Proof of the existence of a complementary subspace by extending a basis

Proof

We extend a basis of $U$ to a basis of $V$ and take the span of the new vectors.

1. By every vector space has a basis, applied to $U$, there is a basis $L$ of $U$. We view $L$ as a set of vectors of $U$.
2. The set $L$ is linearly independent in $U$, hence also in $V$, since linear relations are the same in both spaces.
3. By extension to a basis, there is a basis $B$ of $V$ with $L \subseteq B$.
4. Let $M = B \setminus L$, and let $W = \operatorname{span}(M)$ be its linear span.
5. $W$ is a subspace of $V$. It contains $0$, the empty combination. A sum of two finite linear combinations of $M$, or a scalar multiple of one, is again such a combination.
6. We show $U + W = V$, with the sum of subspaces. Let $v \in V$. Since $B$ spans $V$, $v$ is a finite linear combination of vectors of $B$.
7. Group the terms of this combination: those with vectors in $L$ give a vector $u \in U$, and those with vectors in $M$ give a vector $w \in W$. Then $v = u + w$.
8. So $V \subseteq U + W$. Also $U + W \subseteq V$, since $U$ and $W$ are subspaces of $V$. Hence $U + W = V$.
9. We show $U \cap W = \{0\}$. Let $x \in U \cap W$.
10. Since $L$ spans $U$, $x = a_1 \ell_1 + \cdots + a_p \ell_p$ for distinct $\ell_1, \dots, \ell_p \in L$ and scalars $a_i \in F$.
11. Since $x \in W$, $x = c_1 m_1 + \cdots + c_q m_q$ for distinct $m_1, \dots, m_q \in M$ and scalars $c_k \in F$.
12. Subtracting gives $a_1 \ell_1 + \cdots + a_p \ell_p - c_1 m_1 - \cdots - c_q m_q = 0$.
13. The sets $L$ and $M$ are disjoint, so the vectors $\ell_1, \dots, \ell_p, m_1, \dots, m_q$ are distinct elements of $B$.
14. The basis $B$ is linearly independent, so all $a_i$ and $c_k$ are $0$. Hence $x = 0$.
15. By steps 8 and 14, $U + W = V$ and $U \cap W = \{0\}$. So $W$ is a complement of $U$, and $V = U \oplus W$.