[proofplan] We proceed by induction on $n$, the size of the linearly independent set. At each step, we express $e_{k+1}$ as a linear combination of the current spanning set, identify an original spanning vector $f_j$ with a non-zero coefficient (which must exist to avoid contradicting independence), swap it for $e_{k+1}$, and verify the new set still spans. After $n$ exchanges, the inequality $n \leq m$ follows from the requirement that exchangeable vectors remain available. [/proofplan] [step:Establish the base case $n = 0$] When $n = 0$, the linearly independent set is empty. The claim holds vacuously: $0 \leq m$ and the original spanning set $\{f_1, \ldots, f_m\}$ spans $V$ without modification. [/step] [step:Perform the inductive exchange: replace $f_{k+1}$ by $e_{k+1}$] Suppose the result holds for linearly independent sets of size $k < n$. By the inductive hypothesis, after reordering the $f_i$, the set \begin{align*} \{e_1, \ldots, e_k, f_{k+1}, \ldots, f_m\} \end{align*} spans $V$. Since this set spans $V$, there exist scalars $\alpha_1, \ldots, \alpha_k, \beta_{k+1}, \ldots, \beta_m \in \mathbb{F}$ with \begin{align*} e_{k+1} = \sum_{i=1}^{k}\alpha_i e_i + \sum_{j=k+1}^{m}\beta_j f_j. \end{align*} [claim:Some $\beta_j$ is non-zero] There exists $j \in \{k+1, \ldots, m\}$ with $\beta_j \neq 0$. [/claim] [proof] Suppose for contradiction that $\beta_j = 0$ for all $j \in \{k+1, \ldots, m\}$. Then $e_{k+1} = \sum_{i=1}^{k}\alpha_i e_i$, expressing $e_{k+1}$ as a linear combination of $e_1, \ldots, e_k$. This contradicts the linear independence of $\{e_1, \ldots, e_n\}$ (a subset of a linearly independent set is linearly independent, so $\{e_1, \ldots, e_{k+1}\}$ is independent, but the relation $e_{k+1} - \sum \alpha_i e_i = \mathbf{0}$ with coefficient $1$ on $e_{k+1}$ is non-trivial). [/proof] After reordering, assume $\beta_{k+1} \neq 0$. Solving for $f_{k+1}$: \begin{align*} f_{k+1} = \frac{1}{\beta_{k+1}}\left(e_{k+1} - \sum_{i=1}^{k}\alpha_i e_i - \sum_{j=k+2}^{m}\beta_j f_j\right). \end{align*} This expresses $f_{k+1}$ as a linear combination of $\{e_1, \ldots, e_{k+1}, f_{k+2}, \ldots, f_m\}$. Any $v \in V$ is a linear combination of the previous spanning set $\{e_1, \ldots, e_k, f_{k+1}, \ldots, f_m\}$; replacing each occurrence of $f_{k+1}$ by the expression above shows that \begin{align*} \{e_1, \ldots, e_{k+1}, f_{k+2}, \ldots, f_m\} \end{align*} also spans $V$. This completes the inductive step. [guided] The exchange argument works as follows. We have a spanning set $\{e_1, \ldots, e_k, f_{k+1}, \ldots, f_m\}$ from the previous step and want to "swap in" $e_{k+1}$. Since this set spans $V$, we can write: \begin{align*} e_{k+1} = \sum_{i=1}^{k}\alpha_i e_i + \sum_{j=k+1}^{m}\beta_j f_j. \end{align*} The crucial point is that at least one $\beta_j$ must be non-zero. If all $\beta_j = 0$, then $e_{k+1} = \sum_{i=1}^{k}\alpha_i e_i$, expressing $e_{k+1}$ as a combination of $e_1, \ldots, e_k$. This contradicts the linear independence of $\{e_1, \ldots, e_n\}$ (a non-trivial relation $e_{k+1} - \sum \alpha_i e_i = \mathbf{0}$ with coefficient $1$ on $e_{k+1}$). With $\beta_{k+1} \neq 0$ (after relabelling), we solve for $f_{k+1}$: \begin{align*} f_{k+1} = \frac{1}{\beta_{k+1}}\left(e_{k+1} - \sum_{i=1}^{k}\alpha_i e_i - \sum_{j=k+2}^{m}\beta_j f_j\right). \end{align*} Why does the new set still span? Any $v \in V$ was a combination of the old spanning set. Wherever $f_{k+1}$ appears, substitute its expression in terms of $e_{k+1}$ and the remaining vectors. The result is a combination of the new set, so spanning is preserved. [/guided] [/step] [step:Conclude the inequality $n \leq m$] After $n$ exchanges, the spanning set is $\{e_1, \ldots, e_n, f_{n+1}, \ldots, f_m\}$. At each of the $n$ exchange steps, we consumed one vector from $\{f_1, \ldots, f_m\}$. For the process to be possible, we need at least $n$ vectors $f_j$ available for exchange, which requires $n \leq m$. [/step]