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Let $V$ be the real vector space $\mathbb{R}[X]$ and $M \subset \mathbb{R}$ a set with $d$ elements. Let $$U_2 := \{ f \in \mathbb{R}[X] \mid \deg(f) \leq d − 1\}$$ be a vector space 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 an hoomomorphiism, injective and 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?

How can we show that the mapping is surjective?

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    Please carefully re-read your question statement and fix it rather than expecting us to do so for you. Particular problems: (1) $U_1$ appears to be irrelevant. (2) "It follows that $f = m$". (3) What is $Ab(M, \Bbb R)$? (4) The value $d$ appears to be unbound. – 2017-01-19
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    At "f=m" there is a typo. I meant $f(m)=g(m)$, is this correct? There is no further information about $\text{Ab}(M,\mathbb{R})$, it is the range of the mapping. @JohnHughes – 2017-01-19
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    Presumably, $Ab(M,\Bbb R)$ is the space of $\Bbb R$-valued functions over $M$. – 2017-01-19

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To show that $\Phi$ is injective, it suffices to show that $\Phi(f) = 0$ implies that $f = 0$ (which is to say that $\Phi$ has a trivial kernel).

Now, suppose that $f \in U_2$ is given by $f(x) = a_0 + a_1x + \cdots + a_{d-1}x^{d-1}$. The statement $[\Phi(f)](m) = 0$ (for every $m \in M$) gives us $d$ equations. In particular, if $M = \{m_1,\dots,m_d\}$, then we have $$ a_0 + a_1 m_1 + \cdots + a_{d-1}m_1^{d-1} = 0\\ a_0 + a_1 m_2 + \cdots + a_{d-1}m_2^{d-1} = 0\\ \vdots\\ a_0 + a_1 m_d + \cdots + a_{d-1}m_d^{d-1} = 0 $$ Now, how can we deduce that $f(x) = 0$?

With the appropriate matrix considerations, we could show that this function is also surjective.

However, once you've shown either that the matrix is injective or surjective, it suffices to apply the rank-nullity theorem. In particular, $\Phi$ is a map between two vector spaces of dimension $d$. It follows that either $\Phi$ is both injective and surjective, or it is neither.

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    $f$ has degree $d-1$ and it has $d$ roots. So, it must be the zero-polynomial, right? – 2017-01-19
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    If that's a theorem you're allowed to use here, then that's great. – 2017-01-19
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    Great!! For the surjectivity: We have that $\dim \ker \Phi|_{U_2}+\dim im \Phi_{U_2}=\dim U_2$. We have shown that the kernel is the trrivial and so $\dim \ker \Phi|_{U_2}=0$. So $\dim im \Phi_{U_2}=\dim U_2$. Is this correct? – 2017-01-19
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    That's exactly what I meant. In order to use that, however, it's important to note that both spaces have dimension $d$ – 2017-01-19
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    Ah ok!! So, we have that $\Phi$ is also surjective. So, it is left to show that $\Phi$ is an homomorphism, or not? Or is trivial this is a linear mapping? – 2017-01-19
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    It's pretty easy to show that this is a linear map. In particular, we should note that $$(f+kg)(m)=f(m)+kg(m)$$ by definition – 2017-01-19
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    It is given that $\Phi$ is linear. When we restrict it remains linear, right? – 2017-01-19
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    Yes: if you restrict a linear map to a linear subspace, the result is again a linear map. – 2017-01-19
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    I see!! Thank you!! :-) – 2017-01-19
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    Let $$U_1 := \{ f \in \mathbb{R}[X] | \forall m \in M : f(m) = 0\}$$ an other vector space. Why can we write every polynomial as a sum of an element of $U_1$ and one of $U_2$ ? If we take a polynomial with degree $\leq d-1$ we can write it as $h=0+g$, where $g\in U_2$. What for the other polynomials? – 2017-01-19
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    There are no other polynomials over $M$; $\Phi$ is surjective – 2017-01-19
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    $\Phi$ is surjective when we restruct it to $U_2$, or not? – 2017-01-19
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    It is; that's what we just showed – 2017-01-19