Matrix of the differentiation operation

Question

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?

Answer 1

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.

Answer 2

Let's precise some points.

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

2. 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

Answer 3

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.