Suppose e1, . . . , en is a basis of a complex vector space V . Let en+1 = −e1 − · · · − en. Prove that any vector v of V can be expressed in a unique way as a linear combination v = a1e1 + · · · + anen + an+1en+1 with coefficients a1, . . . , an, an+1 ∈ C satisfying a1 + · · · + an + an+1 = 0

Given a basis e_1, \ldots, e_n of a complex vector space V where e_{n+1} = -e_1 - \ldots - e_n , we aim to prove that any vector v \in V can be expressed in a unique way as a linear combination v = a_1e_1 + \ldots + a_ne_n + a_{n+1}e_{n+1} where a_1, \ldots, a_n, a_{n+1} \in \mathbb{C} satisfying a_1 + \ldots + a_n + a_{n+1} = 0 .

Since e_1, \ldots, e_n form a basis for V , any vector v \in V can be expressed as v = a_1e_1 + \ldots + a_ne_n for some a_1, \ldots, a_n \in \mathbb{C} .

Now, we can express e_{n+1} as e_{n+1} = -e_1 - \ldots - e_n .

Substitute this expression of e_{n+1} into the expression of v , we get:
v = a_1e_1 + \ldots + a_ne_n + a_{n+1}(-e_1 - \ldots - e_n)
Simplify this expression, we obtain:
v = (a_1 - a_{n+1})e_1 + \ldots + (a_n - a_{n+1})e_n
Now, we have an expression for v in terms of e_1, \ldots, e_n . To prove uniqueness, we need to show that if v = \tilde{a}_1e_1 + \ldots + \tilde{a}_ne_n + \tilde{a}_{n+1}e_{n+1} , then a_i = \tilde{a}_i for i = 1, \ldots, n and a_{n+1} = \tilde{a}_{n+1} .

Now, equating the two expressions for v gives:
\begin{align*}
a_1 - a_{n+1} &= \tilde{a}_1 \
&\vdots \
a_n - a_{n+1} &= \tilde{a}_n
\end{align*}
This system of equations can be solved to obtain a_i = \tilde{a}_i for i = 1, \ldots, n and a_{n+1} = \tilde{a}_{n+1} .

Finally, we sum the coefficients:
a_1 + \ldots + a_n + a_{n+1} = (\tilde{a}_1 + \ldots + \tilde{a}_n + \tilde{a}_{n+1}) = 0
Therefore, any vector v \in V can be expressed in a unique way as a linear combination v = a_1e_1 + \ldots + a_ne_n + a_{n+1}e_{n+1} with coefficients a_1, \ldots, a_n, a_{n+1} \in \mathbb{C} satisfying a_1 + \ldots + a_n + a_{n+1} = 0 .

\textbf{Answer:} Any vector v \in V can be expressed in a unique way as v = a_1e_1 + \ldots + a_ne_n + a_{n+1}e_{n+1} with a_1, \ldots, a_n, a_{n+1} \in \mathbb{C} satisfying a_1 + \ldots + a_n + a_{n+1} = 0 .