Mathematics_1_oneside.pdf

# Theorem 418 p roof see problem 418 let a and b be two

• No School
• AA 1
• 110

This preview shows page 28 - 31 out of 110 pages.

Theorem 4.18 P ROOF . See Problem 4.18 . Let A and B be two invertible matrices of the same size. Then AB is Theorem 4.19 invertible and ( AB ) - 1 = B - 1 A - 1 . P ROOF . See Problem 4.19 . Let A be an invertible matrix. Then the following holds: Theorem 4.20 (1) ( A - 1 ) - 1 = A (2) ( A 0 ) - 1 = ( A - 1 ) 0 P ROOF . See Problem 4.20 . 4.5 Block Matrix Suppose we are given some vector x = ( x 1 ,..., x n ) 0 . It may happen that we naturally can distinguish between two types of variables (e.g., endoge- nous and exogenous variables) which we can group into two respective vectors x 1 = ( x 1 ,..., x n 1 ) 0 and x 2 = ( x n 1 + 1 ,..., x n 1 + n 2 ) 0 where n 1 + n 2 = n . We then can write x = x 1 x 2 . Assume further that we are also given some m × n Matrix A and that the components of vector y = Ax can also be partitioned into two groups y = y 1 y 2 where y 1 = ( y 1 ,..., x m 1 ) 0 and y 2 = ( y m 1 + 1 ,..., y m 1 + m 2 ) 0 . We then can parti- tion A into four matrices A = A 11 A 12 A 21 A 22

Subscribe to view the full document.

4.5 B LOCK M ATRIX 23 where A i j is a submatrix of dimension m i × n j . Hence we immediately find y 1 y 2 = A 11 A 12 A 21 A 22 · x 1 x 2 = A 11 x 1 + A 12 x 2 A 21 x 1 + A 22 x 2 . Matrix A 11 A 12 A 21 A 22 is called a partitioned matrix or block matrix . Definition 4.21 The matrix A = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 can be partitioned in numerous Example 4.22 ways, e.g., A = A 11 A 12 A 21 A 22 = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Of course a matrix can be partitioned into more than 2 × 2 submatrices. Sometimes there is no natural reason for such a block structure but it might be convenient for further computations. We can perform operations on block matrices in an obvious ways, that is, we treat the submatrices as of they where ordinary matrix elements. For example, we find for block matrices with appropriate submatrices, α A 11 A 12 A 21 A 22 = α A 11 α A 12 α A 21 α A 22 A 11 A 12 A 21 A 22 + B 11 B 12 B 21 B 22 = A 11 + B 11 A 12 + B 12 A 21 + B 21 A 22 + B 22 and A 11 A 12 A 21 A 22 · C 11 C 12 C 21 C 22 = A 11 C 11 + A 12 C 21 A 11 C 12 + A 12 C 22 A 21 C 11 + A 22 C 21 A 21 C 12 + A 22 C 22 We also can use the block structure to compute the inverse of a parti- tioned matrix. Assume that a matrix is partitioned as ( n 1 + n 2 ) × ( n 1 + n 2 ) matrix A = A 11 A 12 A 21 A 22 . Here we only want to look at the special case where A 21 = 0 , i.e., A = A 11 A 12 0 A 22 We then have to find a block matrix B = B 11 B 12 B 21 B 22 such that AB = A 11 B 11 + A 12 B 21 A 11 B 12 + A 12 B 22 A 22 B 21 A 22 B 22 = I n 1 0 0 I n 2 = I n 1 + n 2 .
S UMMARY 24 Thus if A - 1 22 exists the second row implies that B 21 = 0 n 2 n 1 and B 22 = A - 1 22 . Furthermore, A 11 B 11 + A 12 B 21 = I implies B 11 = A - 1 11 . At last, A 11 B 12 + A 12 B 22 = 0 implies B 12 = - A - 1 11 A 12 A - 1 22 . Hence we find A 11 A 12 0 A 22 - 1 = ˆ A - 1 11 - A - 1 11 A 12 A - 1 22 0 A - 1 22 ! — Summary • A matrix is a rectangular array of mathematical expressions. • Matrices can be added and multiplied by a scalar componentwise. • Matrices can be multiplied by multiplying rows by columns.

Subscribe to view the full document.

• Fall '19

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern