# Prove $n+1$ vectors in $P_n(\Bbb{F})$ with the specific condition is not linearly independent

Let $p_0,p_1,\ldots,p_n$ is polynomials in $P_n(\Bbb{F})$ such that $p_i(2)=0$ for every $i\in\{0,1,\ldots,n\}$. Prove that $p_0,p_1,\ldots,p_n$ is not linearly independent in $P_n(\Bbb{F})$.

I saw that $p_0(x)=0$ if we want $p_0(2)=0$, since one vector is zero vector ,then that vectors is linear dependent. What do you think about this proof?

• Why $p_0 = 0$? Maybe something about $p_j$ is missing? – xbh Sep 2 '18 at 8:23
• if $p0\not=0$ then you have $p0=a$ where $a\in\mathbb R$ if you put p0(2)=a so that does not belong to that vectors that i write $p(2)=0$, – Marko Škorić Sep 2 '18 at 8:29
• Then some assumptions about $p_j$ are missing. Please edit your question and add them. – xbh Sep 2 '18 at 8:30
• Did you see that I put that for every polynomials $p(2)=0$ can you write some polynom such that $p_0(2)=0$? Write that polynom and prove that is zero for 2 – Marko Škorić Sep 2 '18 at 8:33
• @MarkoŠkorić How about $p_0(X)=X^5-32$ (assuming $n\ge 5$)? – Hagen von Eitzen Sep 2 '18 at 8:34

For any $i \in \{0, \ldots, n\}$, since $p_i(2) = 0$ there exists $q_i \in P_{n-1}(\mathbb{F})$ such that $p_i(x) = (x-2)q_i(x)$.

We have $\dim P_{n-1}(\mathbb{F}) = n$ so the set $\{q_1, \ldots, q_n\} \subseteq P_{n-1}(\mathbb{F})$ is linearly dependent. Therefore, there exists scalars $\alpha_0, \ldots, \alpha_n \in \mathbb{F}$ not all equal to $0$ such that $\sum_{i=0}^n \alpha_i q_i(x) = 0$.

Then $$\sum_{i=0}^n \alpha_ip_i(x) = \sum_{i=0}^n \alpha_i (x-2)q_i(x) = (x-2) \left(\sum_{i=0}^n \alpha_i q_i(x)\right) = 0$$

so $\{p_0, \ldots, p_n\}$ is linearly dependent in $P_n(\mathbb{F})$.

Suppose it is linearly independent, then it forms a basis for $P_n$, since $\text{dim}(P_n)=n+1=\Big\vert \{p_0,p_1,...,p_n\} \Big\vert$

Is all members of $P_n$ are in $\text{Span}\{p_0,p_1,...,p_n\}$ ?

Hint: $\dim P_n=n+1$, hence $n+1$ linearly independent vectors would be a bassis and hence span all of $P_n$.