Are the $2\times 2$ symmetric matrices a ring? Ok so I am looking at Rings. I saw somewhere that the $2 \times 2$ symmetric matrices with entries in $\mathbb{R}$ is a ring. But if we look at matrix multiplication I am not convinced:
If $ A = \left( \begin{array}{cc}
a & b \\
b & c\\
\end{array} \right) $
and $ B = 
\left( \begin{array}{cc}
x & y \\
y & z\\
\end{array} \right) $
then $AB = \left( \begin{array}{cc}
ax+by & ay+bz \\
bx+cy & by+cz\\
\end{array} \right) $
which is not symmetric?
What am I missing? Does $a=c$ and $x=z$? If so, why? The matrices aboves are still symmetric without those being equal.
Thank you
 A: As you observed correctly the symmetric $2 \times 2$ matrices are not a ring (with the usual operations) since the set is not closed under multiplications; it is at least an additive subgroup though.
Examples where already given, but let me add one.
$$\begin{pmatrix}
a & b \\
b & d
\end{pmatrix} \begin{pmatrix}
0 & 0 \\
0 & 1
\end{pmatrix} = \begin{pmatrix}
0 & b \\
0 & d
\end{pmatrix}  $$
which is not symmetric (unless $b=0$).
It is however true that the matrices of the form 
$$\begin{pmatrix}
a & b \\
b & a
\end{pmatrix}
$$
form a ring and possibly this was what was meant in your source.
Indeed, the matrices form clearly an additive subgroup, and writing $I$ for the identity matrix and $J$ for the matrix $$\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
$$
each such matrix can be written (uniquely) as $xI+ yJ$ with real $x,y$. Since we have $I^2 = I$, $IJ=JI= J$ and $J^2 = I$, and the distributive law, it follows that the matrix is again of the claimed form.
Furthermore, we get from this consideration that this ring is isomorphic to $\mathbb{R}[X]/(X^2 -1)$. Since $X^2 - 1$ is not irreducible over $\mathbb{R}$ this is not a field, and we get that a further isomorphism to  $\mathbb{R}[X]/(X -1) \times \mathbb{R}[X]/(X +1) \cong \mathbb{R} \times \mathbb{R}$.
In this context it might also be interesting to recall that one way to construct the complex numbers is as matrices of the form 
$$\begin{pmatrix}
a & -b \\
b & a
\end{pmatrix}
$$
There the polynomial we get is $X^2 + 1$, which is irreducible over $\mathbb{R}$.
A: It is very easy to construct a counterexample. Take any 2x2 symmetric matrices and look at their product. Is it symmetric? Try.
I tried with
$$\begin{pmatrix}
1 & 2 \\
2 & 0
\end{pmatrix}$$
and $$\begin{pmatrix}
1 & 1 \\
1 & 1
\end{pmatrix}$$
(just putting random numbers on the entries)
A: It is a ring if both addition and multiplication are performed entrywise. If we use the usual matrix multiplication, however, the set does not form a ring, as you correctly observed.
