Math 307 Spring, 2005 Hentzel Time: 10:00 to 10:50 MWF Room: 205 Carver Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: hentzel@iastate.edu Friday, April 1, 2005 5.1 and 5.2 Eigen values and the characteristic equation Main Idea: How do you tell what a matrix is going to do? Key Words: Characteristic Polynomial, Characteristic Roots, Characteristic Vectors, Characteristic Values, Eigen Values Eigen Vectors. p(x) = Det[ A - xI] Goal: Learn to compute eigen values and eigen vectors. Previous Assignment: Three people work together to build themselves three houses. Here is a table showing how much time each worked on his own house and the other persons houses. Thus A worked 70% of the time on his own house, and 20% of the time on B, and only 10% of the time on C. A B C A .7 .1 .1 B .2 .8 .2 C .1 .1 .7 Who got the better deal? How much better. You are asked to suggest how to even things out. We assume that the value of the house is the value of the labor going into it. We assume the wage of the earner is determined by the value of the house he received. Therefore: We let $A, $B, $C be the values of the three houses. And consequently $A, $B and $C will the value of the three laborers. | .7 .1 .1 | | $A | | $A | | .2 .8 .2 | | $B | = | $B | | .1 .1 .7 | | $C | | $C | <---Labor in house--> = The steady state vector for the matrix is | 1 | | 2 | | 1 | Thus all things considered, B earns twice as much as A and also twice as much as C. ------------------------------------------------------------ RowReduce[1/10{{7,1,1},{2,8,2},{1,1,7}}-IdentityMatrix[3]]; ------------------------------------------------------------ New Material: It is hard to tell what a matrix does by just looking at it. Some matrices though are transparent. | 0 0 0 | | 0 0 0 | just annihilates everything. | 0 0 0 | | 3 0 0 | | 0 3 0 | just triples everything. | 0 0 3 | | 1 0 0 | Fixes the x component, | 0 2 0 | doubles the y component | 0 0 3 | triples the z component. | a | | a | | b |-------->| 2b | | c | | 3c | ===================================================== | 2 0 0 | | 0 0 1 | doubles the x component and | 0 1 0 | switches the y and z components. | a | | 2a | | b |-------->| c | | c | | b | | 0 1 0 | Collapses 3 space into X-Y plane | 0 0 1 | Collapses the X-Y plane to the X axis | 0 0 0 | Collapses the X-axis. Thus in 3 steps, everything goes to zero. ============================================================ One way to study a matrix is to apply it successively to a vector and see what happens to the vector. If you try to watch what is happening, it is usually disappointing because the vector goes off to infinity. The vectors can spiral along an axis as they go off to infinity. Or they can simply collapse towards a common line. ========================================================= The way to examine the action of a matrix is through its eigen values and eigen vectors. Given a matrix A, if AV = cV for some non zero V, then V is called an eigen vector and A is called an eigen value. ==================================================== Suppose c V , c V , c V are eigen values and 1 1 2 2 3 3 vectors of A. Then if W = b V + b V + b B 1 1 2 2 3 3 n n n n A W = b c V + b c V + b c B . 1 1 1 2 2 2 3 3 3 What happens if b1 > b2 > b3, then the n contribution of c quickly over shadows the 1 n n contributions of c and c and so it appears 2 3 n like the A W approaches the line through V . 1 ========================================================== Find the eigen values and eigen vectors of the matrix | 8 9 9 | A = | 3 2 3 |. |-9 -9 -10 | AX = cX means (A-cI)X = 0 and so Det[A-cI] = 0. Thus c is a root of the polynomial Det[A-xI] = 0 ===================================================== | 8-x 9 9 | A-xI = | 3 2-x 3 |. |-9 -9 -10-x| | 8-x 9 9 | Det[ A-xI] = | 3 2-x 3 |. |-9 -9 -10-x| | 8-x 9 0 | Det[ A-xI] = | 3 2-x 1+x |. |-9 -9 -1-x | | 8-x 9 0 | Det[ A-xI] = (1+x) | 3 2-x 1 |. | -9 -9 -1 | | 8-x 9 0 | Det[ A-xI] = (1+x) | 3 2-x 1 |. |-6 -7-x 0 | | 8-x 9 | Det[ A-xI] = -(1+x)|-6 -7-x|. 2 Det[ A-xI] = -(1+x) ( -56 -x+x +54 ) 2 Det[ A-xI] = -(1+x) ( x - x -2 ) Det[A-xI] = -(1+x)(x+1)(x-2) --------------------------------------------- The Eigen Values are -1,-1, 2 We now solve for the eigen vectors. Eigen Value -1 | 9 9 9 | A +I = | 3 3 3 |. |-9 -9 -9 | RCF[A+I] = | 1 1 1 | | 0 0 0 | | 0 0 0 | The eigen vectors are | -1 | | -1 | | 1 | | 0 | | 0 | | 1 | | 8 9 9 | | -1 -1 | | 1 1 | A [V1 V2] = | 3 2 3 | | 1 0 | = | -1 0 | |-9 -9 -10 | | 0 1 | | 0 -1 | ====================================================== Eigen Value 2 | 6 9 9 | A - 2I = | 3 0 3 |. |-9 -9 -12 | === | 1 0 1 | | 0 9 3 | | 0 -9 -3 | === | 1 0 1 | | 0 1 1/3| | 0 0 0 | | -1 | | -1/3 | | 1 | ================================================= | -3 | V = | -1 | 3 | 3 | | 8 9 9 || -3 | | -6 | | -3 | A V = | 3 2 3 || -1 | = | -2 | = 2 | -1 | 3 |-9 -9 -10 || 3 | | 6 | | 3 | ===================================================== | -1 0 0 | A [ V V V ] = [ V V V ] | 0 -1 0 | 1 2 3 1 2 3 | 0 0 2 | Let P = [ V V V ] then 1 2 3 -1 | -1 0 0 | P A P = | 0 -1 0 | | 0 0 2 |