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

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.

• 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 '18 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. – user1101010 Nov 6 '18 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 '18 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. – user1101010 Nov 8 '18 at 15:26
• 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 '18 at 22:14