Prove Basis for symmetric matrix. **Let V be the vector subspace of M$_{2}$ ($\mathbb{R})$ consisting
of all symmetric matrices, That is 
A$^{t}$ = A.
1) Show that
$\clubsuit$= $\left\{ \left(\begin{array}{cc}
1 & -2\\
-2 & 1
\end{array}\right),\left(\begin{array}{cc}
2 & 1\\
1 & 3
\end{array}\right),\left(\begin{array}{cc}
4 & -1\\
-1 & -5
\end{array}\right)\right\} $ is a basis for V. 
2) Find the co-ordinates of Z =  $\left(\begin{array}{cc}
4 & -11\\
-11 & -7
\end{array}\right)$ with respect to this Basis.** 
For Part 1) could we argue that the generalised form of a symmetric
matrix in M$_{2}$ ($\mathbb{R})$ would be 
something like $\left(\begin{array}{cc}
x & z\\
z & y
\end{array}\right)$: x,y,z $\epsilon\mathbb{R}$. Then if $\left(\begin{array}{cc}
1 & -2\\
-2 & 1
\end{array}\right),$$\left(\begin{array}{cc}
2 & 1\\
1 & 3
\end{array}\right)$, $\left(\begin{array}{cc}
4 & -1\\
-1 & -5
\end{array}\right)$ are linearly independent and spanning then it is 
a basis for V , so could we get a set of equations like 
$\left(\begin{array}{cc}
x & z\\
z & y
\end{array}\right)$= A$\left(\begin{array}{cc}
1 & -2\\
-2 & 1
\end{array}\right)$+ B $\left(\begin{array}{cc}
2 & 1\\
1 & 3
\end{array}\right)$+ C$\left(\begin{array}{cc}
4 & -1\\
-1 & -5
\end{array}\right)$ - (1) 
Giving 
x= A + 2B +4C 
y= A + 3B -5C 
z = -2A+B -C 
So if this system of equations has a solution not all zero, then $\clubsuit$
is a basis for V. 
2) For part 2) would we have a similar approach to part one solving a set of values in (1) but replacing the LHS of (1) with Z?.

For 1) it is enough to show that they are  spanning all the space or independent, since as if they are independent and the set of $2$ by $2$ symmetric matrices is a vector space of dimension $3$, then in-dependency results in spanning also. Note that if
$$
T =\left(
\begin{matrix}
1 & 2 &4\\ 1& 3& -5\\ -2 &1 &-1\\
\end{matrix}\right)
$$
then $\det T \ne 0$
For 2) your trend is correct and you should find the coefficients in linear representation, i.e.
$$
\left(
\begin{matrix}
1 & 2 &4\\ 1& 3& -5\\ -2 &1 &-1\\
\end{matrix}\right)
\left(
\begin{matrix}
x\\y\\z
\end{matrix}\right)=
\left(
\begin{matrix}
4\\-7\\-11
\end{matrix}\right).
$$
