A robust solver using a continuation method for Nevanlinna-Pick interpolation with degree constraint

We can assume that all components of x, y, u, v are nonzero. To see why this is so, suppose that the result was proven for this case and we were given four arbitrary nonzero vectors x, y, u, and v. We could transform them via a single nonsingular transformation T such that each component of T x, T y, T u, T v was nonzero (Lemma 2.2). Then for all Hermitian matrices P we would have (T x) T P (T y) = x T (T T PT)y, and hence, that (T x) T P (T y) = 0k(Tu) T P (T v). Then, T x = T u and, thus, x = u or T x = T v and x = v. So, we shall assume that all components of x, y, u, v are nonzero. Suppose that x is not a scalar multiple of u to begin with. Then, for any index i with 1 i n, there is some other index j and two nonzero real numbers c i ; c j such that xi = ciui xj = cjuj; ci 6 = cj: (8) Choose one such pair of indexes i, j. Equating the coefficients of p ii , p jj and p ij , respectively, in the identity x T Py = 0ku T Pv yields the following equations: xiyi = 0 kuivi (9) x j y j = 0 ku j v j If we combine (8) with (9) and (10), we find yi = 0 k c i vi (12) yj = 0 k c j vj: (13) Using (9)–(13), we find c i u i y j + c j u j y i = 0k(u i v j + u j v i). Hence, u i v j (c j 0 c i =c j) = u j v i (c j 0 c i =c i). Recall that c i 6 = c j , so we can divide by cj 0 ci and rearrange the terms to get ci cj = vi vj uj ui : (14) However, using (8), we find c i c j = x i x j u j u i : (15) Combining (14) and (15) yields v i v j = x i x j : (16) Thus, x i = cv i , x j = cv j for some …