[proofplan]
We prove the result by induction on the number $k$ of eigenvectors. The induction step starts from an arbitrary linear relation among $v_1,\ldots,v_k$, applies $T$ to that relation, and subtracts $\lambda_k$ times the original relation. This eliminates the $v_k$ term and produces a linear relation among $v_1,\ldots,v_{k-1}$, so the induction hypothesis forces all earlier coefficients to vanish. The original relation then forces the final coefficient to vanish as well.
[/proofplan]
[step:Prove the one-vector case directly]
For $k=1$, let $a_1 \in F$ satisfy
\begin{align*}
a_1 v_1 = 0.
\end{align*}
If $a_1 \neq 0$, then $a_1$ has an inverse in the field $F$, and multiplying by $a_1^{-1}$ gives $v_1 = 0$, contradicting the hypothesis $v_1 \neq 0$. Hence $a_1 = 0$, so $v_1$ is linearly independent.
[/step]
[step:Use the induction hypothesis after eliminating the last eigenvector]
Assume the theorem has been proved for $k-1$ eigenvectors, where $k \geq 2$. Let $a_1,\ldots,a_k \in F$ be scalars satisfying the linear relation
\begin{align*}
\sum_{i=1}^{k} a_i v_i = 0.
\end{align*}
Since $v_i \in E_{\lambda_i}(T)$, the definition of eigenspace gives
\begin{align*}
T(v_i) = \lambda_i v_i
\end{align*}
for every $i \in \{1,\ldots,k\}$. Applying the linear map $T$ to the relation and using $T(0)=0$ gives
\begin{align*}
0 = T\left(\sum_{i=1}^{k} a_i v_i\right) = \sum_{i=1}^{k} a_i T(v_i) = \sum_{i=1}^{k} a_i \lambda_i v_i.
\end{align*}
Multiplying the original relation by $\lambda_k$ gives
\begin{align*}
\sum_{i=1}^{k} a_i \lambda_k v_i = 0.
\end{align*}
Subtracting this relation from the previous one yields
\begin{align*}
\sum_{i=1}^{k} a_i(\lambda_i-\lambda_k)v_i = 0.
\end{align*}
The $i=k$ term is zero, so this is
\begin{align*}
\sum_{i=1}^{k-1} a_i(\lambda_i-\lambda_k)v_i = 0.
\end{align*}
The eigenvalues $\lambda_1,\ldots,\lambda_{k-1}$ are pairwise distinct, and the vectors $v_1,\ldots,v_{k-1}$ are nonzero with $v_i \in E_{\lambda_i}(T)$. By the induction hypothesis, $v_1,\ldots,v_{k-1}$ are linearly independent. Therefore
\begin{align*}
a_i(\lambda_i-\lambda_k)=0
\end{align*}
for every $i \in \{1,\ldots,k-1\}$. Since $\lambda_i \neq \lambda_k$, the scalar $\lambda_i-\lambda_k$ is nonzero and hence invertible in $F$. Thus $a_i=0$ for every $i \in \{1,\ldots,k-1\}$.
[guided]
Assume the theorem is known for any list of $k-1$ pairwise distinct eigenvalues and corresponding nonzero eigenvectors, with $k \geq 2$. We must show that every linear relation among $v_1,\ldots,v_k$ has all coefficients equal to zero. So let $a_1,\ldots,a_k \in F$ be scalars such that
\begin{align*}
\sum_{i=1}^{k} a_i v_i = 0.
\end{align*}
The goal is to use the induction hypothesis, but that hypothesis only applies to the shorter list $v_1,\ldots,v_{k-1}$. We therefore need to manufacture a relation involving only those first $k-1$ vectors. The standard way to remove $v_k$ is to compare the original relation with the relation obtained after applying $T$.
Because $v_i \in E_{\lambda_i}(T)$, the definition of eigenspace gives
\begin{align*}
T(v_i)=\lambda_i v_i
\end{align*}
for every $i \in \{1,\ldots,k\}$. Applying the $F$-linear map $T:V \to V$ to the original relation gives
\begin{align*}
0 = T(0) = T\left(\sum_{i=1}^{k} a_i v_i\right) = \sum_{i=1}^{k} a_i T(v_i) = \sum_{i=1}^{k} a_i \lambda_i v_i.
\end{align*}
We also multiply the original relation by the scalar $\lambda_k$:
\begin{align*}
0 = \lambda_k \sum_{i=1}^{k} a_i v_i = \sum_{i=1}^{k} a_i \lambda_k v_i.
\end{align*}
Subtracting the second displayed equality from the first gives
\begin{align*}
\sum_{i=1}^{k} a_i(\lambda_i-\lambda_k)v_i = 0.
\end{align*}
This subtraction is designed so that the last coefficient becomes $a_k(\lambda_k-\lambda_k)=0$. Hence the relation reduces to
\begin{align*}
\sum_{i=1}^{k-1} a_i(\lambda_i-\lambda_k)v_i = 0.
\end{align*}
Now the induction hypothesis applies to the vectors $v_1,\ldots,v_{k-1}$: the eigenvalues $\lambda_1,\ldots,\lambda_{k-1}$ are still pairwise distinct, and each $v_i$ is nonzero and lies in $E_{\lambda_i}(T)$. Therefore $v_1,\ldots,v_{k-1}$ are linearly independent. By the definition of linear independence, every coefficient in the displayed relation must be zero:
\begin{align*}
a_i(\lambda_i-\lambda_k)=0
\end{align*}
for every $i \in \{1,\ldots,k-1\}$. Since the eigenvalues are pairwise distinct, $\lambda_i \neq \lambda_k$ for each such $i$. Thus $\lambda_i-\lambda_k$ is a nonzero element of the field $F$, so it is invertible. Multiplying by its inverse gives $a_i=0$ for every $i \in \{1,\ldots,k-1\}$.
[/guided]
[/step]
[step:Return to the original relation to eliminate the final coefficient]
Substituting $a_1=\cdots=a_{k-1}=0$ into the original relation gives
\begin{align*}
a_k v_k = 0.
\end{align*}
Since $v_k \neq 0$, the same field-inverse argument used in the base case implies $a_k=0$. Thus every linear relation among $v_1,\ldots,v_k$ has all coefficients equal to zero, so $v_1,\ldots,v_k$ are linearly independent. By induction, the theorem holds for every $k \in \mathbb{N}$.
[/step]
