2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Why $p_0 = 0$? Maybe something about $p_j$ is missing? $\endgroup$ – xbh Sep 2 '18 at 8:23
  • $\begingroup$ 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 $p(2)=0$, $\endgroup$ – Marko Škorić Sep 2 '18 at 8:29
  • 1
    $\begingroup$ Then some assumptions about $p_j$ are missing. Please edit your question and add them. $\endgroup$ – xbh Sep 2 '18 at 8:30
  • $\begingroup$ 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 $\endgroup$ – Marko Škorić Sep 2 '18 at 8:33
  • $\begingroup$ @MarkoŠkorić How about $p_0(X)=X^5-32$ (assuming $n\ge 5$)? $\endgroup$ – Hagen von Eitzen Sep 2 '18 at 8:34
0
$\begingroup$

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 $P_n(\mathbb{F})$.

$\endgroup$
2
$\begingroup$

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 $\text{Span}\{p_0,p_1,...,p_n\}$ ?

Answer: (Think about constant polynomials)

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.