Show: $\{p_1,...,p_{n+1}\}$ basis of $Pol_n(\mathbb{R}), \Phi$ bijective and $f\in Pol_n(\mathbb{R})$ is zero function if $n+1$ different roots exist

Question

This exercise causes a lot of difficulties for me.

Given:

Let $n \in \mathbb{N}, \space Pol_n(\mathbb{R})$ be the vector space of the polynomial functions with degree $\le n$.

Given real numbers $x_1 \lt x_2 \lt \dots \lt x_{n+1}$, consider $$\Phi: Pol_n(\mathbb{R}) \rightarrow \mathbb{R}^{n+1}, \space f \mapsto \begin{pmatrix} f(x_1)\\f(x_2)\\\vdots\\f(x_{n+1}) \end{pmatrix}$$ and the functions $$p_i:\mathbb{R}\rightarrow\mathbb{R}, \space x \mapsto \prod_{j=1,\space j \ne i}^{n+1}(x-x_j)$$ for all $i \in \{1,\dots,n+1 \}$. The index traverses all $j \in \{1,\dots,n+1 \} \setminus \{i\}$.

With the help of Frank Lu, I have already shown that all $p_i$ are polynomial functions in $Pol_n(\mathbb{R})$ and the set $\{ \Phi(p_1),\dots,\Phi(p_{n+1}) \}$ is basis of $\mathbb{R}^{n+1}$.

With that, I have to

1. show that $\{p_1,\dots,p_{n+1}\}$ is basis of $Pol_n(\mathbb{R})$ and $\Phi$ is bijective, and
2. with 1., show any $f \in Pol_n(\mathbb{R})$ is the zero function if it has at least $n+1$ different roots.

My approaches:

Not much.

For 1., I thought that because of the $j=1, j \ne i$ from the big product of $p_i$, every two different $p_i$ are linearly independent, as for every different $p_i$, a different polynomial is skipped. Also, if $\Phi$ is bijective, $\Phi^{-1}$ has to exist. The problem is that I do not know how that $\Phi^{-1}$ has too look like but I considered $$\Phi^{-1}: \mathbb{R}^{n+1} \rightarrow Pol_n(\mathbb{R}), \begin{pmatrix} \lambda_1 \\ \vdots \\ \lambda_{n+1} \end{pmatrix} \mapsto (x \mapsto \prod_{j=1}^{n+1} (x-\lambda_j)).$$

For 2., I simply don't know how to show the claim.

Thanks in advance for any help.

addition to the bijectivity part:

If this $\Phi^{-1}$ is correct, $\Phi^{-1} \circ \Phi = Id_{Pol_n(\mathbb{R})}$ and $\Phi \circ \Phi^{-1} = Id_{\mathbb{R}^{n+1}}$ would have to apply.

For the first part this would be:

let $f \in Pol_n({\mathbb{R}})$, it follows:

$$\Phi^{-1}(\Phi(f)) = \Phi^{-1} \left( \begin{pmatrix} f(x_1)\\ \vdots \\ f(x_{n+1}) \end{pmatrix} \right) = (x \mapsto \prod_{j=1}^{n+1} (x-f(x_j))=f,$$

so $\Phi^{-1} \circ \Phi = Id_{Pol_n(\mathbb{R})}$ applies

for the second part:

let $\begin{pmatrix} \lambda_1\\ \vdots \\ \lambda_{n+1} \end{pmatrix} \in \mathbb{R}^{n+1}$, it follows:

$$\Phi\left(\Phi^{-1}\begin{pmatrix} \lambda_1\\ \vdots \\ \lambda_{n+1} \end{pmatrix}\right) = \Phi (x \mapsto \prod_{j=1}^{n+1} (x-\lambda_j)=\begin{pmatrix} \lambda_1\\ \vdots \\ \lambda_{n+1} \end{pmatrix},$$

so $\Phi \circ \Phi^{-1} = Id_{\mathbb{R}^{n+1}}$ applies

Is this correct?

Answer 1

For part 1, you need to show that $\{p_1,\dots , p_{n+1}\}$ is linearly independent. To do that, suppose we have $\lambda_1,\ldots,\lambda_{n+1}\in \mathbb{R}$ such that $\lambda_1 p_1+\cdots+\lambda_{n+1} p_{n+1}=0$. Then applying $\Phi$, which is linear -you need to check that-, we get $\lambda_1 \Phi(p_1)+\cdots + \lambda_{n+1} \Phi(p_{n+1})=\Phi(0)=0$. Since $\{ \Phi(p_1),\dots,\Phi(p_{n+1}) \}$ is linearly independent, it follows $\lambda_1=\cdots=\lambda_{n+1}=0$. Hence $\{p_1,\dots , p_{n+1}\}$ is linearly independent.

For part 2, suppose $f\in Pol_n(\mathbb{R})$ has $n+1$ distinct roots, say $x_1 \lt x_2 \lt \dots \lt x_{n+1}$, then $\Phi(f)=0$. Where $\Phi$ is defined using this particular choice of real numbers $x_1 \lt x_2 \lt \dots \lt x_{n+1}$. But by part 1, we know that $\Phi$ is injective. Hence $\Phi(f)=0$ implies $f=0$.