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1 Markets I: The Ecosystem and Exchange-Traded Marketsবাজার অনলাইনইংরেজি 2 Markets II: Rates, FX and Creditবাজার অনলাইনইংরেজি 3 Markets III: Commodities, Energy and Cryptoবাজার অনলাইনইংরেজি 4 Quantitative Methodsপদ্ধতি অনলাইনইংরেজি 5 Derivatives and Volatilityডেরিভেটিভস অনলাইনইংরেজি 6 Rates, Credit, XVA and Riskসুদহার, ঋণ ও ঝুঁকি অনলাইনইংরেজি 7 Research Craft: Predictors, Backtests, Measurement, Portfoliosগবেষণা অনলাইনইংরেজি 8 Strategies I: Equities and Futuresকৌশল অনলাইনইংরেজি 9 Strategies II: Volatility, Relative Value, Macro and the Bank Desksকৌশল অনলাইনইংরেজি 10 Microstructure and Executionএক্সিকিউশন অনলাইনইংরেজি 11 Market Making and High-Frequency Tradingমার্কেট মেকিং অনলাইনইংরেজি 12 Machine Learning for Marketsমেশিন লার্নিং অনলাইনইংরেজি 13 Low-Latency Softwareপ্রযুক্তি অনলাইনইংরেজি 14 Networks, Hardware and Trading Infrastructureপ্রযুক্তি অনলাইনইংরেজি 15 Research, Data and Risk Platformsপ্রযুক্তি অনলাইনইংরেজি 16 The Desk and the Firmপ্রতিষ্ঠান অনলাইনইংরেজি 17 The Industry: Firms, Roles and Careersক্যারিয়ার অনলাইনইংরেজি 18 Interviews I: Maths, Probability and Speedক্যারিয়ার অনলাইনইংরেজি 19 Interviews II: Markets, Trading Games and the Final Roundক্যারিয়ার অনলাইনইংরেজি 20 Interviews III: Codeক্যারিয়ার অনলাইনইংরেজি
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Matemática · পরিভাষা

Posto de uma família কী?

Definição 19.15 Matemática universitária — Graduação 1 · অধ্যায় 19 — Dimensão Finita

O posto de uma família finita de vetores é a dimensão do seu espaço gerado: rk⁡(x1,…,xp)=dim⁡Vect⁡(x1,…,xp)≤min⁡(p,dim⁡E)\operatorname{rk}(x_1, \dots, x_p) = \dim \operatorname{Vect}(x_1, \dots, x_p) \leq \min(p, \dim E), com igualdade a pp se e somente se a família é livre.

উদাহরণ

Exemplo 19.16 (Calcular um posto por eliminação)

Posto de ((1,2,3),(2,3,4),(3,4,5),(1,1,1))\bigl((1,2,3), (2,3,4), (3,4,5), (1,1,1)\bigr) em R3\R^3. O espaço gerado não muda quando se subtrai de um vetor uma combinação dos outros (as duas famílias geram as mesmas combinações): substitua (2,3,4)(2,3,4) por (2,3,4)−(1,2,3)=(1,1,1)(2,3,4) - (1,2,3) = (1,1,1) e (3,4,5)(3,4,5) por (3,4,5)−(1,2,3)=(2,2,2)(3,4,5) - (1,2,3) = (2,2,2). O espaço gerado é agora Vect⁡((1,2,3),(1,1,1),(2,2,2),(1,1,1))=Vect⁡((1,2,3),(1,1,1))\operatorname{Vect}\bigl((1,2,3), (1,1,1), (2,2,2), (1,1,1)\bigr) = \operatorname{Vect}\bigl((1,2,3), (1,1,1)\bigr), e estes dois vetores não são proporcionais: o posto é 22. Este procedimento de subtrair e descartar é sistematizado como eliminação de Gauss no Capítulo 22.

Exemplo 19.17 (Somar por concatenação)

Tome, em R3\R^3,

F=Vect⁡((1,2,3), (1,1,1)),G=Vect⁡((2,3,4)).F = \operatorname{Vect}\bigl((1,2,3),\ (1,1,1)\bigr), \qquad G = \operatorname{Vect}\bigl((2,3,4)\bigr) .

A soma F+GF + G é gerada pela família concatenada dos três geradores, e

(2,3,4)=(1,2,3)+(1,1,1)(2, 3, 4) = (1, 2, 3) + (1, 1, 1)

mostra que o terceiro é redundante: F+G=FF + G = F, de dimensão 22 — equivalentemente G⊆FG \subseteq F, o que a relação exibe. Grassmann confirma: dim⁡(F∩G)=2+1−2=1=dim⁡G\dim(F \cap G) = 2 + 1 - 2 = 1 = \dim G. As somas calculam-se concatenando geradores e depois reduzindo a pilha pelo algoritmo do posto; nenhuma técnica nova é jamais necessária.

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