Linear Algebra theorem I want to prove the following theorem:
Let $F \subseteq E$ be fields, and let $\alpha_1, \alpha_2, . . . , \alpha_m$ be vectors in $F^n$. A vector $v \in F^n$ is in the subspace of $F^n$ spanned by the $α_i$ if and only if v is in the subspace of $E^n$ spanned by the $α_i$.
The "only if" part is easy because of this: 
If $x$ exists in the subspace of $F^n$ spanned by the $\alpha_i$, then $x=c_1\alpha_1+\cdots+c_m\alpha_m$, where $c_i\in F$. But $F \subseteq E$, so $c_i\in E$ and it follows that $x$ exists in the subspace of $E^n$ spanned by the $\alpha_i$.
The "if" part is troubling me. Can anyone help?
Thanks
 A: I don’t think it’s true. Let $F=\mathbb Q$, $E=\mathbb R$, $n=2$, $\alpha_1=(1,0)$ and $\alpha_2=(0,1)$. Then $(\sqrt2,\sqrt2)=\sqrt2\alpha_1+\sqrt2\alpha_2$ is in the span of $\alpha_1,\alpha_2$ in $\mathbb R^2$ but not in the span of the vectors in $\mathbb Q^2$.
A: I don't think this is true either, and intuitively so because it seems the converse is not powerful enough. 
All you know is that $F \subseteq E$. The only assumption you have is that $x \in W$, where $W$ is a subspace of $E^n$. It is easy to construct an example such that not all of $W$ is in the subspace $F^n$ but all of it is in $E^n$ (you can even construct such that some vectors in $E^n$ have unique elements from the field $E$ but not $F$) , thus violating the proof.
A: Here is the "if" part.
Write
$$
v = \begin{bmatrix}v_{1}\\ \vdots \\ v_{n}\end{bmatrix} \in F^{n}
$$
and
$$
\alpha_{i} =
\begin{bmatrix}\alpha_{1i} \\ \vdots \\ \alpha_{ni}\end{bmatrix} \in F^{n}
$$
and consider the two matrices with coefficients in $F$
$$
A = \begin{bmatrix}
\alpha_{11} & \dots & \alpha_{1m}\\
&\ddots\\
\alpha_{n1} & \dots & \alpha_{nm}\\
\end{bmatrix}
$$
and
$$
C = \begin{bmatrix}
v_{1} & \alpha_{11} & \dots & \alpha_{1m}\\
&&\ddots\\
v_{n} &\alpha_{n1} & \dots & \alpha_{nm}\\
\end{bmatrix}.
$$
The hypothesis that you can write $v$ as a linear combination, with coefficients in $E$, of the $\alpha_{i}$, tells you that the (column) ranks of $A$ and $C$ are the same. But then, reversing the argument, this tells you that $v$ is a linear combination of the $\alpha_{i}$ with coefficients in $F$.
