Bachelor in Statistics and Business Universidad Carlos III de Madrid Mathematical Methods II María Barbero Liñán Homework sheet 4: EIGENVALUES AND EIGENVECTORS DIAGONALIZATION (with solutions) Year - Is λ = an eigenvalue of Is 4? Why or why not? 6 9 Solution: Yes, it is because it satisfies det(a ( )I ) = an eigenvector of Solution: Yes, it is because is 5? If so, find the eigenvalue 6 5 6 = The eigenvalue of the eigenvector 6 7 Is an eigenvector of 7? If so, find the eigenvalue 5 6 4 Solution: It is not an eigenvector because 6 7 7 = 5 λ for 5 6 4 any λ R 4 Find a basis for the eigenspace corresponding to each listed eigenvalue: 4 a) A =, λ = 5 4 b) A =, λ =, 7 6 4 c) A =, λ = 5 Solution: (a) V ( 5) = span R (b) V () = span R, V (7) = span R (c) V () = span R
5 Find the eigenvalues and eigenvectors of the following matrices: 4 5 (a) 6 (b) (c) 4 (d) 4 6 7 7 Solution: (a) V () = span R, V (4) = span R 6 7 (b) V (4) = span R, V () = span R, V ( ) = span R 5 (c) V () = span R, V (4) = span R, V ( ) = span R (d) V () = span R 6, V (7) = span R, V ( 4) = span R 8 6 List the real eigenvalues, repeated according to their multiplicities, of the following matrices: 5 5 (a) 6 (b) 6 6 5 5 Solution: (a) 5 has multiplicity of, and have both multiplicity (b),, 6, 5 All have multiplicity 7 It can be shown that the algebraic multiplicity of an eigenvalue λ is always greater than or equal to the dimension of the eigenspace corresponding to λ Find h in the matrix A below such that the eigenspace for λ = 4 is two-dimensional 4 A = h 4 4 Solution: h = 8 Let λ be an eigenvalue of an invertible matrix A Show that λ is an eigenvalue of A (Hint: Suppose a nonzero x satisfies Ax = λ x) Solution: As λ is an eigenvalue of an invertible matrix A, the eigenvalue λ is nonzero and there exist a nonzero vector x such that Ax = λx Multiply the equation by λ A, Thus λ is an eigenvalue of A λ A Ax = (λ λ)a x; λ x = A x
9 Let A = P DP Compute A 4 if P = Solution: A 4 = P D 4 P = 6 5 9 9 5 7 and D = Use the factorization A = P DP to compute A k, where k represents an arbitrary positive integer, if a a A = = (a b) b b Solution: A k = k a b ( = a k (a k b k ) b k ) The matrix A is factored in the form P DP Find the eigenvalues of A and a basis for each eigenspace A = 4 9 = 4 9 Solution: The eigenvalues are (with multiplicity ) and 4 A basis for the eigenspace of eigenvalue is given by, and a basis for the eigenspace of eigenvalue 4 is given by Diagonalize the following matrices and give the matrices P and D in the factorization A = P DP, if possible a) b) 4 c) d) e) 5
f ) Solution: (a) It is not diagonalizable because there is only one eigenvalue,, and it only has one eigenvector associated with (b) It is diagonalizable because it has two different eigenvalues: 5 and D = P = 4 (c) D =, P = 5 (d) D =, P = 5, (e) There is only one eigenvalue which has multiplicity This matrix is not diagonalizable in R (f) It is diagonalizable because the matrix has different eigenvalues D =, P = A is a matrix with two eigenvalues Each eigenspace is one-dimensional Is A diagonalizable? Why? Solution: No, A is not diagonalizable because we cannot have a basis of eigenvectors for R 4 A is a 7 7 matrix with three eigenvalues One eigenspace is two-dimensional, and one of the other eigenspace is three-dimensional Is it possible that A is not diagonalizable? Justify your answer Solution: If the remaining eigenspace has dimension strictly less than, then A is not diagonalizable because we cannot construct a basis of eigenvectors for R 7 5 Show that if A is both diagonalizable and invertible, then so is A Solution: If A is invertible, all the eigenvalues are nonzero If A is diagonalizable, there exist matrices P and D such that A = P DP Take the inverse of this equality, Thus A is diagonalizable A = (P DP ) = (P ) D P = P D P 4
6 Determine if the application matrix of the following linear transformations is diagonalizable a) F (x, y) = (y, x y), b) G(x, y) = (x y, x), c) H(x, y, z) = ((x + z), x z, x + z) Solution: (a) The matrix is diagonalizable with P = and D = (b) The matrix is not diagonalizable because it only has one eigenvalue,, and this eigenvalue has only one eigenvector associated with (c) The matrix is diagonalizable with P = and D = 4 7 Determine a basis for R such that the application matrix of the above linear transformation H(x, y, z) = ((x + z), x z, x + z) in that basis is diagonal Solution: In the basis given by the columns of P in 6(c), the application matrix of H is just D Additional exercises: D C Lay Linear algebra and its applications, Sections 5-54 5