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Regarding this question,

Why do spectral projections give norm approximations?

I have figured out what is meant by spectral projection, and have thus found the answer as well. A spectral projection is the image of $x$ under a step/indicator function defined on its spectrum, which is hence an orthogonal projection on some closed subspace. (Here I am using the Bounded Borel functional calculus.)

Okay, it's a projection. And I said the word "spectrum" in the above paragraph. But surely it's more "spectral" than that. Does the image of the so-obtained projection have anything to do with $x$ of a spectral nature? (This is an open-ended question, so please qualify what sort of spectral significance there is. But some specific interpretations are below):

  1. Suppose $\lambda$ is an eigenvalue of $x$ which hence means that its in the spectrum of $x$. Does the indicator function of the one point $\lambda$ applied to x via the Borel Functional Calculus project onto the eigenspace associated to it? By considering $z*\chi_\lambda$ one can see that the image of this projection is a subset of the eigenspace of $\lambda$. Does the other inclusion hold?

  2. More generally, if $\lambda$ is only a general element of the spectrum, is the indicator function of it applied to $x$ equal to a projection projecting onto some sort of "spectrally significant space for $\lambda$?"

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1 Answer 1

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If $\chi_\lambda$ is the indicator of the single point $\lambda$, then the corresponding spectral projection $\chi_\lambda(A)$ for your normal operator $A$ is indeed the orthogonal projection on the kernel of $A - \lambda I$, i.e. the eigenspace for $\lambda$. If $\lambda$ is not an eigenvalue, that means the projection is $0$.

EDIT: Why is the first statement true? Since $z \chi_\lambda(z) = \lambda \chi_\lambda(z)$ for all complex numbers $z$, $A \chi_\lambda(A) = \lambda \chi_\lambda(A)$ which says the range of $\chi_\lambda(A)$ is contained in the eigenspace. On the other hand, let $g_n(x) = 1/(x-\lambda)$ for $1/(n-1) \ge |x - \lambda| > 1/n$ (ignore the $1/(n-1)$ in the case $n=1$). Then $g_n(x)$ is a bounded Borel function, and $E_n(A) = g_n(A) (A - \lambda I)$ is the spectral projection on $S_n = \{x: 1/(n-1) \ge |x-\lambda| > 1/n\}$. Now if $\psi$ is an eigenvector of $A$ for eigenvector $\lambda$, $E_n(A) \psi = g_n(A) (A - \lambda I) \psi = 0$. The disjoint union of the sets $S_n$ for positive integers $n$ being ${\mathbb C} \backslash \{\lambda\}$, countable additivity for the projection-valued measure implies $$(I - \chi_\lambda(A)) \psi = \sum_{n=1}^\infty E_n(A) \psi = 0$$ i.e. $\psi = \chi_\lambda(A) \psi$ is in the range of $\chi_\lambda(A)$.

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How can I verify this? –  Jeff May 17 '12 at 21:55
    
I mean the first statement. The second one follows from the first, and in fact even from the inclusion for the first that I already know how to prove. –  Jeff May 18 '12 at 0:23

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