Vtyjkyuk

Let $p_0,p_1,ldots,p_n$ is polynomials in $P_n(BbbF)$ such that $p_i(2)=0$ for every $iin0,1,ldots,n$. Prove that $p_0,p_1,ldots,p_n$ is not linearly independent in $P_n(BbbF)$.

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?

For any $i in 0, ldots, n$, since $p_i(2) = 0$ there exists $q_i in P_n-1(mathbbF)$ such that $p_i(x) = (x-2)q_i(x)$.

We have $dim P_n-1(mathbbF) = n$ so the set $q_1, ldots, q_n subseteq P_n-1(mathbbF)$ is linearly dependent. Therefore, there exists scalars $alpha_0, ldots, alpha_n in mathbbF$ 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 $P_n(mathbbF)$.

Suppose it is linearly independent, then it forms a basis for $P_n$, since $textdim(P_n)=n+1=Bigvert p_0,p_1,...,p_n Bigvert$

Is all members of $P_n$ are in $textSpanp_0,p_1,...,p_n$ ?

Answer: (Think about constant polynomials)

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