Linear independence of certain vectors of $\mathbf{C}^2$ over $\mathbf{R}$ Suppose that $\{e_1,e_2\}$ where $e_1=(1,0)$ and $e_2=(0,1)$ is the standard basis of $\Bbb C^2$ as a vector space over $\Bbb C$. Show that $\{e_1,ie_1,e_2,ie_2\}$ is a basis of $\Bbb C^2$ as a vector space over $\Bbb R$ and conclude that $\dim_\Bbb R \Bbb C^2=2\dim_\Bbb C \Bbb C^2$.
I know that in general in order to show that a set of vectors form a basis you must show that they are linearly independent and are a spanning set. So normally I would solve $Ax=0$ and if $x$ was $0$ I would know they were linearly independent. However, I'm not sure how to use that here. 

As another user has described linear independence (and hopefully made it clear that $dim_{\mathbb{R}} \mathbb{C}^2 = 4 = 2\dim_{\mathbb{C}}\mathbb{C}^2$ ), you need now to show that $\{e_1,e_2,ie_1,ie_2\}$ spans $\mathbb{C}^2$. Let $\xi \in \mathbb{C}^2$ be arbitrary. Then, by our assumptions, there exists $x_1,x_2 \in \mathbb{C}$ so that $\xi = x_1e_1+x_2e_2$. Since each $x \in \mathbb{C}$ we can write $x = a + bi$ for $a,b$ real. Hence, \begin{equation}
\xi = x_1e_1+x_2e_2 \\
=(a_1+b_1i)e_1+(a_2+b_2i)e_2 \\
=a_1e_1+b_1ie_1+a_2e_2+b_2ie_2.
\end{equation}
This should complete your proof. 
