Suppose D∈L(P(R),P(R))D \in \mathcal L(\mathcal P(\mathbf R), \mathcal P(\mathbf R)) is such that deg⁡Dp=(deg⁡p)−1\deg Dp = (\deg p) - 1 for every nonconstant polynomial p∈P(R)p \in \mathcal P(\mathbf R). Prove that DD is surjective.

Consider p∈P(R)p \in \mathcal P(\mathbf R) we must show there exists a q∈P(R)q \in \mathcal P(\mathbf R) such that Dq=pDq = p.
Let n=(deg⁡p)+1n = (\deg p) + 1 and consider DD as a map from Pn(R)\mathcal P_n(\mathbf R) to itself.
Since deg⁡(Dp)=(deg⁡p)−1\deg (Dp) = (\deg p) - 1 we have null D=P0(R)\text{null }D = \mathcal P_0(\mathbf R) (the space of constants) meaning dim⁡null D=1\dim \text{null }D = 1. Therefor by 3.22

dim⁡range D=dim⁡Pn(R)−dim⁡null D=(n+1)−1=dim⁡Pn−1(R)\begin{aligned} \dim \text{range }D &= \dim \mathcal P_n(\mathbf R) - \dim \text{null }D \\ &= (n+1) - 1 \\ &= \dim \mathcal P_{n-1}(\mathbf R) \end{aligned}

Thus there exists a q∈Pnq \in \mathcal P_n such that Dq=pDq = p for p∈Pn−1(R)p \in \mathcal P_{n-1}(\mathbf R).

This shows any differential-like operator which takes constants to zero is surjective! I can see how this could be useful in differential equation theory! See 27 for a basic example.