# Eigenvalues of a series of orthogonal projections

In "Introduction to Hilbert spaces - Debnath, Mikusinski", third edition, ch. 4 pag. 194 theorem 4.9.21 it is stated that, if $$P_k:\mathcal{H}\to W_k$$ are pairwise orthogonal projection operators and $$\lambda_k\to 0$$, the only possible eigenvalues of the following operator $$A \doteq \lim_{n\to +\infty} \sum_{k=1}^n \lambda_k P_k$$ are just the $$\lambda_k$$ and $$0$$.

To demonstrate so, it is considered a generic eigevector $$u$$ and it is decomposed in an element of $$A(\mathcal{H})$$ and an element of $$(A(\mathcal{H}))^{\perp\mathcal{H}}$$: the issue is here, I'm not sure decomposition theorem (pag. 130 th. 3.6.6 of the same book) is applicable. My questions are

• Is it actually necessary to decompose an eigenvector? By construction $$A u=\lambda u$$ so it should be $$u\in A(\mathcal{H})$$
• How to demonstrate decomposition theorem is valid here? The single space $$W_k$$ is certainly closed by construction, but I'm not sure $$A(\mathcal{H})$$ is, and I have no idea how to demonstrate it

Thank you for any suggestion, I tried several wrong paths and I don't have any left at the moment

• Have you told us everything you're given? Is there some condition on the $W_k$ being orthogonal, or $P_kp_j=0$ or something??? Commented Jul 10, 2022 at 14:50
• @DavidC.Ullrich Oh yes sorry!! Yes $W_k$ pairwise orthogonal. Corrected Commented Jul 10, 2022 at 15:05

So lets define $$v\in\mathcal{H}\setminus\oplus_k W_k$$ (not $$v\in(\oplus_k W_k)^{\perp\Omega}$$ as stated in the book) in the following way $$u \doteq \sum\limits_k P_k u + v$$ The element $$v$$ certainly exists because by theorem 4.7.11 pag. 177 and theorem 4.7.12 pag. 178 we know that $$\sum\limits_k P_k = P_{\oplus_k W_k}$$ and so we have $$v = u - P_{\oplus_k W_k} u \\ \lVert v\rVert \leq \lVert u\rVert + \lVert P_{\oplus_k W_k} u\rVert \leq 2\lVert u\rVert$$ because for every orthogonal projection application $$P$$ it holds $$\lVert P\rVert\leq 1$$.
An important property of $$v$$ is that $$P_k v=0\,\forall k$$: we actually defined $$v\in\mathcal{H}\setminus\oplus_k W_k$$, but if you are dubious check it out $$P_j u = P_j \lim_{n\to +\infty} \sum_{k=1}^n P_k u + P_j v \\ P_j u = \lim_{n\to +\infty} \sum_{k=1}^n P_j P_k u + P_j v$$ because each $$P_j$$ is a bounded linear application between normed space, and hence is continuos. We now got $$P_j u = P_j^2 u + P_j v$$ because $$W_j\perp W_k\,\forall j\neq k$$ and so $$P_j P_k=0\,\forall j\neq k$$ (pag. 177 th. 4.7.11). Each one of the $$P_j$$ is also idempotent (pag. 176 th 4.7.7) so $$P_j^2=P_j$$ and finally $$P_j v=0\,\forall j$$.
That said, $$u$$ is still an eigenvector of $$A$$ with eigenvalue $$\lambda$$, so we write $$A u = \lambda u \\ \sum\limits_k \lambda_k P_k u = \lambda \sum\limits_k P_k u + \lambda v \\ \sum\limits_k (\lambda_k - \lambda) P_k u = \lambda v$$ So the sum of the series on the first member of the equation is $$\lambda v$$, that exists, meaning that the series converges strongly to $$\lambda v$$, and hence also weakly, so $$\lim_{n\to +\infty} \left\langle \sum_{k=1}^n (\lambda_k - \lambda) P_k u - \lambda v, y \right\rangle = 0\,\forall y\in\mathcal{H}$$ In particular if $$y=P_j u$$ we have $$\lim_{n\to +\infty} \sum_{k=1}^n (\lambda_k - \lambda) \langle P_k u, P_j u\rangle - \lambda \langle v, P_j u\rangle = 0$$ But as stated before $$P_j P_k=0\,\forall j\neq k$$ and because orthogonal projection applications are also self-adjoint $$\langle P_k u, P_j u\rangle=\langle P_j P_k u, u\rangle=\delta_{jk}\lVert P_j u\rVert^2$$, while at the same time $$\langle v, P_j u\rangle=\langle P_j v,u\rangle=0\,\forall j$$ meaning that $$\lambda_j - \lambda = 0 \,\forall j:P_j u\neq 0$$ If exists at least one $$j: P_j u\neq 0$$ then the eigeinvalue $$\lambda$$ must be $$\lambda_j$$. If this $$j$$ doesn't exist then $$P_j u=0\,\forall j$$ and we simply have $$u=v$$, but $$P_j v=0$$ always so $$A v= 0$$: at the same time $$A v=\lambda v$$ meaning that $$0=\lambda v$$ and if $$v\neq 0$$, as requested for an eigenvector, then $$\lambda=0$$.