Proof there is always a base in a v.s such that the coordinates of a vector are the elements of a given set 

Given any non-null vector of a vector space over a field $K$, of finite dimension $n$, and given any ordered set of $n$ elements (not all null), all in $K$, prove that there exists a base such that the elements of the set are the coordinates of the vector in that base.  


So I have to prove that if given a set  $A:= \{ x_1, ..., x_n \}\neq \{0\}$ ($x_1, ..., x_n \in K$) and a vector $v \neq 0$, there exists always a base $B=\{e_1, ..., e_n\}$ such that $v = BX$ ($X= \begin{pmatrix} x_1 \\ \vdots \\ x_n\end{pmatrix} $).
I don't even know where to start.
 A: Initially, you must have an old basis $\{b_1,...,b_n\}$ in which $v=y_1b_1+...+y_nb_n$, then $v=x_1(\frac{y_1}{x_1})b_1+...+x_n(\frac{y_n}{x_n})b_n$, so $\{\frac{y_1}{x_1}b_1,...,\frac{y_n}{x_n}b_n\}$ is the new basis you need. At least in case of all $x_i\neq0$.
A: Extend $\mathbf{v}$ to a basis $\mathcal{B}$ of $V$ and extend $\mathbf{x}$ to a basis $\mathcal{C}$ of $\mathbb{K}^n$. It is relatively well-known that there exists a unique mapping $T:V\rightarrow \mathbb{K}^n$ taking $\mathcal{B}$ to $\mathcal{C}$. It also follows that $T$ is invertible. 
Define a second basis of $V$ as
$$\mathcal{B}' = \{T^{-1}(\mathbf{e}_1),\ \cdots,\ T^{-1}(\mathbf{e}_n)\}.$$
It follows that $T$ maps each vector in $V$ to its coordinate representation under $\mathcal{B}'$. But $T(\mathbf{v}) = \mathbf{x}$ by construction, so it follows that the coordinate representation of $\mathbf{v}$ under $\mathcal{B}'$ is $\mathbf{x}$.
A: You are given $v$ and $x_1,\dots,x_n$. 
The question is asking you to find a base $v_1,\dots,v_n$ such that you have
$$
  v = x_1 v_1 + \dots + x_n v_n.
$$
Suppose you choose any base $e_1,\dots,e_n$. Can you make some simple modification to your base so that you get the correct coefficients?
A: This answer somewhat similar in spirit but somewhat different in technique than that of Eu Yu:
Let's call our vector space $Y$.
Choose a basis $V_1 = V, V_2, . . . , V_n$  of $Y$ and likewise, flesh out $x = (x_1, x_2, . . ., x_n)^T$ to a basis $y_1 = x, y_2, . . . , y_n$.  Define a linear transformation $T:Y \to Y$ via $TV_i = y_i$; then $TV = TV_1 = y_1 = x$.  Furthermore, $T$ is nonsingular, hence invertible, since it takes a basis ($V_i)$ to a basis ($y_i$).  Let $\theta_1, \theta_2, . . . , \theta_n$ be the rows of $T$.  We can think of each $\theta_i$ as a dual vector or linear functional on $Y$, $\theta_i \in Y^*$, since in fact $\theta_i(z) = \sum t_{ij} z_j$ where the matrix of $T$ is $[t_{ij}]$ and $z = (z_1, z_2, . . . , z_n)^T$.  Consider the matrix inverse of $T$, $T^{-1}$.  Let $e_j$, $1 \le j \le n$, be the columns of $T^{-1}$.  Since $TT^{-1} = I$, the $\theta_j \in Y^*$ stand in a dual relationship to the $e_k \in Y$; that is, $\theta_j(e_k) = \delta_{jk}$ for all $1 \le j, k \le n$; $\theta_j$ and $e_k$ are dual bases (of $Y^*$ and $Y$, respectively) to one another.  Thus if $w = \sum w_k e_k$ is the expansion of $w$ in the $e_k$ basis, $\theta_j(w) = \sum w_k \theta _j(e_k) = \sum w_k \delta_{jk} = w_j$; thus $\theta_j$ picks off the $j$-th coefficient in the basis $e_k$, that of $e_j$.  Now since $x = TV$, we have $V = T^{-1}x = \sum x_j e_j$ since the $e_j$ are the columns of $T^{-1}$.  Then $\theta_k(V) = x_k$, by what we have just seen.  Thus the $x_j$ are the components of $V$ in the basis $e_k$.  We have established the existence of a basis ($e_j$) in which the coordinates of $V$ are the $x_j$.  QED.
Hope this helps.  A Happy New Year to One and All,
and as always,
Fiat Lux!!!
