How to find matrix D representing derivative of set of polynomials I'm working on my homework assignment and I came upon this question. 
Going about part a. I easily found $p(x) = c_0 + c_1x + ... + c_nx^n$ and $p = [c_0 ,..., c_n]^T$, $q(x) = c_1 + 2c_2x + ... + nc_nx^(n-1)$, $q = [c_1, ... ,nc_n,  0]^T$. How exactly is $D_2$ found and how could I find $D_n$?
 A: (First, let me note that you might want to use $q = [c_1, ... ,nc_n, 0]^T$ instead of $[c_1, ... , nc_n]^T$. )
You are looking for the matrix $D$ for which $D*p = q$. Let $d_i$ be the i-th row vector of D. Then we get that $q_i = d_i * p$. From looking at the formula you wrote down for q(x) we get that $q_i = i*p_{i + 1}$. For example: $q_2 = 2*c_2 = 2*p_{3}$. So we have to choose the i-th row of D in such a way, that we get $i*p_{i+1} = d_i * p$. This means that $D_{i, i+1} = i$. This last step is probably the most diffucult to understand. If you have difficulties, I suggest you compute $D_2 * p$ to see why this must be the case.
A: You can try turning this matrix equation into system of equations:
$$D_n(1,1)*p(1) + ... D_n(1,n)*p(n) = q(1)$$
$$...$$
$$D_n(n,1)*p(1) + ... D_n(n,n)*p(n) = q(n)$$
and find every element of matrix $D_n$ just from making left side of every equation identically equal to right side, for example first equation of this example that you have:
$$D(1,1)*c_0 + D(1,2)*c_1 + D(1,3)*c_2 = c_1$$
It is obvious that $D(1,2)$ must be 1 and $D(1,1)$ and $D(1,3)$ must be 0
A: Consider the matrix $I_n$, the $n\times n$ identity matrix. Multiplying this matrix to a polynomial keeps all the coefficients and powers the same. Then, shift the elements of $I_n$ to the right by $1$, so that the new matrix is an off-diagonal matrix $J_n$. This reduces the power of each term in the polynomial by $1$ and removes the $c_0$ term. Then, change the $1$'s along the off-diagonal to make a new matrix $K_n$. Multiplying this matrix to a polynomial multiplies the coefficients by the corresponding off-diagonal element of $K_n$. In order to take the derivative of a polynomial, we reduce the power of every term by $1$, remove the $c_0$ term, and multiply each term by it's original power. Thus, we choose a particular matrix of the form $K_n$ where the off-diagonal elements are $1,2,3,\dots$ and so on. This matrix will be $D_{n-1}$.
To see this, let's look at how it works in the case of $D_3$, so that $n=4$. Let $p(x) = c_0 + c_1x + c_2x^2 + c_3x^3$. Obviously, $$ I_4 = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \text{ and } I_4 \begin{bmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \end{bmatrix} $$
which corresponds to transforming $p(x)$ into $p_I(x) = c_0 + c_1x + c_2x^2 + c_3x^3 = p(x)$, i.e. nothing happens.
Then, after shifting each element to the right, the new matrix is
$$ J_4 = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} \text{ and } J_4 \begin{bmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ 0 \end{bmatrix} $$
which corresponds to transforming $p(x)$ into $p_J(x) = c_1 + c_2x + c_3x^2$.
Finally, multiplying the off-diagonal elements by the sequence $1,2,3$ gives the matrix
$$ K_4 = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix} \text{ and } K_4 \begin{bmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} c_1 \\ 2c_2 \\ 3c_3 \\ 0 \end{bmatrix} $$
which corresponds to transforming $p(x)$ into $p_K(x) = c_1 + 2c_2x + 3c_3x^2 = p'(x)$. Thus, $D_3 = K_4$.
