Maximal commutative subring of the ring of $2 \times 2$ matrices over the reals Motivated by complex numbers, I noticed that the set of all elements of the following forms a commutative sub-ring of $M_2(\mathbb{R})$:
\begin{pmatrix}
x & y\\
-y & x
\end{pmatrix}
Is this sub-ring maximal w.r.t commutativity? If this sub-ring is commutatively maximal, are there 'other' such maximally commutative sub-rings?
P.S: 'other'=non-isomorphic.

Maximal dimension of a commutative subalgebra of $n$ by $n$ matrices is 
$$ 1 + \left\lfloor \frac{n^2}{4} \right\rfloor.   $$
In even dimension $2m,$ this is realized by
$$   
\left(
\begin{array}{rr}
\alpha I & A \\
0 & \alpha I
\end{array}
\right),
$$
where $A$ is any $m$ by $m$ matrix and $I$ is the $m$ by $m$ identity.
This is a theorem of Schur, see https://mathoverflow.net/questions/29087/commutative-subalgebras-of-m-n and references in Robin Chapman's answer.
Note
$$   
\left(
\begin{array}{rr}
\alpha I & A \\
0 & \alpha I
\end{array}
\right)   
\left(
\begin{array}{rr}
\beta I & B \\
0 & \beta I
\end{array}
\right)
=   
\left(
\begin{array}{rr}
\alpha \beta I & \beta A + \alpha B \\
0 & \alpha \beta I
\end{array}
\right),
$$
same as
$$   
\left(
\begin{array}{rr}
\beta I & B \\
0 & \beta I
\end{array}
\right)   
\left(
\begin{array}{rr}
\alpha I & A \\
0 & \alpha I
\end{array}
\right)
=   
\left(
\begin{array}{rr}
\alpha \beta I & \beta A + \alpha B \\
0 & \alpha \beta I
\end{array}
\right).
$$
