Foundations of linear algebra by Jonathan S Golan

By Jonathan S Golan

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Z φ/ Q D z0 Pz, and has a nona projection matrix of rank q. 10 in Sect. 9. Putting the above results together we obtain our three expressions for t. z/ D 0. This completes the proof. 4 (Less than full rank) Suppose y D Xβ C ε, where X is n p of rank r (r < p). Instead of introducing identifiability constraints we can focus on what linear restrictions H W a0i β D 0 (i D 1; 2; : : : ; q) we might be able to test, or in matrix terms Aβ D 0 where rankŒA D q. A natural assumption is that the constraints are all estimable, which implies a0i D m0i X (by end of Sect.

P˝ P! /y D y0 P! 4 Contrasts P A contrast of the vector θ is any linear function c0 θ such that i ci D 0. Two contrasts c0 θ and d0 θ are said to be orthogonal if c0 d D 0. For example, Â1 Â2 and Â1 C Â2 C Â3 3Â4 are two orthogonal contrasts. 5 The situation that we often meet in factorial experiments is that we are given a set of independent contrasts a0i θ (i D 1; 2; : : : ; n p) equal to zero and we wish to test whether a further set of q orthogonal contrasts a01i θ (i D 1; 2; : : : ; q), which are orthogonal to the previous set, are also zero.

In P! Xr βr C Xp r βp r / P! 4 with ˝ ? replaced by ! In P! /Xp r with rank p r as CŒXp r  \ ! In P! /Xp r is non-singular. 9) is established. In P! In P! /Xp and it follows from Eq. In P˝ P! In P! In r P! /, that P! In P! 10) which can be used for a Wald test. 1, and we wish to test Aβ D b, where A is q p of rank q. Let β0 be any solution of Aβ D b, put z D y Xβ0 and let γ D β β0 . Then our original model and hypothesis are equivalent to z D Xγ C ε, where ε is Nn Œ0; 2 In , and ! W Aγ D 0. X0 X/ 1 X0 φ, !

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