Question: Well I know the series converges, I just wanted to make sure I did the work properly.
I used the ratio test.

I ended up with $(n+1)^{(k-1)}a/n^k$.
I concluded that the exponent $k>k-1$ and so the denominator is bigger than the numerator making $L<1$ which means the series converges.

Is this correct, or can I not assume the denominator is bigger than the numerator?

and thanks in advance!!:)

Answer: 
Let $A=\pmatrix{a&b\\ c&d}$. Then $M\_A=aM\_{E\_1}+cM\_{E\_2}+bM\_{E\_3}+dM\_{E\_4}$. Here are some entries of $M\_A$:
$$
M\_A=\pmatrix{
0 &b &-c &0\\
c &d-a &0 &-c\\
-b &0 &\ast &\ast\\
0 &-b &\ast &\ast}.\tag{1}
$$
And I will leave the rest to you. Now, the given inner product $\langle X,Y\rangle$ is the usual dot product of $[X]\_{\xi}$ and $[Y]\_{\xi}$ in $\mathbb{R}^4$. Therefore, $L\_A$ is symmetric w.r.t. $\langle\cdot,\cdot\rangle$ if and only if $M\_A$ is a symmetric matrix. More specifically,
\begin{align\*}
L\_A \text{ is symmetric }
&\Leftrightarrow \langle L\_A(X),Y\rangle=\langle X,L\_A(Y)\rangle \text{ for all } X,Y\in M^{2\times2}(\mathbb{R})\\
&\Leftrightarrow [X]\_{\xi}^tM\_A^t [Y]\_{\xi}=[X]\_{\xi}^t M\_A[Y]\_{\xi} \text{ for all } X,Y\in M^{2\times2}(\mathbb{R})\\
&\Leftrightarrow u^tM\_A^tv=u^t M\_Av \text{ for all } u,v\in \mathbb{R}^4\\
&\Leftrightarrow M\_A^t=M\_A.
\end{align\*}

From $(1)$, we see that for $M\_A$ to be symmetric, a necessary condition is $b=c$. If you could determine the unspecified entries of $M\_A$ correctly (note that your $M\_{E\_3}$ is wrong; there's perhaps a typo), you will see that $b=c$ is also a sufficient condition. Hence the result.
"Another" way is the following:

Show that $\langle L\_{E\_2}(X),Y\rangle=\langle X,L\_{E\_3}(Y)\rangle$ and that $\langle L\_{E\_3}(X),Y\rangle=\langle X,L\_{E\_2}(Y)\rangle$ (actually these two are equivalent).  

Therefore for $A=\begin{pmatrix} a\_1&a\_3\\a\_2&a\_4\end{pmatrix}$ we have (note that $M\_{E\_1},M\_{E\_4}$ are symmetric) 
$$
\langle L\_A(X),Y\rangle=\sum\_{i=1}^4a\_i\langle L\_{E\_i}(X),Y\rangle=\\
a\_1\langle X,L\_{E\_1}(Y)\rangle+a\_2\langle X,L\_{E\_3}(Y)\rangle+a\_3\langle X,L\_{E\_2}(Y)\rangle+a\_4\langle X,L\_{E\_4}(Y)\rangle=\\
\langle X,L\_{A^t}(Y)\rangle.
$$
It follows that $L\_A$ is symmetric $\iff \langle X,L\_{A}(Y)\rangle=\langle X,L\_{A^t}(Y)\rangle, \ \forall X,Y\in M^{2\times2}(\mathbb R)\ldots$