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.
 A: Up to conjugacy, there are three maximal abelian subrings in the ring of two by two matrices with real entries:
$$\left\{ \left( \begin{matrix} a & 0 \\ 0 & b \end{matrix} \right) \right\}, \ \left\{ \left( \begin{matrix} a & b \\ -b & a \end{matrix} \right) \right\}, \ \text{and} \ \left\{ \left( \begin{matrix} a & b \\ 0 & a \end{matrix} \right) \right\}.$$
Here is a proof of this. Suppose $R \subseteq M_2(\mathbb{R})$ is a maximal commutative subring. Then by maximality, $R$ contains the scalar matrices and at least one matrix $A$ that is not a scalar matrix. Since $R$ is a subring, $R$ necessarily contains all matrices that can be expressed as polynomials in $A$ with coefficients from $\mathbb{R}$, and since the minimal polynomial of $A$ is of degree $2$, $R$ is of dimension at least $2$. The centralizer of a non-scalar $2$ by $2$ matrix such as $A$ has dimension exactly $2$ (a quick and dirty way to see this: we can assume we are working over $\mathbb{C}$ and $A$ is in Jordan form, so is either a diagonal matrix with distinct diagonal entries, or a single Jordan block, and direct calculation in these two cases does it), therefore $R$ consists exactly of matrices that can be expressed as polynomials in $A$. 
We have reduced your problem to classifying the subrings of $M_2(\mathbb{R})$ that consist of all polynomials in a single non-scalar matrix $A$. Working instead over $\mathbb{C}$, up to conjugacy there are only two of these: the algebra of diagonal matrices and the algebra of upper triangular matrices with equal diagonal entries. But over $\mathbb{R}$, things are slightly more complicated, as the minimal (=characteristic, in our case) polynomial of $A$ may not split. 
If this minimal polynomial is of the form $(x-a)^2$ for some $a$, then since it has real coefficients we must have $a \in \mathbb{R}$ and hence $R$ consists of upper triangular matrices with equal diagonal entries. But if it is of the form $(x-a)(x-b)$ for distinct $a,b \in \mathbb{C}$ (if these are not both real, then they are complex conjugates), then there are two possiblities for $R$: if $a$ and $b$ are both real then $R$ is (up to conjugacy) the algebra of diagonal matrices, while if they are complex conjugate $R$ is (again, up to conjugacy) the algebra of matrices of the form
$$\left\{ \left( \begin{matrix} a & b \\ -b & a \end{matrix} \right) \right\}.$$
In the ring of $n$ by $n$ matrices for $n$ large, things are much more complicated. Maximal commutative subalgebras come in a great profusion, already over an algebraically closed field. For instance, there can be maximal commutative subalgebras of dimension much larger than $n$ (around $(n/2)^2$), and there are also maximal commutative subalgebras of dimension much smaller than $n$ (about $3n^{2/3}$). For examples of the latter---somewhat surprising, in my view---see this paper 
http://projecteuclid.org/euclid.dmj/1077375768
of Courter in Duke Math. J., Volume 32, Number 2 (1965), pp. 225--232.
In order to obtain better behavior you should consider self-normalizing abelian subalgebras instead of maximal ones.
A: 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
$$   
\left(
\begin{array}{rr}
\beta I & B \\
0 & \beta I
\end{array}
\right)   
\left(
\begin{array}{rr}
\alpha I & A \\
0 & \alpha I
\end{array}
\right)
=   
\left(
\begin{array}{rr}
\alpha \beta I & \beta A + \alpha B \\
0 & \alpha \beta I
\end{array}
\right).
$$
