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.

  • $\begingroup$ 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 $n=3$. $\endgroup$ – André Porto Nov 6 '18 at 18:29
  • $\begingroup$ 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. $\endgroup$ – user1101010 Nov 6 '18 at 18:39
  • $\begingroup$ 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? $\endgroup$ – André Porto Nov 8 '18 at 13:54
  • $\begingroup$ @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. $\endgroup$ – user1101010 Nov 8 '18 at 15:26
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    $\begingroup$ 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$$V_1+\cdots+V_k.$$ $\endgroup$ – Alejandro Nasif Salum Nov 13 '18 at 22:14

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