Basis of complex matrix vector space over $\Bbb{R}$ I understand that the basis of the vector space $$Mat_2(\Bbb{R}) = \begin{pmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{pmatrix}$$ over $\Bbb{R}$ is $$e = \left\{ \begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}, \begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix} \right\}$$
However, I can't figure out the basis or the vector space $Mat_2(\Bbb{C}) = \begin{pmatrix}a_{11}+b_{11}i & a_{12}+b_{12}i \\ a_{21}+b_{21}i & a_{22}+b_{22}i\end{pmatrix}$ over $\Bbb{R}$.
Thank You.
 A: That would be $e = \left\{ \begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}, \begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix} \right\}$together with $e' = \left\{ \begin{pmatrix}i & 0\\ 0 & 0\end{pmatrix}, \begin{pmatrix}0 & i\\ 0 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\ i & 0\end{pmatrix},\begin{pmatrix}0 & 0\\ 0 & i\end{pmatrix} \right\}$. 
Think of it this way. You want elements such that you can write matrix of the form
$Mat_2(\Bbb{C}) = \begin{pmatrix}a_{11}+b_{11}i & a_{12}+b_{12}i \\ a_{21}+b_{21}i & a_{22}+b_{22}i\end{pmatrix}$. But you can only multiply the elements with real numbers. Therefore you will need two elements for every position in the matrix. One element takes care of the complex part, one element for the real part. 
Could you find base elements for $\mathbb{C}$ over $\mathbb{R}$? That would be the familiar $1$ and $i$. A square matrix of dimension $n\times m$ has as dimension the dimension of the underlying field to the power $n\times m$. 
A: If you have a basis $(e_1,\dots,e_n)$ of a vector space over $\Bbb C$, $(e_1,\dots,e_n,ie_1,\dots,ie_n)$ a basis of that vector space as a vector space over $\Bbb R$.
This has to do with the fact that a basis of $\Bbb C$ over $\Bbb R$ is $(1,i)$.
