For which positive integer n does there exist a $\Bbb R$-linear ring homomorphism $f:\Bbb C$ $\to M_n(\Bbb R)$? 
For which positive integer $n$ does there exist a $\Bbb R$-linear ring homomorphism $f:\Bbb C \to M_n(\Bbb R)?$

If I send everything to zero matrix that will give trivial ring homomorphism for every $n$ .
Now nontrivial ring homomorphism: for $n=1$ there is no such homomorphism from $\Bbb C$ to $M_n(\Bbb R)$ as there is no ring homomorphism from $\Bbb C$ to $\Bbb R$. Now $M_n(\Bbb R)$ is not an integral domain for $n\ge2$, so $f(1)$ will be an idempotent element. I am not getting  any idea to proceed further.
Thanks in advance.
 A: When $n=2$, you have $f(a+ib) = \begin{bmatrix} a & -b \\ b&a\end{bmatrix}$ as an example of a unital $\mathbb R$-algebra homomorphism.  This 2-by-2 matrix representation of complex numbers is mentioned in Wikipedia.
For $n>2$, if you do not require $f(1)=1$ (as is a common requirement in some contexts), then the easiest way to proceed is to put the $n=2$ example in the top left corner with zeros elsewhere.  
If you do have $f(1)=1$, one restriction comes from $f(i)^2 = -1$, which requires $n$ to be even.  When $n$ is even, the easiest way to proceed is to put the $n=2$ example $n/2$ times into block diagonal matrices.   When $n=4$ this gives a restriction to $\mathbb C$ of a real matrix representation of quaternions.
A: If there is an $\mathbb R$-linear homomorphism $\mathbb C\to M_n(\mathbb R)$, then $\mathbb R^n$ can be turned into a complex vector space compatibly with its usual real vector space structure. Of course, this implies that $n$ is even.
Now, if $n$ is even, then we may identify $\mathbb R^n$ with $\mathbb C^{n/2}$, and there is a obvious may to let $\mathbb C$ act on the latter. Picking a real basis, this gives a map $\mathbb C\to M_n(\mathbb R)$.
