Let $H=(H,(\cdot, \cdot))$ be a Hilbert space, $L:D(L) \subset H \rightarrow H$ be a self-adjoint operator (not necessarily bounded) and $A:H \rightarrow H$ be a bounded and symmetric operator. By Theorem $29.2$ in $[1]$, the spectrum $\sigma(L)$ of $L$ satisfies $\sigma(L) \subset \mathbb{R}$ and by Proposition $6.7$ of $[3]$, we have that $\sigma(A) \subset \mathbb{C}$ is compact . Suppose that both $L$ and $A$ has at least one negative eigenvalue, simbolically $$ {\rm n}(L) \geq 1 \quad \text{and} \quad {\rm n}(A) \geq 1. $$

Questiom. Is true that ${\rm n}(L+A) \geq {\rm n}(L) \cdot {\rm n}(A)$? Or ${\rm n}(L+A) \leq {\rm n}(L)$?

I conjectured this because if $\lambda_1<0$ is negative eigenvalue of $L$ and $\lambda_2<0$ is a negative eigenvalue of $A$, then it's easy to see that $\lambda_1+\lambda_2<0$ is a negative eigenvalue of $L+A$. Thus, the number of possibilities of the quantities of the form $\lambda_1+\lambda_2$ is ${\rm n}(L) \cdot {\rm n}(A)$ and hence $L+A$ has at least ${\rm n}(L) \cdot {\rm n}(A)$ negative eigenvalues. This is in general true?

I tried to use the theory contained in the Section $4.3$, Chapter ${\rm V}$, of $[2]$, but I wasn't successful.

References are welcome.

$[1]$ Bachman, G. and Narici, L. Functional Analysis. New York: Academic Press, $1966$.

$[2]$ Kato, T., Perturbation Theory for Linear Operators, $2$nd edition, Springer, Berlin, $1984$.

$[3]$ Brezis, H. Functional Analysis, Sobolev Spaces and Partial Differential Equations, Springer, New York, $2011$.


1 Answer 1


There is a key error here, namely the claim that if $\lambda_1$ and $\lambda_2$ are eigenvalues of $A$ and $B$, then $\lambda_1+\lambda_2$ is an eigenvalue of $A+B$.

In general, its possible for $n(A),n(B)>0$, while $n(A+B)=0$. For example, take $$A=\begin{bmatrix}10& 0\\ 0& -1 \end{bmatrix},\;\;\;B=\begin{bmatrix}-1& 0\\ 0& 10 \end{bmatrix}.$$ Then $n(A)=n(B)=1$ but $n(A+B)=0$.

  • $\begingroup$ There exists some result that assure ${\rm n}(L+A) \leq {\rm n}(L)$? Since $L+A$ is a perturbation a self-adjoint operator by a bounded operator. $\endgroup$
    – Guilherme
    Commented Sep 14, 2021 at 14:10
  • $\begingroup$ @Guilherme I suspect the only case such a result is roughly when $A\ge 0$ or $L$ has no eigenvalues near zero and $\|A\|$ is small. I would look into Horn's Problem (now theorem) which gives a complete characterization of the possible eigenvalues of $L+A$ in finite dimensions. $\endgroup$
    – Pax
    Commented Sep 14, 2021 at 21:29
  • $\begingroup$ Do you have any reference of the Horn's Theorem? $\endgroup$
    – Guilherme
    Commented Sep 14, 2021 at 21:44
  • 1
    $\begingroup$ @Guilherme This survey seems good math.ucdavis.edu/~kapovich/EPR/eigensurvey.pdf $\endgroup$
    – Pax
    Commented Sep 14, 2021 at 21:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .