Linear algebra and differential equations (Math 54): Lecture 13 - PowerPoint PPT Presentation

Example  0 1 0  Consider the matrix 0 0 1  .  1 0 0  − λ 1 0   = 1 − λ 3 Its characteristic equation is: 0 = det 0 − λ 1  1 0 − λ This has one real solution: λ = 1.   − 1 1 0 Eigenvectors of eigenvalue 1 form the kernel of 0 − 1 1  .  1 0 − 1 This is spanned by (1 , 1 , 1).

Example  0 1 0  Consider the matrix 0 0 1  .  1 0 0  − λ 1 0   = 1 − λ 3 Its characteristic equation is: 0 = det 0 − λ 1  1 0 − λ This has one real solution: λ = 1.   − 1 1 0 Eigenvectors of eigenvalue 1 form the kernel of 0 − 1 1  .  1 0 − 1 This is spanned by (1 , 1 , 1). There is no basis of real eigenvectors.

Example � � 0 1 Consider the matrix . − 1 0

Example � � 0 1 Consider the matrix . − 1 0 Its characteristic equation is:

Example � � 0 1 Consider the matrix . − 1 0 � − λ � 1 = λ 2 + 1. This Its characteristic equation is: 0 = det − 1 − λ has

Example � � 0 1 Consider the matrix . − 1 0 � − λ � 1 = λ 2 + 1. This Its characteristic equation is: 0 = det − 1 − λ has no real solutions, so there are no real eigenvalues or nonzero eigenvectors.

Example � � 0 1 Consider the matrix . − 1 0 � − λ � 1 = λ 2 + 1. This Its characteristic equation is: 0 = det − 1 − λ has no real solutions, so there are no real eigenvalues or nonzero eigenvectors. There is no basis of real eigenvectors.

Example � 1 � 1 Consider the matrix . 0 1

Example � 1 � 1 Consider the matrix . 0 1 Characteristic equation:

Example � 1 � 1 Consider the matrix . 0 1 � 1 − λ � 1 = λ 2 − 2 λ + 1. Characteristic equation: 0 = det 0 1 − λ

Example � 1 � 1 Consider the matrix . 0 1 � 1 − λ � 1 = λ 2 − 2 λ + 1. Characteristic equation: 0 = det 0 1 − λ So the eigenvalues are

Example � 1 � 1 Consider the matrix . 0 1 � 1 − λ � 1 = λ 2 − 2 λ + 1. Characteristic equation: 0 = det 0 1 − λ So the eigenvalues are λ = 1

Example � 1 � 1 Consider the matrix . 0 1 � 1 − λ � 1 = λ 2 − 2 λ + 1. Characteristic equation: 0 = det 0 1 − λ So the eigenvalues are λ = 1 The eigenvectors of eigenvalue 1

Example � 1 � 1 Consider the matrix . 0 1 � 1 − λ � 1 = λ 2 − 2 λ + 1. Characteristic equation: 0 = det 0 1 − λ So the eigenvalues are λ = 1 � 0 � 1 The eigenvectors of eigenvalue 1 are the kernel of . 0 0

Example � 1 � 1 Consider the matrix . 0 1 � 1 − λ � 1 = λ 2 − 2 λ + 1. Characteristic equation: 0 = det 0 1 − λ So the eigenvalues are λ = 1 � 0 � 1 The eigenvectors of eigenvalue 1 are the kernel of . This 0 0 is spanned by

Example � 1 � 1 Consider the matrix . 0 1 � 1 − λ � 1 = λ 2 − 2 λ + 1. Characteristic equation: 0 = det 0 1 − λ So the eigenvalues are λ = 1 � 0 � 1 The eigenvectors of eigenvalue 1 are the kernel of . This 0 0 is spanned by (1 , 0).

Example � 1 � 1 Consider the matrix . 0 1 � 1 − λ � 1 = λ 2 − 2 λ + 1. Characteristic equation: 0 = det 0 1 − λ So the eigenvalues are λ = 1 � 0 � 1 The eigenvectors of eigenvalue 1 are the kernel of . This 0 0 is spanned by (1 , 0). There is no basis of real eigenvectors.

