2007
DOI: 10.13001/1081-3810.1190
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On perfect conditioning of Vandermonde matrices on the unit circle

Abstract: Abstract. Let K, M ∈ N with K < M, and define a square K × K Vandermonde matrix A = A τ, − → n ¡ with nodes on the unit circle: Ap,q = exp (−j2πpnqτ /K) ; p, q = 0, 1, ..., K − 1, where nq ∈ {0, 1, ..., M − 1} and n 0 < n 1 < .... < n K−1 . Such matrices arise in some types of interpolation problems. In this paper, necessary and sufficient conditions are presented on the vector − → n so that a value of τ ∈ R can be found to achieve perfect conditioning of A. A simple test to check the condition is derived and … Show more

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Cited by 21 publications
(19 citation statements)
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“…In fact, if the λ i 's are on the unit circle, then κ 2 (V m ) is, under some additional assumptions (e.g. that the nodes are not tightly clustered), usually moderate [3], [4]. If the underlying operator is nearly unitary then the λ i 's will be close to the unit circle and V m is usually well conditioned.…”
Section: Computation With Vandermonde Matricesmentioning
confidence: 99%
“…In fact, if the λ i 's are on the unit circle, then κ 2 (V m ) is, under some additional assumptions (e.g. that the nodes are not tightly clustered), usually moderate [3], [4]. If the underlying operator is nearly unitary then the λ i 's will be close to the unit circle and V m is usually well conditioned.…”
Section: Computation With Vandermonde Matricesmentioning
confidence: 99%
“…Computing κ(F RF ) is a challenge, especially because the analysis must be valid for the entire range of antennas. Vandermonde matrices with positive real nodes z k ∈ R + are well-known to be ill-conditioned [23] -the condition number grows at least exponentially with the number of nodes K. However, if the nodes are complex-valued, it is possible to lower this growth to polynomial [24] and even achieve perfect conditioning choosing the nodes to be roots of unity [25]. In [26], the authors generalize this result to nodes that are close enough to the unit circle (not necessarily on the unit circle) and not so close to each other, while having N large enough.…”
Section: A Transmitted Powermentioning
confidence: 99%
“…The condition number is small if these knots are (almost) equidistantly distributed on the unit circle, see [4]. There exist several ideas to overcome these difficulties.…”
Section: Solve the Hankel Systemmentioning
confidence: 99%