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?
 A: The numbers in your vectors represent the linear combination of the basis elements needed to form a polynomial. For example, 
\begin{equation*}
v = 
\begin{pmatrix}
3 \\
4 \\
\vdots  \\
6 \\
7  \\
\end{pmatrix}
\end{equation*}
The polynomial represented by this vector is $3+4x+...6x^{n-1}+7x^n$.
Now, for a polynomial like $p(x)=1$, it is represented by 
\begin{equation*}
v = 
\begin{pmatrix}
1 \\
0 \\
\vdots  \\
0 \\
0  \\
\end{pmatrix}
\end{equation*}
This extracts out the first column of $D_{n+1, n+1}$ after left multiplying by $D$, which gives you the zero vector, corresponding to the polynomial $p'(x)=0$. For another example, if $p(x)=x$, the vector representing it is
\begin{equation*}
v = 
\begin{pmatrix}
0 \\
1 \\
\vdots  \\
0 \\
0  \\
\end{pmatrix}
\end{equation*}
After $D$ acts on it, the second column is returned, which is $p'(x)=1$.
So $D$ works as expected.
A: 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
A: The solution is correct. Take a vector from the canonical basis, i.e. $t^k$ which is the vector with all coordinate vanishing except the $k+1$-th one which is equal to $1$.
If you apply the matrix to this vector, the result is the vector with all coordinate vanishing except the $k$-th one which is equal to $k$, ie.e $k t^{k-1}$ as expected.
