Proof that eigenvectors with distinct eigenvalues are linearly independent by induction

Proof

We argue by induction on $k$. For $k = 0$ the empty family is linearly independent, so we treat $k \ge 1$.

1. For $k \ge 1$, let $P(k)$ say: eigenvectors $v_1, \dots, v_k$ of $T$ for pairwise distinct eigenvalues $\lambda_1, \dots, \lambda_k$ are linearly independent.
2. A fact on vector spaces: if $a \in F$, $v \in V$, $v \ne 0$ and $a v = 0$, then $a = 0$. Otherwise $v = a^{-1} (a v) = a^{-1} 0 = 0$.
3. Base case $k = 1$: an eigenvector $v_1$ is not $0$. If $a_1 v_1 = 0$, then $a_1 = 0$ by step 2. So $P(1)$ holds.
4. Induction step: let $k \ge 2$ and assume $P(k-1)$. Let $v_1, \dots, v_k$ be as in $P(k)$, and let $a_1 v_1 + \cdots + a_k v_k = 0$ with $a_i \in F$.
5. Apply the linear map $T$, and use $T(v_i) = \lambda_i v_i$. This gives $a_1 \lambda_1 v_1 + \cdots + a_k \lambda_k v_k = 0$.
6. Multiply the relation of step 4 by $\lambda_k$, and subtract it from step 5. The terms with $v_k$ cancel.
7. This gives $\sum_{i=1}^{k-1} a_i (\lambda_i - \lambda_k) v_i = 0$.
8. The vectors $v_1, \dots, v_{k-1}$ are eigenvectors for the distinct eigenvalues $\lambda_1, \dots, \lambda_{k-1}$. By $P(k-1)$, $a_i (\lambda_i - \lambda_k) = 0$ for $i < k$.
9. For $i < k$, $\lambda_i \ne \lambda_k$, so $\lambda_i - \lambda_k$ is invertible in the field $F$. Hence $a_i = 0$ for $i < k$.
10. Then step 4 becomes $a_k v_k = 0$. Since $v_k \ne 0$, step 2 gives $a_k = 0$.
11. So all $a_i$ are $0$, and $v_1, \dots, v_k$ are linearly independent. This proves $P(k)$, and by induction $P(k)$ holds for all $k \ge 1$.