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Chapter 2 More Properties of Matrices and Matrix Arithmetic Chapter - PowerPoint PPT Presentation

Chapter 2 More Properties of Matrices and Matrix Arithmetic Chapter 2 Distributive and Associative Properties When dealing with matrices, we should already know that multiplication is NOT commutative AB = BA Chapter 2 Distributive and


  1. Chapter 2 More Properties of Matrices and Matrix Arithmetic Chapter 2

  2. Distributive and Associative Properties When dealing with matrices, we should already know that multiplication is NOT commutative AB � = BA Chapter 2

  3. Distributive and Associative Properties However, matrix multiplication DOES satisfy the distributive and associative properties: A ( B + C ) = AB + AC ( distributive ) ( B + C ) A = BA + CA ( distributive ) A ( BC ) = ( AB ) C = ABC ( associative ) Chapter 2

  4. Example: Distributive and Associative Properties Simplify: A ( λ I + A − 1 C ) A ( λ I + A − 1 C ) = A λ I + A ( A − 1 C ) = λ A + C Chapter 2

  5. Properties of the Matrix Transpose There are also few properties that you should be familiar with regarding the matrix transpose operation: ( A + B ) T = A T + B T ( AB ) T = B T A T ( reverse-order) ( ABCD ) T = D T C T B T A T ( reverse-order, any number of factors ) ( A T ) − 1 = ( A − 1 ) T ( Often, we write A − T ) ( α A ) T = α A T ( transpose of scalar is same scalar ) ( A T ) T = A Chapter 2

  6. Properties of the Matrix Transpose There are a few properties that you should be familiar with regarding the matrix transpose operation: ( A + B ) T = A T + B T ( AB ) T = B T A T ( reverse-order ) ( ABCD ) T = D T C T B T A T ( reverse-order, any number of factors ) ( A T ) − 1 = ( A − 1 ) T ( Often, we write A − T ) ( α A ) T = α A T ( transpose of scalar is same scalar ) ( A T ) T = A Chapter 2

  7. Example: Is this matrix symmetric? X T X 1 Is the transpose equal to the original? ( X T X ) T = X T ( X T ) T = X T X yes! I + xx T 2 Is the transpose equal to the original? ( I + xx T ) T = I T + ( xx T ) T = I + xx T yes! If A is symmetric ( A = A T ) and B is symmetric ( B = B T ), is 3 the product AB symmetric? Is the transpose equal to the original? ( AB ) T = B T A T = BA no! Chapter 2

  8. Example: Simplifying Expressions Simplify the expression ( A T B ) T A − 1 ( A T B ) T A − 1 = B T ( A T ) T A − 1 reverse-order law ( A T ) T = A = B T AA − 1 AA − 1 = I = B T . Chapter 2

  9. Check Your Understanding Simplify the following expressions: 1 C − 1 [( A + B )( C T )] T ( X T X ) T ( X T X ) − 1 Determine whether the following matrices are symmetric: 2 XX T A + A T ABA if both A and B are symmetric. Chapter 2

  10. Check Your Understanding - Solution Simplify the following expressions: 1 C − 1 [( A + B )( C T )] T = A T + B T ( X T X ) T ( X T X ) − 1 = I Determine whether the following matrices are symmetric: 2 XX T yes! A + A T yes! ABA if both A and B are symmetric yes!. Chapter 2

  11. Matrix Powers We define/compute powers of a matrix in the same way we do for scalars: A 0 = I A 1 = A A 2 = AA A 3 = AAA When powering a product of matrices, exercise caution: ( AB ) 2 = ( AB )( AB ) = ABAB � = A 2 B 2 Chapter 2

  12. Chapter 2 Arithmetic with Partitioned Matrices Chapter 2

  13. Partitioned Matrices We will often want to consider a matrix as a collection of either rows or columns (or perhaps “blocks”) rather than individual elements. When we write A = ( A 1 | A 2 | . . . | A n ) we are viewing the matrix A as collection of column vectors, A i , in the following way: A 1 A 2 . . . A n   a 11 a 12 . . . a 1 n a 21 a 22 a 2 n     a 31 a 32 . . . a 3 n A = ( A 1 | A 2 | . . . | A n ) =     . . . . . .   . . .   a m 1 a n 2 . . . a mn Chapter 2

  14. Partitioned Matrices Similarly, we can write A as a collection of row vectors:   A 1   a 11 a 12 a 1 n A 1 . . . A 2 a 21 a 22 a 2 n A 2       a 31 a 32 a 3 n A 3 A 3 . . . A = =     . . . .     . . . . . .   .  . . .  .   a m 1 a n 2 a mn A m . . . A m Chapter 2

  15. Partitioned Matrices We could even draw divisions in the matrix to partition it into blocks :  a 11 a 12 a 13 a 14  � B � a 21 a 22 a 23 a 24 C   A =  =   a 31 a 32 a 33 a 34 D F  a 41 a 42 a 43 a 44 Chapter 2

