<s>[INST] 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),$$\left(\begin{array}{cc}
2 & 1\\
1 & 3
\end{array}\right)$, $\left(\begin{array}{cc}
4 & -1\\
-1 & -5
\end{array}\right)$ are linearly independent and spanning then it is 
a basis for V , so could we get a set of equations like 
$\left(\begin{array}{cc}
x & z\\
z & y
\end{array}\right)$= A$\left(\begin{array}{cc}
1 & -2\\
-2 & 1
\end{array}\right)$+ B $\left(\begin{array}{cc}
2 & 1\\
1 & 3
\end{array}\right)$+ C$\left(\begin{array}{cc}
4 & -1\\
-1 & -5
\end{array}\right)$ - (1) 
Giving 
x= A + 2B +4C 
y= A + 3B -5C 
z = -2A+B -C 
So if this system of equations has a solution not all zero, then $\clubsuit$
is a basis for V. 
2) For part 2) would we have a similar approach to part one solving a set of values in (1) but replacing the LHS of (1) with Z?.
 [/INST] For 1) it is enough to show that they are  spanning all the space or independent, since as if they are independent and the set of $2$ by $2$ symmetric matrices is a vector space of dimension $3$, then in-dependency results in spanning also. Note that if
$$
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).
$$
</s>