% problem notes 6.1, spring 2013 % Edited S2013 to include all edits since 2008 %% S2008, 6.1: 10, 22, 30, 40. One stapled package. % problem notes 6.1, Fall 2008 %% F2008: 6.1: 12, 20, 32, 36. One stapled package. ===================================================================== 6.1-10: A:=Matrix([[9,-10],[2,0]]); Find a basis for each eigenspace of A, that is, find the eigenvalues of A, and for each eigenvalue lambda, find a basis of eigenvectors for lambda. 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 the two 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 rref(B) gives the answers u= to the homogeneous problem Bu=0. The reduced system 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 derivative on t1 gives the eigenvector v1: [2] v1 = [ ] [1] lambda2=5: ========= The calculation 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 ] ANSWER: (4,<2,1>), (5,<5/2,1>) are the eigenpairs. The answer can also be written as lambda=4, v1=<2,1> and lambda2=5, v2=<5/2,1>. Because a scalar multiple of a eigenvecotr is also an eigenvector, the answers for v1, v2 can be multiplied by scalars to remove the fractios, which is the common display standard for MAPLE answers. So another set of valid answers is lambda=4, v1=<2,1> and lambda2=5, v2=<5,2>. MAPLE ANSWER CHECK. A:=Matrix([[9,-10],[2,0]]); linalg[eigenvects](A); # Learn how to decode the cryptic maple output. FOURIERS MODEL and AP=PD. The LinearAlgebra package calculates eigenpairs a different way. Output are two matrices, the first a column vector of eigenvalues, and the second a matrix of corresponding eigenvectors. The following code may be used to find the diagonal matrix DD of eigenvalues and the corresponding invertible matrix P of eigenvectors, for matrices A that satisfy Fourier's Model [A is diagonalizable; see section 6.2]: A:=Matrix([[9,-10],[2,0]]); ans1,ans2:=LinearAlgebra[Eigenvectors](A); DD:=LinearAlgebra[DiagonalMatrix](ans1); P:=ans2; Because D is the maple reserved symbol for differentiation, we use symbol DD instead of D. ===================================================================== 6.1-12: A:=Matrix([[13,-15],[6,-6]]); Find a basis for each eigenspace of A, that is, find the eigenvalues of A, and for each eigenvalue lambda, find a basis of eigenvectors for lambda. The work expected parallels 6.1-10, solved above. [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. Frame Sequence for lambda=3. =========================== [10 -15] B = [ ] = A - (3)I [ 6 -9] In your hand solution, please show the steps with combo, swap and mult. [1 -3/2] rref(B) = [ ] [0 0 ] The reduced echelon system for the homogeneous problem Bu=0, u=, 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 ] MAPLE ANSWER CHECK. A:=Matrix([[13,-15],[6,-6]]); linalg[eigenvects](A); ===================================================================== 6.1-20: A:=Matrix([[1,0,0],[-4,7,2],[10,-15,-4]]); Find a basis for each eigenspace of A, that is, find the eigenvalues of A, and for each eigenvalue lambda, find a basis of eigenvectors for lambda. The work expected parallels 6.1-10, solved above. [ 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 expand by cofactors on row 1, which preserves common factors. (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] In your hand solution, please show the steps with combo, swap and mult. [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. The other eigenvectors are found similary, one frame sequence for each distinct eigenvalue. MAPLE ANSWER CHECK. A:=Matrix([[1,0,0],[-4,7,2],[10,-15,-4]]); linalg[eigenvects](A); ===================================================================== 6.1-22: A:=Matrix([[5,-6,3],[6,-7,3],[6,-6,2]]); Find a basis for each eigenspace of A, that is, find the eigenvalues of A, and for each eigenvalue lambda, find a basis of eigenvectors for lambda. The work expected parallels 6.1-10, solved above. Because the matrix A is full, there is no advantage from computing det(A-lambda I) by cofactors, although this is the default method. A shortcut exists by using problem 6.1-40. The result of 6.1-40 is the formula det(A-rI)=(-r)^3+trace(A)(-r)^2+c(-r)+det(A) where c=sum of the minors along the diagonal of A =Minor(A,1,1)+Minor(A,2,2)+Minor(A,3,3) College Algebra finds the roots lambda=-1,-1,2 of the characteristic polynomial det(A - lambda I) = 0 or (used problem 6.1-40) -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.b For lambda = -1. ============== [6 -6 3] B = [6 -6 3] = A - (-1)I [6 -6 3] In your hand solution, please show the steps with combo, swap and mult. [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. For lambda = 2. ============== [3 -6 3] B = [6 -9 3] = A - (2)I [6 -6 0] In your hand solution, please show the steps with combo, swap and mult. [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. MAPLE ANSWER CHECK. A:=Matrix([[5,-6,3],[6,-7,3],[6,-6,2]]); linalg[eigenvects](A); ===================================================================== 6.1-30: A:=Matrix([[0,-12],[12,0]]); Find a basis for each eigenspace of A, that is, find the eigenvalues of A, and for each eigenvalue lambda, find a basis of eigenvectors for lambda. The eigenpairs will be complex. 