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Let $A : H \rightarrow H$ be a bounded linear map where $H$ is a Hilbert space with $\dim H = \infty$. Suppose that $\lambda \in \sigma(A)$ but $\lambda$ is not an eigenvalue. Let $K : H \rightarrow H$ be compact. Show that $\lambda \in \sigma(A + K).$

($\sigma(A)$ is the set $\{\lambda \in \mathbb{C} : (A - \lambda I)$ is not a bijection$\}$)

What I have so far: Suppose $(A-\lambda I)$ is not injective. Then there exist $x_1 \neq x_2 \in H$ so that $(A-\lambda I)x_1 = (A-\lambda I)x_2$ so $(A - \lambda I)(x_1 - x_2) = 0$. But then $\lambda$ is an eigenvalue, a contradiction. Thus $(A - \lambda I)$ must be injective. Since $\lambda \in \sigma(A)$, $(A - \lambda I)$ cannot be a bijection so we know that $(A - \lambda I)$ is not surjective. Letting $H' = \mbox{im}\, (A-\lambda I)$ we have that $H'$ is a proper subset of $H$ and that $(A - \lambda I)^{-1}$ exists on $H'$.

My professor suggested that we assume that $(A - \lambda I)^{-1}$ is unbounded, so there exists in particular a bounded sequence $\{\mu_k\} \subset H'$ with $\lim_{k \rightarrow \infty} (A - \lambda I)^{-1}\mu_k = \infty$. I don't yet see how to use that idea, or why it should be true. Any suggestions?

Daniel Watkins
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2 Answers2

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If we assume that $A-\lambda I +K$ is invertible, then it is a Fredholm operator of index $0$. But any compact perturbation of a Fredholm operator is also Fredholm, and has the same index ( that's not a trivial result). Therefore $A-\lambda I$ is Fredholm of index $0$. However, $A-\lambda I$ is injective, and being Fredholm of index $0$, it implies is also surjective. That's a contradiction. I suppose there exists a more elementary direct proof, but I don't see it right now.

Theo
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  • Very elegant! This problem is from my PDE class, and we're just finishing a short unit on some topics from functional analysis. Hence I have pretty minimal background in functional. I'm hoping to find a more elementary proof, if one exists. – Daniel Watkins Apr 17 '14 at 22:26
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Here is a proof suggested by another student in my class. It is not as compact as Theo's proof, but it requires less of a functional analysis background. I think it's pretty clever.

Let $\lambda \in \sigma(A), \lambda$ not an eigenvalue. Without loss of generality, $\lambda = 0$. (If $\lambda \neq 0$, then set $\tilde{A} = A - \lambda I$. Then $0 \in \sigma(\tilde{A})$ iff $\lambda \in \sigma(A)$, and $0$ is not an eigenvalue of $\tilde{A}$ if $\lambda$ is not an eigenvalue of $A$). If $0 \in \sigma(A)$ then $A$ is not bijective. If $A$ is not injective, then $0$ is an eigenvalue, a condradiction. Thus $A$ is injective but not surjective.

Suppose that $0 \not \in \sigma(A + K)$. Then $A + K$ is bijective. By the Banach theorem, since $A + K$ is a linear bounded operator and $A + K$ is bijective, we have that $(A + K)^{-1}$ exists and is linear bounded. Moreover, since the composition of a linear bounded operator with a compact operator is compact, we have that $K(A + K)^{-1}$ is compact.

Since $Ax = 0$ only if $x \equiv 0$, $A(A+K)^{-1}y \neq 0$ for all $y \not \equiv 0 \in H$. Therefore, $$Iy = (A + K)(A+K)^{-1}y = K(A + K)^{-1}y + A(A+K)^{-1}y \neq K(A+K)^{-1}y $$ In particular, we now have that for all $y \not \equiv 0 \in H$ $$(I - K(A+K)^{-1})y \neq 0$$ or in other words, $$\ker(I - K(A + K)^{-1}) = \{0\}$$ so by the Fredholm alternative, $$ \mbox{im}(I - K(A + K)^{-1}) = H$$ Now, $$(I - K(A+K)^{-1})y = (A + K)(A+K)^{-1}y - K(A+K)^{-1}y = A(A+K)^{-1}$$ so it is clear that $$H = \mbox{im}(I - K(A+K)^{-1}) = \mbox{im}(A(A+K)^{-1}) \subset \mbox{im}(A)$$ hence $ \mbox{im}(A) = H$. Therefore $A$ is surjective. However, this contradicts that $0 \in \sigma(A)$ is an eigenvalue. We then can conclude that $0 \in \sigma(A + K)$.

Daniel Watkins
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