  16. Partitioned Matrices: Arithmetic Why is this useful? As long as two matrices are partitioned conformably , we can actually multiply them as if the partitioned blocks are entries in a new matrix:  a 11 a 12 a 13 a 14   v 1  a 21 a 22 a 23 a 24 v 2         a 31 a 32 a 33 a 34 v 3     a 41 a 42 a 43 a 44 v 4 � B � � u � C = D F w � Bu + Cw � = Du + Fw Chapter 2

  17. Partitioned Matrices: Arithmetic If you work out all of the multiplication and addition from the previous example, you’ll see we aren’t really saving any time in the computation. But, we save time in the notation and often we can represent unique concepts by ordering the rows and columns into special blocks. Chapter 2

  18. Partitioned Matrices: Arithmetic You’ll recall from a previous worksheet that you’ve worked through several views of matrix multiplication. These can always be found in general from partitioned matrix arithmetic. Chapter 2

  19. Partitioned Matrices: Arithmetic Let’s consider matrix multiplication a few ways, starting with our original formulation.  X T  1 X T   2   X T X T =   X = ( X 1 | X 2 | X 3 | . . . | X p ) and 3   .   . .     X T p So, what can we say about the matrix product X T X ? Chapter 2

  20. Partitioned Matrices: Arithmetic  X T  1 X T   2   X T X T X =   ( X 1 | X 2 | X 3 | . . . | X p ) 3   .   . .     X T p Is this partitioning conformable for multiplication? Always want to multiply Row × Column . If we do that, we will be multiplying X T i X j at each step. Do these calculations make sense? X T X j i 1 × n n × 1 Yes! So the partitioning is conformable for multiplication. Chapter 2

  21. Partitioned Matrices: Arithmetic  X T  1 X T   2   X T X T X =   ( X 1 | X 2 | X 3 | . . . | X p ) 3   .   .  .    X T p X T X T X T X T  1 X 1 1 X 2 1 X 3 . . . 1 X p  X T X T X T X T 2 X 1 2 X 2 2 X 3 . . . 2 X p    X T X T X T X T  3 X 1 3 X 3 3 X 3 . . . 3 X p =    . . . .  ... . . . .   . . . .   X T X T X T X T p X 1 p X 2 p X 3 . . . p X p Chapter 2

  22. Partitioned Matrices: Arithmetic X T X T X T X T   1 X 1 1 X 2 1 X 3 . . . 1 X p X T X T X T X T 2 X 1 2 X 2 2 X 3 . . . 2 X p    X T X T X T X T  3 X 1 3 X 3 3 X 3 . . . 3 X p =     . . . . ... . . . .   . . . .   X T X T X T X T p X 1 p X 2 p X 3 . . . p X p Diagonal elements contain the sum of squares for each column (variable). Chapter 2

  23. Partitioned Matrices: Arithmetic Another way that we will end up using partitioned matrices is to represent consider the opposite formulation where the matrix on the right is partitioned into columns and the matrix on the left is partitioned into rows. U m × r V T r × n V T   1 V T  2   V T  ( U 1 | U 2 | U 3 | . . . | U r )   3   . .   .   V T r Chapter 2

  24. Partitioned Matrices: Arithmetic U m × r V T r × n V T   1 V T   2  V T  ( U 1 | U 2 | U 3 | . . . | U r )   3   . .   .   V T r = U 1 V T 1 + U 2 V T 2 + U 3 V T 3 + · · · + U r V T r Let’s back up a minute... Chapter 2

  25. Partitioned Matrices: Arithmetic U m × r V T r × n V T   1 V T  2    V T ( U 1 | U 2 | U 3 | . . . | U r )   3   . .   .   V T r Is this partitioning conformable for multiplication? Again, want to multiply Row × Column . If we do that, we will be multiplying U i V T j at each step. Do these calculations make sense? V T U i j m × 1 1 × n Sure! Each element in the sum will be an m × n matrix! Chapter 2

  26. Check Your Understanding Compute the following matrix product, using block multiplication:  1 − 1 2   − 2 1 − 1  0 2 3 0 1 2     0 0 1 0 0 1 Chapter 2

  27. Check Your Understanding - Solution       1 − 1 2 − 2 1 − 1 − 2 0 − 1  = 0 2 3 0 1 2 0 2 7      0 0 1 0 0 1 0 0 1 � 1 � � − 2 � � 2 � � � − 2 � − 1 1 0 � (1,1)-block: + 0 0 = 0 2 0 1 3 0 2 � 1 � � − 1 � � 2 � � − 1 � − 1 (1,2)-block: + ( 1 ) = 0 2 2 3 7 � � − 2 � 1 � � � � � (2,1)-block: 0 0 + ( 1 ) 0 0 = 0 0 0 1 � � − 1 � � (2,2)-block: 0 0 + ( 1 )( 1 ) = 1 2 Chapter 2

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