Lecture 4: Partitioned Matrices and Determinants

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1 Lecture 4: Partitioned Matrices and Determinants 1

2 Elementary row operations Recall the elementary operations on the rows of a matrix, equivalent to premultiplying by an elementary matrix E: (1) multiplying row i by a nonzero scalar α, denoted by E i (α), (2) adding β times row j to row i, denoted by E ij (β) (here β is any scalar), and (3) interchanging rows i and j, denoted by E ij, (here i j), called elementary row operations of types 1,2 and 3 resp. Illustrations for m = 4: E 2 (α)= α , E42 (β)= β 0 1 Q. Calculate the determinants of these matrices., E13 = a

3 Determinants Definition (Cullen and Gale). The determinant is the function det : C n n C such that (a) det(e i (α)) = α, for all α C, i 1, n, and (b) det(ab) = det(a) det(b), for all A, B C n n. Q. Prove that this definition is equivalent to the one you know. Q. Explain why det E ij = 1, or equivalently, why det E ij A = det A, A. The Binet Cauchy formula. If A C k n, B C n k then det(ab) = I Q k,n det A I det B I. Here Q k,n is the set of increasing sequences of k elements from 1, n, for example: Q 2,3 = {{1, 2}, {1, 3}, {2, 3}}. 3-b

4 Bordered matrices Theorem (Blattner). Let A Cr m n V satisfy and let the matrices U and (a) U C m (m r) (m r) and the columns of U are a basis for N(A ). (b) V C n (n r) (n r) Then the matrix and the columns of U are a basis for N(A). A V is nonsingular and its inverse is A V U O U, (1) O. (2) 5

5 The Cramer rule Given a matrix A and a vector b, A[j b] denotes the matrix obtained from A by replacing the j th column by b. Theorem (Cramer). Let A C n n be nonsingular. Then for any b C n, the solution x = [x j ] of is given by deta[j b] x j =, j 1, n. deta Proof (Robinson). Write Ax = b as and take determinants Ax = b (1) A I n [j x] = A[j b], j 1, n, deta deti n [j x] = deta[j b]. 4-a

6 Bordered matrices (cont d) Proof. Recall: A Cr m n, and the matrices U and V satisfy (a) U C m (m r) (m r) and the columns of U are a basis for N(A ). (b) V C n (n r) (n r) A V U O and the columns of U are a basis for N(A). A V U O = AA + UU AV. V A V V (1) R(U) = N(A ) = R(A) = AA + UU = I n (2) V A = V A AA = V A A A = (AV ) A A = O (3) AV = AV = A(V V V ) = A(V V V ) = AV V V = O, (4) V V = I n r. 6-a

7 A special case Corollary. Let A Cr m n V C n (n r) satisfy and let the matrices U C m (m r) and AV = O, V V = I n r, A U = O, and U U = I m r. Then the matrix A V U, O is nonsingular and its inverse is A U V. O 7

8 Corollary. Let A C m n r MNLSS and let the matrices U and V satisfy (a) U C m (m r) (m r) and the columns of U are a basis for N(A ). (b) V C n (n r) (n r) Consider the linear equation and the columns of U are a basis for N(A). Ax = b. (1) The solution x,y of A V U x = O y b. (2) 0 satisfies x = A b, the minimal norm least squares solution of (1), Uy = P N(A )b, the residual of (1). 8

9 Corollary. Let A C m n r MNLSS and let the matrices U and V satisfy (a) U C m (m r) (m r) and the columns of U are a basis for N(A ). (b) V C n (n r) (n r) Consider the linear equation and the columns of U are a basis for N(A). Ax = b. (1) The minimal norm least squares solution x = [x j ] of (1) is given by A[j b] det U V [j 0] O x j =, j 1, n. det A U V O 9

10 Greville s method Let A C m n, and let A k = A[, 1, k] C m k be partitioned as ] A k = [A k 1 a k (1) Let the vectors d k and c k be defined by Theorem (Greville). d k := A k 1 a k (2) c k := a k A k 1 d k = P N(A k 1 )a k (3) [ A k 1 a k ] = A k 1 d kb k b k, (4) where b k = c k if c k 0, b k = (1 + d kd k ) 1 d ka k 1 if c k = 0. 10

11 Let A be partitioned as A = Schur complement A 11 A 12, A 11 nonsingular, A 21 A 22 and consider the homogeneous equation A 11 x 1 + A 12 x 2 = 0 A 21 x 1 + A 22 x 2 = 0 Eliminating x 1 we get the equation for x 2 (A 22 A 21 A 1 11 A 12)x 2 = 0 The Schur complement of A 11 in A, denoted A/A 11, is A/A 11 := A 22 A 21 A 1 11 A

12 Schur complement (cont d) Let A = A 11 A 12, A 11 nonsingular, A 21 A 22 A/A 11 := A 22 A 21 A 1 11 A 12. (a) If A is square, its determinant is deta = deta 11 det(a/a 11 ). (b) The quotient property. If A 11 is further partitioned as A 11 = E F, E nonsingular, then A/A 11 = (A/E)/(A 11 /E). G H (c) rank A = rank A 11 A/A 11 = O. 12

13 Schur complement (cont d) Let the equation Ax = b be partitioned as Then (1) is consistent if and only if A 11 A 12 x 1 = b 1 (1) A 21 A 22 x 2 (A/A 11 )x 2 = b 2 A 21 A 1 11 b 1 (2a) is consistent, in which case a solution is completed by b 2 x 1 = A 1 11 (b 1 A 12 x 2 ). (2b) Proof. Eliminate x 1 = A 1 11 (b 1 A 12 x 2 ) from the top of (1), and substitute in the bottom to get, (A 22 A 21 A 1 11 A 12)x 2 = b 2 A 21 A 1 11 b 1 13

14 Basic solutions Let A C m n n, b C m, and consider the equation Ax = b (1) Let I(A) be then index set of maximal full rank (nonsingular) submatrices I(A) = {I Q n,m : ranka[i, ] = n} For each I I(A), the I th basic solution of (1) is the vector the solution of the subsystem x I = A[I, ] 1 b[i], A[I, ]x = b[i]. There are at most ( m n) basic solutions. 14

15 Let A C m n n LSS, b C m. Then the LSS x of the equation Ax = b (1) is unique, x = A b and is a convex combination of the basic solutions x = λ I A[I, ] 1 b[i], I I(A) I I(A) λ I = 1, λ I 0, I. Not surprising, but the real surprise is that the weights λ I are proportional to the squares of the determinants of A[I, ], λ I det 2 A[I, ] 15-a

16 The Moore Penrose inverse and basic inverses Theorem. Let A Cn m n. Then A = I I(A) λ I A[I, ] 1, λ I = det 2 A[I, ] det 2 A[K, ] K I(A) Theorem. Let A Cr m n, and let N(A) be the index set of maximal nonsingular submatrices, N(A) := {(I, J) : rank A[I, J] = r} Then A is a convex combination, A = λ IJ = (I,J) N(A) λ IJ A[I, J] 1 det 2 A[I, J] det 2 A[K, L] (K,L) N(A) 16-a

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