Just a nice little problem I saw. The solution is not difficult but here I present a different one which I like.

Let $\mathbb{P}_{1,001}$ be the vector space of polynomials of degree at most 1,000. Let $T:\mathbb{P}_{1,001}\rightarrow \mathbb{P}_{1,001}$ be the linear map defined by:

$\displaystyle T\{p(x)\}=2p'(x)-p(x)$.

Find the eigenvalues and eigenfunctions of the linear map $T$.

Standard Solution: Let $p(x)=\sum_{i=0}^{1,0001}a_ix_i$. Now

$T\{p(x)\}=-a_{1,001}x^{1,001}+\sum_{i=0}^{1,000}(2(i+1)a_{i+1}-a_i)x^i$.

We want to solve the equation

$T\{p(x)\}=2p'(x)-p(x)=\lambda\, p(x)$.

Suppose that $p(x)$ is a degree $N$ polynomial ($N\leq 1,001$). Let us look at the coefficient of $a_N$ in $T\{p(x)\}$ and $\lambda\,p(x)$:

$2(N+1)a_{N+1}-a_N=\lambda\, a_N$.

Now by assumption $a_{N+1}=0$. Hence the only possible eigenvalue is $\lambda =-1$. This means that we have, for all $i=0,1,2,\dots,1,001$:

$a_{i+1}=0$.

In other words the eigenfunctions only have $a_0\neq 0$ (eigenfunctions are non-zero so we have $a_0\neq 0$). The only eigenfunctions are the non-zero constant functions, each of which has eigenvalue -1 $\bullet$

Alternative Solution: Consider now the map $T_1:\mathcal{C}^\infty(\mathbb{R})\rightarrow \mathcal{C}^\infty(\mathbb{R})$ given by

$\displaystyle T_1=2D-I_{\mathcal{C}^\infty(\mathbb{R})}$

where $D$ is the differential operator. Now suppose that $y(x)$ is an eigenfunction of $T_1$ with eigenvalue $\lambda$:

$\displaystyle 2\frac{dy}{dx}-y=\lambda y$

$\displaystyle\Rightarrow 2\frac{dy}{y}=(\lambda+1)\,dx.$

$\displaystyle\Rightarrow 2\ln y=(\lambda+1)x+C_0$

$\displaystyle\Rightarrow y_{(\lambda,\,C)}(x)= C\sqrt{e^{(\lambda+1)x}}= Ce^{(\lambda +1)x/2}$

for some non-zero constant $C$ (eigenfunctions must be non-zero).

These are the eigenfunctions of $T_1$ with eigenvalue $\lambda$. The question is which of these are polynomials?
Suppose that $y_{(\lambda,\,C)}(x)$ is a polynomial of degree $M$. Then the $(M+1)$th derivative must be zero:

$\displaystyle\frac{d^{M+1}y}{dx^{M+1}}=C\left(\frac{\lambda + 1}{2}\right)^{M+1}e^{(\lambda +1)x/2}\overset{!}{=}0$.

Now $C$ is not zero nor is the exponential function so we must have $\lambda=-1$. In this case we have

$y_{(-1,C)}(x)=Ce^0=C$.

So the only eigenfunctions of $T_1$ that are polynomials are the non-zero constant functions. It follows that the only polynomials of $T$ are the non-zero constant functions $\bullet$