# Concerning the linear independence of $p_0,p_1,…,p_m$ when $p_j(2) = 0$

Is my proof of the following theorem correct? If so can it be written better?

Theorem. Fix polynomials $p_0,p_1,...,p_m\in \mathcal{P}_{m}(\mathbf{R})$ such that $p_j(2)=0$ for each $j$. Then prove that the set $\{p_0,p_1,...,p_m\}$ is not linearly independent in $\mathcal{P}_{m}(\mathbf{R})$.

Proof. Let $I_m = \{n\in\mathbf{N}|n\leq m\}$ and assume that $j\in I_m$ . Since $p_j(2) = 0$ it follows that $p_j = q_j+2$ where $q_j\in\operatorname{span}(x,x^2,...,x^m)$ consequently $1<\deg p_j\leq m$ however we have $m+1$ polynomials and thus by the pigenhole principle there must be at least two polynomials having the same degree indicating that $p_0,p_1,...,p_m$ is linearly dependent.

$\blacksquare$

• Don't confuse the condition $p(2)=0$ with $p(0)=2$ – Paolo Leonetti Feb 26 '18 at 1:55

There are two mistakes: 1) $p_j$ cannot be necessarily written as $q_j+2$ (i.e., $p_j(x)=x-2$). 2) If $p_i$ and $p_j$ have the same degree, they are not necessarily linearly independent (e.g., $p_i(x)=x$ and $p_j(x)=x+1$).
Solution 1. $\mathcal{P}_m(\mathbf{R})$ is a vector space of dimension $m+1$, hence, if $p_0,\ldots,p_m$ would be linearly independent, then they would be a basis. This is a contradiction because $p(x)=1$ cannot be expressed as linear combination of $p_0,\ldots,p_m$.
Solution 2. The vector subspace $S$ generated by $p_0,\ldots,p_m$ has dimension $\le m$. Indeed $$S\subseteq V:=\{p \in \mathcal{P}_m(\mathbf{R}): p(2)=0\}$$ and a basis of $V$ is $\{x-2,x^2-2^2,\ldots,x^m-2^m\}$, i.e., $\mathrm{dim}V=m$.
• Would the above reasoning be correct if i said that $p_j$ would have the form $c_j(x-2)^a$ where $1\leq a\leq m$ – Atif Farooq Feb 26 '18 at 1:57
• The correct form is: there exist integers $0\le n_{1,j} < \cdots < n_{k_j,j} \le m$ and reals $c_{1,j},\ldots,c_{k_j,j}$ such that $$p_j(x)=\sum_{t=1}^{k_j} c_t (x^t-2^t)$$ for each $j=0,1,\ldots,m$. – Paolo Leonetti Feb 26 '18 at 2:00