This is cross post to the question at MSE.

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?

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.

  • 1
    $\begingroup$ For the variation of the problem with $\mathbb{R}$ replaced with $\mathbb{C}$ (not sure it helps): consider the set $S$ of all tuples $v_1,...,v_n$, where $v_i\in V_i$. Clearly, $S$ is a vector space, and so we can identify it with a linear subspace of $\mathbb{C}^{n^2}$. Then $\det(v_1,...,v_n)$ is a polynomial of a degree at most $n$ on $S$, who is linearly isomorphic to $\mathbb{C}^{mn}$. Since the zero-set of an analytic function does not separate $\mathbb{C}^{mn}$, it follows that the bases for a connected set. $\endgroup$ – erz Nov 18 '18 at 13:06

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