Prove that $a$ is a linear map and write the matrix in a given basis

Question

Let $n \in \mathbb N$ and let $a:\,R_n[X]\to R_n[X]$ be the derivative linear map, such that $a(P)=P'$.

I am aware that it must satisfy the conditions of:

1. $f(x+y)=f(x)+f(y)$

2. $f(kx)=kf(x)$

However I don't really understand how I should I apply this to it? Also, how can I write the resulting matrix of $a$ in the basis ($1$, $X$, $X^2$ , . . . , $X^n$).

Can someone run me through this example?

Answer 1

You know that the operation of taking derivatives satisfies 1) the sum rule, and 2) you can take out constants (you could think of this as a special case of the product rule, if you must). I don't see what it is you have to show here, the question seems ill-posed.

For your second question, amd's hint is a good one. Think about what matrix you have to premultiply the column vector $\begin{pmatrix} 1 \\ x \\ \vdots \\ x^n\end{pmatrix}$ to get each term sent to its derivative.

Answer 2

Hint:

The $k$-th column vector of the matrix of $a$ are the coordinates of $a(X^k)=kX^{k-1}$ in the standard basis of $R_n[X]$.

Answer 3

Let $A = (a_{ij})$ be the matrix of $a$ in basis $(1, X, \dots, X^n)$.

Here is how to construct $A$:

$$\begin{array}{cc} \begin{array}{cccc} a(1) & a(X) & \dots & a(X^n) \end{array} & \\ \left(\begin{array}{cccc} a_{11} & a_{12} & \dots & a_{1p} \\ a_{21} & a_{22} & \dots & a_{2p} \\ \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & \dots & a_{np} \end{array}\right)& \begin{array}{c} 1\\ X\\ \vdots \\X^n \end{array} \end{array}$$

So $a_{ij}$ is the coordinate of $a(X^{j-1})$ with respect to $X^{i-1}$.

After you've done all the computations, you should get $$A = \begin{pmatrix} 0 & 1 & \dots & 0 \\ 0 & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \dots & n-1 \\ 0 & 0 & \dots & 0 \end{pmatrix}$$