Prove that a mapping from C to M2(R) is injective and a homomorphismm $\mathcal{M}_2(\Bbb R)$ is the ring of $2\times2$ matrices with real entries.
Define a map $\phi:\Bbb C \to \mathcal{M}_2(\Bbb R)$ by 
$$\phi(a+bi) = \begin{pmatrix} a & b \\ -b & a \end{pmatrix}$$
Prove that the mapping is a homomorphism and that it is injective.
Now, in order to show that it is a homomorphism is it enough if I simply show that $\phi(a)+\phi(b) = \phi(a+b)$ and $\phi(a)\phi(b) = \phi(ab)$?
Also, how do I show that it is injective? Does this have something to do with the kernel? If so, how do I find the kernel?
 A: A ring homomorphism from a field to any ring is necessarily injective (assuming that ring homomorphisms map $1$ to $1$), because its kernel is a proper ideal and a field has only $\{0\}$ as proper ideal.
In this case it's also easy to verify it directly, because if $\phi(a+bi)$ is the null matrix, then necessarily $a=b=0$.
The verification that $\phi$ is a homomorphism consists in showing that
$$
\phi(x+y)=\phi(x)+\phi(y),\qquad
\phi(xy)=\phi(x)\phi(y),\qquad
\phi(1)=\begin{bmatrix}1 & 0\\0&1\end{bmatrix}
$$
for all $x,y\in\mathbb{C}$. For multiplication, consider $x=a+bi$ and $y=c+di$; then $xy=(ac-bd)+(ad+bc)i$, so
$$
\phi(xy)=\begin{bmatrix}ac-bd & ad+bc \\ -(ad+bc) & ac-bd \end{bmatrix}
$$
whereas
$$
\phi(x)\phi(y)=
\begin{bmatrix} a & b \\ -b & a \end{bmatrix}
\begin{bmatrix} c & d \\ -d & c \end{bmatrix}
$$
Can you finish doing the matrix product? The check for the addition should be carried on similarly.
Note. In the definition of $\phi$ it is implicitly assumed that $a,b\in\mathbb{R}$ and the same assumption is made in the check, also for $c$ and $d$.
A: Hint:
This is a ''canonical '' isomorphism between $\mathbb{C}$ and  a subring of $M_2(\mathbb{R})$ (here).
to show this fact the first step is to prove that the set $\mathcal{C}$ of matrices of the form 
$$
\begin{bmatrix}
a&b\\
-b&a
\end{bmatrix}
\qquad a,b \in \mathbb{R}
$$
is a field.
The associative and distributive properties of sum and product are inherited  by the ring structure of $M_2(\mathbb{R}$, and so the existence of the neutral elements for the two operations. But we have to prove that any element of $\mathcal{C}$ has an inverse and that the product is commutative. 
Commutativity is easy verified :
$$
\begin{bmatrix}
a&b\\
-b&a
\end{bmatrix}
\begin{bmatrix}
x&y\\
-y&x
\end{bmatrix}
=
\begin{bmatrix}
ax-by&ay+bx\\
-bx-ay&-by+ax
\end{bmatrix}=
\begin{bmatrix}
x&y\\
-y&x
\end{bmatrix}
\begin{bmatrix}
a&b\\
-b&a
\end{bmatrix}
$$
For the invertibility note that, for $A \in \mathcal{C} $, $\det(A)=a^2+b^2 \ne0 \forall a,b \in \mathbb{R} $
and 
$$
\begin{bmatrix}
a&b\\
-b&a
\end{bmatrix}^{-1}=
\dfrac{1}{a^2+b^2}
\begin{bmatrix}
a&-b\\
b&a
\end{bmatrix}
$$
Now (second step) it is easy to see that for the given $\phi$ we have
$$
\phi(1)=\begin{bmatrix}
1&0\\
0&1
\end{bmatrix}
$$
and
$$ \phi(z)=A \Rightarrow \phi(\bar z)=A^T\Rightarrow \phi(z^{-1})=A^{-1}$$
and , obviously, $ \phi(z)=A \Rightarrow \phi(-z)=-A$.
The last step is to show that $ker (\phi)$ is trivial, but this is immediate since 
$$
\phi(a+ib)=\begin{bmatrix}
0&0\\
0&0
\end{bmatrix} \iff
a=b=0
$$
Finally: note that there are many (infinite) representations of $\mathbb{C}$ of this type, but is seems that there only two  such that $\psi(\bar z)=A^T$ ( see  the question: Matrix representation of $\mathbb{C}$ as $^*$Algebra.). 
A: Note that by the definition of $ \phi$, $ \phi(z)=0\iff z=0$. 
Moreover $ \phi$ is isomorphic to the restricted image set of $ \phi$. Hence we can conclude that we can always have a subring of $ \mathcal{M}_2(\Bbb R)$ that is isomorphic to $ \Bbb{C}$.  
Now the only question appears in my mind that how to prove that $ \phi$ is only a monomorphism but may not be an epimorphism.
