Image and Kernel bases of differentiation map

Question

I'm trying to solve e) and f). What I've done for e):

Since we have a basis $(1,x,x^2,x^3)$, the image would be $$(0,1,2x,3x^2)=(0,1,0,0)+x(0,0,2,0)+x^2(0,0,0,3).$$ Hence the base of the image would be

$$(0,1,0,0),(0,0,2,0),(0,0,0,3).$$

f) and then the kernel would be $(0,0,0,0)$.

This should all make sense since by rank-nullity theorem the dimensions of the image and kernel adds up to $4$, which is the dimension of my basis.

Is my approach/answer true?

Answer 1

Close, but is the zero polynomial the only element that, when differentiated, yields the zero polynomial? In fact, any constant function $f(x) = k$, when differentiated, yields the zero element of this vector space ($f(x) = 0$). In vector form, any element $(k,0,0,0)$ will be mapped to $\vec{0}$, so the kernel is any scaled version of the vector $(1,0,0,0)$, and therefore a basis would be $(1, 0, 0, 0)$.

I also have a slight problem with your image basis: you have to be consistent in how you represent a polynomial as a set of coordinates. I assume you are expressing a polynomial $f(x) = a + bx + cx^2 + dx^3$ as $(a, b, c, d)$. You are right that the set $\{1, 2x, 3x^2\}$ is a basis for the image space; however, when expressing this in coordinates, this should be the set $\{(1, 0, 0, 0);(0, 2, 0, 0);(0, 0, 3, 0)\}$.

You will see that this answer is consistent with the rank-nullity theorem as well.

Answer 2

By your same reasoning, the matrix associated to $\frac{d}{dx}$ is $[\frac{d}{dx}]=\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0& 0 & 3 \\ 0 & 0 & 0 & 0 \end{pmatrix}$. So the image is given by the span of the polynomials whose coordinates are the columns of $[\frac{d}{dx}]$; and the kernel is given by solving the associated system of homogeneous equations, so $[\frac{d}{dx}] \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0 \\ 0 \end{pmatrix}$ . By simple inspection the image is given by <$1, x, x^2$> and the kernel is given by the span of the first vector in your basis (since the first column of you matrix is zero), so that would be the <1>. Your mistake was that you chose an arbitrary basis and considered the kernel to be the number of vectors which went to 0 when applying the operator. This is not in general true; you can see this by changing you basis by swapping the constant polynomial 1 (the first vector in you basis) with 1+x. This is still a basis for the vector space however none of the vectors in the basis go to 0. If you calculate the associated matrix in this new basis you’ll find that kernel and image don’t change, since these are properties which are invariant with respect to changes of basis. Note that the way you found the image was correct, since simply considering the span of whatever the operator does to any basis of your vector space is exactly the same as just considering the span of the columns of the matrix associated to said basis.