How to find the matrix of the derivative endomorphism?

[Cathr]

We have $B=(1, X, X^2, X^3)$ as a base of $R3 [X]$ and we have the endomorphisms $d/dX$ and $d^2/dX^2$ so that:

$d/dX (P) = P'$ and $d^2/dX^2 (P) = P''$.

Calculating the matrix in class, the teacher found the following matrix, call it $A$:
\begin{bmatrix}
0 & 1 & 0 & 0 \\

0 & 0 & 2 & 0 \\

0 & 0 & 0 & 3 \\
0 & 0 & 0 & 0
\end{bmatrix}

This is the matrix for the first derivative, and it has the right property that when it's multiplied with itself, it gives the matrix of the second derivative:
\begin{bmatrix}
0 & 0 & 2 & 0 \\

0 & 0 & 0 & 6 \\

0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}

However, when I try to calculate the derivative of a polynomial, say $P=X^3+ X^2 + X + 1$, I don't find the right answer.
I found that the transpose $C$ of the matrix $A$ works, so that if we multiply $P$ with the following matrix:\begin{bmatrix}
0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 3 & 0
\end{bmatrix} *

I find $P= 3X^2 + 2X +1$, which is the derivative.

However, this doesn't work for all vectors and, multiplied by itself, it doesn't give the matrix of the second derivative, so it's still wrong.

May you please give me a hint on finding the solution?

 

[Erland]

You write the basis in the order $B=\{1,X,X^2,X^3\}$, but then you write the polynomial in the order $P=X^3+X^2+X+1$, which indicates that you take the basis elements in the opposite order. I think this is the cause of the confusion, since the transformation matrix depends on the order of the basis elements. If we write $P=1+X+X^2+X^3$ and differentiate by multiplying the transformation matrix by the coordinate vector, you obtain the correct result $P'=1+2X+3X^2$.

 

[mathwonk]

an even nicer basis is obtained by tweaking this one a little to {1, X, X^2/2!, X^3/3!}. Then the non zero entries are all = 1, and the matrix for D is in "Jordan form". The operator (D-c) acting on the similar space {e^ct, te^ct, t^2/2! e^ct, t^3.3! e^ct} has a similar matrix, but also with the constant c on the diagonal. The amazing theorem is that this is, up to "similarity", the most general possible matrix! I.e., given any linear transformation at all on a finite dimensional space, as long as the roots of its characteristic polynomial all lie in the scalar field, then in some basis it has a matrix made up of copies of blocks like this.

So in some sense every linear transformation looks like copies of the derivative operator (D-c) acting on the solution space of the differential equation (D-c)^n = 0. This is discussed in my free linear algebra notes:

http://alpha.math.uga.edu/%7Eroy/laprimexp.pdf