Prove that any vector v of V can be expressed in a unique way as a linear combination Suppose $\{e_1, . . . , e_n\}$ is a basis of a real vector space $V$ . 
Let 
$$e_{n+1} = −e_1 − ... − e_n$$ 
Prove that any vector $v$ of $V$ can be expressed in a unique way as a
linear combination
$$v = a_1e_1 + · · · + a_ne_n + a_{n+1}e_{n+1}$$
with coefficients $a_1, . . . , a_n, a_{n+1} \in \mathbb{R}$ satisfying $a_1 + · · · + a_n + a_{n+1} = 0$
could anyone help with this or even where to start as it is confusing me already!
 A: Suppose $\sum\limits_{i=1}^{n+1}a_ie_i=\sum\limits_{i=1}^{n+1}b_ie_i$. Let $c_i=a_i-b_i$ so that $\sum\limits_{i=1}^{n+1}c_ie_i=0$. This gives $\sum\limits_{i=1}^{n}c_ie_i=-c_{n+1}e_{n+1}=c_{n+1}\sum\limits_{i=1}^{n}e_i$. Hence $\sum\limits_{i=1}^{n}(c_i-c_{n+1})e_i=0$. This implies $c_i=c_{n+1}$ for all $n$. Add this over $i \leq n$ and use the fact that $\sum c_i=0$ to see that $c_i=0$ for all $i$ or $a_i=b_i$ for all $i$.
A: You can also show this by an explicit construction which produces unique values for the $a_i$ coefficients. Since $\{e_1, \dots, e_n\}$ is a basis of $V$ you know that there are unique real co-ordinates $b_i$ such that
$v = b_1e_1 + \dots + b_ne_n$
If you also have
$v = a_1e_n + \dots + a_ne_n + a_{n+1}e+{n+1} = (a_1-a_{n+1})e_1 + \dots + (a_n-a_{n+1})e_n$
then you can equate the co-ordinates to get
$b_1 = a_1 - a_{n+1}\\b_2 = a_2 - a_{n+1}\\ \dots \\b_n = a_n - a_{n+1}$
Adding these $n$ equations gives
$\sum_{i=1}^n b_i =(\sum_{i=1}^n a_i) - na_{n+1}$
You also know that $\sum_{i=1}^n a_i = -a_{n+1}$, so
$\sum_{i=1}^n b_i =-(n+1)a_{n+1} \\ \Rightarrow a_{n+1} = - \frac {\sum_{i=1}^n b_i }{n+1} \\ \Rightarrow a_i = b_i - \frac {\sum_{i=1}^n b_i }{n+1} $
