Endomorphism on polynomial vector space $\mathbb{R}_3[x]$

Question

Let $ \mathbb{R}_3[x] $ be the vector space of the polynomials with the degree $ \le 3$. Given the endomorphism on this vector space, $$ T:\mathbb{R}_3[x] \to \mathbb{R}_3[x], T(f)(x) = f(x+1)-f(x), $$ calculate its eigenvalues, eigenvectors and also its matrix representation in the base $ \{{1,x,x^2, x^3} \} $ of this space. So far I tried to calculate the eigenvalues using the basic formula $T(f)(x) = \lambda x$. This resulted in an equation depending on $\lambda : f(x+1)-f(x)(\lambda-1)=0$. What should I do next? Thank you!

Answer 1

There are two ways to see that only $ 0 $ is an eigenvalue, with eigenvectors the constant polynomials :

• Consider your polynomial $ P = aX^k + R $ where $ k > 0 $ and $ \operatorname{deg} R < k $. We have $$ \begin{align} T(P)&= a((X+1)^k - X^k) + T(R) \\ &= a(X+1 -X)\sum_{i=0}^{k-1}(X+1)^iX^{k-1-i} + T(R)\\ &= a\sum_{i=0}^{k-1}(X+1)^iX^{k-1-i} + T(R) \end{align}$$ which is of degree $ < k $, and thus cannot be equal to $ P $

• Consider your polynomial $ P $ in $ \mathbb{C}[X] $ instead, then suppose $ T(P) = \lambda P $ and $ P $ non-constant. It has one root $ z $ by d'Alembert-Gauss, and the functional equation gives you that $ z + 1 $ is another root. By induction $ P $ has an infinite number of roots and $ P = 0 $.

Calculation of $ T $ on constant polynomials then gives you the desired conclusion.

As for the matrix representation, just use the identity $ (a + b)^n = \sum_{k=0}^n \binom{n}{k}a^kb^{n-k}.$