Eigenvalues/Eigenvectors of Endomorphism on Space of Polynomials I was working on some self-study, when I ran across the following question in Golan's book on Linear Algebra:
Let $V$ be a vector space composed of all polynomial functions from $\mathbb{R}$ to itself, let $a \in \mathbb{R}$, and let $\alpha$ be an endomorphism on $V$ defined by 
$$\alpha(p): x \mapsto (x-a)[p'(x)+p'(a)]-2[p(x)-p(a)]$$
where $p'$ denotes the derivative of $p$.  Find the eigenvalues and eigenvectors of $\alpha$.
I have tried to figure a way to more succinctly write out $\alpha$; for example, I was hoping that this was just something related to a Taylor expansion.  I have also tried, for the sake of intuition, to see what happened when I plugged in the basis vectors $1, x, x^2, x^3$, but I got tired after that, since I didn't really see any sort of useful pattern forming.  I suspect I should get some answers in terms of $a$, but I am quite stuck.  I have asked around a bit in person, but no one seems to know how to proceed with this one.  
Any thoughts, hints or advice would be very much appreciated.
 A: We want to find all eigenvectors of the endomorphism.  That is, we want to find all $p(x)$ such that $\alpha(p) = \lambda(p)$ for some $\lambda$.
Let's see what this map to the basis $\{1,(x-a),(x-a)^2,\dots\}$.  We see that
$$
1 \overset{\alpha}{\mapsto} (x-a)[0] - 2[0] = 0\\
x-a \overset{\alpha}{\mapsto} (x-a)[2]-2[(x-a)] = 0\\
$$
Then, for $n \geq 2$, we have
$$
(x-a)^n  \overset{\alpha}{\mapsto} 
(x-a)[n(x-a)^{n-1} - 0] - 2[(x-a)^n - 0] =
(n-2)(x-a)^n
$$
So, we have a basis of eigenvectors.

Another way to look at this: define the linear map $\tau:p(x) \mapsto p(x-a)$.  It is equivalent to diagonalize the map $\tau^{-1}\alpha \tau$.  We compute
$$
\tau^{-1}\alpha \tau[p(x)] = 
\tau^{-1}\alpha [p(x-a)] = \\
\tau^{-1}\left( 
(x-a)[p'(x-a) + p'(0)] - 2[p(x-a) - p(0)]
\right)=\\
x[p'(x) + p'(0)] - 2[p(x) - p(0)]
$$
We can then check how this map acts on the standard basis $\{1,x,x^2,\dots\}$.
I don't have an intuitive explanation, however as to why these are the eigenvectors, or how to find them other than with a clever guess.
Perhaps it would have been helpful to start by analyzing the map $p(x) \mapsto x\,p'(x)$.
