Spanning $\mathbb {C}^4$ with a system of vectors I have the following set of column vectors:
$$v_1 = (1, 1, 0, 0)^T\\
v_2 = (1, 0, 1, 0)^T\\
v_3 = (0, 0, 1, 1)^T\\
v_4 = (0, 1, 0, 1)^T$$
I was asked:

(a) Are the vectors linearly independent?
(b) Do they span $\mathbb{R}^4$?
(c) Do they span $\mathbb{C}^4$?

I solved parts (a) and (b) be writing the augmented matrix that corresponds to the vector equation $av_1 + bv_2 +cv_3 +dv_4 = 0$. I found that the system had free variables, and thus the system of vectors is not linearly independent because there exist non-trivial solutions to the above vector equation. I used the fact the there did not exist a pivot in every row of the coefficient matrix to reason that the system of vectors does not span $\mathbb{R}^4$.
However, I want to ensure that I understand part (c) correctly: I do not believe that these vectors span $\mathbb{C}^4$ for the same reason that they do not span $\mathbb{R}^4$. Is this correct? I may be missing something. Btw, I am an engineering student.
Best
 A: You're not missing anything. It can definitely be a little subtle to pin down exactly why "dependent in $\Bbb R^4$ implies dependent in $\Bbb C^4$," though. Note that I'm implicitly using dimension here, since four vectors will only span a four-dimensional space if they're linearly independent (that is, not linearly dependent).
It's essentially because $\Bbb R$ is a subfield of $\Bbb C$. Let's call your set of vectors $S = \{v_1, v_2, v_3, v_4\}$, and consider the two spans
$$\operatorname{Span}_\Bbb R(S) = \left\{\sum \alpha_i v_i : \alpha_i \in \Bbb R,\ v_i \in S \right\}$$ and
$$\operatorname{Span}_\Bbb C(S) = \left\{\sum \alpha_i v_i : \alpha_i \in \Bbb C,\ v_i \in S \right\}.$$
Now, since $\Bbb R \subset \Bbb C$, we have $\operatorname{Span}_\Bbb R(S) \subset \operatorname{Span}_\Bbb C(S)$; if you can use real coefficients to achieve a particular linear combination, then you can certainly use complex coefficients (by continuing to use the same real coefficients). 
To say the set $S$ is linearly dependent in $\Bbb R^4$ is to say that $0 \in \operatorname{Span}_\Bbb R(S)$ (requiring at least one $\alpha_i$ to be nonzero), which implies that $0 \in \operatorname{Span}_\Bbb C(S)$ (again, with not all $\alpha_i = 0$), hence the set is linearly dependent in $\Bbb C^4$ as well.
A: The vectors $v_1,v_2,v_3,v_4$ are, indeed, linearly dependent. For example, 
$$v_1=\begin{bmatrix}
1\\
1\\
0\\
0
\end{bmatrix}= v_2- v_3+v_4=
\begin{bmatrix}
1\\
0\\
1\\
0
\end{bmatrix}
-\begin{bmatrix}
0\\
0\\
1\\
1
\end{bmatrix}+
\begin{bmatrix}
0\\
1\\
0\\
1
\end{bmatrix}
.$$
So,  $v_1,v_2,v_3,v_4$ cannot span $R^4$, a fortiori, they cannot span $C^4.$
$v_2,v_3,v_4$ are linearly independent because, as far as the first component, there are no constants $\gamma$ and $\delta$ such that $\gamma0+\delta0=1.$
As a result $v_2,v_3,v_4$ span a three dimensional real or complex space.
