_j$ form again a basis of $\mathbb{R}^n$, andf if $\epsilon_j = 1$, then any continuous path in $V_j$ connecting $u_j$ and $w_j$ goes through $0$ so that $\gamma(t)$ cannot be a basis for all $t$.

Edited:

As a first step beyond $m=1$ it seems reasonable to study the case $n= 3, m=2$.

Morerover, if the assertion is true for $m=2$, then it also true for all $m =2,\dots,n-1$ (because for each $m > 2$ there exist two-dimensional subspaces $V'_j \subset V_j$ such that $u_j,w_j \in V'_j$).

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  • $\begingroup$ Thanks for answering my question. Could you give an example for the dimension of subspace $m \ge 2$? $\endgroup$ – user1101010 Oct 19 '18 at 15:56
  • $\begingroup$ I do not have an example. However, I add a remark to my answer. $\endgroup$ – Paul Frost Oct 19 '18 at 17:12
  • $\begingroup$ Thanks. I have upvoted your answer. I do want to see the more general case. Do you mind me changing the question to be $m>2$ explicitly? $\endgroup$ – user1101010 Oct 19 '18 at 17:29
  • $\begingroup$ Absolutely okay! $\endgroup$ – Paul Frost Oct 19 '18 at 17:32

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