proof of the full exchange lemma Let V be spanned by $\{v_1,...,v_k\}$ and let $\{u_1,...,u_k\}$ be a linearly independent subset of V, then:
1) $k\leq n$
2) $\exists$ a spanning set $\{w_1,...,w_n\}$ for V where $w_i = u_i$ for $1 \leq i \leq k$
and $w_i \in \{v_1...v_n\}$ for $k < 1$
proof:
by induction on k,
for k = 1, proved by the little exchange lemma (done before),
assume true for $k-1$, then we have constructed a set $\{w_1,...,w_n\}$ which spans with $w_i = u_i$ $i \leq k-1$ and $w_i \in \{v_1,...,v_n\}$ for $k-1 > i$
we can express $u_k = \displaystyle \sum_{r=1}^n\lambda_rw_r$ as $\{w_1,...,w_n\}$ spans,
since $\{u_1,...,u_k\}$ is LI, $u_k \not = 0 $ for some $\lambda_r \not = 0$ (?)
Claim that $\lambda_r \not = 0 $ for some $k \leq t$ Otherwise, $u_k = \sum_{r=1}^{k-1}\lambda_ru_r$ (?) which is a dependence relationship and contradicts that $\{u_1...u_k\}$ is LI
Choose r, $k \leq r \leq n $ s.t. $\lambda_r \not = $, by the simplified little exchange lemma, $\{u_1,...,u_{k-1},u_k,w_{k+1}...w_n\}$ is a spanning set, hence by swapping $\{u_1...u_{k-1}, w_{k+1}...,u_k,...,w_n\}$ still spans V.(?)
I have questions where the question marks are,
firstly I understand that $u_k \not = 0$, but what does "$u_k \not= 0$ for some $\lambda_r \not=0$" mean? What has lambda got to with it
secondly if $\lambda_r = 0$ why does that give a dependence relationship, and why do we need for $k \leq t$?
Thirdly, why does rearranging it prove anything? I don't even know why the proof is finished - is it proved for k+1 now? have we even proved $k \leq n$?
 A: I think you can work your way through this much easierif you first prove the useful, easy
Proposition: Let $\;B:=\{v_1,...,v_k\}\;$ be a linearly independent set in a vector space $\;V\;$ and let $\;x\in V\;$ . Then, 
$$x\in\text{Span}\,(B)\iff \{v_1,...,v_k,x\}\;\text{is linearly dependent}$$
Can you see how to use the above in the different versions of the exchange theorem?
A: Comments on the proof (by question marks): 


*

*Since $u_k\neq 0$ there must be a coefficient $\lambda_r$ which is not zero.

*Since $\{u_1,..,u_k\}$ forms a linearly independent set, there must be some $\lambda_r \neq 0$ for $r > k$.

Another proof which is more intuitive:
Start from $U_k=span(\{u_1,..,u_k\})$. Then $U_k$ has dimension $k$. If $k=n$ we are done; else $k<n$ and therefore $U_k$ cannot contain all $v_i,\ i=1..n$. Thus there exists an index $i_1$ such that $v_{i_1}\notin U_k$ and define $U_{k+1}=span(U_k \cup \{v_{i_1}\})$.
If $k+1<n$ then again $U_{k+1}$ cannot contain all $v_i$ and there exists an index $i_2$ such that $v_{i_2} \notin U_{k+1}$. Using the same construction we do this until we span the whole space. 
