I have a question about the following proof of the statement that for every subspace $U$ of a finite dimensional vector space $V$ there is a subspace $W$ of $V$ such that $V = U \oplus W$.
Proof:
Because $V$ is finite-dimensional, so is $U$. Thus there is a basis $(u_1,\ldots,u_m)$ of U. Of course $(u_1,\ldots,u_m)$ is a linearly independent list of vectors in $V$, and thus it can be extended to a basis $(u_1,\ldots,u_m, w_1, \ldots,w_n)$ of $V$. Let $W = \text{span}(w_1,\ldots,w_n).$
To prove that $V = U \oplus W$, we need to show that $$V = U + W \text{ and } U \cap W = \{0\}.$$
To prove the first equation, suppose that $v \in V$. Then, because the list $(u_1,\ldots,u_m, w_1, \ldots,w_n)$ spans $V$, there exist scalars $a_1,\ldots,a_m,b_1,\ldots,b_n \in \mathbb{F}$ such that $$v = \underbrace{a_1u_1 + \dotsb + a_mu_m} + \underbrace{b_1w_1 + \dotsb + b_nw_n}.$$
In other words, we have $v = u + w$, where $u \in U$ and $w \in W$ are defined as above. Thus $v \in U + W$.
I understand everything up until this point, the next part of the proof is the portion that I wish to clarify.
To show that $U \cap W = \{0\}$, suppose $v \in U \cap W$. Then there exist scalars $a_1,\ldots,a_m,b_1,…,b_n \in \mathbb{F}$ such that $$v = a_1u_1 + \dotsb+a_mu_m = b_1w_1 + \dotsb + b_nw_n. $$
Halt! See the above. How does the notion that $v \in U\cap W$ lead to the equality $v = a_1u_1 + \dotsb+a_mu_m = b_1w_1 + \dotsb + b_nw_n$? Or how does the containment of $v$ in $U$ and $W$ yield the equality $v = a_1u_1 + \dotsb+a_mu_m = b_1w_1 + \dotsb + b_nw_n$?
Best Answer
Since $v$ is in $U$, it can be written as a linear combination of basis elements of $U$ , and the same goes for $W$.
While you're at it, and since this wasn't too thrilling, you can go ahead and prove that
$({\rm i})$ $U\oplus W=V$
$(\rm ii)$ If $B'$ is a basis of $U$ and $B''$ a basis of $W$ then $B=B'\cup B''$ is a basis of $V$
are equivalent statements.