If I have a complex matrix ${\bf W} \in {\Bbb C}^{M\times N}$. Why can this matrix be written as follow?
$$ {\bf W} = \begin{bmatrix} {\bf W}_r & -{\bf W}_i \\ {\bf W}_i & {\bf W}_r \end{bmatrix} \in {\Bbb R}^{2M\times 2N} $$
I appreciate your answers!
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.