By Phoebus J. Dhrymes
This publication offers with a few mathematical themes which are of significant significance within the examine of classical econometrics. there's a long bankruptcy on matrix algebra, which takes the reader from the main ordinary points to the partitioned inverses, attribute roots and vectors, symmetric, and orthogonal and confident (semi) sure matrices. The e-book additionally covers pseudo-inverses, recommendations to structures of linear equations, strategies of vector distinction equations with consistent coefficients and random forcing capabilities, matrix differentiation, and permutation matrices. Its novel positive factors comprise an creation to asymptotic expansions, and examples of functions to the general-linear version (regression) and the overall linear structural econometric version (simultaneous equations).
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Let A be m × n, B be n × r. The i th row of C = AB is given by n ci· = aiq bq· , i = 1, 2, . . , m. q=1 Proof: Obvious from the deﬁnition of matrix multiplication. 3. Let A, B be m × n, and n × r, respectively. Then, C =BA, where C = AB. 3. RANK AND INVERSE OF A MATRIX 17 Proof: The typical element of C is given by n cij = ais bsj . s=1 By deﬁnition, the typical (i, j) element of C , say cij , is given by n cij = cji = ajs bsi . e. ajs is the (s, j) element of A , say asj , and bsi is the (i, s) element of B , say bis .
Deﬁne the vector x = Bξ and note that Ax = ABξ = Dξ = z; this means that z ∈ C(A), which shows C(D) ⊂ C(A). 7) But Eqs. 7) together imply C(A) = C(D). d. 7. Let A be p × q and B q × r, and put D = AB. Then r(D) ≤ min[r(A), r(B)]. Proof: Since D = AB, we note that if x ∈ N (B) then x ∈ N (D); hence, we conclude N (B) ⊂ N (D), and thus that n(B) ≤ n(D). 4. HERMITE FORMS AND RANK FACTORIZATION 23 we ﬁnd, in view of Eq. 8), r(D) ≤ r(B). 9) Next, suppose that y ∈ C(D). This means that there exists a vector, say, x ∈ Vr , such that y = Dx or y = ABx = A(Bx), so that y ∈ C(A).
D. 20. Let A be a square matrix of order m, and suppose all elements in its r th row are zero. Then, |A| = 0. Proof: By the deﬁnition of a determinant, we have |A| = (−1)s a1j1 a2j2 a3j3 · · · amjm , and it is clear that every term above contains an element from the i th row, say aiji . Hence, all terms vanish and thus |A| = 0. d. 16. It is clear that, in any of the propositions regarding determinants, we may substitute “column” for “row” without disturbing the conclusion. 15 and the example following.