Sachchidanand Prasad

21-07-2024

Problem: Let $\mathcal{P} _n(\mathbb{C} )$ denotes the set of all polynomials of degree at most $n$. Let $p_1(t), p_2(t),\dots, p_{n+1}(t) \in \mathcal{P} _n(\mathbb{C} )$ such that \[ p_1(1) = \dots = p_{n+1} (1) = 0. \] Show that the set $\left\{ p_1, p_2,\dots, p_{n+1} \right\} $ is linearly dependent.
linear-algebra
Solution: We will provide two different solutions to this problem.
  • Define a map \[ f: \mathcal{P} _n(\mathbb{C} ) \to \mathbb{C} ,\quad p \mapsto p(1). \] We have the following properties of the map $f$.
    • $f$ is linear.
    • Given any complex number $\alpha $, define $p(t) = \alpha $ for any $t \in \mathbb{C} $. Then $f(p) = p(1) = \alpha $. Hence, $f$ is surjective.
    • The kernel of $f$ is \begin{align*} \ker f & = \left\{ p\in \mathcal{P} _n(\mathbb{C} ): f(p) = 0 \right\} \\ & = \left\{ p \in \mathcal{P} _n(\mathbb{C} ): p(1) = 0 \right\}. \end{align*}
    Therefore, using the Rank-Nullity theorem, the nullity of the map $f$ is $n$. Hence, \[ \dim \operatorname{span}\left\{ p_1, p_2, \dots, p_{n+1} \right\} \leq n. \] Therefore, the given set is linearly dependent.
  • Consider the polynomials \[ q_i(t) = \frac{p_i(t)}{t - 1},\quad i = 1,2,\dots, n+1. \] Since $\deg q_i\leq n-1$, the set $\left\{ q_1, q_2,\dots, q_{n+1} \right\} $ must be linearly dependent. Therefore, there exists $c_1, c_2, \dots, c_{n+1} $ not all zero such that \begin{align*} \sum_{i=1}^{n+1} c_i q_i(t) = 0 & \implies (t -1)\sum_{i=1}^{n+1} q_i(t) = 0 \\ & \implies \sum_{i=1}^{n+1} c_i p_i(t) = 0. \end{align*} Hence, the set $\left\{ p_1, p_2, \dots, p_{n+1} \right\} $ is linearly dependent.