Is $\Phi|_{U_2}$ a Vector Space Isomorphism?

[mathmari]

Hey!

Let $V$ be the real vector space $\mathbb{R}[X]$ and $M \subset \mathbb{R}$ a set with $d$ elements. Let $$U_1 := \{ f \in \mathbb{R}[X] | \forall m \in M : f(m) = 0\}, \ \ U_2 := \{ f \in \mathbb{R}[X] \mid \deg(f) \leq d − 1\}$$ be two vector spaces of $V$. Let $\Phi: V\rightarrow Ab(M,\mathbb{R})$ be a linear mapping that is defined by $\Phi (f)(m):=f(m)$.

I want to show that $\Phi\mid_{U_2}:U_2\rightarrow \text{Ab}(M,\mathbb{R})$ is a vector space isomorphism.

So, we have to show that $\Phi$ is injective ans surjective.

To show that the mapping is injective, we take to elements of $U_2$, say $f,g\in U_2$.

For them it holds that $f(m)=g(m)=0$ for every $m\in M$.

Suppose that $\Phi (f)=\Phi (g)$ then it follows that $f(m)=g(m)$.

Is this correct? So, $\Phi$ is injective, right? (Wondering)

How can we show that the mapping is surjective?

 

[Euge]

Hi mathmari,

Based on context, I assume $\operatorname{Ab}(M,\Bbb R)$ is the space of real-valued functions on $M$, with addition and scalar multiplication defined pointwise. (Although, the symbol $\operatorname{Ab}$ used in such a way would indicate homomorphisms of abelian groups.)

Since $\Phi$ is a linear transformation, so is $\Phi \big|_{U_2}$. So to show $\Phi|_{U_2}$ is injective, it suffices to show $\operatorname{Ker}(\Phi\big|_{U_2}) = 0$. If $f\in \operatorname{Ker}(\Phi\big|_{U_2})$, then $f(m) = 0$ for all $m\in M$. Since $f\in U_2$, $f$ has degree less than $d$, and hence has less than $d$ roots. (Recall: A polynomial of degree $n$ over a field has no more than $n$ roots.) As $M$ has $d$ elements, we deduce $f = 0$. Hence, $\Phi|_{U_2}$ is injective.

To see that $\Phi|_{U_2}$ is surjective, let $M = \{m_1,\ldots, m_d\}$, take an element $g : M \to \Bbb R$, and define $f(x) = \sum\limits_{j = 1}^d g(m_j)\, f_j(x)$, where

$$f_j(x) = \prod_{{k=1 \atop k\neq j}}^d \frac{x-m_k}{m_j - m_k}$$

Show that $f\in U_2$ and $\Phi(f) = g$.