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Matriks Operator Pertemuan 9

Matriks Operator Pertemuan 9. Matakuliah : Matrix Algebra for Statistics Tahun : 2009. Partitioned Matrices. Purpose: To obtain expressions for the inverse and determinant of an m x m matrix A that is partitioned into the 2 x 2 block

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Matriks Operator Pertemuan 9

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  1. Matriks OperatorPertemuan 9 Matakuliah : Matrix Algebra for Statistics Tahun : 2009

  2. Partitioned Matrices Purpose: To obtain expressions for the inverse and determinant of an m x m matrix A that is partitioned into the 2 x 2 block where A11is m1xm1. A12 is m1xm2. A21 is m2xml, and A22is m2xm2

  3. Contoh: Let Amxm and Bmxm partitioned as A, A11, and A22 are nonsingular matrices B = A-1 and partition B as

  4. Then, • B11 =(A11-A12A22-1A21)-1 =A11 +A11A12B22A21A11, b) B22= (A22– A21-1A12)-1 = A22-1 + A22-1 A21B11A12A22-1 c) B12= -A11-1A12B22, d) B21 =-A22-1A21 B11

  5. Matrix equation Yields, A11B11 +A12B21= Im1 A21B12 +A22B22= 1m2 A11B12 +A12B22= (0) A21B11 + A22B21= (0)

  6. Example 1 Consider the regression model y = Xᵦ + є, where y is Nx1, X is Nx(k + 1), β is (k + 1) x1, and є is N x 1. Suppose that β and X are partitioned as β= (β 1T, β 2T)T and X = (X1, X2) so that the product X1β1is defined

  7. Example 2 Let the m x m matrix A be partitioned If A22= Im2 and A12 = (0) or A21= (0), then IAI = IA11I· To find the determinant use the cofactor expansion formula for a determinant

  8. where B is the (m2- 1) x m1 matrix obtained by deleting the last row from A21 Repeating this process another (m2- 1) times yields IA I = IA11l. In a similar way we obtain IA I = IA11I. when A21 = (0). by repeatedly expanding along the last row.

  9. Nonnegatif Vector An m x n matrix A is a nonnegative matrix, indicated by A ≥ (0), if each element of A is nonnegative. Similarly, A is a positive matrix, indicated by A > (0), if each element of A is positive. We will write A ≥ B and A> B to mean that A-B ≥ (0) and A-B > (0), respectively.

  10. Any matrix A can be transformed to a nonnegative matrix by replacing each of its elements by its absolute value. This will be denoted by abs(A); that is, if A is an mxn matrix, then abs(A) is also an mxn matrix with (i,j)th element given by Ia ij I.

  11. Let A be an mxm matrix and x be an mx1 vector. If A ≥(0) and x > 0, then with similar inequalities holding when minimizing and maximizing over columns instead of rows

  12. Theorm 1 Let A be an m x m positive matrix. Then ρ(A) is positive and is an eigenvalue of A. In addition, there exists a positive eigenvector of A corresponding to the eigenvalue ρ(A).

  13. g Theorm 2 Let A be an mxm positive matrix and suppose that  is an eigenvalue of A satisfying = ρ(A). If x is any eigenvector corresponding to , then A abs(x) = ρ(A)abs(x)

  14. Theorm 3 If A is an mxm positive matrix, then the dimension of the eigenspace corresponding to the eigenvalue ρ(A) is one. Further, if A is an eigenvalue of A and Aρ(A), then II < ρ(A).

  15. VEC OPERATOR There are situations in which it is useful to transform a matrix to a vector that has as its elements the elements of the matrix. One such situation in statistics involves the study of the distribution of the sample covariance matrix S The operator that transforms a matrix to a vector is known as the vec operator

  16. If the mxn matrix A has aias its i th column, then vec(A) is the mnx1 vector given by

  17. Example A is 2x3 matrix, If a is mx1 and b is nx1, then abTis mxn and vec(abT) = vec([b1a, b2a, ... , bna]) =

  18. Theorm Let a and b be any two vectors, while A and B are two matrices of the same size. Then a) vec(a) = vec(aT) = a, b) vec(αbT) = b a, c) vec( αA + (βB) = α vec(A) + β vec(B), where α and β are scalars.

  19. Example Suppose that we are interested in the distribution of the sample covariance matrix or the distribution of the sample correlation matrix computed from a sample of observations on three different variables. The resulting sample covariance and correlation matrices would be of the form

  20. So that vec(S) = (S11 S12, S13, S12, S22, S23, S13, S23, S33)T, vec(R) = (1, rl2, r13, r12, 1, r23, r13, r23, 1)T Since both S and R are symmetric, there are redundant elements in vec(S) and vec(R). The elimination of these results in v(S) and v(R) given by v(S) = (S11, S12, S13, S22, S23, S33)T, v(R) = (I, rl2, r13, 1, r23, 1)T

  21. Eliminating the nonrandom 1 s from v(R), we obtain which contains all of the random variables in R.

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