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.
**Note:** I am sorry this is a little far from your notations. But lots of people would define this inner product as $\mbox{Tr}(X^\*Y)$. This idea of using the trace to create bilinear objects is frequent and natural (see the [Killing form](http://en.wikipedia.org/wiki/Killing_form), to mention one famous example). Last but not least, observe that **the argument below applies to $M\_n(\mathbb{R})$ and even $M\_n(\mathbb{C})$ for any $n$.** This is my main motivation for doing it the following way. If you are only interested in the $2\times 2$ case, then you should keep your notations and do like @user1551 did.

Observe that the inner product you are given on the matrices has a nice and tractable [matrix trace](http://en.wikipedia.org/wiki/Trace_%28linear_algebra%29) expression:

> 
> $$
> (X,Y)=\mbox{Tr}(X^\*Y)
> $$
> 
> 
> 

where $X^\*$ denotes the adjoint of $X$, i.e. the transpose in the real case.

Now, using the commutativity of the trace $\mbox{Tr}(MN)=\mbox{Tr}(NM)$, straightforward computations show that
$$
(L\_A(X),Y)=\mbox{Tr}(A^\*(YX^\*-X^\*Y))
$$
while
$$
(X,L\_A(Y))=\mbox{Tr}(A(YX^\*-X^\*Y)).
$$

**Easy direction:** it follows at once that if $A=A^\*$, then $(L\_A(X),Y)=(X,L\_A(Y))$ for every $X,Y$. Hence $L\_A=L\_A^\*$, i.e. $L\_A$ is symmetric.

**Conversely:** assume that $L\_A=L\_A^\*$. Then the matrix $B=A-A^\*$ satisfies $$\mbox{Tr}(B(YX^\*-X^\*Y))=\mbox{Tr}((BY-YB)X^\*) =0$$ for every $X,Y$.
First, this implies, with $X=BY-YB$, 
$$
\|BY-YB\|^2=\mbox{Tr}((BY-YB)^\*(BY-YB))=\mbox{Tr}((BY-YB)(BY-YB)^\*) =0.
$$
Where the norm is the one induced by your inner product. It follows that $BY-YB=0$ for every $Y$, i.e. $B$ commutes with every other matrix. It is well-know that the only such matrices are the scalar matrices (nice exercise, just compute the products with matrices of the canonical basis $E\_{ij}$). So $B=A-A^\*=\lambda I\_2$. But taking the trace of the latter, we find $2\lambda=\mbox{Tr}(A)-\mbox{Tr}(A^\*)=0$. Finally, $B=A-A^\*=0$, so $A=A^\*$ is symmetric.