Announcements Monday, November 06

Transcription

1 Announcements Monday, November 06 This week s quiz: covers Sections 5 and 52 Midterm 3, on November 7th (next Friday) Exam covers: Sections 3,32,5,52,53 and 55

2 Section 53 Diagonalization

3 Motivation: Difference equations Now do multiply matrices Many real-word (linear algebra problems): Start with a given situation (v 0) and want to know what happens after some time (iterate a transformation): v n = Av n = = A n v 0 Ultimate question: what happens in the long run (find v n as n ) Old Example Recall our example about rabbit populations: using eigenvectors was easier than matrix multiplications, but Taking powers of diagonal matrices is easy! Working with diagonalizable matrices is also easy We need to use the eigenvalues and eigenvectors of the dynamics

4 Powers of Diagonal Matrices If D is diagonal Then D n is also diagonal, the diagonal entries of D n are the nth powers of the diagonal entries of D Example D = D 2 = ( ) ( ) ( ) D n 2 n 0 = 0 3 n 0 0 M = 0 0 2, M 2 = 0 0 4, ( ) n 0 0 M n = n n

5 Powers of Matrices that are Similar to Diagonal Ones When is A is not diagonal? Example Let A = ( 2 4 ) Compute A n Using that A = PDP where P = From the first expression: ( ) 2 and D = ( ) A 2 = (PDP )(PDP ) = PD(P P)DP = PDIDP = PD 2 P A 3 = (PDP )(PD 2 P ) = PD(P P)D 2 P = PDID 2 P = PD 3 P A n = PD n P Plug in P and D: ( ) ( A n 2 2 n = n Closed formula in terms of n: easy to compute ) ( ) = 2 ( 2 n+ 3 n 2 n n 2 n 3 n 2 n n )

6 Diagonalizable Matrices Definition An n n matrix A is diagonalizable if it is similar to a diagonal matrix: A = PDP for D diagonal Important d 0 0 If A = PDP 0 d 22 0 for D = then 0 0 d nn d k 0 0 A k = PD K P 0 d22 k 0 = P P 0 0 dnn k So diagonalizable matrices are easy to raise to any power

7 Diagonalization The Diagonalization Theorem An n n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors In this case, A = PDP for λ 0 0 P = v v 2 v n 0 λ 2 0 D =, 0 0 λ n where v, v 2,, v n are linearly independent eigenvectors, and λ, λ 2,, λ n are the corresponding eigenvalues (in the same order) Important If A has n distinct eigenvalues then A is diagonalizable Fact 2 in 5 lecture notes: eigenvectors with distinct eigenvalues are always linearly independent If A is diagonalizable matrix it need not have n distinct eigenvalues though

8 Diagonalization Example Problem: Diagonalize A = The characteristic polynomial is ( ) 2 4 f (λ) = λ 2 Tr(A) λ + det(a) = λ 2 5λ + 6 = (λ 2)(λ 3) Therefore the eigenvalues are 2 and 3 Let s compute some eigenvectors: ( ) 2 (A 2I )x = 0 x = 0 rref ( ) 2 x = The parametric form is x = 2y, so v = ( 2 ) is an eigenvector with eigenvalue 2 ( ) 2 2 (A 3I )x = 0 x = 0 rref ( ) x = The parametric form is x = y, so v 2 = ( ) is an eigenvector with eigenvalue 3 The eigenvectors v, v 2 are linearly independent, so the Diagonalization Theorem says A = PDP for P = ( ) 2 D = ( )

9 Diagonalization Example Problem: Diagonalize A = 2 0 The characteristic polynomial is f (λ) = det(a λi ) = λ 3 + 4λ 2 5λ + 2 = (λ ) 2 (λ 2) Therefore the eigenvalues are and 2, with respective multiplicities 2 and First compute the -eigenspace: (A I )x = x = 0 rref x = x 0 The parametric vector form is y = y + z 0 z 0 Hence a basis for the -eigenspace is B = { } 0 v, v 2 where v =, v 2 = 0 0

10 Diagonalization Example 2, continued Now let s compute the 2-eigenspace: (A 2I )x = x = 0 rref x = The parametric form is x = 3z, y = 2z, so an eigenvector with eigenvalue 2 is 3 v 3 = 2 Note that v, v 2 form a basis for the -eigenspace, and v 3 has a distinct eigenvalue Thus, the eigenvectors v, v 2, v 3 are linearly independent and the Diagonalization Theorem says A = PDP for P = 0 2 D = In this case: there are 3 linearly independent eigenvectors and only 2 distinct eigenvalues

11 Diagonalization Procedure How to diagonalize a matrix A: Find the eigenvalues of A using the characteristic polynomial 2 Compute a basis B λ for each λ-eigenspace of A 3 If there are fewer than n total vectors in the union of all of the eigenspace bases B λ, then the matrix is not diagonalizable 4 Otherwise, the n vectors v, v 2,, v n in your eigenspace bases are linearly independent, and A = PDP for λ 0 0 P = v v 2 v n 0 λ 2 0 and D =, 0 0 λ n where λ i is the eigenvalue for v i

12 Diagonalization A non-diagonalizable matrix Problem: Show that A = The characteristic polynomial is ( ) is not diagonalizable 0 f (λ) = det(a λi ) = (λ ) 2 Let s compute the -eigenspace: ( ) 0 (A I )x = 0 x = ( ) A basis for the -eigenspace is ; solution has only one free variable! 0 Conclusion: ( ) All eigenvectors of A are multiples of 0 So A has only one linearly independent eigenvector If A was diagonalizable, there would be two linearly independent eigenvectors!

