By Moody T. Chu, Gene H. Golub
Inverse eigenvalue difficulties come up in a striking number of functions and linked to any inverse eigenvalue challenge are primary questions-the theoretic factor on solvability and the sensible factor on computability. either questions are tough and not easy. during this textual content, the authors talk about the basic questions, a few identified effects, many functions, mathematical houses, quite a few numerical options in addition to a number of open difficulties. this is often the 1st ebook within the authoritative Numerical arithmetic and medical Computation sequence to hide numerical linear algebra, a large region of numerical research. Authored via world-renowned researchers, this publication is geared toward graduates and researchers in utilized arithmetic, engineering and laptop technology and makes a great graduate textual content.
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Additional resources for Inverse Eigenvalue Problems: Theory, Algorithms, and Applications
Example text
Cm ]T ∈ Fm . An IEP being parameterized in this way will be called a linear PIEP (LiPIEP). 1) and scalars {λk }nk=1 , find values of c := [c1 , . . , cm ]T ∈ Fm such that σ(A(c)) ⊂ {λk }nk=1 . 32 PARAMETERIZED INVERSE EIGENVALUE PROBLEMS Note that the basis matrices {Ai }m i=0 need not have any inherent structure in themselves. They are just the constituent elements that make up A(c). 1). That is, assuming that the basis matrices {Ai }m i=0 are linearly independent, then all matrices A(c) lie on a specific m-dimensional affine subspace.
4) n |Ak |, S := k=1 d(λ) := min |λi − λj |. 5) The following theorem gives rise to one of the sufficient conditions for real solvability. 3. (Biegler-K¨onig, 1981b) Given n real numbers λ = {λk }nk=1 , and n + 1 matrices Ai ∈ Rn×n , i = 0, 1, . . , n, assume that (i) diag(Ak ) = ek , k = 1, . . , n, (ii) π(S) < 1, (iii) the gap d(λ) is sufficiently large, that is, d(λ) ≥ 4 π(S) λ − diag(A0 ) ∞ + π(A0 ) . 1 − π(S) Then the LiPIEP (with m = n) has a real solution c ∈ Rn . Proof The original proof is quite insightful, so we outline the proof below.
By the implicit function theorem, there exists a neighborhood N of 0 ∈ Cm and a function z(c) analytic in N such that f (z(c), c) = 0. 13), it can be checked that ˜ A(c) 1 1 ˜12 z(c)) .