Properties of Transpose Transpose has higher precedence than - PDF document

3.2, 3.3 Inverting Matrices P. Danziger Properties of Transpose Transpose has higher precedence than multiplica- tion and addition, so AB T = A and A + B T = A + � B T � � B T � As opposed to the bracketed expressions ( AB ) T and ( A + B ) T Example 1 � � � � 1 2 1 1 0 1 Let A = and B = . 2 5 2 1 1 0 Find AB T , and ( AB ) T . �   1 1 � T � � � � 1 2 1 1 0 1 1 2 1 AB T = = 0 1   2 5 2 1 1 0 2 5 2   1 0 � � 2 3 = 4 7 Whereas ( AB ) T is undefined. 1

3.2, 3.3 Inverting Matrices P. Danziger Theorem 2 (Properties of Transpose) Given ma- trices A and B so that the operations can be pre- formed 1. ( A T ) T = A 2. ( A + B ) T = A T + B T and ( A − B ) T = A T − B T 3. ( kA ) T = kA T 4. ( AB ) T = B T A T 2

3.2, 3.3 Inverting Matrices P. Danziger Matrix Algebra Theorem 3 (Algebraic Properties of Matrix Multiplication) 1. ( k + ℓ ) A = kA + ℓA (Distributivity of scalar multiplication I) 2. k ( A + B ) = kA + kB (Distributivity of scalar multiplication II) 3. A ( B + C ) = AB + AC (Distributivity of matrix multiplication) 4. A ( BC ) = ( AB ) C (Associativity of matrix mul- tiplication) 5. A + B = B + A (Commutativity of matrix ad- dition) 6. ( A + B ) + C = A + ( B + C ) (Associativity of matrix addition) 7. k ( AB ) = A ( kB ) (Commutativity of Scalar Mul- tiplication) 3

3.2, 3.3 Inverting Matrices P. Danziger The matrix 0 is the identity of matrix addition. That is, given a matrix A , A + 0 = 0 + A = A. Further 0 A = A 0 = 0, where 0 is the appropriately sized 0 matrix. Note that it is possible to have two non-zero ma- trices which multiply to 0. Example 4 � � � � � � � � 1 − 1 1 1 1 − 1 1 − 1 0 0 = = − 1 1 1 1 − 1 + 1 − 1 + 1 0 0 The matrix I is the identity of matrix multiplica- tion. That is, given an m × n matrix A , AI n = I m A = A Theorem 5 If R is in reduced row echelon form then either R = I , or R has a row of zeros. 4

3.2, 3.3 Inverting Matrices P. Danziger Theorem 6 (Power Laws) For any square ma- trix A , A r A s = A r + s and ( A r ) s = A rs Example 7 2 4 2       0 0 1 0 0 1 1.   1 0 1 = 1 0 1             2 2 0 2 2 0 2. Find A 6 , where � � 1 0 A = 1 1 A 2 � 2 . A 6 = A 2 A 4 = A 2 � � � 1 0 Now A 2 = , so 2 1 � 2 � � � � � � � A 2 � 2 = 1 0 1 0 1 0 1 0 A 2 � = 2 1 2 1 2 1 3 1 � � 1 0 = 5 1 5

3.2, 3.3 Inverting Matrices P. Danziger Inverse of a matrix Given a square matrix A , the inverse of A , denoted A − 1 , is defined to be the matrix such that AA − 1 = A − 1 A = I Note that inverses are only defined for square ma- trices Note Not all matrices have inverses. If A has an inverse, it is called invertible . If A is not invertible it is called singular . 6

3.2, 3.3 Inverting Matrices P. Danziger Example 8 � � � � 1 2 5 − 2 A − 1 = 1 . A = 2 5 − 2 1 � � � � � � 1 2 5 − 2 1 0 Check: = 2 5 − 2 1 0 1 � � 1 2 2 . A = Has no inverse 2 4     1 1 1 3 − 1 − 1 A − 1 = 3 . A = 1 2 1 − 1 1 0         1 1 2 − 1 0 1       1 1 1 3 − 1 − 1 1 0 0 Check: 1 2 1 − 1 1 0  = 0 1 0            1 1 2 − 1 0 1 0 0 1   1 2 1 4 . A = 2 1 3 Has no inverse     3 3 4 7

3.2, 3.3 Inverting Matrices P. Danziger Inverses of 2 × 2 Matrices Given a 2 × 2 matrix � � a b A = c d A is invertible if and only if ad − bc � = 0 and � � 1 d − b A − 1 = − c a ad − bc The quantity ad − bc is called the determinant of the matrix and is written det( A ), or | A | . Example 9 2 � � � � � � − 1 1 2 3 − 2 A − 1 = 1 3 A = = − 1 − 3 3 3 − 3 1 1 3 � � � � � � 3 − 2 1 2 − 3 0 1 = − 1 Check: = I − 3 3 − 3 1 3 3 0 − 3 8

3.2, 3.3 Inverting Matrices P. Danziger Algebra of Invertibility Theorem 10 Given an invertible matrix A : 1. ( A − 1 ) − 1 = A , 2. ( A n ) − 1 = ( A − 1 ) n � = A − n � , 3. ( kA ) − 1 = 1 k A − 1 , 4. ( A T ) − 1 = ( A − 1 ) T , 9

