[proofplan] We prove the equivalence in both directions by translating between the affine relation $\sum_{i=0}^n t_i a_i = 0$ with $\sum_{i=0}^n t_i = 0$ and the linear relation $\sum_{i=1}^n \lambda_i (a_i - a_0) = 0$. The substitution $\lambda_i = t_i$ for $i \geq 1$, $t_0 = -\sum_{i \geq 1} t_i$ gives a bijection between the two types of dependence relations, so the triviality of one is equivalent to the triviality of the other. The algebra in each direction is a single straightforward expansion. [/proofplan]

[step:Recall the definitions and set up the bijection between relations] Recall that $a_0, \ldots, a_n \in \mathbb{R}^m$ are **[affinely independent](/page/Affine%20Independence)** if the only solution $(t_0, \ldots, t_n) \in \mathbb{R}^{n+1}$ to the simultaneous system \begin{align*} \sum_{i=0}^n t_i a_i = 0 \quad \text{and} \quad \sum_{i=0}^n t_i = 0 \end{align*} is $t_0 = t_1 = \cdots = t_n = 0$. Vectors $v_1, \ldots, v_n \in \mathbb{R}^m$ are **[linearly independent](/page/Linear%20Independence)** if the only solution $(\lambda_1, \ldots, \lambda_n) \in \mathbb{R}^n$ to \begin{align*} \sum_{i=1}^n \lambda_i v_i = 0 \end{align*} is $\lambda_1 = \cdots = \lambda_n = 0$. Given coefficients $(t_0, t_1, \ldots, t_n)$ satisfying $\sum_{i=0}^n t_i = 0$, set $\lambda_i := t_i$ for $i = 1, \ldots, n$, so $t_0 = -\sum_{i=1}^n \lambda_i$. Conversely, given $(\lambda_1, \ldots, \lambda_n)$, set $t_i := \lambda_i$ for $i \geq 1$ and $t_0 := -\sum_{i=1}^n \lambda_i$; then $\sum_{i=0}^n t_i = 0$. This correspondence is bijective between \begin{align*} \{(t_0, \ldots, t_n) \in \mathbb{R}^{n+1} : \textstyle\sum_{i=0}^n t_i = 0\} \quad \text{and} \quad \mathbb{R}^n, \end{align*} and sends the zero tuple on one side to the zero tuple on the other. The two conditions "all $t_i = 0$" and "all $\lambda_i = 0$" are equivalent under this bijection. [/step]

[step:Prove the forward direction: affine independence implies linear independence] Assume $a_0, \ldots, a_n$ are affinely independent. Let $(\lambda_1, \ldots, \lambda_n) \in \mathbb{R}^n$ satisfy \begin{align*} \sum_{i=1}^n \lambda_i (a_i - a_0) = 0. \end{align*} Expanding the left-hand side by distributing the scalars, \begin{align*} \sum_{i=1}^n \lambda_i (a_i - a_0) = \sum_{i=1}^n \lambda_i a_i - \left(\sum_{i=1}^n \lambda_i\right) a_0. \end{align*} Define $t_0 := -\sum_{i=1}^n \lambda_i$ and $t_i := \lambda_i$ for $i = 1, \ldots, n$. The expansion becomes \begin{align*} \sum_{i=1}^n \lambda_i (a_i - a_0) = \sum_{i=1}^n t_i a_i + t_0 a_0 = \sum_{i=0}^n t_i a_i. \end{align*} So $\sum_{i=0}^n t_i a_i = 0$. Moreover \begin{align*} \sum_{i=0}^n t_i = t_0 + \sum_{i=1}^n t_i = -\sum_{i=1}^n \lambda_i + \sum_{i=1}^n \lambda_i = 0. \end{align*} The pair $(t_0, \ldots, t_n)$ therefore satisfies both affine-dependence equations. By affine independence, $t_0 = t_1 = \cdots = t_n = 0$, which forces $\lambda_i = t_i = 0$ for $i = 1, \ldots, n$. Hence $a_1 - a_0, \ldots, a_n - a_0$ are linearly independent. [/step]

[step:Prove the reverse direction: linear independence implies affine independence]Assume $a_1 - a_0, \ldots, a_n - a_0$ are linearly independent. Let $(t_0, t_1, \ldots, t_n) \in \mathbb{R}^{n+1}$ satisfy \begin{align*} \sum_{i=0}^n t_i a_i = 0 \quad \text{and} \quad \sum_{i=0}^n t_i = 0. \end{align*} From the second equation, $t_0 = -\sum_{i=1}^n t_i$. Substituting into the first, \begin{align*} 0 = \sum_{i=0}^n t_i a_i = t_0 a_0 + \sum_{i=1}^n t_i a_i = -\left(\sum_{i=1}^n t_i\right) a_0 + \sum_{i=1}^n t_i a_i = \sum_{i=1}^n t_i (a_i - a_0). \end{align*} This is a linear relation among the vectors $a_i - a_0$ for $i = 1, \ldots, n$. By linear independence, $t_i = 0$ for $i = 1, \ldots, n$. Substituting back, $t_0 = -\sum_{i=1}^n t_i = 0$ as well. Hence all $t_i = 0$, and $a_0, \ldots, a_n$ are affinely independent.[/step]

[guided]We already saw in Step 1 that the map $(t_0, \ldots, t_n) \leftrightarrow (\lambda_1, \ldots, \lambda_n)$ with $\lambda_i = t_i$ ($i \geq 1$) and $t_0 = -\sum_i \lambda_i$ is a bijection between affine relations (with $\sum t_i = 0$) and linear relations in the shifted vectors. We now use this bijection going the opposite direction: given an affine relation, produce a linear relation. Starting from $\sum_{i=0}^n t_i a_i = 0$ with $\sum_{i=0}^n t_i = 0$, eliminate $t_0$ using the sum constraint: $t_0 = -\sum_{i=1}^n t_i$. Substitute: \begin{align*} 0 = \sum_{i=0}^n t_i a_i = t_0 a_0 + \sum_{i=1}^n t_i a_i = -\sum_{i=1}^n t_i \cdot a_0 + \sum_{i=1}^n t_i a_i. \end{align*} The crucial algebraic move is combining the two sums by factoring: \begin{align*} \sum_{i=1}^n t_i a_i - \sum_{i=1}^n t_i \cdot a_0 = \sum_{i=1}^n t_i (a_i - a_0). \end{align*} This is now a linear relation among the shifted vectors $(a_i - a_0)_{i \geq 1}$. Linear independence of those vectors forces $t_i = 0$ for all $i \geq 1$, and then the sum constraint $\sum_{i=0}^n t_i = 0$ forces $t_0 = 0$ as well. So the only affine dependence is the trivial one, proving affine independence. Notice that the two directions of the proof are computationally nearly identical — this reflects the fact that Step 1's bijection between affine and linear dependence relations is the genuine content of the theorem, and the two directions of the implication are just the two directions of that bijection.[/guided]