Suppose $T \in \mathcal L(\mathcal P(\mathbf R))$ is such that $T$ is injective and $\deg Tp \le \deg p$ for every nonzero polynomial $p \in \mathcal P(\mathbf R)$.

1. Prove that $T$ is surjective.
2. Prove that $\deg Tp = \deg p$ for every nonzero $p \in \mathcal P(\mathbf R)$.

Fix $n \in \mathbf{N}$ and consider $T$ as an operator on $\mathcal P_n(\mathbf F)$ (this is possible since $\deg Tp \le \deg p$). because $T$ is injective 3.69 implies $T$ is surjective on $\mathcal P_n(\mathbf F)$, and since $n$ was arbitrary this shows $T$ is surjective on $\mathcal P(\mathbf F)$.

To show $\deg Tp = \deg p$ consider $1,x,\dots$ and apply induction:

Clearly $T(1)$ must be a constant by injectivity and $\deg Tp \le \deg p$.

If $T(x)$ were a constant then $T$ would fail to be injective since we could find nonzero $a_0,a_1$ such that $T(a_0 + a_1x) = 0$. Thus $\deg T(x) = 1$.

Continue like this to show $\deg Tx^n = n$ for all $n \in \mathbf{N}$.

todo: find better proof of (2)