What is the dimension of $M(n,\mathbb{C})$ over $\mathbb{R}$?

Question

Suppose $M(n,\mathbb{C})$ is the vector space of all $n\times n$ matrices with elements from $\mathbb{C}$, the elements are of the form $a_{ij}+\iota b_{ij} $. It seems the dimension of $M(n,\mathbb{C})$ over $\mathbb{R}$ is $2n^2$ since there are $2n\times n$ real numbers in the matrix.

But consider the space of all endomorphisms on $\mathbb{C} ^n$ denoted by $\operatorname{End}(\mathbb{C} ^n)$ which has dimension of $2n \times 2n = 4n^2$ over $\mathbb{R}$ and it is isomorphic to $M(n,\mathbb{C})$. This gives dimension of $M(n,\mathbb{C})$ to be $4n^2$.

What am I missing ?

Answer 1

The key is that $\mathrm{End}_{\Bbb C} (\Bbb C^n)$ is not isomorphic to $M(n, \Bbb C)$ over $\Bbb R$ (only over $\Bbb C$).

The first result $2n^2$ is the valid one.

Answer 2

Given any finite dimensional vector space $V$ over $\Bbb C$ one has $$ \dim_{\Bbb R}V=2\dim_{\Bbb C}V. $$ If $\{v_1,...,v_n\}$ is a basis over $\Bbb{C}$ then$\{v_1,iv_1...,v_n,iv_n\}$ is a basis over $\Bbb{R}$.