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Then, it is not difficult to verify that x is an isometric reflection vector.
Proposition 2 Let X be a Minkowski plane, x ∈ S X be an isometric reflection vector.
Then, x ∈ H X, and it is not an isometric reflection vector.
In the rest of the proof, we show that x is not an isometric reflection vector.
Then, it follows from Lemma 1 that T x, x * is an isometry and x is an isometric reflection vector.
For any isometric reflection vector x, there is a unique isometric reflection functional x* corresponding to it (cf. [8]).
Similar(48)
The roots placement of reflection vectors is studied.
The presented paper deals with the development of robust control algorithm based on reflection vectors methodology.
If the stable simplex (or polytope) of reflection vectors is not given then a simple search procedure is needed.
For the relation between isometric reflection vectors and Roberts orthogonality, Chan He et al. proved the following lemma.
This procedure is quite straightforward because for a special family of polynomials the linear cover of so-called reflection vectors is stable.
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