% problem notes 6.1, Fall 2008 %%6.1: 12, 20, 32, 36. One stapled package. 6.1-10: The eigenvalues lambda1, lambda2 of A are scale factors that represent a system of units along two independent axes v1, v2. The latter are called the eigenvectors in Fourier's model A(c1 v1 + c2 v2) = c1 lambda1 v1 + c2 lambda2 v2 Find eigenvalues from the determinant equation det(A - lambda I)=0 The expansion of the determinant is a quadratic polynomial in the variable lambda. The two roots of this quadratic are the answers lambda1=4 and lambda2=5. lambda^2 -9 lambda + 20 = 0 The eigenvectors v1, v2 are found from frame sequence calculations that end with the last frame algorithm, which takes partial derivatives on invented symbols t1, t2, ... The partial derivative answers are a basis for the solution space of the system. The computed vector basis answers are the eigenvectors. lambda1=4: Compute B = A - lambda1 I = matrix A with 4 subtracted from its diagonal [ 5 -10 ] B = [ ] [ 2 -4 ] This is frame 1. Apply combo, swap, mult to find rref(B). [ 1 -2 ] rref(B)= [ ] [ 0 0 ] The reduced echelon system for rref(B) is x -2y=0 0=0 The free variable is y. Assign symbol t1 to the free variable, then back-sub to get the general solution x = 2 t1 y = t1 The partial on t1 gives the eigenvector v1: [2] v1 = [ ] [1] The caclulation for v2 is similar. Subtract 5 from the diagonal of A to get matrix B, then form a frame sequence to rref(B), and apply the last frame algorithm to obtain [5/2] v2 = [ ] [ 1 ] ===================================================================== 6.1-12: The work expected parallels 6.1-10. [13 -15] A = [ ] [ 6 -6] Find eigenvalues from the determinant equation det(A - lambda I)=0 lambda^2 - 7 lambda + 12 = 0 (lambda - 3)(lambda - 4) = 0 The expansion of the determinant is a quadratic polynomial in the variable lambda. The two roots of this quadratic are the answers lambda1=3 and lambda2=4. The eigenvectors v1, v2 are found from frame sequence calculations that end with the last frame algorithm, which takes partial derivatives on invented symbols t1, t2, ... The partial derivative answers are a basis for the solution space of the system. The computed vector basis answers are the eigenvectors. Sequence for lambda=3. [10 -15] B = [ ] = A - (3)I [ 6 -9] [1 -3/2] rref(B) = [ ] [0 0 ] The reduced echelon system for rref(B) is x -3y/2=0 0=0 The free variable is y. Assign symbol t1 to the free variable, then back-sub to get the general solution x = 3 t1/2 y = t1 The partial on t1 gives the eigenvector v1: [3/2] v1 = [ ] [ 1 ] The caclulation for v2 is similar. Subtract 4 from the diagonal of A to get matrix B, then form a frame sequence to rref(B), and apply the last frame algorithm to obtain [5/3] v2 = [ ] [ 1 ] ===================================================================== 6.1-20: The work expected parallels 6.1-10. [ 1 0 0] A =[-4 7 2] [10 -15 -4] College Algebra finds the roots lambda=1,1,2 of the characteristic polynomial det(A - lambda I) = 0 by cofactors on row 1 (1-lambda)(lambda^2-3 lambda + 2) = 0 (1-lambda)(lambda-1)(lambda-2) = 0 Frame sequence calculations must appear as follows. For lambda = 1. [ 0 0 0] B =[-4 6 2] = A - (1)I [10 -15 -5] [1 -3/2 1/2] rref(B) = [0 0 0 ] [0 0 0 ] Apply the last frame algorithm. The general solution is then x1 = 3 t1/2 - t2/2, x2 = t1, x3 = t2. Partials on symbols t1 and t2 give two basis vectors v1, v2 which are the eigenvectors for lambda = 1. For lambda = 2. [-1 0 0] B =[-4 5 2] = A - (2)I [10 -15 -6] [1 0 0 ] rref(B) = [0 1 2/5] [0 0 0 ] Apply the last frame algorithm. The general solution is then x1 = 0, x2 = -2 t1/5, x3 = t1. Partial on symbol t1 gives one basis vector v3, which is the eigenvector for lambda = 2. ===================================================================== 