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**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?.

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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. $$ \left( \begin{matrix} 1 & 2 &4\\ 1& 3& -5\\ -2 &1 &-1\\ \end{matrix}\right) \left( \begin{matrix} x\\y\\z \end{matrix}\right)= \left( \begin{matrix} 4\\-7\\-11 \end{matrix}\right). $$

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