Is this tiny conjecture true concerning $\alpha$ true? As part of a problem that i am trying to solve i would like to know if the following conjecture is true.

Conjecture. Given a non-zero vector $\alpha$ in a finite dimensional vector space $W$ there exist a basis for $W$ say
  $w_1,w_2,...,w_m$ such that $\alpha =
 a_1w_1+a_2w_2+\cdot\cdot\cdot+a_mw_m$ where $a_j\neq 0$ $\forall j\in\{1,2,...,m\}$ and $m = \dim W$. The Field of Scalars is either $\mathbf{R}$ or $\mathbf{C}$.

 A: Yes. Intuitively, since we're over $\mathbb{F} = \mathbb{R}$ or $\mathbb{C}$, you can rotate coordinate axes so that $\alpha$ doesn't point along any coordinate axis. 
To prove the result, suppose that $\alpha$ has coordinate vector $a$ in a given basis. Its coordinate vector $a'$ in a new basis is $a' = Q a$, where $Q$ is an invertible matrix, the inverse of the "change of basis" matrix $P = Q^{-1}$. Moreover, every invertible $Q$ corresponds to a change of basis.
Now, consider first a basis with $w_1 = \alpha$. (This exists as $\alpha \neq 0$.) In this basis, we have $a = (1, 0, \dots, 0)$. That means that changing bases according to $Q$ gives $a' = $ first column of $Q$. We want $a'$ to have nonzero entries. So the question becomes "does there exist an invertible matrix $Q$ with first column having no vanishing entries?" And the answer is yes. Pick your favorite vector $q_1 \in \mathbb{F}^m$ with no vanishing entries to be the first column of $Q$; then extend $q_1$ to a basis $q_1, \dots, q_m$ of $\mathbb{F}^m$. If we use these for the columns of $Q$, then $Q$ is invertible, and we have $a' = Q(1, 0, \dots, 0) = q_1$, which is has no vanishing entries. We've found a basis for $W$ for which the coordinate vector $a'$ of $\alpha$ has no vanishing entries.
A: Hint
Let $(a_k)$ be any scalars (not all $0$).
If you complete $\alpha=\alpha_1$ into a basis $(\alpha_1,\alpha_2, \ldots, \alpha_n)$ of $W$ a "stronger" affirmation is that there exist a basis $(w_1,w_2,\ldots, w_n)$ such that:
$$\alpha_1=\sum_{k=1}^n a_k w_k$$
$$\alpha_i=w_i$$
But such a base can be explictely computed from $(\alpha_1, \ldots, \alpha_n)$ and :
$$\begin{pmatrix} a_1& a_2&a_3 &\ldots &a_n \\
0&1&0&\ldots &0 \\
0&0&1&\ldots&0\\
\vdots&&&\ddots&&\\
0&0&0&\ldots&1\end{pmatrix}$$
