I want to see if I am understanding the proof of the replacement theorem correctly.
Let $V$ be a vector space that is spanned by a set $G$ containing $n$ vectors.
Let $L \subseteq V$ be a linearly independent subset containing $m$ vectors.
Then $m\leq n$ and there exists a subset H of G containing exactly $n-m$ vectors s.t. $L \cup H$ generates $V$.
The proof in the book says to start with $m=0$. In that case $L=\varnothing$, the empty set.
Let's assume this theorem is true for some integer $m \geq 0$.
Let $L = \{v_j\}_{j=1}^{m+1}$ and define it as a linearly independent subset of $V$ consisting of $m+1$ vectors.
Since any subset of a linearly independent set is linearly independent as well ($S_1 \subseteq S_2 \subseteq V$), then $\{v_j\}_{j=1}^m$ is linearly independent also.
It then says to use the induction hypothesis to say that $m \geq n$. Well, ok, but that seems sort of pulled out of nowhere to me. The next step is to say that there is therefore another subset, $\{u_k\}_{k=1}^{n-m}$ of G such that $\{v_j\}_{j=1}^m \cup \{u_k\}_{k=1}^{n-m}$ spans $V$.
That being the case there are scalars $(a_j)_{j=1}^m$ and $(b_k)_{k=1}^{n-m}$ which you can multiply by the vectors $v_j$ and $u_k$. You can then add the two sets of vectors, yielding $$\sum_{j=1}^m a_j v_j + \sum_{k=1}^{n-m} b_k u_k = v_{m+1}$$
At this point I am not sure I understand things. Because we are assuming $n-m>0$ — otherwise $v_{m+1}$ is linearly dependent. But then it says not only is $n>m $ but $n>m+1$, and that is where I am losing the plot to get to the next step.
Best Answer
I'm assuming you're working through the proof in Friedberg et al. as its very similar. I'd really encourage you to double check what you've written because there are many typos above and probably isn't coherent enough for someone without a copy of the text.
Okay, and I hope you realize that this also proves the inequality $m \leq n$ for the base case.
So we're going to use induction on $m$ to prove the theorem. Also, it should be $L = \{v_1, \ldots, v_{m+1} \}$.
That's correct. I'll call this set $L'$.
It isn't pulled out of nowhere because it follows from the induction hypothesis. As we have assumed it for some $m \geq 0$, we can apply the theorem to $\{v_1, \ldots, v_m\}$ which gives us the inequality $m \leq n$ and the set $H' = \{u_1, \ldots, u_{n-m}\} \subset G$ such that $L' \cup H'$ generates $V$.
We can do this as $L' \cup H'$ generates $V$.
What follows is this:
Do you understand why it must be $n - m >0$? Well if we had $n - m = 0$ then $H' = \emptyset$ and this would mean that $(\text{*})$ is only composed of vectors of $L'$ and therefore, $L' \cup \{v_{m+1}\} = L$ would be linearly dependent as $v_{m+1} \in \operatorname{span}(L')$, but that would be a contradiction.
Why? Because another way of expressing $n > m$, as $n$ and $m$ are integers, is $n \geq m +1$. That's all there is to it.
Of course. If $b_i$ were all zero in $(\text{*})$ it would again only be composed of vectors of $L'$ and we know that that can't be the case for the same reason as above.
This should be clear.
To see why you can apply Theorem 1.5 you must remember that if $S \subseteq V$ then $\operatorname{span}(S)$ is a subspace of $V$ and therefore, by Theorem 1.5, if $W \subseteq \operatorname{span}(S)$ then $\operatorname{span}(W) \subseteq \operatorname{span}(S)$. So we have that $V = \operatorname{span}(L' \cup H') \subseteq \operatorname{span}(L \cup H)$ and since obviously, $\operatorname{span}(L \cup H) \subseteq V$ we have $\operatorname{span}(L \cup H) = V$.
This should be clear.
Hope this helps!