Approximate spectral decomposition

A detailed attempt below.

I am interested in effective and constructive computations for finding approximate spectral decompositions in some suitable format.

Namely, let $A: H \rightarrow H$ be a Hermitian operator on an $n-$dimensional Hilbert space $H$ with the spectrum $\{\lambda_1, \ldots \lambda_m\}, m \leq n$. Then, $A$ can be decomposed as:

$$ A = \sum_{i=1}^{m}\lambda_i P_i,$$

where $P_i, i=1,\ldots m$ are orthogonal projections with $P_i, P_j = 0, i \neq j$ onto the eigenspaces $H_i = \ker \{ \lambda_i I – A \}$ such that:

$$ H= \displaystyle \underset{i=1}{\overset{m}{\oplus}} H_i.$$

In an approximate format, the theorem can be stated as follows (p. 380):

for any $\varepsilon > 0$, there exist projections $P_i, i=1, \ldots n$ with $P_i, P_j = 0, i \neq j$, and real numbers $\alpha_1, \ldots \alpha_n$ such that $\big|\big| A – \displaystyle \sum_{i=1}^{n} \alpha_i P_i \big|\big| \leq \varepsilon$.

What about the approximate eigenspaces?

A particular example is this article, but it addresses exact spectral decomposition at the cost of additional input (cardinality of spectrum).


The attempt

Fix $\varepsilon>0$. Let $\{\lambda_1, \ldots \lambda_n\}$ be the $n$ roots of the characteristic polynomial $ \det \{ \lambda I – A \} = 0 $. These roots are computable. However, it is undecidable whether $\lambda_i = \lambda_j \lor \lambda_i \neq \lambda_j, i \neq j$. Therefore, the multiplicities of the eigenvalues are uncomputable.

On the other hand, for any real numbers $x, y, z$ such that $y < z$, it is decidable whether $ x < z $ or $ x > y $. This enables us to find “fuzzy” eigenvalues in the following sense: there are $k \leq n$ intervals of length $\varepsilon$ each of which contains a known number of roots of the characteristic polynomial.

Following the advice of TrialAndError, we may use the Kato’s theorem 6.17 on page 178 to compute projectors via contour integrals of the resolvent (also called Riesz projectors):

$$ P_j = \frac{1}{2\pi i}\oint_{\mathcal{C}\left(\alpha_j, \frac{\varepsilon}{2}\right)}(z I – A)^{-1} dz, j = 1 \ldots k, $$

where $\mathcal{C}\left(\alpha_j, \frac{\varepsilon}{2}\right)$ is a circle of radius $\frac{\varepsilon}{2}$ whose center is the center of $j$-th interval. This formula is also a consequence of the Stone’s formula that seems to admit a constructive proof. In our case, we only have “agglutinate” projectors and we can’t constructively show that they equal sums of exact projectors. At this point, we have:

$$ \Big|\Big| A – \displaystyle \sum_{j=1}^{k} \alpha_j P_j \Big|\Big| \leq k \varepsilon $$

If we could compute the bases of subspaces, that $P_j$’s project onto, we could take these as the approximate eigenvectors in the sense:

$$\| A x – \alpha_j x \| \leq \varepsilon \|x\|, $$

where $x$ is a vector from the column space of $P_j$, i. e., $\text{range}\{P_j\}$.

Wanted: an explicit proof, without resorting to the spectral theorem, that:

$$ \text{tr} \left \{ \frac{1}{2\pi i}\oint_{\mathcal{C}\left(\alpha_j, \frac{\varepsilon}{2}\right)}(z I – A)^{-1} dz \right \} = \text{# of roots of } \det \{ \lambda I – A \} = 0 \text{ in } \mathcal{C}\left(\alpha_j, \frac{\varepsilon}{2}\right) $$

(winding number properties?)

We can explicitly show that $P_j = P_j^2$ meaning that $P_j$ is an orthogonal projection, and that is commutes with $A$.

It can be constructively shown (Theorem 2.3) that if $P$ is an orthogonal projection, then it can be transformed into the form

$$ \begin{pmatrix}
I_r & 0 \\
0 & 0
\end{pmatrix} $$

for some integer $r$ via $T P T^{-1}$ with $T$ being the transformation operator.

How does this $r$ relate to the trace of $P$?

Assuming that $r$ equals the trace, and that $T$ effectively represents a change of basis, we can compute the explicitly compute the column space, can we?

Solutions Collecting From Web of "Approximate spectral decomposition"

This is not an answer, and don’t take it as such. From a theoretical point of view, you have $A=\sum_{k=1}^{n}\lambda_k P_k$ where the $P_k$ are orthogonal projections. If you start iterating in $A$, or perform various functions of $A$, the problem is that $f(A)=\sum_{k=1}^{n}f(\lambda_k)P_k$. You can get down below the level of a projection. You can form functions of $A$ in such that way that $f_k(A)=P_k$, but that’s the smallest granularity. Anything else starts with a vector and, even then, the best you can do is $P_kv$ for some vector $v$.

If you know where an eigenvalue is located, then you can use Complex Analysis to assert that
$$
P_{k} = \frac{1}{2\pi i}\oint_{|\lambda-\lambda_k|=\delta}(\lambda I-A)^{-1}d\lambda.
$$
where the circular contour is positively oriented, centered at $\lambda_k$ and of radius $\delta > 0$ chosen small enough that there are no other eigenvalues on or inside the circle of radius $\delta$ centered at $\lambda_k$. What’s nice about such a representation is this: if you perturb $A$ so that the eigenvalues don’t change by much, even if there are bifurcations in the eigenvalues, so long as the eigenvalues do not cross the boundary of your circle of integration, then $P_k$ remains a projection–it is the projection onto the direct sum of the eigenspaces associated with all eigenvalues in the circle. This tells you something about what happens under perturbations. What’s nice about this is that you don’t really care whether the eigenvalues are very close or whether they’re exactly on top of each other–this integral will give you the sum of all the projections associated with eigenvalues in the circle. And the sum of all such projections is an orthogonal projection. I suspect from your point of view of computability, this may be interesting, but I don’t know.

If you know the positions of eigenvalues to a certain accuracy, then you could theoretically perform several such integrations in order to obtain $\lambda_1,\cdots,\lambda_k$ and contour-integration-derived projections $P_l$ such that $\|\sum_{l=1} \lambda_l P_l-A\| < \epsilon$. The projection $P_l$ would be the sum of all the orthogonal projections for eigenvalues within a certain tolerance of $\lambda_l$. This approximation would in the operator norm, with the exact difference being
$$
\sum_{l}\sum_{\{m : |\lambda_m-\lambda_l| < \epsilon\}}\lambda_m Q_m -\sum_{l}\sum_{\{ m : |\lambda_m-\lambda_l| < \epsilon\} }\lambda_l Q_m \\
= \sum_{l}\sum_{\{m : |\lambda_m-\lambda_l|\}}(\lambda_m-\lambda_l)Q_m
$$
The projection $P_l$ is then given by
$$
P_l = \sum_{\{ m : |\lambda_m-\lambda_l| \}}Q_m.
$$
These are orthogonal projections, and $P_l P_l’ = 0$ if $l\ne l’$.