# Represent a complex-valued matrix into real-valued matrix

If I have a complex matrix $${\bf W} \in {\Bbb C}^{M\times N}$$. Why can this matrix be written as follow?

$${\bf W} = \begin{bmatrix} {\bf W}_r & -{\bf W}_i \\ {\bf W}_i & {\bf W}_r \end{bmatrix} \in {\Bbb R}^{2M\times 2N}$$

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.
• +1) For the nice explanation. This is a really good answer. Just one thing how do you write the matrix $T$? I know its a transformation matrix I just did not get its structure. Can you please explain little more on it? Commented Nov 14, 2016 at 1:04
• $T$ is just an elementary matrix for swapping rows/columns. $T^{-1}=T$ in fact. To see $T^{-1}AT$ you just swap the 2nd and 3rd row and 2nd and third column of $A$. Write down the matrix for 3 by 3 and then see what you have to swap. The columns of $T$ are just the standard basis elements in the other order. Commented Nov 14, 2016 at 1:42