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.

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 $x_i\neq0$.

• @ janmarqz: what happens if $y_i \ne 0$ but $x_i = 0$? – Robert Lewis Dec 27 '13 at 20:13
• this is a homework for hallplay835, my friend – janmarqz Dec 27 '13 at 20:22
• @ janmarqz: indeed it may be, though not tagged as such . . . I noticed you edited your answer to address the case in point. – Robert Lewis Dec 27 '13 at 20:24
• ok and yes, 'cuz this is in working – janmarqz Dec 27 '13 at 20:26

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 $\mathbf{x}$.

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?

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!!!