Find the matrix for d/dx acting on the vector space 
Im not sure if Im going about this problem the right way and any help would be appreciated.  
I took the derivative of the vector space and ended up with  $c_1e^{2x}+2c_1e^{2x}+ 2c_2e^{2x}$.


If the original ordered basis only has two components does that mean that the final matrix only has two rows? Will my matrix be in $\Bbb{R}^2$?



Thanks
 A: Make your linear map act on the given basis and write the outcome as a linear combination of the same basis:
$$\begin{align*}&\frac d{dx}(xe^{2x})=e^{2x}+2xe^{2x}=&\color{red}2\cdot xe^{2x}+\color{red}1\cdot e^{2x}\\{}\\
&\frac d{dx}(e^{2x})=2e^{2x}=&\color{red}0\cdot xe^{2x}+\color{red}2\cdot e^{2x}\end{align*}$$
Thus the matrix for the linear operator $\;\frac d{dx}\;$ with respect to the given basis $\;B\;$ is
$$A:=\begin{pmatrix}2&0\\1&2\end{pmatrix}$$
A: First apply the $\frac{d}{dx}$operator on the basis to obtain the transformed basis:
$$D 
  \left[ {\begin{array}{c}
   1  \\       0  \\      \end{array} } \right] = \left[ {\begin{array}{c}
   2  \\       1  \\      \end{array} } \right]$$
$$D 
  \left[ {\begin{array}{c}
   0  \\       1 \\      \end{array} } \right] = \left[ {\begin{array}{c}
   0  \\       2  \\      \end{array} } \right]$$
Hence, the transformation of the Identity matrix, and hence the matrix representation of the transformation is given by
$$D 
  \left[ {\begin{array}{cc}
   1 & 0  \\       0 & 1  \\      \end{array} } \right] = DI = D = \left[ {\begin{array}{c}
   2 & 0  \\       1 & 2  \\      \end{array} } \right]$$
A: All you need is how it acts on the basis. If $A$ is your operator then $A(e_i)=\sum_{j}a_{ji}e_j$. So the matrix is the $[A]_{ij}=a_{ij}$.
As user296113 points out, if $e_1=xe^{2x}$ and $e_2=e^{2x}$, letting A=$d\over{dx}$
${d\over{dx}}(xe^{2x})=2e_1+e_2$ so $a_{11}=2$ and $a_{21}=1$.
Similarly,
${d\over{dx}}(e^{2x})=0e_1+2e_2$ so $a_{12}=0$ and $a_{22}=2$.
Altogether $\operatorname{mat}(A)=\begin{pmatrix} 2 & 0\\ 1 & 2 \end{pmatrix}$
You can check by acting the matrix on your basis, either $(1,0)^T$ or $(0,1)^T$
The peculiarity that we get $a_{ji}$ when we act with the operator $A$ comes about because of the way we define matrix multiplication, and so when you act on $(1,0)^T$ you get back the first column of $\operatorname{A}$, hence the sum over $j$ is down the rows $a_{ji}$.
