Rewriting a complex matrix as a real matrix and changing basis to $(z,\bar z)$ If we have a complex $n\times n$ matrix $A$ and we rewrite it as a real $2n\times 2n$ matrix $B$ by using the identity $z=x+iy$. I don't get how if we change basis from $(x,y)$ to $(z,\bar z)$ that is $(x+iy,x-iy)$ we are then changing a basis via a matrix $M$ as $M^{-1}BM$. And how then $M^{-1}BM=\begin{bmatrix} A&0\\0&\bar A\end{bmatrix}$, where $\bar A$ is the complex conjugate of $A$?
 A: The complex $1$ dimensional  case is the well known one: if $w=a+ib$ and $z=x+iy$
$$wz=ax-by+i(ay+bx)$$
So via the usual $(x+iy)\in\mathbb{C}\cong \mathbb{R}^2\ni(x,y)$ identification, multiplication becomes
$$B\begin{pmatrix}x \\ y\end{pmatrix}:=\begin{pmatrix}a & -b \\ b& a\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}.$$
If one would choose the identification $(x+iy)\in\mathbb{C}\hookrightarrow  \mathbb{C}^2\ni(x+iy,x-iy)$ then multiplication would appear as 
$$wz=ax-by+i(ay+bx)\hookrightarrow (wz,\overline{wz})=(wz,\bar{w}\bar{z})$$
and so in matrix notation as 
$$\begin{pmatrix}w & 0 \\ 0& \bar{w}\end{pmatrix}\begin{pmatrix}z \\ \bar{z}\end{pmatrix}.$$
The generalization to the $n$ dimensional case is similar, where in the second case the identification is made as
$$(z_1,\dots,z_n)\hookrightarrow  (z_1,\dots,z_n,\bar{z_1},\dots,\bar{z_n}),$$
and so one obtains
$$Az\hookrightarrow(Az,\overline{Az})=(Az,\overline{A}\overline{z})$$
In the first case, using the identification
$$(z_1=x_1+iy_1,\dots,z_n=x_n+iy_n)\hookrightarrow(x_1,y_1,\dots, x_n,y_n),$$
one gets the representation of $A=((a_{ij}+ib_{ij}))_{ij}$ as
$$B=\begin{pmatrix}a_{11} & -b_{11} & a_{12} & -b_{12} & \dots\\ b_{11} & a_{11} & b_{12} & a_{12} & \dots
\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix},$$
i.e. one replaces each entry $a+bi$ of $A$ with a $2\times 2$ block $\begin{pmatrix}a & -b \\ b& a\end{pmatrix}$ analogously to the $2$ dimensional case.
Going back to the $2$ dimensional case note that the change of basis matrix is given by 
$$M:=\begin{pmatrix}1 & i \\ 1 & -i\end{pmatrix}.$$
One can then check that the above found representations agree with the one given by the change of basis, as 
$$M^{-1}=\frac{i}{2}\begin{pmatrix}-i & -i \\ -1 & 1\end{pmatrix}=\frac{1}{2}\begin{pmatrix}1 & 1 \\ -i &i\end{pmatrix}$$
and so
$$M^{-1}BM=\begin{pmatrix}a+ib & 0 \\ 0& a-ib\end{pmatrix}=\begin{pmatrix}w & 0 \\ 0& \bar{w}\end{pmatrix}$$
