Prove joining bases of two disjoint bases spans the entire space I have a question I need to prove or disprove. I think it's correct, but I don't know how to prove it formally.
Let V be a vector space above F, and U be a subspace of V. Let $v_1,...v_n$ (vectors in V)  be the basis of V, and $u_1,...,u_k$ (vectors in U) be the basis of U. we know that $k<n$. Let's define $W = span(v_{k+1},...,v_n)$. Obviously, W is a subspace of V, and $v_{k+1},...,v_n$ is its basis. 
We also know, that for every vector in $u_1,...,u_k$ the vector isn't in W, i.e   $u_1,...,u_k \not\in W $, and also that for every vector in $v_{k+1},...,v_n$ the vector isn't in U, i.e   $v_{k+1},...,v_n \not\in U $. Prove or disprove that $u_1,...,u_k,v_{k+1},...,v_n$ is the basis of V.
I think this is true. I wanted to prove that $u_1,...,u_k,v_{k+1},...,v_n$ is linear independent, and since it's a linear independent group in V, the size of dim V, it's also spans V and therefore the basis of V. I can prove that $u_1,...,u_k,v_{k+1}$ is linear independent, but I don't manage for $u_1,...,u_k,v_{k+1}, v_{k+2}$. is this even the right approach?
 A: We want a linearly independent set of $n$ vectors to form a basis of $V$. We already have $u_i$ for $1 \leq i \leq k$. We then can add $v_{k+1}$ to get a set of $k+1$ independent elements since $v_{k+1} \notin U$. However, now our set spans not just $U$, but $U+\langle v_{k+1} \rangle$. Could $v_{k+2}$ be in this set? Well, let's say it is. Then we have that $v_{k+2}=u+cv_{k+1}$ for some $u \in U$, so $u=v_{k+2}-cv_{k+1} \in W$. However, this could still happen because even if all of the $u_i$ are not in $W$, one of their linear combinations could be.
Let's say $V=\Bbb{R}^4$, $v_1=(1 \ 0 \ 0 \ 0)^t$, $v_2=(0 \ 0 \ 1 \ 0)^t$, $v_3=(0 \ 1 \ 0 \ -1)^t$, $v_4=(0 \ 1 \ 0 \ 1)^t$, $u_1=(0 \ 1 \ -1 \ 0)^t$, and $u_2=(0 \ 1 \ 1 \ 0)^t$. Now, $v_3,v_4 \notin U$ and $u_1,u_2 \notin W$, but $u_1,u_2,v_3,v_4$ does not span $V$ because no linear combination of them can create $v_1$, so the statement is false. The key behind this counterexample is that if we choose $u=u_1+u_2$, then we have $u=v_3+v_4 \in W$, which satisfies the equation at the end of the first paragraph.
