MATH3200, Lecture 8: Properties of Matrix Multiplication

Transcription

1 Lecture 8: Properties of Matrix Multiplication Winfried Just, Ohio University September 7, 206

2 Some familiar-looking properties Let A = [a ij ] k m, B = [b ij ] n p, and C = [c ij ] q r be matrices. Then A(BC) = (AB)C = ABC We need here: m = n and p = q. The order of ABC will then be k r. (Associativity Law), A(B + C) = AB + AC (Left Distributivity Law), We need here: m = n = q and p = r. The order of A(B + C) will then be k r. (B + C)A = BA + CA (Right Distributivity Law), We need here: k = p = r and n = q. The order of (B + C)A will then be q m. Homework 23: Prove the Left Distributivity Law. Ohio University Since 804

3 Applications of the Associativity Law We have used the Associativity Law A(BC) = (AB)C already when we considered compositions of transformations: T A (T B ( v)) = A(B v) = (AB) v = (T A T B )( v). Our next application of the Associativity Law will rely on the observation that sums of vectors can be expressed as inner products: x [ [ ] x 2 n ].... = x n = [ ] x x 2... x n l=. x n Ohio University Since 804

4 Calculating grades: The instructor s spreadsheet again a a 2... a n a 2 a a 2n Let A =... = [a ij] m n a m a m2... a mn Assume m is the number of students in the class, n is the number of gradable items, and the entry a ij represents the score of student i on gradable item number j. Let x be the vector of total scores for each student and let y be a vector of mean scores for each item. Question: (a) How can you express x as the product of A and a vector v? Would the result of this operation give you a row vector or a column vector x? (b) How can you express y as the product of A and a vector w? Would the result of this operation give you a row vector or a column vector y? Ohio University Since 804

5 The answer: a a 2... a n a 2 a a 2n Let A =... = [a ij] m n a m a m2... a mn Assume m is the number of students, n is the number of gradable items, and a ij the score of student i on gradable item number j. Let x be the vector of total scores for each student and let y be a vector of mean scores for each item. Then x = A. and y = [ m m... m] A The row vector has m entries and the column vector has n entries. Ohio University Since 804

6 Another use of the Associativity Law: Homework again Let A be the instructor s spreadsheet. Method of calculating the mean total score, adding up the total scores of all students and dividing the result by m, can be expressed as the inner product m [ ]... A. Similarly, Method 2 of calculating the mean total score, adding up the mean scores of all gradable items, can be expressed as ([ m... ] ) m A. = ([ ] )... A m. The Associativity Law implies that the two products are equal. Ohio University Since 804

7 Commutativity fails Even for square matrices A = [a ij ] n n and B = [b ij ] n n we may have AB BA. Consider the example: [ ] [ ] [ ] c c AB = = c 2 c 22 [ ] [ ] [ ] d d BA = = d 2 d 22 Ohio University Since 804

8 Commutativity fails Even for square matrices A = [a ij ] n n and B = [b ij ] n n we may have AB BA. Consider the example: [ ] [ ] [ ] c2 AB = = c 2 c 22 c = a b + a 2 b 2 = = 6. [ ] [ ] [ ] d2 BA = = d 2 d 22 d = b a + b 2 a 2 = = 0. Since c d, the matrices AB and BA differ in at least one element and we conclude that AB BA. Ohio University Since 804

9 The transpose of a product Let A = [a ij ] k n and B = [b ij ] n p. Is then (AB) T = A T B T? Not in general. The expression on the right may not even be defined. But we always have: (AB) T = B T A T Since B T has order p n and A T has order n k, the product B T A T is defined. (We are assuming here that all elements of these matrices are numbers, which we will from now on always do when we work with the transpose.) Homework 24: Test the above claims for [ ] A = B = [ ] 2 Ohio University Since 804

10 Multiplication of symmetric matrices is commutative Let A = [a ij ] n n and B = [b ij ] n n be symmetric, that is, A T = A and B T = B. Homework 25: Assume A and B are symmetric and of the same order. Prove that the product AB is also a symmetric matrix. That is, prove the (AB) T = AB. Now let A, B be symmetric. Then: BA = B T A T = (AB) T = AB. The first equality follows from the definition of symmetry. The second equality follows from reading (AB) T = B T A T backward. The third equality follows from the result of Homework 25. Thus multiplication of symmetric matrices is commutative. Ohio University Since 804