If you mean that the spectrum is finite, a necessary and sufficient condition is that the operator is algebraic over C i.e. satisfies a polynomial \prod_i (X - \lambda_i).
thank you very much for your useful answer. please, can you provided me with document can use as a reference in my thesis.
No but it is easy to prove:
Suppose A is normal and has finite spectrum (\lambda_1, ....\lambda_n) i.e we can write A = \sum \lambda_i E_i for orthogonal projections E_i with E_iE_j = \delta_{ij}E_i and \sum_j E_j = 1. Then
\prod (A - \lamdba_i) = \sum_j \prod_ i (A - \lambda_i) E_j
= \sum_j (\prod_i (\lambda_j - \lambda_i)) E_j
= \sum_j 0 E_j
= 0
Conversely suppose A is normal and there is a polynomial P(X) \in C[X] such that
P(A) = 0.
Since A is normal there is a spectral decompostition
A = \int_C \lambda E(d\lambda)
for a unique orthogonal projection valued measure E on C with \int E(d\lambda) = 1 whose support is the spectrum of A (see any book on functional analysis). Now if f(\lambda) is a continuous complex valued function we have
\int f(\lambda) P(A) E(d\lambda) = \int f(\lambda)P(\lambda) E(d\lambda) = 0
Since f is arbitrary, the projection valued measure must have support on the zero's of P, i.e. the spectrum is finite.
thank you for your help. the statement E_iE_j = \delta_{ij}E_i what is exactly means?
and \prod (A - \lamdba_i) = \sum_j \prod_ i (A _\lambda_i) E_j according what i can do this ?
Helllo Noor,
The Kronecker \delta_{ij} = 0 iff i \ne j, and 1 iff i = j. Hence E_iE_j = \delta_{ij}E_i is just a shorthand for E_i E_j = 0 iff i \ne j and E_i^2 = E_i which just expresses that the eigenspaces for different eigenvalues are orthogonal, and that the E_i are projections.
The second relation is a direct insertion of the "decomposition of the identity"
\sum_j E_j = 1, i.e. the completeness of the spectral decomposition:
\prod_i (A - \lamdba_i) = (\prod_i (A - \lambda_i)) 1
= \prod_i (A - \lambda_i) \sum_j E_j
= \sum_j \prod_ i (A _\lambda_i) E_j
Hello Rogier,
i am really thank you very much for your helpful information. please, do you mean by C{X} the space of continuous functions? . In fact, i am study the idea of paper and looking for the answer about representation of normal operator the researcher represent the normal operator in summation form but as known the representation of normal operator by using integral on infinite dimension Hilbert space. please , can you see the attach file (theorem 3.2) and consider the problem?
Hello Noor.
glad to help.
The space C[X] is the space of complex polynomials in the (formal) variable X, so I just say that P is a polynmial in the variable X, and P(A) is the normal operator obtained by substituting A for X.
Indeed I am writing Latex (or at least something that is supposed to be Latex) so \prod_i is the product over i = 1, ...n, the index set of the distinct eigenvalues, in other words
\prod_i (A - \lamdba_i) = (A -\lambda_1)(A-\lambda_2)...(A-\lambda_n)
the sum \sum_j is likewise a sum over j =1, ...,n i.e. all the different eigenvalues.
Hello Rogier,
the problem that lead me to ask the question is : on the following paper the researcher using some trick in proof to represent normal operator in summation form but not suppose that the normal operator is algebraic over C. please, can you check the attachment file theorem (3.2) .
In the proof theorem 3.2 the author does not assume that the operator N has finite spectrum. Instead, he takes the the spectral decomposition
N = \int_C \lambda E(d\lamda) = \int_\sigma(N) \lamda E(d\lamda).
Then he approximates the function f(\lamda) = \lambda by a "stair function"
s(\lambda) = \sum \lambda_i chi__{\sigma_i}
where the \chi_\sigma is the indicator function of the Borel set \sigma \subset (C) and \lamda_i \in \sigma_i. For each \sigma_i there is an orhogonal projection E(\sigma_i) = E_i (which we may assume non zero!). Without actually stating this he uses that this E_i commutes with A, i.e. E_i \in {A}' (I think this is a well known fact, it follows e.g from the fact that
E_i = E(\sigma_i) = \int_C \chi_{\sigma_i} (\lambda) E(d\lambda) = \chi_{\sigma_i}(N)
and you can approximate \chi_{\sigma_i} by polynomials on the compact spectrum of N. It is clear that if N commutes with A then so does every polynomial in N ) .
