Q: Matrix Representation of a Linear Operator? Related Previous Question
I have a follow-up question related to my previous submission, specifically in regards to solving for the matrix representation $A$ of the linear operator $T$
I am looking to solve for $A$ in the following:
$$A[u]_{\beta _U} = [v]_{\beta _V}$$ 
where I chose the basis of $U$ to be:
$$\beta _U = \{1, x^2, x^4, x^6\}$$
and the basis for $V$ to be:
$$\beta _V = \{x, x^3, x^5\}$$
Now, formulating the equation:
$$
\begin{bmatrix}
    c_0 & c_1 & c_2 & c_3 \\
    c_4 & c_5 & c_6 & c_7 \\
    c_8 & c_9 & c_{10} & c_{11}
\end{bmatrix} [u]_{\beta _U} = [v]_{\beta _V}
$$
where A is known to be a 3x4 matrix and $[u]_{\beta _U}$ and $[v]_{\beta _V}$ are directly related to the coefficients of:
$$P = \{1, x^2, x^4, x^6 \} $$ where $U = \text{span}(P)$
and $$Tu = v = \frac {du}{dx}$$
with $$v = \{2x, 4x^3, 6x^5\}$$
My question is:
How do you solve for the matrix $A$ when the bases are not common?
My first thought would be the first row of matrix $A$ would consist of:
$$
\begin{bmatrix}
    2 & 0 & 0 & 0
\end{bmatrix}
$$
but this doesn't make sense to me because $c_0$ would be related to the first element of the basis $\beta _U$, which is not the same as the first element of basis $\beta _V$
Any clarity would be greatly appreciated!
Thanks!
 A: AS you said the vector $(1,x^2,x^4,x^6) \mapsto (2x,4x^3,6x^5) $.
Usually when we consider a certain base for a vector space with $n$ dimension say the basis are $\{ e_1 , e_2 ,\cdots , e_n\}$ then we mean by a vector in $V$ as $(a_1 , a_2 , \cdots , a_n) = \sum_{i=1}^{n} a_i e_i $. We write each base vector to where it maps. The basis for $U$ are $(1,0,0,0) $ represent $1$ , $(0,1,0,0)$ represent $x^2$, $(0,0,1,0)$ represent $x^4$ and $(0,0,0,1)$ represent $x^6$. 
The first vector maps to $0$ vector the second $(0,1,0,0)\mapsto (2,0,0)$ since $x^2 \mapsto 2x$ and $x$ in $V$ is the base $(1,0,0)$ so $2x$ is $(2,0,0)$. We get the following 
$$\begin{bmatrix}
    c_0 & c_1 & c_2 & c_3 \\
    c_4 & c_5 & c_6 & c_7 \\
    c_8 & c_9 & c_{10} & c_{11}
\end{bmatrix} \begin{bmatrix} 1 \\ 0 \\0 \\0 \end{bmatrix}= \begin{bmatrix} 0\\0\\0 \end{bmatrix}$$
$$\begin{bmatrix}
    c_0 & c_1 & c_2 & c_3 \\
    c_4 & c_5 & c_6 & c_7 \\
    c_8 & c_9 & c_{10} & c_{11}
\end{bmatrix} \begin{bmatrix} 0 \\ 1 \\0 \\0 \end{bmatrix}= \begin{bmatrix} 2\\0\\0 \end{bmatrix}$$ 
$$\begin{bmatrix}
    c_0 & c_1 & c_2 & c_3 \\
    c_4 & c_5 & c_6 & c_7 \\
    c_8 & c_9 & c_{10} & c_{11}
\end{bmatrix} \begin{bmatrix} 0 \\ 0 \\1 \\0 \end{bmatrix}= \begin{bmatrix} 0\\4\\0 \end{bmatrix}$$ 
$$\begin{bmatrix}
    c_0 & c_1 & c_2 & c_3 \\
    c_4 & c_5 & c_6 & c_7 \\
    c_8 & c_9 & c_{10} & c_{11}
\end{bmatrix} \begin{bmatrix} 0 \\ 0 \\0 \\1 \end{bmatrix}= \begin{bmatrix} 0\\0\\6 \end{bmatrix}$$ 
Each base maps into a certain vector. 
