Error Analysis
The original proof lets $\mu_1, \dots, \mu_n$ be the zeros of $q$, then for each $k \in {1, \dots, n}$, substitutes $\mu_k$ into $rp + sq = 0$ to obtain $r(\mu_k) = 0$. It then concludes that $r$ is a polynomial of degree at most $n - 1$ with $n$ zeros, so $r = 0$ by Theorem 4.8.
However, this argument implicitly assumes that $\mu_1, \dots, \mu_n$ are distinct. If $q$ has repeated roots, say $q$ has only $t < n$ distinct zeros, then substitution only yields $r(\mu_k) = 0$ at $t$ distinct points. Since $\deg r \leq n - 1$ and $t < n$, Theorem 4.8 does not apply, and we cannot conclude $r = 0$.
Proof
(a) Note that $m, n \geq 1$ since $p, q$ are non-constant. Let $\lambda_1, \dots, \lambda_m$ be the zeros of $p$ and let $\mu_1, \dots, \mu_t$ be the distinct zeros of $q$ with multiplicities $d_1, \dots, d_t$, so that $d_1 + \cdots + d_t = n$. By assumption we have $p(\mu_i) \neq 0$ and $q(\lambda_k) \neq 0$ for all $i, k$.
Suppose that
$$r \in \mathcal{P}_{n-1}(\mathbf{C}) \quad \text{and} \quad s \in \mathcal{P}_{m-1}(\mathbf{C})$$
are such that $rp + sq = 0$. Then $q \mid rp$. For each $i \in \lbrace 1, \dots, t \rbrace$, since $(x - \mu_i)^{d_i} \mid q(x)$ and $p(\mu_i) \neq 0$, the polynomials $(x - \mu_i)^{d_i}$ and $p(x)$ are coprime, so
$$(x - \mu_i)^{d_i} \mid r(x).$$
Since the factors $(x - \mu_i)^{d_i}$ are pairwise coprime, we have
$$\prod_{i=1}^{t}(x - \mu_i)^{d_i} \mid r(x).$$
The left side has degree $d_1 + \cdots + d_t = n$, while $r$ is a polynomial of degree at most $n - 1$. Thus $r = 0$. A similar argument with the $\lambda_k$'s shows that $s = 0$. Thus null $T = \lbrace 0 \rbrace$, i.e. $T$ is injective.
Error Analysis
The original proof lets$\mu_1, \dots, \mu_n$ be the zeros of $q$ , then for each $k \in {1, \dots, n}$ , substitutes $\mu_k$ into $rp + sq = 0$ to obtain $r(\mu_k) = 0$ . It then concludes that $r$ is a polynomial of degree at most $n - 1$ with $n$ zeros, so $r = 0$ by Theorem 4.8.
However, this argument implicitly assumes that$\mu_1, \dots, \mu_n$ are distinct. If $q$ has repeated roots, say $q$ has only $t < n$ distinct zeros, then substitution only yields $r(\mu_k) = 0$ at $t$ distinct points. Since $\deg r \leq n - 1$ and $t < n$ , Theorem 4.8 does not apply, and we cannot conclude $r = 0$ .
Proof
(a) Note that$m, n \geq 1$ since $p, q$ are non-constant. Let $\lambda_1, \dots, \lambda_m$ be the zeros of $p$ and let $\mu_1, \dots, \mu_t$ be the distinct zeros of $q$ with multiplicities $d_1, \dots, d_t$ , so that $d_1 + \cdots + d_t = n$ . By assumption we have $p(\mu_i) \neq 0$ and $q(\lambda_k) \neq 0$ for all $i, k$ .
Suppose that
are such that$rp + sq = 0$ . Then $q \mid rp$ . For each $i \in \lbrace 1, \dots, t \rbrace$ , since $(x - \mu_i)^{d_i} \mid q(x)$ and $p(\mu_i) \neq 0$ , the polynomials $(x - \mu_i)^{d_i}$ and $p(x)$ are coprime, so
Since the factors$(x - \mu_i)^{d_i}$ are pairwise coprime, we have
The left side has degree$d_1 + \cdots + d_t = n$ , while $r$ is a polynomial of degree at most $n - 1$ . Thus $r = 0$ . A similar argument with the $\lambda_k$ 's shows that $s = 0$ . Thus null $T = \lbrace 0 \rbrace$ , i.e. $T$ is injective.