Proof of The Basis Theorem in Linear Algebra So I saw the following proof of the Basis Theorem in Leon Simon's "An Introduction to Multivariable Mathematics". I was wondering if anyone could help explain what is happening in it. I understand part a) about 80% and part b) about 1%. I was wondering if anyone could help explain what is going on in the proof. The proof is as follows:
The Basis Theorem:  
Suppose $V$ is a non trivial subspace of $\mathbb R^n.$ Then: 

(a) $V$ has a basis 
(b) If $u_1,...,u_k$ are l.i. vectors in $V$, then there is a basis for $V$ which contains $u_1,...,u_k$; more precisely, there is a basis $v_1, . . . , v_q$ for $V$ with $q \geq k$ and $v_j = u_j$ for each $j = 1, . . . , k$ 
Proof of (a): 
Define:  
$\qquad \qquad \qquad \qquad $ $q =$ max {$l∈${$1,...,n$}$: ∃$ l.i.vectors $w_1,...,w_l ∈V$},  
and choose l.i. vectors $v_1,...,v_q ∈ V$. (Thus, roughly speaking, $v_1,...,v_q$ are chosen to give “a maximum number of linearly independent vectors in $V$.”) We claim that such $v_1, . . . , v_q$ must span $V$.To see this let $v$ be an arbitrary vector in $V$ and consider the vectors $v_1,...,v_q,v$. If $q = n$ this is a set of $n+1$ vectors in $\mathbb R^n=$ span{$e1,...,en$} and so must be l.d. by the Linear Dependence Lemma. On the other hand, if $q < n$ then $q + 1 ≤ n$ and so $v_1,...,v_q,v$ must again be l.d., otherwise $v_1,...,v_q,v$ is a set of $q+1∈${$1,...,n$} l.i. vectors in $V$ contradicting the definition of $q$. Thus, in either case $(q=n,q<n)$, the vectors $v_1,...,v_q,v$ are l.d. Thus,$c_0v+c_1v_1+···+c_qv_q =0$ for some $c_0, . . . , c_q$ not all zero. But of course then $c_0 \neq 0$ because otherwise this identity would tell us that $c_1v_1 + · · · + c_q v_q = 0$ with not all $c_1, . . . , c_q$ zero, contradicting the linear independence of $v_1, . . . , v_q$. Thus, $v = −c_0^{-1} (c_1v_1 + · · · + c_q v_q )$, which completes the proof. 

Proof of (b): 
The proof of (b) is almost the same as the proof of (a), except that we start by defining   
$\qquad$ $q =$ max{$l∈${$k,...,n$}$: ∃$ l.i. vectors $w_1,...,w_l ∈V$
with $w_j = u_j$ for each $j =$ {$1, . . . , k$},   
so that we can select l.i. vectors $v_1,...,v_q ∈V$ with $v_j =u_j$ for $j =1,...,k$. The remainder of the proof is identical, word for word, with the proof of (a), and yields the conclusion that $v_1, . . . , v_q$ span $V$ , and hence $v_1, . . . , v_q$ are a basis for $V$ with $q ≥ k$ and $v_j = u_j$ for each $j = 1, . . . , k$.
 A: Here's what's going down.
In part (a) they're setting $q$ equal to the dimension of $V$. A basis for a vector space $V$ is defined to be a set of linearly independent that span $V$. They set the proof up so that $\{v_{1}, ..., v_{q}\}$ is maximally linearly independent by definition. Then they go on to show that any set which is constructed this way must by necessity span $V$.
In part (b) they doing the same thing by letting $q = dim(V)$. What this implies is that $q \geq k$. If $q = k$ then $\{u_{1}, ... , u_{k}\}$ is a basis for $V$, and the proof of this is exactly the same as in part (a). If $q > k$, then they're assuming that there exists a set of vectors $\{x_{1}, ... x_{q-k}\}$ such that the vectors in this set are linearly independent with each other and with each of the vectors in $\{u_{1}, ... , u_{k}\}$. Once you assume the existence of this set, you can go on to prove that the set $\{u_{1}, ... , u_{k}, x_{1}, ... , x_{q-k}\}$ spans $V$ using the exact same method that's used in part (a).
