How do you complexify vector spaces? Suppose you have a linear map $T: \mathbb{R}^2 \to \mathbb{R}^2$ given by $$Tv = \begin{bmatrix}
a &b \\ 
c & d
\end{bmatrix}v$$
So, if I understand correctly complexification (of the domain and the target space and the map) means $\tilde{T}: \mathbb{R}^2 \otimes \mathbb{C} \to \mathbb{R}^2 \otimes \mathbb{C}$ where $$\tilde{T}(v \otimes z) = T(v) \otimes z$$ as I can not think of any other way to define a map from $\mathbb{R}^2 \otimes \mathbb{C} \to \mathbb{R}^2 \otimes \mathbb{C}$. This will now be linear map from a 4 dimensional space to a 4 dimensional space. And if we use the standard tensor product basis then $\tilde T$  is just a block diagonal matrix with each block identical to T. This seems pretty pointless.
Now if we look at a map $f: \mathbb{C} \to \mathbb{C}$ which is given by $f = u+iv$ or $f = (u,v)$ thinking of $\mathbb{C}$ as $\mathbb{R}^2$. Then the derivative $f^*: T_p \mathbb{R}^2 \to T_{f(p)} \mathbb{R}^2$ is given by $$\begin{bmatrix}
u_x &u_y \\ 
v_x & v_y
\end{bmatrix}$$ if we let the basis of both tangent spaces to be $\partial_x, \partial_y$. Now, if we play the same game and complexify these tangent spaces and $f^*$ and write it in terms of the standard tensor product basis then we should get a block diagonal matrix with each block identical to that of $f^*$. Now if we write the matrix corresponding to the complexification with respect to the basis $$\partial_z = \partial_x \otimes1 - \partial_y \otimes i \\ \partial_{\bar z} = \partial_x \otimes1 + \partial_y \otimes i$$ (what are the other two basis vectors) why do we get the following matrix?
$$\begin{bmatrix}
f_z &f_{\bar z} \\ 
\bar f_z & \bar f _{\bar z}
\end{bmatrix}$$ i.e. a 2 by 2 and not a 4 by 4 matrix.
 A: $\newcommand{\dd}{\partial}\newcommand{\Number}[1]{\mathbf{#1}}\newcommand{\Cpx}{\Number{C}}\newcommand{\Reals}{\Number{R}}$If $B = (v_{1}, \dots, v_{n})$ is an ordered basis of a real vector space $V$, then

*

*$B = (v_{1}, \dots, v_{n})$ is an ordered complex basis of the complexification $V \otimes \Cpx$;

*$B_{\Cpx} := (v_{1}, \dots, v_{n}, iv_{1}, \dots, iv_{n})$ is an ordered real basis of $V \otimes \Cpx$. Note that we're suggestively writing $v_{k} \otimes 1 = v_{k}$ and $v_{k} \otimes i = iv_{k}$. (Caution: This is not the only ordering used in the literature.)

Correspondingly, if the linear operator $T:V \to V$ has (real $n \times n$) matrix $A$, then the induced complex-linear operator $T$ on $V \otimes \Cpx$

*

*(also) has matrix $A$ with respect to $B$;

*(as you say) has block diagonal matrix $A \oplus A$ with respect to $B_{\Cpx}$.

Particularly, if we choose $(\dd_{x}, \dd_{y})$ as a basis of $\Reals^{2}$, then $(\dd_{x}, \dd_{y})$ is a complex basis of $\Reals^{2} \otimes \Cpx = \Cpx^{2}$, and $(\dd_{x}, \dd_{y}, i \dd_{x}, i \dd_{y})$ is a real basis of $\Reals^{2} \otimes \Cpx = \Cpx^{2}$. As Brevan Ellefsen notes in the comments, there is a straightforward change of basis calculation, using the complex bases $(\dd_{x}, \dd_{y})$ and
$$
\dd_{z} = \tfrac{1}{2}(\dd_{x} - i \dd_{y}),\qquad
\dd_{\bar{z}} = \tfrac{1}{2}(\dd_{x} + i \dd_{y}).
$$
(The factors of $\frac{1}{2}$ aren't essential for the calculation, but are needed to make these vectors dual to the complex $1$-forms $dz = dx + i\, dy$ and $d\bar{z} = dx - i\, dy$.)
