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

*Let be the vector space of polynomials of degree at most 1,000. Let be the linear map defined by:*

*.*

*Find the eigenvalues and eigenfunctions of the linear map .*

*Standard Solution:* Let . Now

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We want to solve the equation

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Suppose that is a degree polynomial (). Let us look at the coefficient of in and :

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Now by assumption . Hence the only possible eigenvalue is . This means that we have, for all :

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

*Alternative Solution*: Consider now the map given by

where is the differential operator. Now suppose that is an eigenfunction of with eigenvalue :

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

These are the eigenfunctions of with eigenvalue . The question is which of these are polynomials?

Suppose that is a polynomial of degree . Then the th derivative must be zero:

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Now is not zero nor is the exponential function so we must have . In this case we have

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So the only eigenfunctions of that are polynomials are the non-zero constant functions. It follows that the only polynomials of are the non-zero constant functions

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