# Infinite number of eigenvalues

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 $$f(2x+1) = \sum_{i=0}^n a_i (2x+1)^i = \sum_{i=0}^n a_i \sum_{k=0}^i \binom{i}{k} 2x^{i-k}$$.

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.)
• 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. – Tim Jun 11 '19 at 11:33
• @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. – Arthur Jun 11 '19 at 12:00
• 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? – Tim Jun 12 '19 at 7:01
• @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. – Arthur Jun 12 '19 at 7:16