Finding a matrix for a differentiation map

Question

How do I find a matrix for a differentiation map, that is, a map defined by $Tp=p'$ with $T∈L(\mathbb P_4(\mathbb R),\mathbb P_3(\mathbb R))$ under the standard basis $\{1, x, x_2,x_3\}$? I know that $(x^n)'=n(x^{n-1})$, so wouldn't the transformation matrix be such that, when the subspace is applied to it, the resulting vector is $(0,1,2x,3x^2)$?

Answer 1

"applying a subspace to a matrix" has no meaning. Anyway if you want to know the matrix you have to know how the matrix of a linear appplication is determined. You chose the canonical base for both the domain and codomain ( they are the ordered sets $\{1,x,x^2,x^3,x^4 \}$ for $\mathbb{P}_4(\mathbb{R})$, $\{1,x,x^2,x^3 \}$ for $\mathbb{P}_3(\mathbb{R})$ ). Once chosen the bases, the first column of T is rapresented by the coordinates of $Tp=p'$ in the codomain base. Since $T1=0$ the first column of $T$ will be $\begin{pmatrix} 0\\0\\0\\0 \end{pmatrix}$. The second column will be the coordinates of $Tx=1$, meaning $\begin{pmatrix} 1\\0\\0\\0 \end{pmatrix}$. The same way you have $Tx^2=2x$, with coordinates $\begin{pmatrix} 0\\2\\0\\0 \end{pmatrix}$, the coordinates of $3x^2$ $\begin{pmatrix} 0\\0\\3\\0 \end{pmatrix}$ and of $4x^3$ $\begin{pmatrix} 0\\0\\0\\4 \end{pmatrix}$. Then the matrix is: $$\begin{pmatrix} 0& 1& 0& 0&0 \\0&0&2&0&0\\0&0&0&3&0\\0&0&0&0&4 \end{pmatrix} $$