Eigenvalues of a polynomial transformation

[zwingtip]

Homework Statement

Let V be the linear space of all real polynomials p(x) of degree < n. If p $$\in$$ V, define q = T(p) to mean that q(t) = p(t + 1) for all real t. Prove that T has only the eigenvalue 1. What are the eigenfunctions belonging to this eigenvalue?

Homework Equations

Not sure there are any.

The Attempt at a Solution

I am completely lost on this. I know how to solve this for polynomials of a specific degree, but n is unspecified. Here's an example I worked for a polynomial of degree 2:

$$p(t)=a_{0}+a_{1}t+a_{2}t^2$$

can be written as

$$\left[ \begin{array}{ccc} a_0 & 0 & 0 \\ 0 & a_1 & 0 \\ 0 & 0 & a_2 \end{array} \right] <br /> \left[ \begin{array}{c} 1 \\ t \\ t^2 \end{array} \right] = \left[ \begin{array}{c} a_0 \\ a_{1}t \\ a_{2}t^2 \end{array} \right]$$

then
$$T(p)=a_{0}+a_{1}(t+1)+a_{2}(t+1)^2<br /> <br /> \left[ \begin{array}{ccc} a_0 & 0 & 0 \\ a_1 & a_1 & 0 \\ a_2 & 2a_2 & a_2 \end{array} \right] <br /> \left[ \begin{array}{c} 1 \\ t \\ t^2 \end{array} \right] = \left[ \begin{array}{c} a_0 \\ a_{1}(t+1) \\ a_{2}(t+1)^2 \end{array} \right]$$

But taking the characteristic polynomial of T(p) for p of degree 3, I'm getting distinct eigenvalues equal to the coefficients $$a_0, a_1, a_2$$ and not 1. Unless this has to do with the eigenfunction somehow. Any ideas?

 

[lanedance]

so for a general polynomial you can write the requirement as
$$q(t) = \sum_n a_n t^n = \sum_n b_n (t+1)^n = p(t+1)$$

now consider given powers of t

another hint is that any constant is clearly an eigenfunction - what is it eigenvalue?

 

[zwingtip]

Thanks, that made sense. Question, though. I found that only a constant can be the eigenfunction. Is this true?

 

[lanedance]

yeah i agree, i haven't worked it thorugh, but i think it was heading that way

P(t+1) is basically translating the function by -1, so as no basis polynomials will overlie themselves after the translation, the constant is the only eigenfunction