Let's say I have a closed form expression for $\lambda_n$ which is the n-th eigenvalue, how do I solve for the sequence of eigenvectors of increasing length ? That is,





here the n-th eigenvalue is

$\begin{array}{ll} \lambda_m & = W_{m, m}\\ & = \sum_{n = 1}^{\infty} n^{- m - 1} (n + 1)^{- m - 1}\\ & = \sum_{n = 0}^{\frac{m}{2} + \frac{1}{4} - \frac{(- 1)^m}{4}} \zeta (2 n) \frac{2 (- 1)^{m + 1 - 2 n} \Gamma (2 m - 2 n + 2)}{\Gamma (m + 2 - 2 n) \Gamma (m + 1)}\\ & = \sum_{n = 0}^{\frac{m}{2} + \frac{1}{4} - \frac{(- 1)^m}{4}} \zeta (2 n) 2 (- 1)^{m + 1 - 2 n} \binom{2 m + 2 n + 1}{m} \end{array}$

but the question is for any general countably infinite set of eigenvalues . The solution $A$ is a triangular matrix.

This is what I've got so far..

The eigenvector equation to be solved is \begin{equation} \sum_{n = 0}^{\infty} W_{m n} v_n = \lambda v_m \end{equation} where $v$ is an eigenector with components $\{ v_n \}$ so the $k$-th eigen-equation is \begin{equation} \sum_{n = 0}^{\infty} W_{m n} v_{n k} = \lambda_k v_{m k} \end{equation} the $k$-th polynomial eigenfunction is then given by \begin{equation} e_k (x) = \sum_{n = 0}^{\infty} v_{n k} x^n \end{equation}.

The eigenfunctions and eigenvalues are related by \begin{equation} \frac{[\mathcal{L}_w e_n (x)]}{e_n (x)} = \lambda_n \end{equation}


$[\mathcal{L}_w f] (x) = \sum_{m = 0}^{\infty} \frac{g^{(m)} (0)}{m!} x^m = \sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \frac{f^{(k)} (0)}{k!} \left( \frac{n - x}{n (n + 1)} \right)^k$

is the transfer operator of the harmonic sawtooth map

$w (x) = \lfloor \frac{1}{x} \rfloor \left( x \lfloor \frac{1}{x} \rfloor + x - 1 \right)$

whose iterates are the coefficients of a modified Engel expansion. See See Why can the transfer operator of the Gauss (continued fraction) map be used to prove the Riemann hypothesis but not the map for the Engel expansion?

The 1st eigenfunction of $[\mathcal{L}_w f] (x)$ is just the identity map with eigenvalue $1$ \begin{equation} e_0 (x) = x \end{equation} the 2nd eigenfunction is \begin{equation} e_1 (x) = 1 + x 2 \frac{\zeta (2) - 1}{\zeta (2) - 2} \end{equation}

so that

${\frac {{\it Lw} \left( x\mapsto {\frac { \left( 2\,{\pi }^{2}-12 \right) x}{{\pi }^{2}-12}}+1,x \right) }{3-1/3\,{\pi }^{2}} \left( 2 \,{\frac { \left( {\pi }^{2}-6 \right) x}{{\pi }^{2}-12}}+1 \right) ^{ -1}}=-1/3\,{\frac {2\,{\pi }^{4}x+{\pi }^{4}-30\,{\pi }^{2}x-21\,{\pi }^{2} +108\,x+108}{ \left( {\pi }^{2}-12 \right) \left( 3-1/3\,{\pi }^{2} \right) } \left( 2\,{\frac { \left( {\pi }^{2}-6 \right) x}{{\pi }^{2 }-12}}+1 \right) ^{-1}}=1$

