Eigenvectors and Eigenvalues Repeated application 1 1 0 0 A - PowerPoint PPT Presentation

Eigenvectors and Eigenvalues

Repeated application 1   1 0 0 A = − 1 0 0   0 0 2     1 0 0 1 0 0 A 2 n + 1 = A 2 n = − 1 0 0 0 1 0     2 2 n + 1 2 2 n 0 0 0 0

General LA 2   − 9 14 4 A = − 2 3 0   − 18 22 9 A n =?

General LA 3       − 9 14 4 2 2 3 2 0 0  = − 2 − 3 3 0 1 0 1 0 0      − 18 22 9 2 3 4 0 0 1   − 1 − 2 3 2 − 2 − 1   − 3 2 2

General LA 4       − 9 14 4 2 2  = 2 3 0 − 2 1 1      − 18 22 9 2 2       − 9 14 4 2 2  = − 3 − 2 3 0 0 0      − 18 22 9 3 3       − 9 14 4 3 3  = 1 − 2 3 0 1 1      − 18 22 9 4 4

Division of polynomials 5 Theorem Let a ( x ) and b ( x ) be polynomials such that b ( x ) � = 0 ( x ) , then there are quotient q ( x ) and remainder r ( x ) polynomials such that a ( x ) = q ( x ) b ( x ) + r ( x ) satisfying deg r ( x ) < deg b ( x ) .

Corollaries 6 Corollary Let a ( x ) and b ( x ) = x − λ be polynomials, then a ( x ) = q ( x )( x − λ ) + a ( λ ) . Corollary a ( x ) has root λ if and only if x − λ divides a ( x ) .

Fundamental Theorem of Algebra 7 Theorem Every non-constant polynomial with complex coefficients has at least one complex root.

Eigenvalues and eigenvectors 8 Definition (eigenvalues and eigenvectors) Let A be a square matrix. A non-zero vector � u is an eigenvector for A if A � u = λ� u for some λ . The value λ is called eigenvalue for the eigenvector � u .

Eigenvalues and eigenvectors 9 A � u = λ� u is A � u = λ I � u u = � ( A − λ I ) � 0 Want linearly dependent columns!

Example 10 from       − 9 14 4 2 2  = 2 − 2 3 0 1 1      − 18 22 9 2 2 to       − 11 14 4 2 0  = − 2 − 2 3 1 0      − 18 22 7 2 0

Example 11 from       − 9 14 4 2 2  = − 3 − 2 3 0 0 0      − 18 22 9 3 3 to       − 6 14 4 2 0  = − 2 3 3 0 0      − 18 22 12 3 0

Example 12 from       − 9 14 4 3 3  = − 2 3 0 1 1      − 18 22 9 4 4 to       − 10 14 4 3 0  = − 1 − 2 3 1 0      − 18 22 8 4 0

Goal 13 Given   · · · a 11 a 12 a 1 n a 21 a 22 a 2 n   A =  . .  ... . .   . .   a n 1 a n 2 · · · a nn are the columns of A linearly dependent.

Computing Eigenvalues and Eigenvectors 14 1. compute the determinant of A − λ I 2. compute the roots λ 1 , . . . , λ n of the resulting polynomial, the n (possibly repeated) roots are the eigenvalues 3. for each eigenvalue i find the corresponding x = � eigenvectors by computing ( A − λ i I ) � 0

Example 15   − 9 14 4 A = − 2 3 0   − 18 22 9 det( A − λ I ) = λ 3 − 7 λ + 6 = ( λ − 2 ) · ( λ − 1 ) · ( λ + 3 ) so λ 0 = 2 λ 1 = 1 λ 2 = − 3

A − λ 0 I 16       − 9 − 11 14 4 1 0 0 14 4  − 2  = = 3 0 − 2 0 1 0 3 − 2 − 2     − 18 − 18 22 9 0 0 1 22 7 solve         − 11 14 4 0 2    | α ∈ C 3 − 2 − 2  � x = 0 ⇒  α 1     − 18 22 7 0 2 

A − λ 1 I 17       − 9 − 10 14 4 1 0 0 14 4  − 1  = − 2 − 1 − 2 3 0 0 1 0 3     − 18 − 18 22 9 0 0 1 22 8 solve         − 10 14 4 0 3    | α ∈ C  � 3 − 1 − 2 x = 0 ⇒  α 1     − 18 22 8 0 4 

