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


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:


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


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