Represent a complex-valued matrix into real-valued matrix If I have a complex Matrix $\boldsymbol{W}\in C ^{M\times N}$, Why this matrix can be written as 
$$W=
\begin{bmatrix}
W_r & -W_i \\ W_i &W_r
\end{bmatrix}\in R ^{2M\times 2N}$$
I appreciate your answers! 
 A: Do this first for 1-by-1 complex matrix.  Try
$$
\begin{bmatrix} a & -b \\ b & a \end{bmatrix}
$$
Try multiplication/addition of such matrices and then notice that this is exactly the same formulas as multiplication of complex numbers, where the matrix above represents $a+ib$.  And transpose is then complex conjugate.  Really it is a representation of the complex number field inside the ring of 2-by-2 real matrices.  So in a certain sense ${\mathbb C} \subset M_2({\mathbb R})$ as a subring.
A way to philosophically think of this is to notice that multiplication is something that ought to be bilinear.  Then you just need to think of what $1$ and $i$ do.  In this identification we have
$$
1 = 
\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
,
\qquad
i = 
\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} ,
$$
which are the identity, and the matrix that rotates counterclockwise by 90 degrees.
Now let's get to larger matrices.  Take an $n$-by-$n$ complex matrix and replace every entry by a 2-by-2 matrix as above.  In
$$
\begin{bmatrix}
z_{11} & z_{12} & z_{13} \\
z_{21} & z_{22} & z_{23} \\
z_{31} & z_{32} & z_{33}
\end{bmatrix}
$$
replace each complex number $z_{ij}$ with the 2-by-2 matrix representing the complex number.  You get a 3-by-3 block matrix, or in other words a 6-by-6 real matrix.  Matrix multiplication for block matrices works in the familiar way, you multiply blocks as you would the entries.  Therefore the 6-by-6 real matrix really represents the 3-by-3 complex matrix.  When you want to multiply the matrix by a vector, you need to make the vector into a block matrix again, in this case a vector is a 3-by-1 complex matrix, or a 6-by-2 real matrix.
Now finally, how to write it in the form as you did?  Let's start with a 2-by-2 complex matrix
$$
\begin{bmatrix}
a_{11} + ib_{11} & a_{12} + ib_{12} \\
a_{21} + ib_{21} & a_{22} + ib_{22} \\
\end{bmatrix}
$$
and do the above procedure (and let's call it $A$)
$$
A = \begin{bmatrix}
a_{11} & -b_{11} & a_{12} & -b_{12} \\
b_{11} & a_{11} & b_{12} & a_{12} \\
a_{21} & -b_{21} & a_{22} & -b_{22} \\
b_{21} & a_{21} & b_{22} & a_{22} \\
\end{bmatrix}
$$
Let's apply a real-linear change of coordinates $T$ to $A$ (we reorder the 2nd and 3rd entry)
$$
T = \begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 1
\end{bmatrix}
$$
So
$$
T^{-1} A T
=
\begin{bmatrix}
a_{11} & a_{12} & -b_{11} & -b_{12} \\
a_{21} & a_{22} & -b_{21} & -b_{22} \\
b_{11} & b_{12} & a_{11} & a_{12} \\
b_{21} & b_{22} & a_{21} & a_{22} \\
\end{bmatrix}
$$
Which is the form that you want.  It is just in a different bases (really just reordering the vectors of the basis by writing all the real parts first and then the imaginary parts.
