Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: hentzel@iastate.edu http://www.math.iastate.edu/hentzel/class.307.ICN Text: Linear Algebra With Applications, Second Edition Otto Bretscher ============================================================ Friday, Apr 23 Chapter 8.2 Page 381 Problems 18,20,22 Main Idea: Use the Eigen vectors for the x and y axes. Key Words: Rotate the axes. Goal: Learn how to put a quadratic equation in standard form. ============================================================== Previous Assignment Wednesday, April 23 Chapter 7.4 Page 337 Problems 16,20,34.36 --------------------------------------------------------- Page 337 Problem 16 Diagonalize: | 4 0 -2 | A = | 0 1 0 | | 1 0 1 | | 4-x 0 -2 | Det[A-x I] = | 0 1-x 0 | | 1 0 1-x | = (1-x)( 4-5x+x^2 +2) = (1-x)( x^2 - 5 x + 6) = (1-x)(x-2)(x-3) Eigen values 1,2,3. x=1 | 3 0 -2 | | 1 0 0 | | 0 | A-I = | 0 0 0 | = | 0 0 1 | V1 = | 1 | | 1 0 0 | | 0 0 0 | | 0 | x=2 | 2 0 -2 | | 1 0 -1 | | 1 | A-2I = | 0 -1 0 | = | 0 1 0 | V2 = | 0 | | 1 0 -1 | | 0 0 0 | | 1 | x=3 | 1 0 -2 | | 1 0 -2 | | 2 | A-2I = | 0 -2 0 | = | 0 1 0 | V3 = | 0 | | 1 0 -2 | | 0 0 0 | | 1 | P = | 0 1 2 | | 1 0 0 | | 0 1 1 | -1 | 0 1 2 | | 4 0 -2 | | 0 1 2 | P^(-1) A P = | 1 0 0 | | 0 1 0 | | 1 0 0 | | 0 1 1 | | 1 0 1 | | 0 1 1 | | 0 1 0 | | 0 2 6 | = | -1 0 2 | | 1 0 0 | | 1 0 -1 | | 0 2 3 | | 1 0 0 | = | 0 2 0 | It checks | 0 0 3 | Page 337 Problem 20 Diagonalize: | 1 0 1 | A = | 1 1 1 | | 1 0 1 | | 1-x 0 1 | Det[A-x I} = | 1 1-x 1 | | 1 0 1-x | = (1-x)( 1 - 2 x + x^2 -1) = (1-x)(x^2-2x) = (1-x)x(x-2) eigen values 0,1,2. x = 0 | 1 0 1 | | 1 0 1 | | -1 | A-0I = | 1 1 1 | ~ | 0 1 0 | V1 = | 0 | | 1 0 1 | | 0 0 0 | | 1 | x = 1 | 0 0 1 | | 1 0 0 | | 0 | A-I = | 1 0 1 | ~ | 0 0 1 | V2 = | 1 | | 1 0 0 | | 0 0 0 | | 0 | x = 1 |-1 0 1 | | 1 0 -1 | | 1 | A-2I = | 1-1 1 | ~ | 0 -1 2 | V2 = | 2 | | 1 0-1 | | 0 0 0 | | 1 | | -1 0 1 | P = | 0 1 2 | | 1 0 1 | -1 | -1 0 1 | | 1 0 1 | | -1 0 1 | P^(-1) A P = | 0 1 2 | | 1 1 1 | | 0 1 2 | | 1 0 1 | | 1 0 1 | | 1 0 1 | |-1/2 0 1/2 | | 0 0 2 | | -1 1 -1 | | 0 1 4 | | 1/2 0 1/2 | | 0 0 2 | | 0 0 0 | | 0 1 0 | It checks | 0 0 2 | ======================================================== Page 337 Problem 34 | 1/2 1/4 | | 1/2 3/4 | Det[A-xI] = | 1/2-x 1/4 | = 3/8 -5/4 x + x^2 -1/8 | 1/2 3/4-x | = x^2 -5/4 x + 1/4 = (x-1)(x-1/4) eigen values 1, 1/4 x^n = (x^2 - 5/4 x + 1/4) q(x) + a x + b 1 = a + b 1/4^n = a/4 + b 1-1/4^n = 3/4 a a = 4/3( 1-1/4^n) 1-4/4^n = -3 b b = -1/3(1-4/4^n) A^n = 4/3(1-1/4^n) A - 1/3(1-4/4^n)I 4 A -I -4 A +4 I A^n = ---------- + 1/4^n ----------- 3 3 | 1 1 | | 2 -1 | A^n = | 2 2 | + 1/4^n | -2 1 | ------------ -------------- 3 3 n | 1 | | 1 | A | 2 | = | 2 | -------------------------------------------------- Page 337 Problem 36 Is matrix | -1 6 | similar to | 3 0 | | -2 6 | | 0 2 | Same trace and same determinant so it is possible. In fact, since the eigen values are distinct, it can be diagonalized and the diagonal form must have a 3 and a 2 on the diagonal. By putting the vector for 3 first and the vector for 2 second, we will end up with | 3 0 |. Therefore, they | 0 2 | are similar. Or, doing it the long hard way ... Det[A-x I ] = | -1-x 6 | | -2 6-x | = -6 -5x+x^2+12 = x^2 - 5 x + 6 = (x-2)(x-3) x=2 A-2I = | -3 6 | ~ | 1 -2 | V1 = | 2 | | -2 4 | | 0 0 | | 1 | x=3 A-3I = | -4 6 | ~ | 1 -3/2 | V2 = | 3/2 | | -2 3 | | 0 0 | | 1 | P = | 3 2 | | 2 1 | -1 P^(-1) A P = | 3 2 | | -1 6 | | 3 2 | | 2 1 | | -2 6 | | 2 1 | = - | 1 -2 | | 9 4 | |-2 3 | | 6 2 | = - | -3 0 | | 0 -2 | = | 3 0 | it checks. | 0 2 | ------------------------------------------------ New Material: Rotation of Axes. Page 379 Example 4. 8 x^2 - 4 x y + 5 y^2 = 1 Write the equation in matrix form as | x y | | 8 -2 | | x | = 1 |-2 5 | | y | Note that the xy term is split into the xy and yx entries of the matrix. ================================================== The theory needs this fact: If A is symmetric, then we can find a complete set of eigen vectors. Moreover, we can find a complete set of mutually orthogonal eigenvectors. Therefore you can find an orthogonal matrix P such that -1 P A P is diagonal. ===================================================== Now given that A is symmetric, let P be an orthogonal matrix of eigen vectors of A. In the equation X^T A X = 1. We substitute X = PW and the equation becomes (PW)^T A PW = 1 W^T (P^T A P) W = 1 W^T (P^(-1) A P) W = 1 Since P^(-1) A P is diagonal, the new equation is a u^2 + b v^2 + c w^2 = 1 ------------------------------------------------------ We proceed with the example: (x,y) | 8 -2 | | x | = 1 |-2 5 | | y | Det[A-xI] = | 8-x -2 | = 40 -13 x + x^2 - 4 | -2 5-x | = x^2 -13 x + 36 = (x-4)(x-9) x = 4 A-4I = | 4 -2 | V1 = | 1 | | -2 1 | | 2 | x = 9 A-9I = | -1 -2 | V2 = | 2 | | -2 -4 | |-1 | V1 and V2 are going to be orthogonal. They will make an orthonormal basis of the space. P = 1/Sqrt[5] | 1 2 | | 2 -1 | So P^(-1) is P^(T). Make the substitution |x | = 1/Sqrt[5] | 1 2 | | u | |y | | 2 -1 | | v | [u v] 1/Sqrt[5] | 1 2 | | 8 -2 | 1/Sqrt[5] | 1 2 | | u | = 1 | 2 -1 | | -2 5 | | 2 -1 | | v | 1/5 [u,v] | 4 8 | | 1 2 | | u | = 1 |18 -9 | | 2 -1 | | v | 1/5 [u,v] | 20 0 | | u | = 1 | 0 45 | | v | [u,v] | 4 0 | | u | = 1 | 0 9 | | v | 4 u^2 + 9 v^2 = 1 <== Same graph but equation is for a different basis. u^2 v^2 ---- + ---- = 1 (1/2)^2 (1/3)^2 v ... . | . . 1/3 . . | . --:--1/2---+--1/2---:----- u . 1/3 . . | . ... To find the location with respect to the original axis. We have to locate the new axes. |x | = 1/Sqrt[5] | 1 2 | | u | |y | | 2 -1 | | v | u in direction | 1 | . / | 2 | . / . / . / . | / . | / .' ' .| / . . |. / . . | ./ . . | / . -----------------------.----+----.--------------------------- . .|\ . .. | \ . .. | \ . . .| \ . . . \ . . | . . \ | \ direction | 2 | \ |-1 | ------------------------------------------------------------------------ ans = Solve[8 x^2 - 4 x y + 5 y^2 == 1,y]; f[x_] = y /. ans[[1]]; g[x_] = y /. ans[[2]]; U = 1/Sqrt[5] {1,2}; V = 1/Sqrt[5] {2,-1}; a = Plot[{f[x],g[x]},{x, -1,1}]; b = ParametricPlot[ t U {1, 2},{t,0,1},PlotStyle->{RGBColor[1,0,0}]; c = ParametricPlot[ t V,{t,0,1},PlotStyle->{RGBColor[1,0,0}]; bns = Solve[4 u^2 + 9 v^2 == 1, v]; h[u_] = v /. bns[[1]]; k[u_] = v /. bns[[2]]; d = ParametricPlot[ h[u] U + k[u] V,{u,-1/3,1/3},PlotStyle->{RGBColor[1,0,0]}] e = Show[a,b,c,d,PlotLabel->" 8 x^2 - 4 x y + 5 y^2 = 1"); ---------------------------------------------------------------------------- Friday, Apr 25 Chapter 8.2 Page 381 Problems 18,20,22 Page 381 Problem 18 9 x^2 - 4 x y +6 y^2 = 1 (a) Express this in matrix Form. (b) What is the u-v equation? (c) What are U and V? ---------------------------------------------------- eq = 9 x^2 - 4 x y +6 y^2 ans = Solve[eq==1,y]; f1[x_] = y /. ans[[1]]; f2[x_] = y /. ans[[2]]; a = Plot[ f1[x],{x,-10,10},PlotPoints->100]; b = Plot[ f2[x],{x,-10,10},PlotPoints->100]; c = Show[a,b,AspectRatio->Automatic,PlotLabel->"9 x^2 - 4 x y +6 y^2 = 1"]; Display["18.ps",c]; ------------------------------------------------------- Page 381 Problem 20 -3x^2 + 6 x y + 5 y^2 = 1 (a) Express this in matrix Form. (b) What is the u-v equation. (c) What are U and V? ------------------------------------------------------ eq = -3x^2 + 6 x y + 5 y^2 ans = Solve[eq==1,y]; f1[x_] = y /. ans[[1]]; f2[x_] = y /. ans[[2]]; a = Plot[ f1[x],{x,-10,10},PlotPoints->100]; b = Plot[ f2[x],{x,-10,10},PlotPoints->100]; c = Show[a,b,AspectRatio->Automatic,PlotLabel->"-3x^2 + 6 x y + 5 y^2 = 1"]; Display["20.ps",c]; ------------------------------------------------------- Page 381 Problem 22 -x^2 + y^2 -z^2 + 10 x z = 1 (a) Express this in matrix Form. (b) What is the u-v equation. (c) What are U and V? ----------------------------------------------------- eq = -x^2 + y^2 -z^2 + 10 x z ; ans = Solve[eq==1,z]; f1[x_,y_] = z /. ans[[1]]; f2[x_,y_] = z /. ans[[2]]; a = ParametricPlot3D[ {1/3 r Cos[t], r Sin[t],f1[ 1/3 r Cos[t], r Sin[t]]}, {r, 0 ,2},{t, 0,2 Pi},PlotPoints->50]; b = ParametricPlot3D[ {1/3 r Cos[t], r Sin[t], f2[1/3 r Cos[t] ,r Sin[t] ] }, {r, 0, 2},{t, 0,2 Pi}, PlotPoints->100]; c = Show[a,b,AspectRatio->Automatic,PlotLabel->"-x^2 + y^2 -z^2 + 10 x z =1"]; Display["22.ps",c]; ------------------------------------------------------- In general the equations are x^2 y^2 --------- + -------- = 1 ellipse a^2 b^2 x^2 y^2 ------- - -------- = 1 hyperbola a^2 b^2 x^2 y^2 - ------ + -------- = 1 hyperbola. a^2 b^2 ----------------------------------------------------- In each case make the box. o \* | * / o o \ | / o o \ * | * / o \ * | * / o o \ ____*____ / o |\ . '|` . /| o | ' b . | o |' \ | / .| o . \ | / . o. \|/ . o ------------:--a--+--a--:------------------- o. /|\ . o / | \ o |` / b \ '| o | ` | ' | o |/_` ..._'_\| o o / * \ o / * | * \ o o / * | * \ o o / | \ o / * | *\ o Notice that the sign pattern of the equation can be determined from the determinant. If the diagonal form of the matrix is | c1 0 | | 0 c2 | Then it is an ellipse if c1 c1 > 0 and is a hyperbola if c1 c2 < 0. It is a parabola if c1 c2 = 0. Since c1 c2 is the determinant, we know immediately which type of curve we have since a x^2 + b x y + c y^2 has determinant >0 it is an ellipse. a c - b^2/4 Thus if ac-b^2/4 <0 it is a hyperbola. =0 it is a parabola. ---------------------------------------------------------------