Combination of two sets of basis Given two sets of basis $V = \{v_1,...,v_n\}$ and $U = \{u_1,...,u_n\}$, each spanning $\mathbb{R}^n$, is it possible to select a subset of the two sets $T$ such that T = $\{v_1\, ...,v_m\} \cup \{u_1\, ...,v_o\}$ with $m+o = n$ such that $T$ spans $\mathbb{R}^n$?
 A: First, we need to impose the condition that $m\neq 0$ and $o\neq 0$, or the question becomes trivial because we can always choose $T=V$ or $T=U$.
Now suppose $m,o\neq 0$. We want to know if we can swap any $u_i$ with any $v_j$. We claim that this is possible. Consider $u_1$. If we can swap $u_1$ with $v_n$, that is, $\{u_1,v_1,\cdots,\cdots,v_{n-1}\}$ is linearly independent, then we are done. If not, $u_1$ is a linear combination of $v_1,\cdots,v_{n-1}$. Now do the same thing for $u_2,\cdots,u_n$. If we can swap any of these $u_i$ with $v_n$, then we are done. If not, each of $u_2,\cdots,u_n$ is a linear combination of $v_1,\cdots,v_{n-1}$. Since $\{u_1,\cdots,u_n\}$ spans $\mathbb{R}^n$, this implies $\{v_1,\cdots,v_{n-1}\}$ spans $\mathbb{R}^n$, which is a contradiction. Therefore, we must be able to swap $v_n$ with some $u_i$ and create a new basis.
A: You can do this many ways. Start with any subset of $U$. Then go through the elements of $V$ one at a time in any order you like. Add each to your growing basis whenever it is independent of what you have so far. Since $V$ by itself spans the space, when you are done you will have a basis of the space consisting of some elements of $U$ and some of $V$.
