By Adi Ben-Israel

The sphere of generalized inverses has grown a lot because the visual appeal of the 1st variation in 1974, and remains to be becoming. This publication money owed for those advancements whereas conserving the casual and leisurely type of the 1st version. New fabric has been additional, together with a bankruptcy on functions, an appendix at the paintings of E.H. Moore, new workouts and purposes.

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99) (c) If A, A11 , and A22 are nonsingular, then A−1 = (A/A22 )−1 −1 −A22 A21 (A/A22 )−1 −1 −A−1 11 A12 (A/A11 ) . −1 (A/A11 ) (100) 5. DETERMINANTS AND VOLUME 31 Proof. (a) follows from A11 A21 A12 A22 I O A11 −A−1 11 A12 = A21 I O . A/A11 (b) is left for the reader, after noting that A11 /E is nonsingular (since A11 and E are nonsingular, and det A11 = det E det(A11 /E) by (97)). (c) is veriﬁed by multiplying A and RHS(100). Ex. 62. The set S = {x1 , . . , xk } are linearly independent if, and only if, the Gram matrix G(S) is nonsingular.

Stein’s Theorem. A square matrix is convergent if and only if there exists a PD matrix H such that H − A∗ HA is also PD (Stein [776], Taussky [799]). 22 0. PRELIMINARIES 3. 1. Elementary operations. 2 Applying an elementary row operation to the identity matrix Im results in an elementary matrix of the same type. We denote these elementary matrices also by E i (α), E ij (β), and E ij . Elementary matrices of types 1, 2 have only one row that is diﬀerent from the corresponding row of the identity matrix of the same order.

3. ) Exercises Ex. 51. Transforming a matrix into Hermite normal form. Let A ∈ Cm×n and let T0 = [A Im ]. A matrix E transforming A into a Hermite normal form EA can be found by Gaussian elimination on T0 where, after the elimination is completed, ET0 = [EA E], E being recorded as the right-hand m × m submatrix of ET0 . We illustrate this procedure for the matrix 0 2i i 0 4 + 2i 1 −6 −3 − 3i , (80) A = 0 0 0 −3 0 2 1 1 4 − 4i 1 4. THE HERMITE NORMAL FORM AND RELATED ITEMS marking the pivots by boxes, 0 T0 = 0 0 2i i 0 4 + 2i 1 0 0 −3 −6 −3 − 3i 1 1 4 − 4i 1 2 31 T1 = E (−2)E 0 = 0 0 1 1 0 0 1 0 0 0 0 , 1 1 ( 2i )T0 1 1 2 0 1 − 2i − 12 i 0 0 −3 −6 −3 − 3i 0 0 1 2 1+i T2 = E 32 (−1)E 2 (− 13 )T1 0 = 0 0 ..