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: 
**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.
"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 $\iff \langle X,L\_{A}(Y)\rangle=\langle X,L\_{A^t}(Y)\rangle, \ \forall X,Y\in M^{2\times2}(\mathbb R)\ldots$