Complexification of v.s. of all endos on $V$ is isomorphic to v.s. of all endos on complexification of $V$ Let $V$ be a real vector space. By calculating their dimension, $\text{End}(V)_\mathbb{C}\cong \text{End}(V_\mathbb{C})$ when $\dim_\mathbb{R} V< \infty$.
I hope to find exactly an isomorphism between them, and wonder if this isomorphism is true if $\dim V=\infty$.
Let me clarify some concepts...
If $V$ is a vector space over field $\mathbb{F}$, then $\text{End}(V)$ denote the set of all endomorphisms on $V$. It is proved that $\text{End}(V)$ is also a vector space over $\mathbb{F}$, and $\dim_\mathbb{F} \text{End}(V) = n^2$ if $\dim_\mathbb{F} V = n< \infty$.
If $V$ is a vector space over real numbers $\mathbb{R}$, then the complexification is the set $V\times V = \{(u,v)\colon u,v\in V\}$, denoted $V_\mathbb{C}$, with addition
$$
(u_1,v_1) + (u_2,v_2) = (u_1+u_2,v_1+v_2),
$$
and scalar multiplication over complex field $\mathbb{C}$
$$
(a+bi)(u,v) = (au-bv,av+bu),
$$
when $a,b\in \mathbb{R}$. It is proved that $V_\mathbb{C}$ is a vector space over $\mathbb{C}$, and if $\dim_\mathbb{R} V = n<\infty$, then $\dim_\mathbb{C} V_\mathbb{C} =  n$ (if $v_1,\dots,v_n$ is a basis of $V$, then $(v_1,0),\dots,(v_n,0)$ is a basis of $V_\mathbb{C}$).
Therefore, when $\dim_\mathbb{R} V=n$, where $V$ is a real vector space, then $\dim_\mathbb{C} \text{End}(V)_\mathbb{C} = \dim_\mathbb{C} \text{End}(V_\mathbb{C}) = n^2$, so they are isomorphic in finite dimensional case.
If you are interested...
If $V$ is a real vector space and $T\in \text{End}(V)$, then the complexification of the operator $T$, denoted $T_\mathbb{C}$, is the map $V_\mathbb{C}\to V_\mathbb{C}$, defined by
$$
T_\mathbb{C}(u,v) = Tu + iTv,
$$
where $u,v\in V$. Trivially $T_\mathbb{C}\in \text{End}(V)$. Notice that if $\dim V<\infty$, then the minimal polynomial of $T_\mathbb{C}$ (complex) is equal to that of $T$ (real), so the set of all complexification of some $T\in \text{End}(V)$ does not span $\text{End}(V_\mathbb{C})$. Let this set be $S$:
$$
S = \{R\in \text{End}(V_\mathbb{C})\colon \text{There exists $T\in \text{End}(V)$ s.t. $T_\mathbb{C} = R$}\}.
$$
If we view $\text{End}(V_\mathbb{C})$ as a real vector space (then its dimension is doubled), by routine check, $S$ is a subspace of $\text{End}(V_\mathbb{C})$. Then what is dimension of $S$?
 A: Yes, they are naturally isomorphic and the obvious choice of isomorphism works. $\text{End}(V)_{\mathbb{C}}$ naturally acts on $V_{\mathbb{C}}$ via the obvious map, namely
$$(T + iS)(v + iw) = (Tv - Sw) + i(Sv + Tw)$$
and this gives an isomorphism $\text{End}(V)_{\mathbb{C}} \cong \text{End}(V_{\mathbb{C}})$ with no dimension hypotheses. Abstractly this comes from understanding complexification as an extension of scalars $V_{\mathbb{C}} \cong V \otimes_{\mathbb{R}} \mathbb{C}$ and then computing that
$$\text{Hom}_{\mathbb{C}}(V \otimes_{\mathbb{R}} \mathbb{C}, V \otimes_{\mathbb{R}} \mathbb{C}) \cong \text{Hom}_{\mathbb{R}}(V, V \otimes_{\mathbb{R}} \mathbb{C}) \cong \text{Hom}_{\mathbb{R}}(V, V) \otimes_{\mathbb{R}} \mathbb{C}$$
where on the right we need that $\mathbb{C}$ is finite-dimensional over $\mathbb{R}$ and on the left we've used the tensor-hom adjunction.
Concretely what this is saying is that a $\mathbb{C}$-linear map $V_{\mathbb{C}} \to V_{\mathbb{C}}$ is, by $\mathbb{C}$-linearity, uniquely and freely determined by the $\mathbb{R}$-linear map $V \to V_{\mathbb{C}}$ given by restricting it to the real subspace of $V_{\mathbb{C}}$. Then this map splits up into real and imaginary parts (since $V_{\mathbb{C}}$ splits up like this) and each of these is an $\mathbb{R}$-linear map $V \to V$. But this is exactly how $\text{End}(V)_{\mathbb{C}}$ splits up.
As for your second question, complexification is injective so $S$ is just a copy of $\text{End}(V)$ and has dimension $n^2$ in the finite-dimensional case.