Example � 1 � 0 Consider the matrix . 0 1

Example � 1 � 0 Consider the matrix . 0 1 Characteristic equation:

Example � 1 � 0 Consider the matrix . 0 1 � 1 − λ � 0 = λ 2 − 2 λ + 1. Characteristic equation: 0 = det 0 1 − λ

Example � 1 � 0 Consider the matrix . 0 1 � 1 − λ � 0 = λ 2 − 2 λ + 1. Characteristic equation: 0 = det 0 1 − λ So the eigenvalues are

Example � 1 � 0 Consider the matrix . 0 1 � 1 − λ � 0 = λ 2 − 2 λ + 1. Characteristic equation: 0 = det 0 1 − λ So the eigenvalues are λ = 1

Example � 1 � 0 Consider the matrix . 0 1 � 1 − λ � 0 = λ 2 − 2 λ + 1. Characteristic equation: 0 = det 0 1 − λ So the eigenvalues are λ = 1 The eigenvectors of eigenvalue 1

Example � 1 � 0 Consider the matrix . 0 1 � 1 − λ � 0 = λ 2 − 2 λ + 1. Characteristic equation: 0 = det 0 1 − λ So the eigenvalues are λ = 1 � 0 � 0 The eigenvectors of eigenvalue 1 are the kernel of . 0 0

Example � 1 � 0 Consider the matrix . 0 1 � 1 − λ � 0 = λ 2 − 2 λ + 1. Characteristic equation: 0 = det 0 1 − λ So the eigenvalues are λ = 1 � 0 � 0 The eigenvectors of eigenvalue 1 are the kernel of . This 0 0 is all vectors in R 2 .

Example � 1 � 0 Consider the matrix . 0 1 � 1 − λ � 0 = λ 2 − 2 λ + 1. Characteristic equation: 0 = det 0 1 − λ So the eigenvalues are λ = 1 � 0 � 0 The eigenvectors of eigenvalue 1 are the kernel of . This 0 0 is all vectors in R 2 . There is a basis of real eigenvectors.

Example d Consider the linear transformation dx : P n → P n .

Example d Consider the linear transformation dx : P n → P n . Constant polynomials are eigenvectors of eigenvalue zero.

Example d Consider the linear transformation dx : P n → P n . Constant polynomials are eigenvectors of eigenvalue zero. There are no others:

Example d Consider the linear transformation dx : P n → P n . Constant polynomials are eigenvectors of eigenvalue zero. There are no others: the derivative lowers the degree of a polynomial

Example d Consider the linear transformation dx : P n → P n . Constant polynomials are eigenvectors of eigenvalue zero. There are no others: the derivative lowers the degree of a polynomial so any eigenvector must have eigenvalue zero

Example d Consider the linear transformation dx : P n → P n . Constant polynomials are eigenvectors of eigenvalue zero. There are no others: the derivative lowers the degree of a polynomial so any eigenvector must have eigenvalue zero hence be a constant.

Example Or, we pick a basis, say 1 , x , x 2 , . . .

Example Or, we pick a basis, say 1 , x , x 2 , . . . and write the matrix of d dx .

Example Or, we pick a basis, say 1 , x , x 2 , . . . and write the matrix of d dx . I’ll write the case P 4 :

Example Or, we pick a basis, say 1 , x , x 2 , . . . and write the matrix of d dx . I’ll write the case P 4 :  0 1 0 0 0  0 0 2 0 0     0 0 0 3 0     0 0 0 0 4   0 0 0 0 0

Example Or, we pick a basis, say 1 , x , x 2 , . . . and write the matrix of d dx . I’ll write the case P 4 :  0 1 0 0 0  0 0 2 0 0     0 0 0 3 0     0 0 0 0 4   0 0 0 0 0 The characteristic equation of this matrix is − λ 5 = 0.

Example Or, we pick a basis, say 1 , x , x 2 , . . . and write the matrix of d dx . I’ll write the case P 4 :  0 1 0 0 0  0 0 2 0 0     0 0 0 3 0     0 0 0 0 4   0 0 0 0 0 The characteristic equation of this matrix is − λ 5 = 0. So 0 is the only eigenvalue.

Example Or, we pick a basis, say 1 , x , x 2 , . . . and write the matrix of d dx . I’ll write the case P 4 :  0 1 0 0 0  0 0 2 0 0     0 0 0 3 0     0 0 0 0 4   0 0 0 0 0 The characteristic equation of this matrix is − λ 5 = 0. So 0 is the only eigenvalue. The corresponding eigenspace is

Example Or, we pick a basis, say 1 , x , x 2 , . . . and write the matrix of d dx . I’ll write the case P 4 :  0 1 0 0 0  0 0 2 0 0     0 0 0 3 0     0 0 0 0 4   0 0 0 0 0 The characteristic equation of this matrix is − λ 5 = 0. So 0 is the only eigenvalue. The corresponding eigenspace is in this basis

Example Or, we pick a basis, say 1 , x , x 2 , . . . and write the matrix of d dx . I’ll write the case P 4 :  0 1 0 0 0  0 0 2 0 0     0 0 0 3 0     0 0 0 0 4   0 0 0 0 0 The characteristic equation of this matrix is − λ 5 = 0. So 0 is the only eigenvalue. The corresponding eigenspace is in this basis spanned by (1 , 0 , 0 , 0 , 0).

Example Or, we pick a basis, say 1 , x , x 2 , . . . and write the matrix of d dx . I’ll write the case P 4 :  0 1 0 0 0  0 0 2 0 0     0 0 0 3 0     0 0 0 0 4   0 0 0 0 0 The characteristic equation of this matrix is − λ 5 = 0. So 0 is the only eigenvalue. The corresponding eigenspace is in this basis spanned by (1 , 0 , 0 , 0 , 0). These are the constant polynomials.

When is there a basis of eigenvectors?

When is there a basis of eigenvectors? Theorem Let T : V → V be a linear transformation,

When is there a basis of eigenvectors? Theorem Let T : V → V be a linear transformation, and suppose v 1 , . . . , v n are nonzero eigenvectors, T v i = λ i v i

When is there a basis of eigenvectors? Theorem Let T : V → V be a linear transformation, and suppose v 1 , . . . , v n are nonzero eigenvectors, T v i = λ i v i If the eigenvalues λ i are distinct, then the eigenvectors v i are linearly independent.

When is there a basis of eigenvectors? Proof.

When is there a basis of eigenvectors? Proof. We check n = 2.

When is there a basis of eigenvectors? Proof. We check n = 2. Assume v 1 , v 2 are linearly dependent, i.e. c 1 v 1 + c 2 v 2 = 0 where c 1 , c 2 are constants not both zero. Applying T , we have λ 1 c 1 v 1 + λ 2 c 2 v 2 = 0 as well.

When is there a basis of eigenvectors? Proof. We check n = 2. Assume v 1 , v 2 are linearly dependent, i.e. c 1 v 1 + c 2 v 2 = 0 where c 1 , c 2 are constants not both zero. Applying T , we have λ 1 c 1 v 1 + λ 2 c 2 v 2 = 0 as well. If c 1 = 0,

When is there a basis of eigenvectors? Proof. We check n = 2. Assume v 1 , v 2 are linearly dependent, i.e. c 1 v 1 + c 2 v 2 = 0 where c 1 , c 2 are constants not both zero. Applying T , we have λ 1 c 1 v 1 + λ 2 c 2 v 2 = 0 as well. If c 1 = 0, then we can conclude v 2 = 0,

When is there a basis of eigenvectors? Proof. We check n = 2. Assume v 1 , v 2 are linearly dependent, i.e. c 1 v 1 + c 2 v 2 = 0 where c 1 , c 2 are constants not both zero. Applying T , we have λ 1 c 1 v 1 + λ 2 c 2 v 2 = 0 as well. If c 1 = 0, then we can conclude v 2 = 0, which is a contradiction since we assumed the v i � = 0.

When is there a basis of eigenvectors? Proof. We check n = 2. Assume v 1 , v 2 are linearly dependent, i.e. c 1 v 1 + c 2 v 2 = 0 where c 1 , c 2 are constants not both zero. Applying T , we have λ 1 c 1 v 1 + λ 2 c 2 v 2 = 0 as well. If c 1 = 0, then we can conclude v 2 = 0, which is a contradiction since we assumed the v i � = 0. So c 1 � = 0.

When is there a basis of eigenvectors? Proof. We check n = 2. Assume v 1 , v 2 are linearly dependent, i.e. c 1 v 1 + c 2 v 2 = 0 where c 1 , c 2 are constants not both zero. Applying T , we have λ 1 c 1 v 1 + λ 2 c 2 v 2 = 0 as well. If c 1 = 0, then we can conclude v 2 = 0, which is a contradiction since we assumed the v i � = 0. So c 1 � = 0. If λ 2 = 0, then λ 1 c 1 v 1 = 0