Math 307 Spring, 2005 Hentzel Time: 10:00 to 10:40 MWF Room: 205 Carver Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: hentzel@iastate.edu http://orion.math.iastate.edu/hentzel/class.307.05 Text: Linear Algebra and its Applications, Third Edition David C. Lay Wednesday, February 9 2.5 Page 150 Problems 24, 25, 26 Main Idea: It is just finding the RCF Key Words: LU-decomposition, Back Substitution Goal: Learn the LU-decomposition. Previous Assignment Assignment Page 139 Problem 10, 13, 15, 19 Page 139 Problem 10 | I 0 0 | | I 0 0 | The inverse of | C I 0 | is | Z I 0 | | A B I | | X Y I | Find X,Y,Z. | I 0 0 | | C+Z I 0 | |A+BZ+X B+Y I | C+Z = 0 so Z = -C B+Y = 0 so Y = -B A+BZ+X = 0 so X = -A-BZ = -A+BC. | I 0 0 | | I 0 0 | The inverse of | C I 0 | is | -C I 0 | | A B I | |-A+BC -B I | --------------------------------------------------------- Page 139 Problem 13 | B 0 | Let A = | | Show that A is invertible if and | 0 C | only if B and C are invertible. If A is invertible then ... | B 0 | | X Y | | B X B Y | | | | | = | | | 0 C | | Z W | | C Z C W | So BX = I CW = I thus B and C are invertible. And whenever B and C are invertible, then the inverse is obviously | -1 | | B 0 | | | | -1 | | 0 C | =================================================== Page 139 Problem 15 Suppose A11 is invertible. Find X and Y such that | A11 A12 | | I 0 | | A11 0 | | I Y | | | = | | | | | | | A21 A22 | | X I | | 0 S | | 0 I | | A11 A12 | | A11 0 | | I Y | | | = | | | | | A21 A22 | | X A11 S | | 0 I | | A11 A12 | | A11 A11 Y | | | = | | | A21 A22 | | X A11 X A11 Y + S | -1 A12 = A11 Y so A11 A12 = Y -1 A21 = X A11 so A21 A11 = X -1 -1 A22 = X A11 Y + S So S = A22 - X A11 Y = A22 - A21 A11 A11 A11 A12 -1 S = A22 - A21 A11 A12 -------------------------------------------------------------- Page 139 Problem 19 Assume A-s In is invertible and view (8) as a system of two matrix equations. Solve the top equation for X and substitute into the bottom equation. The result is an equation of the form W(s) U = y, where W(s) is a matrix that depends on s. W(s) is called the transfer function of the system because it transforms the input U into the output Y. Find W(s) and describe how it is related to the partioned system matrix on the left side of (8). | A - sI B | | X | | 0 | | | | | = | | | C I | | U | | Y | (A-sI)X + BU = 0 -1 X = -(A-sI) B U C X + U = Y -1 - C (A-sI) B U + U = Y _ _ | -1 | | - C (A-sI) B + I | U = Y |_ _| <----- W(s)--------> ================================================================= In Class Assignment: ------------------------------------------------------------------------------- || || e|| || d || || *********************** **************************** || || f|| || c || || ----------------------- ------------------------------ | * | | * | | * | | ========== * =========== | | a * b | | * | | * | | * | | * | If you measure the cars passing through each of a,b,c,d,e,f can you determine the number of cars that went straight, turned right, or turned left from each approach. NE E NW W WS ES NE + E = c NW + W = e WS + ES = a NE + NW = b W + WS = d E + ES = f | 1 1 0 0 0 0 || NE | | c | | 0 0 1 1 0 0 || E | | e | | 0 0 0 0 1 1 || NW | = | a | | 1 0 1 0 0 0 || W | | b | | 0 0 0 1 1 0 || WS | | d | | 0 1 0 0 0 1 || ES | | f | | 1 1 0 0 0 0 c | | 0 0 1 1 0 0 e | | 0 0 0 0 1 1 a | | 1 0 1 0 0 0 b | | 0 0 0 1 1 0 d | | 0 1 0 0 0 1 f | A = { { 1, 1, 0, 0, 0, 0, c }, { 0, 0, 1, 1, 0, 0, e }, { 0, 0, 0, 0, 1, 1, a }, { 1, 0, 1, 0, 0, 0, b }, { 0, 0, 0, 1, 1, 0, d }, { 0, 1, 0, 0, 0, 1, f }}; NE E NW W WS ES |1 0 0 0 0 -1 -a + b + d - e| |0 1 0 0 0 1 a - b + c - d + e| |0 0 1 0 0 1 a - d + e| |0 0 0 1 0 -1 -a + d| |0 0 0 0 1 1 a| |0 0 0 0 0 0 -a + b - c + d - e + f| No solution unless -a+b-c+d-e+f = 0 that is a+c+e = b+d+f the number of cars entering the intersection equals the number of cars leaving the intersection. | NE | | -a+b +d-e | | 1 | | E | | a-b+c-d+e | | -1 | | NW | = | a -d+e | + t| -1 | | W | | -a +d | | 1 | | WS | | a | | -1 | | ES | | 0 | | 1 | The null space is NE+ W+ES - E-NW-WS So the counting arrangement cannot tell the difference if six cars switch directions. Naturally the numbers have to be non negative, but there is quit a range of possibilities that could produce the same counting pattern. ===================================================== The LU decomposition when you are lucky. Using the LU decomposition to solve AX = B. A = QR QR Factorization T A = U D V Singular Value Decomposition -1 A = P D P Spectral Factorization ---------------------------------------------------