13 Poll Poll Which of the following matrices are diagonalizable, and why? ( ) ( ) ( ) ( ) A B C D Matrix D is already diagonal! Matrix B is diagonalizable because it has two distinct eigenvalues Matrices A and C are not diagonalizable: Same argument as previous slide: ( ) All eigenvectors are multiples of 0

14 Non-Distinct Eigenvalues Definition Let λ be an eigenvalue of a square matrix A The geometric multiplicity of λ is the dimension of the λ-eigenspace Theorem Let λ be an eigenvalue of a square matrix A Then (the geometric multiplicity of λ) (the algebraic multiplicity of λ) Note: If λ is an eigenvalue, then the λ-eigenspace has dimension at least but it might be smaller than what the characteristic polynomial suggests The intuition/visualisation is beyond the scope of this course Multiplicities all one If there are n eigenvalues all with algebraic multiplicity (so does the geometric multiplicities), then their corresponding eigenvectors are linearly independent Therefore A is diagonalizable

15 Non-Distinct Eigenvalues (Good) examples From previous exercises we know: Example The matrix A = 2 0 has characteristic polynomial f (λ) = (λ ) 2 (λ 2) ( ) 2 The matrix B = has characteristic polynomial 4 f (λ) = ( λ)(4 λ) + 2 = (λ 2)(λ 3) Matrix A Geom M Alg M λ = 2 2 λ = 2 Thus, both matrices are diagonalizable Matrix B Geom M Alg M λ = 2 λ = 3

16 Non-Distinct Eigenvalues (Bad) example Example ( ) The matrix A = has characteristic polynomial f (λ) = (λ ) 2 0 We showed before that the -eigenspace has dimension and A was not diagonalizable The geometric multiplicity is smaller than the algebraic Eigenvalue Geometric Algebraic λ = 2 The Diagonalization Theorem (Alternate Form) Let A be an n n matrix The following are equivalent: A is diagonalizable 2 The sum of the geometric multiplicities of the eigenvalues of A equals n 3 The sum of all algebraic multiplicities is n And for each eigenvalue, the geometric and algebraic multiplicity are equal

17 Applications to Difference Equations Let D = ( ) 0 0 /2 Start with a vector v 0, and let v = Dv 0, v 2 = Dv,, v n = D n v 0 Question: What happens to the v i s for different starting vectors v 0? Answer: Note that D is diagonal, so ( ) ( D n a n = b If we start with v 0 = ( ) a, then b 0 0 /2 n the x-coordinate equals the initial coordinate, the y-coordinate gets halved every time ) ( ) ( ) a a = b b/2 n

18 Applications to Difference Equations Picture D ( ) ( ) ( ) ( ) a 0 a a = = b 0 /2 b b/2 v 0 v v 2 v 3 v 4 e 2 e -eigenspace /2-eigenspace So all vectors get collapsed into the x-axis, which is the -eigenspace

19 Applications to Difference Equations More complicated example Let A = ( ) 3/4 /4 /4 3/4 Start with a vector v 0, and let v = Av 0, v 2 = Av,, v n = A n v 0 Question: What happens to the v i s for different starting vectors v 0? Matrix Powers: This is a diagonalization question Bottom line: A = PDP for ( ) ( ) 0 P = D = 0 /2 Hence v n = PD n P v 0 Details: The characteristic polynomial is f (λ) = λ 2 Tr(A) λ + det(a) = λ λ + ( 2 = (λ ) λ ) 2 We compute eigenvectors with eigenvalues and /2 to be, respectively, ( ) ( ) w = w 2 =

20 Applications to Difference Equations Picture of the more complicated example A n = PD n P acts on the usual coordinates of v 0 in the same way that D n acts on the B-coordinates, where B = {w, w 2} /2-eigenspace -eigenspace v 0 v v 2 v 3 v 4 w w 2 So all vectors get collapsed into the -eigenspace

21 Extra: Proof Diagonalization Theorem Why is the Diagonalization Theorem true? A diagonalizable implies A has n linearly independent eigenvectors: Suppose A = PDP, where D is diagonal with diagonal entries λ, λ 2,, λ n Let v, v 2,, v n be the columns of P They are linearly independent because P is invertible So Pe i = v i, hence P v i = e i Av i = PDP v i = PDe i = P(λ i e i ) = λ i Pe i = λ i v i Hence v i is an eigenvector of A with eigenvalue λ i So the columns of P form n linearly independent eigenvectors of A, and the diagonal entries of D are the eigenvalues A has n linearly independent eigenvectors implies A is diagonalizable: Suppose A has n linearly independent eigenvectors v, v 2,, v n, with eigenvalues λ, λ 2,, λ n Let P be the invertible matrix with columns v, v 2,, v n Let D = P AP De i = P APe i = P Av i = P (λ i v i ) = λ i P v i = λ i e i Hence D is diagonal, with diagonal entries λ, λ 2,, λ n Solving D = P AP for A gives A = PDP