Let $V$ be a finite-dimensional vector space. Let $U$ be a subspace of $V$. If I extend a basis $\{v_1, ... , v_m\}$ for $U$ to a basis $\{v_1, ... , v_n\}$ for $V$, does this then imply that $V=U\bigoplus Span(v_{m+1}, ... , v_n)$?
My attempt: I feel like it should because if $v\neq 0 \in V$ then either $v\in U$ or not. If so then $v\notin Span(v_{m+1}, ... , v_n)$ and vice versa if $v$ is not in $U$. Is that right?
The proof as it is now is not correct. The statement(which you also give no justification for) “either $v\in U$ or not” is not true. Consider the subspaces $$U_1 = \{(x,0) \in R^2\} $$ $$U_2 = \{(0,y) \in R^2\}$$ It can be easily shown that $$R^2 = U_1 \oplus U_2$$ Now notice that the vector $(1.1) \in R^2$ is neither in $U_1$ nor in $U_2$. But rather in their sum.
Another giveaway that the proof is wrong is the fact that you never use your hypothesis. Presumably this problem comes from a source that understands the necessity of the hypothesis.
What you should do is just use your hypothesis that $v_1,\dots, v_n$ is linearly independent. Because they are linearly independent, we have that $$U\cap span(v_{m+1}, \dots, v_n)=\{0\}$$ as if that were not true, then the list $v_1,\dots,v_n$ would be linearly dependent.
Now use the(easy to prove) theorem that the sum of two subspaces $U$ and $W$ is a direct sum if and only if $U\cap W =\{0\}$ to conclude that $$U+span(v_{m+1}, \dots, v_n)$$ is a direct sum.