# Number of negative eigenvalue of a perturbation by a bounded operator

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, $$2011$$.

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 $$n(A+B)=0$$.
• 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. Commented Sep 14, 2021 at 14:10
• @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.