Finding Matrix Representation of a Polynomial Linear Transformation The question gives the transformation $D:P_3 \to P_3$, where $D$ is the first derivative. I need to find the matrix representation of the transformation, and determine whether it is invertible (if it is invertible, find the matrix for inverse transformation).
I guess my question is, if $D$ is the first derivative, how is the transformation going from $P_3$ to $P_3$? Wouldn't it have to go to $P_2$?
 A: When we write $D:P_3\to P_3$ we're simply claiming that if a polynomial has degree at most 3 then its derivative has degree at most 3. Of course we can actually make the stronger claim that $\text{image}(D)= P_2\subsetneq P_3$, and therefore it makes sense to define a new function with restricted target space, $\tilde{D}:P_3\to P_2$, where the 'rule' of $\tilde{D}$ is the same as that of $D$.
Actually, the observation above already tells you $D:P_3\to P_3$ is not invertible, because $\text{image}(D)= P_2\subsetneq P_3$, so the mapping is not surjective (hence not invertible).
You should probably take a closer look at the distinction between the target space/codomain of a function and the image of a function.
A: I assume that $P_3$ is the space of all polynomials of degree three at most. You are correct in thinking that, say, $$(at^3+b)'=3at^2$$
can be viewed as an element of $P_2$. But since $P_3$ is the space of all polynomials of degree three at most, the map $D$ can without trouble be defined as taking values in $P_3$ (i.e. mapping into $P_3$).
Hence it doesn't "have to go" into $P_2$, but we can find a copy of $P_2$ properly within $P_3$ (see below). If you read peek-a-boo's answer, this point of view becomes helpful to solve your question of invertibility.
$$P_3 \ni 0t^3+3at^2+0t+0 \quad\leftrightarrow\quad 3at^2+0t+0 \in P_2.$$
$$P_3 \ni at^3+0t^2+0t+b \quad\not\leftrightarrow\quad \text{ any element } \in P_2.$$