We know the eigenvalues of $\mathcal{L}_w$ which are $\lambda_w = W_{w, w}$ and also the off-diagonal elements of $W_{m,n}$ and would like to solve for the non-zero elements in the triangular matrix of polynomial eigenfunction coefficients $A_{n,m}$

\begin{equation} \left( \begin{array}{cccccccc} \lambda_1 & W_{1, 2} & W_{1, 3} & W_{1, 4} & W_{1, 5} & W_{1, 6} & W_{1, \ldots} & \ldots\\ 0 & \lambda_2 & W_{2, 3} & W_{2, 4} & W_{2, 5} & W_{2, 6} & W_{2, \ldots} & \ldots\\ 0 & 0 & \lambda_3 & W_{3, 4} & W_{3, 5} & W_{3, 6} & W_{3, \ldots} & \ldots\\ 0 & 0 & 0 & \lambda_4 & W_{4, 5} & W_{4, 6} & W_{4, \ldots} & \ldots\\ 0 & 0 & 0 & 0 & \lambda_5 & W_{5, 6} & W_{5, \ldots} & \ldots\\ 0 & 0 & 0 & 0 & 0 & \lambda_n & W_{6, \ldots} & \ldots\\ 0 & 0 & 0 & 0 & 0 & 0 & \lambda_{n + 1} & \ldots\\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \end{array} \right) \left( \begin{array}{c} A_{n, 1}\\ A_{n, 2}\\ A_{n, 3}\\ A_{n, 4}\\ A_{n, 5}\\ A_{n, 6}\\ A_{n, 7}\\ \ldots \end{array} \right) = \lambda_n \left( \begin{array}{c} A_{n, 1}\\ A_{n, 2}\\ A_{n, 3}\\ A_{n, 4}\\ A_{n, 5}\\ A_{n, 6}\\ A_{n, 7}\\ \ldots \end{array} \right) \end{equation}

  • $\begingroup$ Typo: the "kth polynomial" eqn should show an $x^n$ not $x^k$. $\endgroup$
    – Linas
    Apr 16, 2018 at 7:47

1 Answer 1


What follows is perhaps a non-answer, but rather an attempt to straighten out some of the confusion. First, if you know the eigenvalues, you can trivially write down the operator and its eigenvectors, simply by listing the eigenvalues along the diagonal. The eigenvectors are the columns of the identity matrix. But as this is trivial, its not informative.

Next, you imply that the eigenvectors are polynomials -- presumably orthogonal polynomials over the unit interval. However, given some arbitrary operator, there is no reason to presume that the operator even acts on the unit interval ("act" in the sense of "group action"). Operators can act on arbitrary spaces, sets, things; in general, there is no reason to assume that an operator acts on the unit interval, or that it can be represented by such an action ("representation" in the sense of a matrix, possibly infinite-dimensional, that transforms in the same way that the operator transforms; viz "group representation")

But lets assume that such a representation exists. There is no reason to believe that the eigenfunctions will be polynomials; they could be anything. For example, $\cos (n\pi x)$ gives a certain kind of basis on the unit interval, but these are clearly not polynomials.

Next, you state that the operator is (upper) triangular; but this is exactly the definition of a "solvable operator"; its called "solvable" precisely because the you can take linear combinations of successive columns to "solve" it. In general, operators need not be solvable (viz, no such triangular form exists).

These three assumptions (that the operator can be represented by its action on the unit interval; that the operator is solvable; that the operator has polynomial eigenfunctions) -- this already is forcing a lot of details to work out in a very special kind of way; one is lucky to have problems that are "this easy".

By the end of the question, it seems as if you've answered your own question. The operator is not just any operator, but a transfer operator; you have an explicit form for it; you claim that its upper-triangular (although this is not obvious to me). If it really is upper-triangular, then go solve it: the first eigenvector is just the first column; the second eigenvector is the second column, minus a multiple of the first, leaving the off-diagonal entry zero. The third eigenvector is just the third column, minus some linear combination of the first two columns, leaving the two off-diagonal entries zero. And so one marches down the line, with all but the diagonal entries zero. The columns of the identity operator are then the eigenvectors.

  • $\begingroup$ Apparently the method I need to use is described at mathfaculty.fullerton.edu/mathews/n2003/… $\endgroup$
    – crow
    Apr 18, 2018 at 0:05
  • $\begingroup$ Yes, that shows how to solve triangular matricies. $\endgroup$
    – Linas
    Apr 19, 2018 at 12:05

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