# Constructing 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 $$m < n$$. 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 $$m \ge 2$$.

• How do you mean this 'whether $U$ is connected with $W$? – Berci Oct 18 '18 at 23:39
• @Berci: I edited the question to make it more clear. Thanks. – user1101010 Oct 19 '18 at 2:09
• 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. – Paul Frost Oct 19 '18 at 10:12

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.

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 $$u_j,w_j \in V'_j$$).

• Thanks for answering my question. Could you give an example for the dimension of subspace $m \ge 2$? – user1101010 Oct 19 '18 at 15:56
• I do not have an example. However, I add a remark to my answer. – Paul Frost Oct 19 '18 at 17:12
• 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? – user1101010 Oct 19 '18 at 17:29
• Absolutely okay! – Paul Frost Oct 19 '18 at 17:32