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Let $V$ be the real vector space $\mathbb{R}[X]$ and $M⊂\mathbb{R}$ with $d$ elements.

Furthermore let $U := \{ f \in \mathbb{R}[X]\ |\ \deg(f) ≤ d − 1 \} $ and $\Phi : V → \operatorname{Func}(M, \mathbb{R})$ which is a linear function given via $Φ(f)(m) := f(m)$.

How could I show step by step that $Φ|_U : U → \operatorname{Func}(M, \mathbb{R})$ is a vector space isomorphism?

My idea is to show that:

(1) It is a homomorphism (as it is a linear function, it is also a homomorphism, so that one is already proved, right?).

(2) It is injective.

(3) It is surjective.

Can you help me how is it possible to show those three circumstances?

(Can you please give me an explanation for every step, that I can understand everything?)

Chill2Macht
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Code-G
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  • What does Abb mean? – Tobias Kildetoft Jan 22 '17 at 12:01
  • What is $Abb(M,R)$ the set of functions from $M$ to $\mathbb{R}$? (Abb standing for the German Abbildung)? Also, if what you are asking for is a proof that any linear transformation is determined by its action on a basis, then this might help you: https://wiki.math.ntnu.no/users/ehrnstro/teaching/linearmethods/linearmappings – Chill2Macht Jan 22 '17 at 12:01
  • William, thanks for your remark, I wanted to write $Func(M,R)$ I've edited it. (I am solving this problem from a german problem collection, and I forgot to change this one) – Code-G Jan 22 '17 at 12:04
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    1 is immediate because the definition of homomorphism between vector spaces is a linear function. – Chill2Macht Jan 22 '17 at 12:06
  • Any chance that you and Mary Star are in the same class? http://math.stackexchange.com/questions/2104452/vector-space-isomorphism#comment4327265_2104452 . Perhaps you should work on these problems together... – John Hughes Jan 22 '17 at 12:39
  • More then likely, I guess :) – Code-G Jan 22 '17 at 12:43

1 Answers1

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As I noted in the comments, (1) is immediate from the comments, so I will focus on (2) and (3). Note that this only shows that $\Phi$ is a bijective vector space homomorphism, but every bijective vector space homomorphism is even a vector space isomorphism, because the inverse of any linear function is again linear (the analogous statement is not true, e.g., for continuous functions).

Also I should note that I am being lazy and, unlike what you asked for, I am not really showing every step in what follows, just giving pointers so that you can find resources which will explain the answers to you.

(2) We want to show that, $\Phi(f) = \Phi(g) \implies f=g$. This is basically the statement that any polynomial of degree less than or equal to $(d-1)$ is determined uniquely by its values at $d$ points -- this can be shown, for example, using Lagrange interpolation.

(3) We want to show that any function from $M$ to $\mathbb{R}$ can be described by the action of a polynomial. In other words, given data points $(x_1, y_1), \dots, (x_d,y_d)$, we want to find a polynomial such that $f(x_i)=y_i$ for all $i=1,\dots,d$. That this is possible is again discussed and proved in the theory of Lagrange interpolation.

Since there is such an extensive literature out there discussing Lagrange interpolation and other polynomial interpolation methods (e.g. Newton interpolation), I figured it would be best if I don't try to reinvent the wheel and explain all of it again here myself. If you have followup questions please of course let me know.

Chill2Macht
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    The (2) is clear. But as $Func(M,R)$ means all of the functions from $M$ to the real numbers, it means $(x^5+x^3+1)∈Func(M,R)$. But let's say $d=3$. So, it means that $(X^5+X^3+1)∉U$.

    From that it is not true: $\forall f\in Func(M,R),,\exists g\in U:;;Φ(g)=f.$

    Which means $Φ|_U→Func(M,R)$ is not surjectiv.

     – Code-G Jan 22 '17 at 12:34
  • @ÁrminBéda Let's take $d=3$, say that $M={0,1,2}$. Then what you are calling $(x^5+x^3+1)$ is just the set of points $(0,1)(1,3)(2,41)$. The claim is that there is a quadratic polynomial which interpolates those three points, $ax^2 + bx +c$, thus restricted to $M$ they are the same function. Make sure not to confuse polynomials as symbols versus polynomials as functions -- there are many more polynomials as symbols than functions when the domain is finite http://math.stackexchange.com/questions/815440/polynomials-vs-polynomial-functions – Chill2Macht Jan 22 '17 at 14:33
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    I think I start tp understand now. So if $M={{(0,1,2)}}$ then restricted to M is $f=x^5+x^3+1=18x^2-16x+1$? – Code-G Jan 22 '17 at 14:44
  • Well arguably from a pedantic point of view, $f$ is the following subset of $M \times \mathbb{R}: {(0,1), (1,3), (2,41) }$. One has that both of the rules of assignment you mention generate this function, i.e. one writes $f: x \mapsto x^5 + x^3 +1$ or $f: x \mapsto 18x^2 -16x +1$ (see here: http://math.stackexchange.com/questions/2004895/does-f-colon-x-mapsto-2x3-mean-the-same-thing-as-fx-2x3/2004944#2004944). Both rules of assignment correspond to the same function because of our choice of domain. Only the quadratic rule of assignment is an element of $U$. – Chill2Macht Jan 22 '17 at 17:51
  • I might be doing you a disservice by using $f$ to denote both the function as well as the polynomials which can represent $f$ when restricted to $M$, since what I am trying to do is explain how the two things are technically different. Generally the distinction only becomes important when we are dealing with finite sets (like $M$ or a finite field $\mathbb{Z}/(p\mathbb{Z})$. See these related questions: http://math.stackexchange.com/a/470847/327486, http://math.stackexchange.com/questions/427501/difference-between-polynomial-functions-and-polynomials-and-why-these-two-polyno – Chill2Macht Jan 22 '17 at 17:59
  • Also relevant (from glancing over them): https://www.quora.com/Which-is-the-difference-between-polynomials-and-polynomial-functions, https://www.reddit.com/r/learnmath/comments/234dmf/whats_the_difference_between_functions_and/ -- I hope this helps, even though the answers/explanations all mention abstract algebra which isn't relevant to this case ($M$ isn't a ring/field -- just a subset of one, namely $\mathbb{R}$). – Chill2Macht Jan 22 '17 at 18:03