Find $[T_{c}]^{\alpha}_{\alpha}$ For $c \in R$, let $T_{c}: M_{2x2}(R) \rightarrow M_{2x2}(R)$ be the linear transformation defined by $T_{c}(A) = A + cA^{T}$ for every $A \in M_{2x2}$.
For the basis $\alpha = \left\{\begin{bmatrix} 1 & 0\\ 0 & 0\\\end{bmatrix} \begin{bmatrix} 0 & 0\\ 0 & 1\\\end{bmatrix} \begin{bmatrix} 0 & 1\\ 1 & 0\\\end{bmatrix} \begin{bmatrix} 0 & 1\\ -1 & 0\\\end{bmatrix}\right\}$ of $M_{2x2}(R)$, find $[T_{c}]^{\alpha}_{\alpha}$ and for what values of c is $T_{c}$ an isomorphism?
My working:
$T_{c}\{\begin{bmatrix} 1 & 0\\ 0 & 0\\\end{bmatrix}\} = \begin{bmatrix} 1+c & 0\\ 0 & 0\\\end{bmatrix}$
$T_{c}\{\begin{bmatrix} 0 & 0\\ 0 & 1\\\end{bmatrix}\} = \begin{bmatrix} 0 & 0\\ 0 & 1+c\\\end{bmatrix}$
$T_{c}\{\begin{bmatrix} 0 & 1\\ 1 & 0\\\end{bmatrix}\} = \begin{bmatrix} 0 & 1+c\\ 1+c & 0\\\end{bmatrix}$
$T_{c}\{\begin{bmatrix} 0 & 1\\ -1 & 0\\\end{bmatrix}\} = \begin{bmatrix} 0 & c-1\\ c-1 & 0\\\end{bmatrix}$
So $[T_{c}]^{\alpha}_{\alpha} = \begin{bmatrix} 1+c & 0 & 0 & 0\\ 0 & 0 & 0 & 1+c\\ 0 & 1+c & 1+c & 0\\ 0 & c-1 & c-1 & 0\\ \end{bmatrix}$
$T_{c}$ is an iso-morphism when $c \neq 1$ and $c \neq -1$
Would you say what I did is correct or am I going in the wrong direction?

You are on the right way, but you made mistakes in your computation.
First:
$$
T_c\begin{pmatrix}0&1\\-1&0\end{pmatrix}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}+c\begin{pmatrix}0&-1\\1&0\end{pmatrix}=\begin{pmatrix}0&1-c\\c-1&0\end{pmatrix}.
$$
Second: Since
$$
T_c\begin{pmatrix}1&0\\0&0\end{pmatrix}=(1+c)\begin{pmatrix}1&0\\0&0\end{pmatrix}\\
T_c\begin{pmatrix}0&0\\0&1\end{pmatrix}=(1+c)\begin{pmatrix}0&0\\0&1\end{pmatrix}\\
T_c\begin{pmatrix}1&0\\0&1\end{pmatrix}=(1+c)\begin{pmatrix}1&0\\0&1\end{pmatrix}\\
T_c\begin{pmatrix}0&1\\-1&0\end{pmatrix}=(1-c)\begin{pmatrix}0&1\\-1&0\end{pmatrix}
$$
you get
$$
[T_c]^\alpha_\alpha=\begin{pmatrix}1+c&0&0&0\\0&1+c&0&0\\0&0&1+c&0\\0&0&0&1-c\end{pmatrix}.
$$
But your statement that $T_c$ is an isomorphism if $c\neq 1$ and $c\neq -1$ holds.
If we write $\alpha=\{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}$ 
it seems, you wrote the elements of $T_c\alpha_i$ in the $i$-th row of $[T_c]_\alpha^\alpha$. 
But you have to write $T_c\alpha_i=a_1\alpha_1+a_2\alpha_2+a_2\alpha_3+a_4\alpha_4$ where $a_1,a_2,a_3,a_4\in\mathbb{R}$. Then your $i$-th column of $[T_c]_\alpha^\alpha$ is $\begin{pmatrix}a_1\\a_2\\a_3\\a_4\end{pmatrix}$