3.2, 3.3 Inverting Matrices P. Danziger Theorem 11 Given two invertible matrices A and B ( AB ) − 1 = B − 1 A − 1 . Proof: Let A and B be invertible matricies and let C = AB , so C − 1 = ( AB ) − 1 . Consider C = AB . Multiply both sides on the left by A − 1 : A − 1 C = A − 1 AB = B. Multiply both sides on the left by B − 1 . B − 1 A − 1 C = B − 1 B = I. So, B − 1 A − 1 is the matrix you need to multiply C by to get the identity. Thus, by the definition of inverse B − 1 A − 1 = C − 1 = ( AB ) − 1 . 10

3.2, 3.3 Inverting Matrices P. Danziger A Method for Inverses Given a square matrix A and a vector b ∈ R n , consider the equation A x = b This represents a system of equations with coeffi- cient matrix A . Multiply both sides by A − 1 on the left, to get A − 1 A x = A − 1 b . But A − 1 A = I n and I x = x , so we have x = A − 1 b . Note that we have a unique solution. The as- sumption that A is invertible is equaivalent to the assumption that A x = b has unique solution. 11

3.2, 3.3 Inverting Matrices P. Danziger During the course of Gauss-Jordan elimination on the augmented matrix ( A | b ) we reduce A → I and � � b → A − 1 b , so ( A | b ) → I | A − 1 b . If we instead augment A with I , row reducing will produce (hopefully) I on the left and A − 1 on the � I | A − 1 � right, so ( A | I ) → . The Method: 1. Augment A with I 2. Use Gauss-Jordan to obtain ( I | A − 1 ) . 3. If I does not appear on the left, A is not in- vertable. Otherwise, A − 1 is given on the right. 12

3.2, 3.3 Inverting Matrices P. Danziger Example 12 1. Find A − 1 , where   1 2 3 A = 2 5 5     3 5 8 13

3.2, 3.3 Inverting Matrices P. Danziger Augment with I and row reduce:   1 2 3 1 0 0 R 2 → R 2 − 2 R 1 2 5 5 0 1 0   R 3 → R 3 − 3 R 1   3 5 8 0 0 1   1 2 3 1 0 0 0 1 − 1 − 2 1 0 R 3 → R 3 + R 2     0 − 1 − 1 − 3 0 1   1 2 3 1 0 0 R 3 → − 1 0 1 − 1 − 2 1 0 2 R 3     0 0 − 2 − 5 1 1   1 2 3 1 0 0 R 1 → R 1 − 3 R 3 0 1 − 1 − 2 1 0   R 2 → R 2 + R 3   0 0 1 5 / 2 − 1 / 2 − 1 / 2   1 2 0 − 13 / 2 3 / 2 3 / 2 0 1 0 1 / 2 1 / 2 − 1 / 2 R 1 → R 1 − 2 R 2     0 0 1 5 / 2 − 1 / 2 − 1 / 2   1 0 0 − 15 / 2 1 / 2 5 / 2 0 1 0 1 / 2 1 / 2 − 1 / 2     0 0 1 5 / 2 − 1 / 2 − 1 / 2 14

3.2, 3.3 Inverting Matrices P. Danziger So   − 15 1 5 A − 1 = 1 1 1 − 1   2   5 − 1 − 1 To check inverse multiply together:     1 2 3 − 15 1 5  1 AA − 1 = 2 5 5 1 1 − 1     2    3 5 8 5 − 1 − 1   2 0 0 1 = 0 2 0  = I   2  0 0 2 2. Solve A x = b in the case where b = (2 , 2 , 4) T .     − 15 1 5 2 A − 1 b = 1 = 1 1 − 1 2 x     2     5 − 1 − 1 4     − 18 − 9 1 =  = 0 0     2    4 2 15

3.2, 3.3 Inverting Matrices P. Danziger 3. Solve A x = b in the case where b = (2 , 0 , 2) T .     − 15 1 5 2 A − 1 b = 1 = 1 1 − 1 0 x     2     5 − 1 − 1 2     − 20 − 9 1 = 0  = 0     2    8 4 4. Give a solution to A x = b in the general case where b = ( b 1 , b 2 , b 3 )     − 15 1 5 b 1 1 = 1 1 − 1 x b 2     2     5 − 1 − 1 b 3   − 15 b 1 + b 2 + 5 b 3 1 = b 1 + b 2 − b 3   2   5 b 1 − b 2 − b 3 16

3.2, 3.3 Inverting Matrices P. Danziger Elementary Matrices Definition 13 An Elementary matrix is a matrix obtained by preforming a single row operation on the identity matrix. Example 14 1.   2 0 0 0 1 0 ( R 1 → 2 R 1 )     0 0 1 2.   1 0 0 3 1 0 ( R 2 → R 2 + 3 R 1 )     0 0 1 3.   1 0 0 0 0 1 ( R 1 ↔ R 2 )     0 1 0 17

3.2, 3.3 Inverting Matrices P. Danziger Theorem 15 If E is an elementary matrix ob- tained from I m by preforming the row operation R and A is any m × n matrix, then EA is the matrix obtained by preforming the same row operation R on A . Example 16   1 1 1 A = 2 1 0     3 2 1 1.       2 0 0 1 1 1 2 2 2 0 1 0 2 1 0  = 2 1 0  ∼ 2 R 2 on A           0 0 1 3 2 1 3 2 1 2.       1 0 0 1 1 1 1 1 1  ∼ R 2 → R 2 + 3 R 1 3 1 0 2 1 0  = 5 4 3       on A     0 0 1 3 2 1 3 2 1 3. 18