The dimension of a complex vector space seen as a real space I have problem with the following: Let $V_{n}(\mathbb{C})$ complex vector space whose dimension is "$n$" and whose basis is $B=${$\vec{x_1},...,\vec{x_n}$}


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*What is the dimension of $V$ if $V$ was seeing like a real vector space, calculate a basis of it.

*Let $V$ the endomorphism that is defined like $J: V_n(\mathbb{C}) \longrightarrow V_n(\mathbb{C})\ / J(\vec{V})=i\vec{v}$. Calculate its change matrix basis if $V$ is seeing like a real vector space in the last basis calculated. 


I don't know how I can prove it.

To answer the first, question, consider the more general situation. 
Suppose $\Bbb K$ is a subfield of a field $\Bbb F$. Then $\Bbb F$ may be regarded as a vector space over $\Bbb K$. Now, if $V$ is a vector space over $\Bbb F$, then $V$ may also be regarded as a vector space over $\Bbb K$. The dimensions of these vector spaces are related by the formula
$$ 
\DeclareMathOperator{Span}{Span}
\dim_{\Bbb K}(V)=\dim_{\Bbb F}(V)\cdot\dim_{\Bbb K}(\Bbb F)\tag{1}
$$
Exercise. Prove this assuming all the dimensions are finite.
Now, in our case we have
\begin{align*}
\Bbb K &= \Bbb R & \Bbb F &= \Bbb C & V &= \Span_{\Bbb C}\{v_1,\dotsc,v_n\}
\end{align*}
What does (1) tell us about $\dim_{\Bbb R}V$?
