0
$\begingroup$

Let $F$ be a field, unique $ a_0, a_1, ... ,a_n \in F $, and let $ b_0, b_1, ... ,b_n $ be some elements in $ F $ (not necessarily different).

A) Prove there exists a unique $ f \in\mathbb{F}[x] $, with $ deg f \le n $, and $ f(a_i) = b_i $ for all $ 0\le i\le n $.

B) Let $ F = \mathbb C $, and let $ b_i = \overline{a_i} $. Prove there exists $ f \in\mathbb{R}[x] $ of some degree, such that $ f(a_i) = b_i $ for all $ i $.

Hint: consider the set $ A = {{a_i}}\cup{{\overline{a_i}}} $, and by using A), prove that $ f = \overline{f} $.

I've proven A), But not B). I tried to define a linear transformation, but since $ a_i $ might be equal to $ \overline{a_i} $, I'm unsure of what N-tuple I'd pick. Am I looking at this from the wrong direction?

$\endgroup$
6
  • $\begingroup$ There is a problem in the hint, as what you would want to prove is that $f=\tau\circ f\circ\tau$ where $\tau(z):=\overline{z}$. The identity $f=\overline{f}(=\tau\circ f)$ for a polynomial $f\in\mathbb{C}[x]$ implies $f=0$. And I don't see why it would be a problem if $a_i=\overline{a_i}$, in this case you assign to it the value $b_i=\overline{b_i}$ (part A works for any $a_1,\ldots,a_n$, not necessarily distinct, if you add the extra condition that $b_i=b_j$ whenever $a_i=a_j$). $\endgroup$
    – imtrying46
    Commented Mar 17, 2023 at 18:03
  • $\begingroup$ @imtrying46 are you sure you're thinking of elements in the polynomial ring $\mathbb C[x]$? My guess is you are thinking of polynomial functions from complex analysis, in which case the implication isn't $f=0$ per se but constant $f$ (and evidently real). The mention of $f=\tau\circ f\circ\tau$ also looks like complex analysis; the notion of function composition $f\circ \tau$ doesn't register since $f$ is not a function but an element in $\mathbb C[x]$. $\endgroup$ Commented Mar 17, 2023 at 22:39
  • $\begingroup$ Yeah. The problem I'm having is that f should be in the Reals, I'm unsure how to approach that. $\endgroup$
    – FNB
    Commented Mar 18, 2023 at 7:22
  • $\begingroup$ @user8675309 You’re right, of course if $f=\overline{f}$ then $f$ is constant real, not necessarily $0$. But I don’t get your problem about distinguishing polynomials from say holomorphic functions. If $R$ denotes the ring of holomorphic functions, then there is an injective ring homomorphism $\mathbb{C}[x]\to R$, assigning to a polynomial the associated function. So it makes perfect sense to regard polynomials as holomorphic functions, and use complex analysis on them. For example, the fact that every polynomial in $\mathbb{C}[x]$ has a root is typically proven by using complex analysis. $\endgroup$
    – imtrying46
    Commented Mar 19, 2023 at 9:54
  • $\begingroup$ @imtrying46 The issue is that $f\in \mathbb C[x]$ which is a field with a single element "$x$" adjoined to it hence a polynomial ring; "x" is compatible w/ linear combinations w/ itself and elements in $\mathbb C$ but there is no additional element $\bar x$. You can do a homomorphism $\phi$ into the ring of entire functions if you like but $\phi\circ f\neq f$. Yet $\bar f \in \mathbb C[x]$ which is to say $\bar f$ is $f$ with all scalars $c_k \in \mathbb C$ conjugated, so the official hint is fine. Put differently: you are working in a different ring than the OP which causes discrepancies. $\endgroup$ Commented Mar 19, 2023 at 16:56

2 Answers 2

1
$\begingroup$

An explicit way to do (A) and (B) is via an Vandermonde matrix. For (B), let $S:= \big\{a_1,\dots, a_n\big\}\cup\big\{\overline a_1,\dots, \overline a_n\big\}$ with $r:=\big \vert S\big \vert$ and consider the degree $r$ polynomial $c_0 + c_1 x + \dots c_r x^r = p(z) \in \mathbb C[x]$.
Examine the equation
$V\mathbf c = \mathbf b$ where the $r\times r$ Vandermonde matrix has the moment curve $[1, s, s^2, ..., s^{r-1}]$ for $s\in S$ on each row, each of which is unique, each row of $\mathbf b$ contains the value of $p$ when evaluated at $s$ (technically a substitution homomorphism $\phi:\mathbb C[x]\rightarrow \mathbb C$) and $\mathbf c$ has the coefficients of $p$ to be solved for. $V$ is invertible since all rows are unique hence $\mathbf c$ is uniquely specified. But for for some permutation matrix $P$
$\big(P V\big)\overline {\mathbf c} =\overline V\overline {\mathbf c} = \overline{\mathbf b}=P\mathbf b\implies V\overline {\mathbf c}=\mathbf b$
$\implies \big(\overline {\mathbf c}-\mathbf c\big)\in \ker V\implies \mathbf c\in \mathbb R^r$, i.e. the coefficients of $p$ are real since $V$ is injective

$\endgroup$
0
$\begingroup$

Let $\# A=2n-m.$ That means $a_k=\overline{a_k}$ for $m$ values of $k.$ By (A) there is a unique polynomial $p(z)$ of degree $2n-m-1$ such that $p(a_i)=\overline{a_i}$ and $p(\overline{a_i})=a_i.$ Consider the polynomial $q(z)$ with the coefficients that are complex conjugate of the coefficients of $p(z),$ i.e. $q(z)=\overline{p(\overline{z})}.$ Then $q(a_i)=\overline{a_i}$ and $q(\overline{a_i})=a_i$ for any $i.$ By the unicity we get $q(z)=p(z).$ Hence the coefficients of $p(z)$ are real numbers.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .