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Everyone knows that the eigenvalues of an operator $L$ are the values $\lambda$ such that $Lx=\lambda x$.

Yet in all expositions of Sturm-Liouville theory I've seen (including Wikipeda's), the eigenvalues of a Sturm-Liouville problem with Sturm-Liouville operator $S$ are the numbers $\lambda$ such that $Sx+\lambda x=0$. Surely we can forgive the untutored mathematician seeing this material for the first time who protests: no! If $Sx+\lambda x=0$, "the eigenvalues," which is surely elliptical for "the eigenvalues of the operator $S$," are $-\lambda$ (i.e. the set of values $-\lambda$ such that a solution $x$ exists).

Why, oh why, is this the case? Yes, mathematical notation varies in all manner of ways, but this departure strikes me as supremely ridiculous, even if it means the eigenvalues according to the usual definition will usually come out to be negative. I ask because I haven't seen this material in a while and just spent ten minutes trying to convince a student that the eigenvalues of an S-L problem of the above form must be $-\lambda$.

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    It could be said, perhaps, that the eigenvalues of the Sturm-Liouville problem are the eigenvalues of $-S$. – 2017-02-15
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    @Omnomnomnom: well, yes, that's equivalent but just as ridiculous. *Why* don't we just speak of the eigenvalues of $S$? – 2017-02-15
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    Probably a combination of historical reasons and because we have a huge bias towards the positive numbers: when written that way the eigenvalues usually turn out to be positive numbers (they can be ordered such that they are eventually positive). – 2017-02-15
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    Typically because the equation is quadratic, and solving each case is easier if you are looking at positive terms. Plus, it's nice to have a convention when running into PDEs. – 2017-02-15
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    The real problem is that second-order differential operators are positive when they have *negative* coefficients, i.e. $-DpD$ in the Sturm–Liouville case. It's the same reason that the Laplacian is normally taken with a sign that makes $-\Delta$ positive. (And I'm not suggesting it's a good reason!) – 2017-02-15
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    Not all expositions adopt the convention you state. Also, keep in mind that the parameter $\lambda$ in Sturm-Liouville comes from Fourier's separation of variables, which predates the study of matrix eigenfunction/eigenvalue decompositions. Finite-dimensional eigenvalue and eigenvector decompositions evolved out of studying Sturm-Liouville problems, not the other way around. So you're also reversing History. – 2017-03-04

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