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Partitioned (Or Block) Matrices This Version: 29 Nov 2018 Partitioned (or Block) Matrices This version: 29 Nov 2018 Intermediate Econometrics / Forecasting Class Notes Instructor: Anthony Tay It is frequently convenient to partition matrices into smaller sub-matrices. e.g. 2 3 2 1 3 2 3 2 1 3 4 1 1 0 7 4 1 1 0 7 A B (2×2) (2×3) 3 1 1 0 0 = 3 1 1 0 0 = C I 1 3 0 1 0 1 3 0 1 0 (3×2) (3×3) 2 0 0 0 1 2 0 0 0 1 The same matrix can be partitioned in several different ways. For instance, we can write the previous matrix as 2 3 2 1 3 2 3 2 1 3 4 1 1 0 7 4 1 1 0 7 a b0 (1×1) (1×4) 3 1 1 0 0 = 3 1 1 0 0 = c D 1 3 0 1 0 1 3 0 1 0 (4×1) (4×4) 2 0 0 0 1 2 0 0 0 1 One reason partitioning is useful is that we can do matrix addition and multiplication with blocks, as though the blocks are elements, as long as the blocks are conformable for the operations. For instance: A B D E A + D B + E (2×2) (2×3) (2×2) (2×3) (2×2) (2×3) + = C I C F 2C I + F (3×2) (3×3) (3×2) (3×3) (3×2) (3×3) A B d E Ad + BF AE + BG (2×2) (2×3) (2×1) (2×3) (2×1) (2×3) = C I F G Cd + F CE + G (3×2) (3×3) (3×1) (3×3) (3×1) (3×3) | {z } | {z } | {z } (5×5) (5×4) (5×4) 1 Intermediate Econometrics / Forecasting 2 Examples (1) Let 1 2 1 1 2 1 c 1 4 2 3 4 2 3 h i A = = = a a a and c = c 1 2 3 2 3 0 1 3 0 1 c 0 1 3 0 1 3 3 c1 h i then Ac = a1 a2 a3 c2 = c1a1 + c2a2 + c3a3 c3 The product Ac produces a linear combination of the columns of A. (2) If 0 1 2 1 1 2 1 a1 b1 0 4 2 3 4 2 3 a2 b2 A = = = and b = 0 3 0 1 3 0 1 a3 b3 0 0 1 3 0 1 3 a4 b4 1 4 3 0 where a1 = 2 , a2 = 2 , a3 = 0 , a4 = 1 1 3 1 3 0 a1 0 0 h i a2 0 0 0 0 then b A = b b b b = b a + b a + b a + b a 1 2 3 4 0 1 1 2 2 3 3 4 4 a3 0 a4 The product b0A gives a linear combination of the rows of A. (3) Transposing partitioned matrices: 0 A B A0 C0 (2×2) (2×3) (2×2) (2×3) = C I B0 I (3×2) (3×3) (3×2) (3×3) Intermediate Econometrics / Forecasting 3 (4) If 0 1 2 1 1 2 1 a1 0 4 2 3 4 2 3 a2 A = = = 0 3 0 1 3 0 1 a3 0 0 1 3 0 1 3 a4 1 4 3 0 h i 0 then A = 2 2 0 1 = a1 a2 a3 a4 1 3 1 3 (5) Consider 1 2 1 1 4 3 0 0 4 2 3 A A = 2 2 0 1 3 0 1 1 3 1 3 0 1 3 We can write this as: 0 a1 0 4 0 h i a2 X 0 A A = a a a a = a a 1 2 3 4 0 i i a3 i=1 0 a4 Determinants and Inverses of Partitioned Matrices Suppose we have a square matrix partitioned so that the diagonal blocks are also square, e.g., A B (n×n) (n×m) C D (m×n) (m×m) Suppose for the time being that C is an (m × n) zero matrix 0. Just like usual lower (or upper) triangular matrices, the determinant of a lower (or upper) block triangular matrix is the product of the determinants of the diagonal blocks: A B (n×n) (n×m) = |A| |D| 0 D (m×n) (m×m) Like a matrix of elements, elementary row operations on block matrices do not change its determinant. We can use this fact to derive a formula for the case where C 6= 0. Subtracting Intermediate Econometrics / Forecasting 4 CA−1 times the first row from the second row produces a triangular matrix, thus A B A B (n×n) (n×m) (n×n) (n×m) −1 = = |A| D − CA B −1 C D 0 D − CA B (m×n) (m×m) (m×n) (m×m) This of course assumes that A−1 exists. The inverse for the block matrix A B (n×n) (n×m) C D (m×n) (m×m) is −1 AB A−1 + A−1B(D − CA−1B)−1CA−1 −A−1B(D − CA−1B)−1 = CD −(D − CA−1B)−1CA−1 (D − CA−1B)−1 assuming the necessary inverses exist. This can be shown by multiplying the two matrices together, or by applying the following row operations: A B I 0 C D 0 I A B I 0 −1 → (row2) − CA (row1) 0 D − CA−1B −CA−1 I −1 −1 −1 −1 −1 A 0 I + B(D − CA B) CA −B(D − CA B) −1 −1 → (row1) − B(D − CA B) (row2) 0 D − CA−1B −CA−1 I I 0 A−1 + A−1B(D − CA−1B)−1CA−1 −A−1B(D − CA−1B)−1 → multiplying by appropriate inverses 0 I −(D − CA−1B)−1CA−1 (D − CA−1B)−1 If the matrix is block diagonal, i.e., B = 0 and C = 0, then −1 A 0 A−1 0 = 0 D 0 D−1 Exercises 1. Show that an alternative formula for the determinant of a (suitably) partitioned matrix is A B (n×n) (n×m) −1 = |D| A − BD C C D (m×n) (m×m) Intermediate Econometrics / Forecasting 5 2. Show that an alternative formula for the inverse of a suitably partitioned matrix is −1 AB (A − BD−1C)−1 −(A − BD−1C)−1BD−1 = CD −D−1C(A − BD−1C)−1 D−1 + D−1C(A − BD−1C)−1BD−1 assuming the necessary inverses exist..
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