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! MAPLE ANSWER CHECK. A:=Matrix([[0,-12],[12,0]]); linalg[eigenvects](A); ===================================================================== 6.1-32: A:=Matrix([[0,-4],[36,0]]); Find a basis for each eigenspace of A, that is, find the eigenvalues of A, and for each eigenvalue lambda, find a basis of eigenvectors for lambda. The eigenpairs will be complex. 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! MAPLE ANSWER CHECK. A:=Matrix([[0,-4],[36,0]]); linalg[eigenvects](A); ===================================================================== 6.1-36: Show that the eigenvalues of triangular nxn matrix are its diagonal elements. This is a proof and as such it does not have an answer check. Read the following to get an idea of how the proof was discovered and written. 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. Present the case for matrix A upper triangular. PLAN. Then matrix A has 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. IDEA. The way to discover the proof is to look a lower dimension, like 3x3, and then generalize the proof from the special case to the setting nxn. We consider an upper triangular 3x3 matrix like A:=Matrix([[1,2,3],[0,4,5],[0,0,6]]); The plan is to compute by hand for this special case the eigenvalues of A. Subtract lambda from the diagonal of A and take the determinant. Only the triangular rule is needed because A-lambda I is triangular (no swap, combo, mult required). What you get is product of disg elements = 0 (1-lambda)(4-lambda)(6-lambda)=0 lambda = a diag element of matrix A GENERALIZATION. To generalize from 3x3 to nxn involves inventing symbols and writing dots (...) in displays. For example, if the ideas were formed from 3x3, then write the proof with symbols a_{ij} inserted, e.g., A:=Matrix([[1,2,3],[0,4,5],[0,0,6]]); original example A:=Matrix([[a11,a12,a13],[0,a22,a23],[0,0,a33]]); Replace constants by symbols A:=Matrix([[a11,a12,...,a1n],[0,a22,...,a2n],...,[0,0,...,ann]]); Insert ... and change 3 to n throughout. ===================================================================== 6.1-40: A:=Matrix([[32,-67,47],[7,-14,13],[-7,15,-6]]); Find the characteristic equation, eigenvalues and eigenvectors of the 3x3 matrix A. The result of this problem is the 3x3 formula det(A-rI) = (-r)^3 + trace(A)(-r)^2 + (minor(A,1,1)+minor(A,2,2)+minor(A,3,3))(-r)+det(A) DETAILS. ======= For A:=Matrix([[32,-67,47],[7,-14,13],[-7,15,-6]]) this formula reads det(A-rI)=-r^3+12r^2+(84-13(15)-32(6)+7(47)-14(32)+7(67)](-r)+det(A) =-r^3+12r^2+(-111+137+21)(-r)+60 =-r^3+12r^2-47r+60 The maple solve command can be used to find the roots. Also possible is the rational root theorem, checking the factors of 60 divided by the factors of 1, giving a finite but rather large list. Guessing is allowed, we could guess r=3 is a root, then divide by r-3 to reduce the problem of roots to the quadratic formula, applied to the quotient. MAPLE ANSWER CHECK. solve(-r^3+12*r^2-47*r+60=0,r); r = 4, 5, 3 A:=Matrix([[32,-67,47],[7,-14,13],[-7,15,-6]]); LinearAlgebra[Eigenvectors](A); # lambda1=4, lambda2=5, lambda3=3 # v1=[5/7, -1/2, 3], v2=[1, 1/2, 2], v3=[1, 1, 1]] Solution using the book's hint. ============================== The results of problems 37-38 give for x=lambda the polynomial equation -x^3 + [32-14-6](-x)^2 + c (-x) + det(A)=det(A- x I) Substitute the sample value x=1 to get an equation for c, then solve it. Evaluate det(A) and det(A-I) to finish the problem. Answers: det(A)=60, A-I=Matrix([[31, -67, 47], [7, -15, 13], [-7, 15, -7]]) det(A-I)=24 Equation for c1: -1+12-c+det(A)=det(A-I) -1+12-c+60=24 c=71-24=47 REMARK. It is known that c equals the the sum of the minors of A along the diagonal (However, c1 in the problem statement equals -c) c = minor(A,1,1)+minor(A,2,2)+minor(A,33) To verify that fact, it is necessary to learn about differentiation of determinants. This is because (d/dx)det(A-xI) at x=0 equals -c. %%===end of 6.1 problem notes======