Matrix of the differentiation operation Exercise: Find the matrix of the derivative operation $D$ related to the base $\{1, t, t^2,..., t^n\}$ $$D: \mathcal P_{n} \to \mathcal P_{n}$$
I found a possible solution to this exercise, given that $D(t^k)=kt^{k-1}$ $$ \begin{equation*}
D_{n+1,n+1} = 
\begin{pmatrix}
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 2 & \cdots & 0 \\
\vdots  & \vdots  & \vdots &\ddots & \vdots  \\
0 & 0 & 0 &\cdots & n \\
 0 & 0 & 0 &\cdots & 0 \\
\end{pmatrix}
\end{equation*}$$Nevertheless, it doesn't convince me at all, because when multiplying the matrix with the vectors in $\mathcal P_{n}$, the exponent remains the same. Is this solution correct?

Let's precise some points.


*

*$D$ is an operator on  $\mathbb{R}^n$ so it isn't a vector, which means $D\neq kt^{k-1}$

*I think what you mean is that for every $k \in [0,n]$ integer, $$D(t^k)=kt^{k-1}$$
So from here we can build de matrix of $D$ of dimension $n+1$ given :
The i-th column of $D$ is $D(t^k)=kt^{k-1}$ decomposed on the basis $\mathcal{B}=(1,...,t^n)$ 
$$D(1)=0\\
D(t^k)=ke_{k-1}$$
It is why your result is true