A − λ 2 I 18       − 9 − 6 14 4 1 0 0 14 4  − ( − 3 )  = − 2 − 2 3 0 0 1 0 3 3     − 18 − 18 22 9 0 0 1 22 12 solve         − 6 14 4 0 2    | α ∈ C  � 3 3 − 2 x = 0 ⇒  α 0     − 18 22 12 0 3 

Example 19 � 4 � − 5 A = 2 − 3 det( A − λ I ) = λ 2 − λ − 2 = ( λ − 2 ) · ( λ + 1 ) ◮ characteristic polynomial : λ 2 − λ − 2 ◮ characteristic equation : λ 2 − λ − 2 = 0 λ 0 = 2 λ 1 = − 1

Example 20 � 4 � 1 � 2 � � � − 5 − 5 0 A − λ 0 I = − 2 = 2 − 3 0 1 2 − 5 solve � 2 � 0 � 5 � � � � � − 5 � x = ⇒ | α ∈ C α − 5 2 0 2

Example 21 � 4 � 1 � 5 � � � − 5 − 5 0 A − λ 1 I = − ( − 1 ) = 2 − 3 0 1 2 − 2 solve � 5 � 0 � 1 � � � � � − 5 � x = ⇒ | α ∈ C α − 2 2 0 1

Example double root 22 � 2 � 0 A = 0 2 det( A − λ I ) = λ 2 − 4 λ + 4 = ( λ − 2 ) 2 so λ 0 = 2 λ 1 = 2

Example double root 23 � 2 � 1 � 0 � � � 0 0 0 A − λ 0 I = − 2 = 0 2 0 1 0 0 solve � 0 � 0 � 1 � 0 � � � � � � 0 � x = ⇒ α 0 + α 1 | α 0 , α 1 ∈ C 0 0 0 0 1

Example double root again 24 � 2 � 0 A = 1 2 det( A − λ I ) = λ 2 − 4 λ + 4 = ( λ − 2 ) 2 so λ 0 = 2 λ 1 = 2

Example 25 � 2 � 1 � 0 � � � 0 0 0 A − λ 0 I = − 2 = 1 2 0 1 1 0 solve � 0 � 0 � 0 � � � � � 0 � x = ⇒ α | α ∈ C 1 0 0 1

Zero is an eigenvalue 26 � 4 � − 4 A = 2 − 2 det( A − λ I ) = λ 2 − 2 λ = ( λ − 2 ) · λ ◮ characteristic polynomial : λ 2 − 2 λ ◮ characteristic equation : λ 2 − 2 λ = 0 λ 0 = 2 λ 1 = 0

Example 27 � 4 � 1 � 2 � � � − 4 − 4 0 A − λ 0 I = − 2 = 2 − 2 0 1 2 − 4 solve � 2 � 0 � 2 � � � � � − 4 � x = ⇒ | α ∈ C α − 4 2 0 1

Example 28 � 4 � 1 � 4 � � � − 4 − 4 0 A − λ 1 I = − ( 0 ) = 2 − 2 0 1 2 − 2 solve � 4 � 0 � 1 � � � � � − 4 � x = ⇒ | α ∈ C α − 2 2 0 1

Characteristic polynomial 29 Definition Let T ∈ M n × n . The characteristic polynomial of T is det ( T − λ I ) . The characteristic equation of T is det ( T − λ I ) = 0 . The characteristic polynomial of a linear transformation φ is the characteristic polynomial of any matrix representation R B → B ( φ )

Existence 30 Theorem Let φ be any linear transformation on a non-trivial vector space. Then φ has at least one eigenvalue.

Eigenspace 31 Definition The eigenspace of a transformation φ associated with the eigenvalue λ is � � � ζ | φ ( � ζ ) = λ� V λ = ζ . Theorem An eigenspace is a subspace.

Algebraic vs geometric multiplicity 32 Theorem Let a linear transformation φ has a characteristic polynomials p ( λ ) = ( x − λ 1 ) m 1 ( x − λ 2 ) m 2 . . . ( x − λ k ) m k Then eigenvalue λ k has algebraic multiplicity m k and geometric multiplicity dim V λ k

Linear independence 33 Theorem A set of eigenvectors, corresponding to distinct eigenvalues is linearly independent