Difference between $\mathbb{R}^4$ and $\mathbb{C}^4$ in subspace spanned by some vectors. This is a problem in Hoffman / Kunze, Linear Algebra:
Let
$$\alpha_1=(1,1,-2,1), \quad \alpha_2=(3,0,4,-1), \quad \alpha_3=(-1,2,5,2).$$
Let
$$\alpha=(4,-5,9,-7), \quad \beta=(3,1,-4,4), \quad \gamma=(-1,1,0,1).$$
(a) Which of the vectors $\alpha$, $\beta$, $\gamma$ are in the subspace of $\mathbb{R}^4$ spanned by the $\alpha_i$?
(b) Which of the vectors $\alpha$, $\beta$, $\gamma$ are in the subspace of $\mathbb{C}^4$ spanned by the $\alpha_i$?
(c) Does this suggest a theorem?
I did (a) by solving the linear system. Got that $\alpha=-3\alpha_1+2\alpha_2-\alpha_3$ but $\beta$ and $\gamma$ resulted in systems that are insolvable. So only $\alpha$ is in the subspace spanned by the $\alpha_i$'s.
It looks to me that the same result holds in $\mathbb{C}^4$ but that would be non-sense to ask so I think that I'm wrong. What is the difference between $\mathbb{R}^4$ and $\mathbb{C}^4$ in this case? What theorem does this suggest?
Thanks!
 A: You are right about (a) and (b). They do have the same results.
(a) The vectors $\alpha$, $\beta$, and $\gamma$ are in the subspace of $\Bbb{R}^4$ spanned by the $\alpha_i$ if and only if we can write them as linear combinations of the $\alpha_i$ with real coefficients. 
$$\left[
        \begin{matrix}
        1 & 3 & -1 \\
        1 & 0 & 2 \\
        -2 & 4 & 5 \\ 
        1 &-1 & 2 \\
        \end{matrix}
\right]\rightarrow\left[
        \begin{matrix}
        1 & 3 & -1&y_1 \\
        0 & -3 & 3&y_2-y_1 \\
        0 & 0 & -39&4y_1-10y_2-3y_3 \\ 
        0 &0 & 0&3y_1−14y_2+y_3+13y_4 \\
        \end{matrix}
\right]$$
so $(y_1, y_2, y_3, y_4) \in \Bbb{R}^4$ is in the subspace of $\Bbb{R}^4$
spanned by the $\alpha_i$ if and only if $3y_1−14y_2+y_3+13y_4=0$
Checking this for $\alpha$, $\beta$, and $\gamma$ we see that only $\alpha$ is in the subspace.
(b) The same row reduction is operated here. 
$(y_1, y_2, y_3, y_4) \in \Bbb{C}^4$ is in the subspace of $\Bbb{C}^4$
spanned by the $\alpha_i$ if and only if $3y_1−14y_2+y_3+13y_4=0$. Again only $\alpha$ is in the subspace.
(c) 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 $α_i$.
