Find the eigenvalues of the linear transformation I was trying to solve this problem but I got a little stuck in the second point:


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*Let T be the linear transformation T: $R[x] \rightarrow R[x]$,the polynomials of x with real coefficients, such that:
$T(p(x))=p(3x)$.
a) Show that T is bijective.
b) Find the eigenvalues of T.
c) Prove that does not exit a polynomial $p(x) \in R$ such that: $T^{-1}=p(T)$
It is easy to prove a), but when I tried to prove b) I got some problems, because $R[x]$ is not a finite dimensional space so I can not proceed as I always do finding the matrix of the transformation. So I don't really know how to find the eigenvalues of T when the space does not have a finite basis.
I would really appreciate any hint or advice you could give me.
Thanks.
 A: Suppose $\;\lambda\;$ is an eigenvalue with eigenvector $\;h(x)=\sum\limits_{k=0}^n a_kx^k\in\Bbb R[x]\;$ , then
$$h(3x)=T(h(x))=\lambda h(x)\iff \sum_{k=0}^n a_k3^kx^k=\sum_{k=0}^n a_k\lambda x^k\iff$$
$$\iff\,\forall\,k=0,1,...,n\;,\;\;a_k3^k=a_k\lambda\iff\lambda=3^k\;\;\text{for at least one}\;\;k$$
and the only possibility the above is a constant is if $\;k=0\implies \lambda=1\;$. So all the constant polynomials are eigenvectors of $\;T\;$ belonging to the eigenvalue $\;1\;$ .
A: You should immediately note from $p(3x)=\lambda p(x)$ that $p(x)$ cannot be a polynomial having any $x$ of different degrees. Let $p(x)=ax^k $ with $a\in \mathbb{R}$. Then
$$T(p(x))=ap(3x)=a(3x)^k=a3^kx^k=3^k(ax^k)=3^kp(x)$$
Thus $\lambda = 3^k$. We deduce that $T$ has an infinite number of eigenvectors that come from the set $\{x^k|k\in \mathbb{N}_0\}$ for the eigenvalues $\{3^k|k\in \mathbb{N}_0\}$.
A: Nobody has said anything about part c), so here are my two cents:
Hint: Note that the eigenvalues of $T^{-1}$ are $\{3^{-i}\}$ where $i\geq 0$ is an integer.  The eigenvalues of $p(T)$, on the other hand, are $\{p(3^i)\}$ where $i \geq 0$ is an integer.
Note that for any polynomial $p$, we have
$$
\lim_{i \to \infty} |p\left( 3^i\right)| = \infty
$$
but the eigenvalues of $T^{-1}$ form a bounded set.
