2x2 symmetric matrix is a subspace of vector space. Can you kindly check my proof of the problem and correct if possible. 
The following
$S=\{A\in M_{2,2} | AA^T=A^TA\}$
is a subspace of 
$V=M_{2,2}$ all real $2\times2$ matrices.
My proof:
S should be  of symmetric matrices. (I am mostly confused at this point,if other matrices that satisfy $S=\{A\in M_{2,2} | AA^T=A^TA\}$ exist, my proof is not general solution anymore)
Therefore, let the three matrices be:
$$
A_1=\begin{bmatrix}
a & b\\
b & a\\
\end{bmatrix}$$
$$
A_2=\begin{bmatrix}
c & d\\
d & c\\
\end{bmatrix}$$
$$
null=\begin{bmatrix}
0 & 0\\
0 & 0\\
\end{bmatrix}$$


*

*Taking null matrix leads that space is non-empty - proved

*Symmetric matrices with random values


$$A_1+A_2=\begin{bmatrix}
a+c & b+d\\
b+d & a+c\\
\end{bmatrix} 1st.equation
$$
$$
(A_1+A_2)^T=\begin{bmatrix}
a+c & b+d\\
b+d & a+c\\
\end{bmatrix} 2nd.equation
$$
From 1st and 2nd equations $(A_1+A_2)(A_1+A_2)^T=(A_1+A_2)^T(A_1+A_2)$ - proved.


*

*let $Q=\alpha A_1$


$QQ^T=\alpha A_1 \alpha A_1^T$ 
$\quad\quad=\alpha ^2 A_1 A_1^T$ (since $A_1 A_1^T=A_1^T A_1$ for symmetric matrices)
$\quad\quad=\alpha ^2 A_1^T A_1$
$\quad\quad=\alpha A_1^T \alpha A_1^T$
$\quad\quad=Q^TQ$ - proved
 A: Let me discuss as much as I have understood.Please correct me if there is anything wrong.
Taking $A\in S$ where $A=\begin{bmatrix}a & b\\c & d\\\end{bmatrix}$ we take $AA^T=A^TA$ which (on some direct calculation through the matrix multiplication) gives $b=\pm c$ and $ac+bd=ab+cd$. So if $b=c$ then it is fine (i.e $A$ is symmetric) and in case $b=-c$(with both non zero,otherwise $b=c=0$ which is done already)  then we have $c(a-d)=c(d-a)$ which is possible only when $a=d$.

So we conclude : If $A\in S$ then either $A$ is symmetric or $A$ has the form $\begin{bmatrix}a & b\\-b & a\\\end{bmatrix}$  

Now if we take $A,B\in S$ such that only one of them(say $A$) is symmetric then clearly we can take $A=\begin{bmatrix}a & b\\b & d\\\end{bmatrix}$ and $B=\begin{bmatrix}e & f\\-f & e\\\end{bmatrix}$ to conclude that $A+B=\begin{bmatrix}a+e & b+f\\b-f & d+e\\\end{bmatrix}$ can not be in $S$ in general as $A+B$ is neither symmetric nor has the specified form in general,but it is possible for $A+B$ to be in $S$ only when $a=d,b=0$

So my conclusion is that $S$ can not be a subspace in general.

Now if $S$ is considered to be a subset of symmetric matrices then as OP proved, it is indeed a subspace.
(Please note that to prove scalar multiplication remains in $S$ we donot need symmetric nature,$AA^T=A^TA$ is sufficient to prove so)

To extend the discussion: If we consider $S$ such that every element of $S$ has that specified form 

Then we see that if $A=\begin{bmatrix}a & b\\-b & a\\\end{bmatrix}$ and $A=\begin{bmatrix}c & d\\-d & c\\\end{bmatrix}$ then $A+B=\begin{bmatrix}x & y\\-y & x\\\end{bmatrix}$ where $x=a+c,y=b+d$ and then clearly $(A+B)^T(A+B)=(A+B)(A+B)^T=\begin{bmatrix}x^2+y^2 & 0\\0 & x^2+y^2\\\end{bmatrix}$ and thus $S$ will be a subspace.(scalar multiplication definitely remains in $S$ as usual)
Hope this helps.Please share any further suggestion/correction.Thank You.
