Eigenvalue of a linear transformation substituting $t+1$ for $t$ in polynomials.

Question

Let $V$ be the linear space of all real polynomial $p(x)$ of degree $\leq n$. 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?

It is obvious that constant polynomials are eigenfunctions with eigenvalues 1. But how to prove that this is the only eigenvalue.

Answer 1

You need to find constants $\lambda$ and polynomials $p$, not identically zero, such that $$\qquad\qquad\qquad p(t+1)=\lambda p(t)\quad\hbox{for all $t\in\Bbb R$}.\qquad\qquad\qquad(*)$$ Firstly, if $p$ is a constant $c\ne0$, then $c=\lambda c$ and $\lambda=1$.

There are no solutions with degree $1$ because then we need $$a(t+1)+b=\lambda(at+b)$$ with $a\ne0$, and equating coefficients gives no solution.

If there is a solution with degree $m>1$, then differentiating $(*)$ shows that $p'$ is also a solution, and it has degree $m-1$. Repeated differentiation leads to a solution of degree $1$, which is impossible.

Therefore the only solution is $\lambda=1$ and $p$ constant.

Answer 2

Hint: if you write the matrix of the transformation in the standard basis $\left(1, x, \ldots , x^{n} \right)$, you will see you get an upper triangular $\left( n+1\right) \times \left( n+1 \right)$ matrix with the main diagonal with $A_{ii}=1$ for any $i$ and such that $A - I$ has rank $n$.

Hence the eigenspace has dimension $1$ and you have already shown a generator.

Answer 3

From the binomial formula it is clear that $T$ preserves the degree and the leading coefficient of polynomials. Since multiplication by a scalar$~\lambda$ also multiplies the leading coefficient by$~\lambda$, it is now clear that $\lambda=1$ is the only possible eigenvalue. Again by the binomial formula, $T-I$ decreases the degree of non-constant polynomials by$~1$, multiplying the leading coefficient by the degree (while constant polynomials are mapped to$~0$). Note that multiplying the leading coefficient by the degree never makes it$~0$, because we are in characteristic$~0$ (real coefficients). It follows that the eigenspace $\ker(T-I)$ for $\lambda=1$ consists precisely of the constant polynomials.