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Let $\varphi: \mathbb{R}[x] \to \mathbb{R}[x],f(x) \mapsto f(2x+1)$. To determine the eigenvalues, I have:

Let $b_i = x^{i}$ be the i. basis-vector of $\mathbb{R}[x]$. $\varphi(b_i) = (2x+1)^i = \sum_{k=0}^i \binom{i}{k} 2x^{i-k} = b_i \cdot \sum_{k=0}^i \binom{i}{k} 2x^{-k}$. How can I determine the eigenvalues? I know, that $f(2x+1)=\lambda \cdot f(x)$, but how can I get rid of $x$ in $\sum_{k=0}^i \binom{i}{k} 2x^{-k}$, as $f(2x+1) = \sum_{i=0}^n a_i (2x+1)^i = \sum_{i=0}^n a_i \sum_{k=0}^i \binom{i}{k} 2x^{i-k}$.

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Say $f$ is a degree-$n$ eigenpolynomial of $\varphi$. Then $$ f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0\\ \varphi(f)(x) = a_n(2x+1)^n + a_{n-1}(2x+1)^{n-1} + \cdots + a_1(2x+1) + a_0 $$ By looking at the leading term of $\varphi(f)$, we see that the coefficient has changed from $a_n$ to $2^na_n$. So the eigenvalue corresponding to the degree-$n$ eigenpolynomial $f$ is $2^n$.

So the remaining question becomes: Is there an eigenpolynomial of each degree? I claim that $(x+1)^n$ is such an eigenpolynomial. Indeed, we have: $$ \varphi((x+1)^n) = (2x+1+1)^n = 2^n(x+1)^n $$

(I found this eigenpolynomial by solving explicitly for $n = 0, 1, 2$ and then confirming once for $n = 3$, before starting to expand the binomials to try to actually prove that $(x+1)^n$ worked in general. I was stuck for a minute staring at the sums, then I came to the realization above, which made the proof very simple.)

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  • $\begingroup$ Thank you very much. Why is it sufficient to only look at the eigenpolynomials to determine the eigenvalue? I thought the equality $\varphi(f) = \lambda f$ has to be fulfilled for all real polynoms. $\endgroup$ – Tim Jun 11 '19 at 11:33
  • $\begingroup$ @Tim No. Only the identity transformation (and constant multiples of it) has that property for all $f$. For most linear transformations, most vectors aren't eigenvectors, and thus do not fulfill anything like $\varphi(f) = \lambda f$. Consider $\varphi(x)$. Is that of the form $\lambda\cdot x$ for some real number $\lambda$? The eigenpolynomials are special. $\endgroup$ – Arthur Jun 11 '19 at 12:00
  • $\begingroup$ Alright, I think I mixed it up … So your last equation shows that $(x+1)^n$ is the eigenvector with eigenvalue $2^n$. How can I be sure, that this is the only eigenvalue? $\endgroup$ – Tim Jun 12 '19 at 7:01
  • $\begingroup$ @Tim That's what the first paragraph is for: any degree-$n$ eigenpolynomial has eigenvalue $2^n$. Alternately, for any linear transformation of a finite dimensional vector space, there can only be as many eigenvalues as there are dimensions. And this linear transformation restricts nicely to the $k$-dimensional subspaces of "Polynomials of degree less than $k$", for any $k\in\Bbb N$. And on any of these spaces, the eigenpolynomials $(x+1)^n$ for $0\leq n<k$ gives the $k$ distinct eigenvalues $2^n$, so there is no room for more. $\endgroup$ – Arthur Jun 12 '19 at 7:16
  • $\begingroup$ (For that last argument, it's important to point out that any polynomial in the large space is actually contained in some of the subspaces considered, so in particular, any eigenpolynomial is included in this discussion. If you had allowed "infinite-degree" polynomials, then those would not be considered by that argument.) $\endgroup$ – Arthur Jun 12 '19 at 7:21

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