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  "body": "\n## Proof\n\nWe argue by induction on $k$. For $k = 0$ the empty family is linearly independent, so we treat $k \\ge 1$.\n\n1. 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|0EVHJFTG9Z]].\n2. A fact on [[vector spaces|05EGHEQZAW]]: if $a \\in F$, $v \\in V$, $v \\ne 0$ and $a v = 0$, then $a = 0$. Otherwise $v = a^{-1} (a v) = a^{-1} 0 = 0$.\n3. Base case $k = 1$: an [[eigenvector|0B5H4208Z9]] $v_1$ is not $0$. If $a_1 v_1 = 0$, then $a_1 = 0$ by step 2. So $P(1)$ holds.\n4. 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 $a_i \\in F$.\n5. Apply the [[linear map|0NRYHJPQG2]] $T$, and use $T(v_i) = \\lambda_i v_i$. This gives $a_1 \\lambda_1 v_1 + \\cdots + a_k \\lambda_k v_k = 0$.\n6. Multiply the relation of step 4 by $\\lambda_k$, and subtract it from step 5. The terms with $v_k$ cancel.\n7. This gives $\\sum_{i=1}^{k-1} a_i (\\lambda_i - \\lambda_k) v_i = 0$.\n8. 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 $i \u003c k$.\n9. For $i \u003c k$, $\\lambda_i \\ne \\lambda_k$, so $\\lambda_i - \\lambda_k$ is invertible in the field $F$. Hence $a_i = 0$ for $i \u003c k$.\n10. Then step 4 becomes $a_k v_k = 0$. Since $v_k \\ne 0$, step 2 gives $a_k = 0$.\n11. 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 $k \\ge 1$.\n\n## Idea\n\nApply $T - \\lambda_k \\id_V$ to a linear relation. It multiplies each $v_i$ by $\\lambda_i - \\lambda_k$, so it removes $v_k$ and keeps the other terms with nonzero factors. The induction hypothesis then kills the other coefficients, and the last one follows.\n",
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