MATH 221, Spring Homework 10 Solutions

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1 MATH 22, Spring 28 - Homework Solutions Due Tuesday, May Section 52 Page 279, Problem 2: 4 λ A λi = and the characteristic polynomial is det(a λi) = ( 4 λ)( λ) ( )(6) = λ 6 λ 2 +λ+2 The solutions to the equation λ 2 + λ + 2 = are λ =, λ = 2 Page 279, Problem 4: 8 λ 2 A λi = and the characteristic polynomial is det(a λi) = (8 λ)( λ) ()(2) = λ λ 2 λ + 8 The solutions to λ 2 λ + 8 = are λ = 9, λ = 2 Page 272, Problem 7: [ 5 λ A λi = 4 4 λ and the characteristic polynomial is det(a λi) = (5 λ)(4 λ) ()( 4) = λ 2 9λ + 2 The solutions to λ 2 9λ + 2 = are found using the quadratic formula λ = 9± 9 2 4()(2) 2() λ = 9 2 ± Because expression involves complex roots, there are no REAL eigenvalues Page 279, Problem 8: 4 λ A λi = and the characteristic polynomial is det(a λi) = ( 4 λ)( λ) ()(2) = λ 2 λ 2 +λ The solutions to λ 2 + λ = are λ = 5, λ = 2 Page 28, Problem 25a: /7 6 /7 /7 Because we know that v = is an eigenvector, compute Av 4/7 = = So, λ = must 4 7 4/7 4/7 be the eigenvalue corresponding to v 6 λ To find the other eigenvector, find the eignevalues of the matrix: A λi =, so the characteristic 4 7 λ polynomial is λ 2 λ + and the solutions to λ 2 λ + = are λ = and λ = Thus, the other eigenvector must correspond to λ = To find the other eigenvector, solve (A I)x = for the general solution: x = 4 4 x 2 Therefore, an eigenvector corresponding to λ = is Because eigenvectors corresponding to different eigenvalues are {[ linearly independent } (and two non-zero linearly independent vectors in R 2 must also span R 2 ), the set {v, v 2 } =, is a basis for R /7 4/7 2

2 Page 28, Problem 25b: Solve for c: x = v + cv 2 x v = cv 2 So, and x = v 4 v 2 [ 5 5 [ /7 4/7 = [ /4 /4 = 4 [ = 4 v 2 So, c = 4 Page 28, Problem 25c: To begin, realize that x k = A k x = A k (v 4 v 2) = A k v A k 4 v 2 = A k v 4 Ak v 2 Then, x = Av 4 Av 2 Remember the definition of an eigenvector: if v is an eigenvector corresponding to λ, then Av = λv Because v is an eigenvector corresponding [ to λ = and [ v 2 is an eigenvector [ corresponding to λ =, this equation /7 /4 9/2 can be rewritten as x = v 4 (v 2) = + = 4/7 /4 /2 Similarly, x 2 = A 2 v 4 A2 v 2 = [ A(Av ) [ 4 A(Av 2) = A(v [ ) 4 A(v 2) = Av 4 Av 2 = v 4 (v 2) = /7 9/4 87/2 v 4 ()2 v 2 This is equal to + = 4/7 9/4 /2 It is clear to see that that the formula for x k = v 4 ()k v 2 As k gets larger (tends to infinity), () k tends to Therefore, as k, x k v Section 5 Page 286, Problem 6: A matrix A of the form A = P DP where D is a diagonal matrix consisting of the eigenvalues of A has vectors that form a basis for the eigenspace in the column of P that correspond to the eigenvalue in D Therefore, the eigenvalues of A are and 4 The vectors corresponding to λ = that forms a basis for the eigenspace are columns and of the matrix P :, The vector corresponding to λ = 4 that forms a basis for the eigenspace is column 2 of the matrix P : Page 286, Problem 7: To diagonalize the matrix, first find the eigenvalues: det(a λi) = ( λ)( λ) 6() = λ 2 = λ = ± Then, find a basis for each eigenspace {} / / When λ =, (A I)x = x = x = x 2 So, is a basis {} 2 When λ =, (A + I)x = x = x 6 2 So, is a basis Then, these bases form the columns [ of P with the [ associated eigenvalue in the corresponding column of D (this is very important!): P =, D = 2

3 Page 287, Problem 2: Because the eigenvalues are given, we just need to find a basis for each eigenspace Note: Because there are only 2 distinct eigenvalues, the sum of the dimensions of the eigenspaces must equal in order for A to be diagonalizable When λ = 2, (A 2I)x = x = x 2 + x So,, is a basis When λ = 5, (A 5I)x = x So, is a basis Then, these bases form the columns of P with the associated eigenvalue in the corresponding column of D (this is very important!): P =, D = Page 287, Problem 2: Because the matrix is triangular, the eigenvalues are the entries on the diagonal: λ = 2, λ = (each with multiplicity 2) Note: Because there are only 2 distinct eigenvalues, the sum of the dimensions of the eigenspaces must equal 4 in order for A to be diagonalizable When λ = 2, (A 2I)x =, is a basis When λ =, (A I)x = basis x = x 2 x = x 4 + x So, So, is a Because the dimension of the basis corresponding to λ = is and the basis corresponding to λ = 2 is 2 and + 2 = 4, the matrix is not diagonalizable Page 287, Problem 2a: True or False: A is diagonalizable if A = P DP for some matrix D and some invertible matrix P FALSE: The matrix D needs to be a diagonal matrix (the notation D does not automatically denote a diagonal matrix) Page 287, Problem 2b: True or False: If R n has a basis of eigenvectors of A, then A is diaognalizable TRUE: Because A is an n n matrix (stated in the directions), this statement is true and follows from the Diagonalization Theorem on page 282 x =

4 Section 6 Page 6, Problem 2: w w = () + ( ) + 5( 5) = = 5 x w = 6() + 2( ) + ( 5) = = 5 x w w w = 5 5 = 7 Page 6, Problem 7: w = w w = 5 Page 6, Problem : First, compute the norm of the vector: 6( 6) + 4(4) + ( ) = = 6 6/ 6 Then, normalize the vector (multiply by the scalar 6 : 4/ 6 / 6 Page 6, Problem 4: First find u z = use the formula dist(u, z) = u z = (u z) (u z) = 4(4) + 4( 4) + 6( 6) = 68 = 2 7 Page 6, Problem 6: Vectors are orthogonal if the dot product of the vectors equals zero Compute u v = 2(2) + ( ) + 5() = So, the vectors are orthogonal Page 6, Problem 7: u v = ( 4) + 2() + 5( 2) + (6) = So, the vectors are orthogonal Page 7, Problem 2a: True or False: u v v u = TRUE: By Theorem, u v = v u, so by substitution u v u v = Page 7, Problem 2b: True or False: For any scalar c, cv = c v FALSE: As stated on page, cv = c v Page 7, Problem 2c: True or False: If x is orthogonal to every vector in a subspace W, then x is in W TRUE: This statement follows from the definition of Orthogonal Complements on page 4 (here, the set that spans W is W itself) Page 7, Problem 2d: 4

5 True or False: For any scalar c, u 2 + v 2 = u + v 2, then uand v are orthogonal TRUE: This statement is part of Theorem 2 in this section (the Pythagorean Theorem) Page 7, Problem 2e: True or False: For an m n matrix A, vectors in the null space of A are orthogonal to vectors in the row space of A TRUE: This statement is part of Theorem of this section Page 7, Problem 2: u v = 2( 7) + 5( 4) + (6) = u 2 = u u = 2(2) + 5( 5) + ( ) = v 2 = v v = 7( 7) + 4( 4) + 6(6) = u + v 2 = (u + v) (u + v) = 5( 5) + 9( 9) + 5(5) = Page 7, Problem : Suppose x is in both W and W Because W spans W and x W, xis orthogonal to every vector in W (by definition of orthogonal complements) Because x is orthogonal to every vector in W, that means x x =, which implies x = (by Theorem ) 5