I'm hoping to get some feedback on my attempt for this exercise in Axler's Linear Algebra Done Right.
Suppose $T \in \mathcal{L}(\mathcal{P}(\mathbb{R})$ is such that $T$ is injective and $\deg Tp \leq \deg p$ for every non-zero polynomial $p \in \mathcal{P}(\mathbb{R})$.
a) Prove that $T$ is surjective.
b) Prove that $\deg Tp = \deg p$ for every nonzero $p \in \mathcal{P}(\mathbb{R})$.
Axler uses $\mathcal{P}(\mathbb{R})$ to denote the vector space of polynomials with real coefficients, $\mathcal{P}_m (\mathbb{R})$ to denote the vector space of polynomials of degree less than or equal to $m$ (where the zero polynomial is defined to have degree $-\infty$), and $\mathcal{L}(\mathcal{P}(\mathbb{R})$ to denote the set of all linear maps from $\mathcal{P}(\mathbb{R})$ to $\mathcal{P}(\mathbb{R})$. I'm going to quote the following theorem from the text (paraphased):
(Theorem 3.69) Suppose $V$ is finite-dimensional and $T: V \to V$ a linear map. Then all of the following are equivalent:
a) $T$ is invertible;
b) $T$ is injective;
c ) $T$ is surjective.
Here is my attempt at a proof of the exercise.
a) As $T$ is a linear map, $T(0) = 0$, so it suffices to check that each non-zero polynomial possesses some preimage. Let $p \in \mathcal{P}(\mathbb{R}) - \{0\}$. Define $m := \deg p$. As $p \neq 0$, $m \geq 0$ (i.e., $m \neq -\infty$, which is the degree of the zero polynomial). Define $\varphi_m: \mathcal{P}_m(\mathbb{R}) \to \mathcal{P}_m(\mathbb{R})$ by $\varphi_m (q) = Tq$. This map is well-defined because, for every $q \in \mathcal{P}_m$, we have $\deg \varphi_m q = \deg Tq \leq m$, i.e., $Tq \in \mathcal{P}_m (\mathbb{R})$. As $T$ is injective and $\varphi_m (p) = Tp$ for every $p \in \mathcal{P}_m (\mathbb{R})$, $\varphi_m$ is also injective. By Theorem 3.69, we conclude that $\varphi_m$ is also surjective. That is, there exists $r \in \mathcal{P}_m (\mathbb{R})$ such that $\varphi_m (r) = p = Tr$. Therefore, $T$ is surjective.
(b) Suppose for the sake of contradiction that for some $p \in \mathcal{P}(\mathbb{R}) - \{0\}$, we have $\deg Tp < \deg p$. If $\deg p = 0$, then $\deg Tp = - \infty$, hence $Tp = 0$, meaning that $\text{null} T$ is non-trivial, contradicting the fact that $T$ is injective. Hence, $\deg p > 0$. Define $m := \deg p$. Define $\varphi_{m-1}: \mathcal{P}_{m-1} (\mathbb{R}) \to \mathcal{P}_{m-1}$ by $\varphi_{m-1} (q) = Tq$ for each $q \in \mathcal{P}_{m-1} (\mathbb{R})$. As $T$ is injective, $\varphi_{m-1}$ is injective, hence it is surjective by Theorem 3.69. As $p \in \mathcal{P}_m (\mathbb{R})$ and $\deg Tp < \deg p$, we have $Tp \in \mathcal{P}_{m-1} (\mathbb{R})$. By surjectivity of $\varphi_{m-1}$, there exists $q \in \mathcal{P}_{m-1}$ such that $Tq = \varphi_{m-1} (q) = Tp$. As $T$ is injective, we must have $p = q$. However, $\deg p = m > \deg q = m - 1$, hence $p \neq q$, so this is a contradiction. Therefore, we must have $\deg Tp = \deg p$ for every non-zero $p$.