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The rational first-variation space has dimension set by defect

Statement

Let $r = p / q \in R_{m, n}^{+} (I)$ be nonzero, with its reduced pair and defect $d = d_{m, n} (r)$ , and put $N = m + n - d$ . Then ${u q - p v : u \in P_{m}, v \in P_{n}} = P_{N} .$ For $r = 0$ with the convention $p = 0, q = 1, d = n, N = m$ , the same equality holds.

Facts & Assumptions

Given: A nondegenerate interval $I$ , nonnegative degrees $m, n$ , and a normalized admissible rational function $r = p / q$ with the reduced convention of Weighted rational minimax families and defect.

[F1]

Weighted rational minimax families and defect: for $r \neq 0$ , $p, q$ are coprime, $\deg p \leq m$ , $\deg q \leq n$ , and $d = \min (m - \deg p, n - \deg q)$ ; for $r = 0$ , $(p, q, d) = (0, 1, n)$ .

[F2]

Bézout identity and the Euclidean algorithm for polynomials over a field: if $p, q$ are coprime, there are polynomials $A, B$ with $A p + B q = 1$ .

[F3]

Rank-nullity: $\dim_{F} V = nullity T + rank T$ : domain dimension is kernel dimension plus image dimension for a linear map of finite-dimensional vector spaces.

[F4]

Over an integral domain, degrees add under multiplication of nonzero polynomials: for nonzero polynomials over $R$ , product degrees add.

[F5]

Formal real polynomials, evaluation, degree, leading coefficient, and monic polynomials: $P_{k}$ has basis $1, x, \dots, x^{k}$ and dimension $k + 1$ .

Proof

1.1F1F5construct

Define $Φ : P_{m} \times P_{n} \to R [x]$ by $Φ (u, v) = u q - p v$ . It is linear. If $r = 0$ , [F1] gives $p = 0, q = 1, d = n, N = m$ , so $Φ (u, v) = u$ and its image is exactly $P_{m} = P_{N}$ .

2.1F1F2F4step 1.1algebra

Suppose $r \neq 0$ . If $(u, v) \in \ker Φ$ , then $u q = p v$ . Multiply the Bézout identity in [F2] by $u$ : $u = A p u + B q u = A p u + B p v = p (A u + B v)$ , so $u = h p$ for $h : = A u + B v$ . Substitution into $u q = p v$ and cancellation of the nonzero polynomial $p$ give $v = h q$ . Conversely, every $(h p, h q)$ lies in the kernel.

3.1F1F3F4F5step 2.1algebra

The degree bounds on $u = h p$ and $v = h q$ , together with [F4], are equivalent to $\deg h \leq \min (m - \deg p, n - \deg q) = d$ ; the zero polynomial $h$ is included. Thus the kernel has basis $(p, q), (x p, x q), \dots, (x^{d} p, x^{d} q)$ and dimension $d + 1$ . By [F3] and [F5], the image dimension is $(m + n + 2) - (d + 1) = m + n + 1 - d = N + 1$ .

4.1F1F4F5step 3.1algebra∎

For all $u, v$ , the degree of $u q - p v$ is at most $\max (m + \deg q, n + \deg p) = m + n - d = N$ by [F1] and [F4]. Hence the image is a subspace of $P_{N}$ , whose dimension is $N + 1$ by [F5]. Its dimension from step 3.1 is also $N + 1$ , so the two spaces are equal.

Depends on

Used by

Weighted rational minimizers satisfy the defect-sensitive alternation criterion Theorem

Dependency tree · two levels

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Sources

András Kroó and Allan Pinkus, Strong Uniqueness, Lemma 6.2 (standard reference, not scraped)
Lloyd N. Trefethen, Approximation Theory and Approximation Practice, Chapter 24 (standard reference, not scraped)