# Suppose T is an element of L ( P 3 </msub> ( <mi mathvariant="do

Suppose $T$ is an element of $L\left({P}_{3}\left(\mathbb{R}\right),{P}_{2}\left(\mathbb{R}\right)\right)$ is the differentiation map defined by $Tp={p}^{\prime }$. Find a basis of ${P}_{3}\left(\mathbb{R}\right)$ and a basis of ${P}_{2}\left(\mathbb{R}\right)$ such that the matrix of T with respect to these basis is
$\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\end{array}\right)$
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Raven Higgins
With ${e}_{1}=\frac{1}{3}{x}^{3},{e}_{2}=\frac{1}{2}{x}^{2},{e}_{3}=x,{e}_{4}=1$ represented by vectors
$\left(\begin{array}{c}1/3\\ 0\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1/2\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 0\\ 1\end{array}\right)$
relative to the standard basis, and a basis for the target space as usual just ${x}^{2},x,1$ represented by the usual basis vectors.
The map in this basis is exactly what you're looking for.
More generally, it's pretty straightforward to write linear transformations in different bases, $\alpha$ of the domain and $\beta$ of the range: start by writing the domain's basis in terms of the standard basis and getting a matrix ${T}_{\alpha }$ which changes the standard basis into the basis, $\alpha$. Next find a matrix ${T}_{\beta }$ which changes the standard basis on the domain space into the basis you want. Finally find a matrix ${T}_{0}$ which does the job on the standard basis.
Then your map is $T={T}_{\beta }{T}_{0}{T}_{\alpha }^{-1}$, which you can verify by testing it on a basis, ${T}_{\alpha }^{-1}$ takes an alpha basis vector, turns it into a standard one, ${T}_{0}$ does the work of the transformation, and ${T}_{\beta }$ writes it in the alternate basis.