Complex and real span of a set of complex vectors I have a set of vectors $\mathcal{V}=\{ v_1, \dots ,v_n \}$ in $\mathbb{C}^d$ and I call $V \in \mathbb{C}^{d{\times}n}$ the matrix having these vectors as columns.
I am interested in the dimension of the linear span of these vectors on the fields $\mathbb{R}$ or $\mathbb{C}$.
In particular, I suspect that 
$$\mathrm{dim} \left[ \mathrm{span}_{\mathbb{C}} \mathcal{V}  \right]= \mathrm{dim} \left[  \mathrm{span}_{\mathbb{R}} \mathcal{V} \right] \iff \mathrm{Im}\left(V^\dagger V \right) = \mathbb{0}_n$$,
i.e. the dimension of the real vector space of linear combinations with real coefficients and the dimension of the complex vector space spanned by linear combination with complex coefficients are equal if and only if the Gram matrix of the vectors is a real symmetric matrix.
I don't know if this is trivial or standard statement, but if anybody knows the answer, has some reference to look at or has a hint at the proof it would be greatly appreciated.
 A: The statement isn't true in general. Consider $v_1, v_2 \in \mathbb{C}^2$ given by$$v_1 = \begin{bmatrix} 1 \\ 1\end{bmatrix}, v_2 = \begin{bmatrix} i \\ 0\end{bmatrix}$$
Then $\dim_{\mathbb{R}}\big[\operatorname{span}_{\mathbb{R}}\{v_1, v_2\}\big] = 2 = \dim_{\mathbb{C}}\big[\operatorname{span}_{\mathbb{C}}\{v_1, v_2\}\big]$ but
$$V^*V = \begin{bmatrix} 2 & i \\ -i & 0\end{bmatrix}$$
isn't real.

The other direction on the other hand is true. Assume that $V^*V$ is real. WLOG assume that $v_1, \ldots, v_r$ are linearly independent over $\mathbb{R}$ for some $1 \le r \le d$. We claim that they are also linearly independent over $\mathbb{C}$.

Assume $\sum_{i=1}^r \alpha_iv_i = 0$ for some scalars $\alpha_i \in \mathbb{C}$. Scalar multiplying this by $v_j$ we get
$$\sum_{i=1}^r\alpha_i\underbrace{\langle v_i, v_j\rangle}_{\in\mathbb{R}} = 0, \quad\forall j =1, \ldots, r$$
In particular $$\sum_{i=1}^r(\operatorname{Im}\alpha_i)\langle v_i, v_j\rangle = 0, \quad\forall j =1, \ldots, r$$
Multiplying this by $\operatorname{Im}\alpha_j$ we obtain
$$\left\langle\sum_{i=1}^r(\operatorname{Im}\alpha_i)v_i, (\operatorname{Im}\alpha_j)v_j \right\rangle = \sum_{i=1}^r(\operatorname{Im}\alpha_i)(\operatorname{Im}\alpha_j)\langle v_i, v_j\rangle = 0, \quad\forall j =1, \ldots, r$$
Summing this over $j=1, \ldots, r$ we get
$$\left\|\sum_{i=1}^r(\operatorname{Im}\alpha_i)v_i\right\|^2 = \left\langle \sum_{i=1}^r(\operatorname{Im}\alpha_i)v_i, \sum_{i=1}^r(\operatorname{Im}\alpha_j)v_j\right\rangle = 0$$
so $\sum_{i=1}^r(\operatorname{Im}\alpha_i)v_i = 0$ which implies $\operatorname{Im}\alpha_i = 0$ for $i=1, \ldots, r$ since $v_1, \ldots, v_r$ are linearly independent over $\mathbb{R}$. Hence $\alpha_i \in \mathbb{R}$ and so $\alpha_i = 0$ for $i=1, \ldots, n$.
Therefore
\begin{align}
\dim_{\mathbb{R}}\big[\operatorname{span}_{\mathbb{R}}\{v_1, \ldots, v_n\}\big] &= \dim_{\mathbb{R}}\big[\operatorname{span}_{\mathbb{R}}\{v_1, \ldots, v_r\}\big] \\
&= r\\
&= \dim_{\mathbb{C}}\big[\operatorname{span}_{\mathbb{C}}\{v_1, \ldots, v_r\}\big] \\
&= \dim_{\mathbb{C}}\big[\operatorname{span}_{\mathbb{C}}\{v_1, \ldots,v_n\}\big]
\end{align}