6.1-22: The work expected parallels 6.1-10: College Algebra finds the roots lambda=-1,-1,2 of the characteristic polynomial det(A - lambda I) = 0 or -lambda^3+3 lambda + 2 = 0 The rational root theorem and Decartes rule of signs combine to detect one positive root from the list L={1,-1,2,-2}. The positive root that works is lambda=2. Divide the cubic polynomial by lambda-2, which must divide perfectly by the root-factor theorem. This factors the cubic into a linear factor (lambda-2) and a quadratic factor. Factor the quadratic to find the two roots -1,-1. Frame sequence calculations must appear as follows. For lambda = -1. [6 -6 3] B = [6 -6 3] = A - (-1)I [6 -6 3] [1 -1 1/2] rref(B) = [0 0 0 ] [0 0 0 ] Apply the last frame algorithm. The general solution is then x1 = t1 - t2/2, x2 = t1, x3 = t2. Partials on symbols t1 and t2 give two basis vectors v1, v2 which are the eigenvectors for lambda = -1. [3 -6 3] B = [6 -9 3] = A - (2)I [6 -6 0] [1 0 -1 ] rref(B) = [0 1 -1 ] [0 0 0 ] Apply the last frame algorithm. The general solution is then x1 = t1, x2 = t1, x3 = t1. Partial on symbol t1 gives one basis vector v3, which is the eigenvector for lambda = 2. ===================================================================== 6.1-30: Details expected include college algebra to solve the characteristic equation det(A - lambda I)=0 for the two complex roots lambda = 12i, -12i A frame sequence for B = A -(12i)I is constructed to give [1 -i] rref(B) = [ ] [0 0] The last frame algorithm applies to find the general solution x1 = (i) t1, x2 = t1. Take the partial on symbol t1 to find eigenvector v1. Use i^2 = -1 to re-write the vector in a standard form, if needed. Eigenvector v2 for lambda = -12i is by the theory equal to the complex conjugate of v1 [replace i by -i in the formula for v1]. A second frame sequence is not needed! ===================================================================== 6.1-32: [0 -4] A = [ ] [36 0] Details expected include college algebra to solve the characteristic equation det(A - lambda I)=0 for the two complex roots lambda = 12i, -12i A frame sequence for B = A -(12i)I is constructed to give [1 -i] rref(B) = [ ] [0 0] The last frame algorithm applies to find the general solution x1 = (i/3) t1, x2 = t1. Take the partial on symbol t1 to find eigenvector v1. Use i^2 = -1 to re-write the vector in a standard form, if needed. Eigenvector v2 for lambda = -12i is by the theory equal to the complex conjugate of v1 [replace i by -i in the formula for v1]. A second frame sequence is not needed! ===================================================================== 6.1-36: Show that the eigenvalues of triangular nxn matrix are its diagonal elements. Background: Determinant theory, four rules (1) Triangular, (2) swap, (3) combo, (4) mult. Expected details: Write a matrix display for det(A - lambda I)=0, using symbols for the entries of A, such as found in display (4), section 6.1. Assume that A is upper triangular. Then all entries below the diagonal are zero. Apply the triangular rule to evaluate the determinant. Find all roots of the equation. The lower triangular case follows by citing the theorem on transposes, det(B)=det(B^T), applied to B=A - lambda I, B^T = A^T - lambda I. ===================================================================== 6.1-40: The results of problems 37-38 give for x=lambda the polynomial equation -x^3 + [32-14-6]x^2 + c1 x + det(A)=det(A- x I) Substitute the sample value x=1 to get an equation for c1, then solve it. Evaluate det(A) and det(A-I) to finish the problem. Its known that c1 equals the negative of the sum of the minors of A along the diagonal: - c1 = minor(A,1,1)+minor(A,2,2)+minor(A,33) %%===end of 6.1 problem notes======