Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Following Coddington-Levinson's book Theory of ordinary differential equations, chapter 7: "Self-adjoint problems on finite intervals", let us consider the eigenvalue problem

$$\pi(l):\begin{cases} Lx(t)= lx(t) & t \in [a, b] \\ Ux=0 \end{cases}$$

where $Lx=p_0(t)x^{(n)}+p_1(t)x^{(n-1)}+\ldots + p_n(t)x(t)$ (with $p_0(t)\ne 0$, the problem is not singular) and $Ux=0$ stands for the boundary conditions

$$U_jx=\sum_{k=1}^n(M_{jk}x^{(k-1)}(a)+N_{jk}x^{(k-1)}(b)),\qquad j=1\ldots n.$$

Also let $\pi$ be self-adjoint, meaning that $\int_a^b Lu\overline{v}\, dt=\int_a^bu\overline{Lv}\, dt$ for all $u, v \in C^n$ satisfying boundary conditions $Uu=Uv=0$.

We say that $l\in \mathbb{C}$ is an eigenvalue of $\pi$ if $\pi(l)$ admits non trivial solutions. Coddington-Levinson's theorem 2.1 asserts that all eigenvalues are real and that they have no finite cluster point. What is interesting for this question is the proof: the authors start taking a fundamental system $\{\varphi_j, j=1\ldots n\}$ of solutions of the linear equation $Lx=lx$, observing that each $\varphi_j$ depends analytically on $l$. Then they point out that the generic solution

$$x=\sum_{j=1}^nc_j \varphi_j$$

of $Lx=lx$ is an eigenvalue of $\pi$ if and only if

$$\tag{1} \sum_{j=1}^n c_j U_k\varphi_j=0 \qquad k=1\ldots n, $$

which is a system of $n$ homogeneous linear equations in $n$ unknowns $c_1 \ldots c_n$. The determinant $\Delta$ of $(1)$ is an entire function of $l$ and it vanishes exactly at the eigenvalues of $\pi$. At this point the authors finish off their proof, while we proceed to our question.

Question This $\Delta$, being entire and vanishing at eigenvalues, might be regarded as an infinite-dimensional analogue of the characteristic polynomial of a matrix. Is there any relationship between the multiplicity of its zeros and the geometric multiplicity of the corresponding eigenvalues (i.e. the dimension of the associated eigenspaces)?

Thank you.

share|cite|improve this question
up vote 2 down vote accepted

Yes, there is a relationship between these two multiplicities. If $\lambda$ is a zero of multiplicity $m$, then the multiplicity of the corresponding eigenvalue is $\le m$. There are even better (in terms of equalities) relationships between these multiplicities, but in order to describe them you have to take into account associated functions (i.e. not only eigenfunctions). You can find all these results together with their complete proofs in

  • Naimark, M. A. Linear differential operators. Part I: Elementary theory of linear differential operators. Frederick Ungar Publishing Co., New York, 1967 xiii+144 pp.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.