[proofplan]
Existence is exactly the defining property of the affine span: a point in the affine span is an affine combination of the given points with coefficients summing to $1$. For uniqueness, compare two such affine combinations after choosing $p_0$ as an origin. Their equality becomes a linear relation among the displacement vectors $p_i-p_0$, and affine independence forces all nonzero-index coefficients to agree; the coefficient at $p_0$ then agrees because both coefficient sums are $1$.
[/proofplan]
[step:Use membership in the affine span to obtain barycentric coefficients]
Let
\begin{align*}
q \in \operatorname{Aff}\{p_0,\ldots,p_m\}.
\end{align*}
By the definition of affine span, there exist scalars $\lambda_0,\ldots,\lambda_m \in k$ such that
\begin{align*}
\lambda_0+\cdots+\lambda_m=1
\end{align*}
and
\begin{align*}
q=\lambda_0p_0+\cdots+\lambda_mp_m.
\end{align*}
Thus at least one barycentric expression exists.
[/step]
[step:Compare two barycentric expressions by translating to the model vector space]
Suppose $\lambda_0,\ldots,\lambda_m \in k$ and $\mu_0,\ldots,\mu_m \in k$ are two coefficient tuples satisfying
\begin{align*}
\lambda_0+\cdots+\lambda_m=1
\end{align*}
and
\begin{align*}
\mu_0+\cdots+\mu_m=1,
\end{align*}
with
\begin{align*}
q=\lambda_0p_0+\cdots+\lambda_mp_m=\mu_0p_0+\cdots+\mu_mp_m.
\end{align*}
Choose $p_0$ as an origin and subtract $p_0$ from both affine combinations. In the model [vector space](/page/Vector%20Space) $V$, this gives
\begin{align*}
q-p_0=\sum_{i=1}^m \lambda_i(p_i-p_0)
\end{align*}
and
\begin{align*}
q-p_0=\sum_{i=1}^m \mu_i(p_i-p_0).
\end{align*}
Subtracting these two equalities in $V$ yields
\begin{align*}
\sum_{i=1}^m(\lambda_i-\mu_i)(p_i-p_0)=0.
\end{align*}
[guided]
The point of choosing $p_0$ as an origin is that affine combinations of points become ordinary linear combinations of displacement vectors in $V$. Suppose that the same point $q$ has two affine expressions
\begin{align*}
q=\lambda_0p_0+\cdots+\lambda_mp_m
\end{align*}
and
\begin{align*}
q=\mu_0p_0+\cdots+\mu_mp_m,
\end{align*}
where $\lambda_i,\mu_i \in k$ for each $i \in \{0,\ldots,m\}$ and both coefficient sums are equal to $1$.
Because the coefficients in each expression sum to $1$, translating the affine combination by the origin $p_0$ gives an equality in the vector space $V$:
\begin{align*}
q-p_0=\sum_{i=0}^m \lambda_i(p_i-p_0).
\end{align*}
The term with $i=0$ is $\lambda_0(p_0-p_0)=0$, so this reduces to
\begin{align*}
q-p_0=\sum_{i=1}^m \lambda_i(p_i-p_0).
\end{align*}
Applying the same translation to the second expression gives
\begin{align*}
q-p_0=\sum_{i=1}^m \mu_i(p_i-p_0).
\end{align*}
Both right-hand sides are vectors in $V$ and both equal the same vector $q-p_0$. Therefore subtracting the second equality from the first in the vector space $V$ gives the linear relation
\begin{align*}
\sum_{i=1}^m(\lambda_i-\mu_i)(p_i-p_0)=0.
\end{align*}
This is the crucial reduction: uniqueness of affine coefficients has been converted into uniqueness of coefficients in a linear relation among displacement vectors.
[/guided]
[/step]
[step:Use affine independence to force equality of all coefficients]
By the stated characterization of affine independence, the displacement vectors
\begin{align*}
p_1-p_0,\ldots,p_m-p_0
\end{align*}
are linearly independent in $V$. Applying this [linear independence](/page/Linear%20Independence) to
\begin{align*}
\sum_{i=1}^m(\lambda_i-\mu_i)(p_i-p_0)=0
\end{align*}
gives
\begin{align*}
\lambda_i-\mu_i=0
\end{align*}
for every $i \in \{1,\ldots,m\}$. Hence $\lambda_i=\mu_i$ for every $i \in \{1,\ldots,m\}$.
It remains to compare the coefficient at $p_0$. Since both coefficient sums are $1$,
\begin{align*}
\lambda_0=1-\sum_{i=1}^m\lambda_i
\end{align*}
and
\begin{align*}
\mu_0=1-\sum_{i=1}^m\mu_i.
\end{align*}
The already proved equalities $\lambda_i=\mu_i$ for $i \geq 1$ imply $\lambda_0=\mu_0$. Therefore the two coefficient tuples are equal, so the barycentric expression of $q$ is unique.
[/step]