So now we use lemma 3. 1 (which really has to include the assumption that the P_i are non trivial) to conclude
|max(\lambda_i)| \le |\sum \lamda_i E_i + [A,X]|
Taking the limit we get
esssup|\lambda| = sup|\sigma(N)| = ||N||
Hello Rogier,
thank you for the effort that you do for me. i will understand the proof perfectly and i will ask you if there is any misunderstand step in the proof.
Hello Dr. Rogier,
i was reading the proof and i understand as in the following attachment. please, can you check the folder and let me know your opinion.
Hello Noor,
I am afraid you still miss something important. The function f(\lambda) = \lambda cannot in general be written as f(\lambda) = \sum_{i
Hello Dr. Rogier,
in the equation
\tilde N = \sum \lambda_i \int_\sigma(n) \chi_{\sigma_i}(\lambda) E(d\lambda)
it came from you substitute that \tilde f(\lambda) = \sum \lambda_i \chi_{\sigma_i} and you put the infinite sum \lambda_i out of \int_\sigma(n) because \int_\sigma(n) does not depend on \lambda_i since \lambda_i it is constant? . \lambda_i represents elements in spectrum of normal operator?
and the following step of proof
= \sum \lambda_i E(\sigma_i)
came from the properties of indicator function?. please, can you explain to me the causes.
Hallo Noor
You asked: in the equation
\tilde N = \sum \lambda_i \int_\sigma(n) \chi_{\sigma_i}(\lambda) E(d\lambda)
it came from you substitute that \tilde f(\lambda) = \sum \lambda_i \chi_{\sigma_i} and you put the infinite sum \lambda_i out of \int_\sigma(n) because \int_\sigma(n) does not depend on \lambda_i since \lambda_i it is constant? .
Answ: the small n is a typo (see improved version of my answer above) and the sum over the \lambda_i is finite (ditto), but otherwise you are right: the \lambda_i's are just constants that can be pulled out.
You asked: \lambda_i represents elements in spectrum of normal operator?
Answ: Yes. The \lambda_i must be chosen from the spectrum \sigma(N). Of course in general \sigma(N) contains other elements. This is because of the way we started by partioning \sigma(N) = \union_{i =1}^N in Borel sets, and we chose \lambda_i \in \sigma_i
You asked: and the following step of proof
= \sum \lambda_i E(\sigma_i)
came from the properties of indicator function?. please, can you explain to me the causes.
Answer: yes just as more or less by definition the measure of a set is the integral (with respect to the same measure) of the indicator function, for the projection E(\sigma_i) we have
E(\sigma_i) = \int \chi_{\sigma_i} (\lambda) E(d\lambda)
In fact if you consider the innerproduct for arbitrary x (which determines (E(\sigma_i) completely) the second statement is a direct consequence of the first. Note that it does not matter whether we take the integral over C or \sigma(N) since the complement of \sigma(N) has measure zero.
Hello Dr. Rogier,
thank you very much for your explaining. in the proof we assume that the partitioning \sigma(N) = \union_{i =1}^n in borel sets , can we always partition the spectrum as a union of finite or infinite sets of borel sets as we need in the proof ? or under some conditions on the space ?
Hello Noor,
yes the partitioning in Borel sets of the spectrum can always be done. That is because the Borel sets is a sigma-algebra constructed so that continuous functions are measurable, i.e. the positive (negative) part of the real and imaginary part are the sup (inf) of simple functions. It is then standard that they can can be approximated (in L_1 on C and even in the sup norm on the compact spectrum) by simple functions.
In fact we only have to approximate the identity function f(\lambda) = \lambda so we can be very explicit how we chose \tilde f by just dividing the complex plane in a sufficiently small grid of squares and taking the \lambda_i's as the centre of the squares. More formally: suppose we want
sup_[\lambda \in \sigma(N)} |\tilde f - f| 0 such that for all \lambda \in \sigma(N) ,
|Re(\lambda)|
Hello Dr. Rogier,
i am grateful very much. thank you for your impressive answers.
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