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I'm having trouble understanding this proof about the uniqueness of bases, specifically, a line in the forward direction (highlighted in the image). I feel like I'm missing something obvious, and any clarification would be appreciated.

the proof in question

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  • $\begingroup$ This is almost exactly the same as the proof that prime factorisations, if they exist, are unique. $S$ is analogous to the primes, $a_1 u_1$ is analogous to $p_1^{n_1}$, $+$ is analogous to $\times$, and $\bar{0}$ is analogous to $1$. So if you understand that, then you can translate it over to this setting. $\endgroup$ – Patrick Stevens Jan 19 at 13:12
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Suppose $v = a_1u_1+...+a_nu_n = b_1w_1+...+b_kw_k$ with $u_i,w_i \in S$ and $a_i,b_i \in \mathbb{F}\setminus\{0\}$. Then we have that $$ a_1u_1+...+a_nu_n -b_1w_1-...-b_kw_k = 0 $$ Suppose $a_1 \neq b_l$ and $u_1 \neq w_l$ for all values of $l$. This implies the existence of a nontrivial relation, since $a_1 \neq 0 $ by assumption. Since $u_1,...,u_n, w_1,...,w_k \in S$ and $S$ is linearly independent, this yields a contradiction. This lets you 'pair off' $a_1u_1$ with $b_lw_l$ for some value of $l$, also implying that $n = k$. As with so many things, you can formalize this with induction.

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Since $S$ is linearly independent, the linear dependence relation would normally force all those scalars to be zero; however, by assumption, these scalars are all nonzero. So, $a_1 = b_l$ for some $l$.

Another thing that follows, for the same reason, is that $m = k$. This should’ve been mentioned in the proof, as it is not obvious.

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