- #1
zwingtip
- 20
- 0
Homework Statement
Let V be the linear space of all real polynomials p(x) of degree < n. If p [tex]\in[/tex] 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:
[tex]p(t)=a_{0}+a_{1}t+a_{2}t^2[/tex]
can be written as
[tex]\left[ \begin{array}{ccc} a_0 & 0 & 0 \\ 0 & a_1 & 0 \\ 0 & 0 & a_2 \end{array} \right]
\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][/tex]
then
[tex]T(p)=a_{0}+a_{1}(t+1)+a_{2}(t+1)^2
\left[ \begin{array}{ccc} a_0 & 0 & 0 \\ a_1 & a_1 & 0 \\ a_2 & 2a_2 & a_2 \end{array} \right]
\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][/tex]
But taking the characteristic polynomial of T(p) for p of degree 3, I'm getting distinct eigenvalues equal to the coefficients [tex]a_0, a_1, a_2[/tex] and not 1. Unless this has to do with the eigenfunction somehow. Any ideas?