Diagonalization Hung-yi Lee
Review If Av = λv (v is a vector, λ is a scalar) v is an eigenvector of A excluding zero vector λ is an eigenvalue of A that corresponds to v Eigenvectors corresponding to λ are nonzero solution of (A I n )v = 0 Eigenspace of λ: Eigenvectors corresponding to λ = Null(A I n ) 0 eigenspace A scalar t is an eigenvalue of A Eigenvectors corresponding to λ + 0 det A ti n = 0
Review Characteristic polynomial of A is det A ti n Factorization multiplicity = t λ 1 m 1 t λ 2 m 2 t λ k m k Eigenvalue: Eigenspace: (dimension) λ 1 λ 2 dk d 1 d 2 λ k m 1 m 2 m k
Outline An nxn matrix A is called diagonalizable if A = PDP 1 D: nxn diagonal matrix P: nxn invertible matrix Is a matrix A diagonalizable? If yes, find D and P Reference: Textbook 5.3
Diagonalizable An nxn matrix A is called diagonalizable if A = PDP 1 D: nxn diagonal matrix P: nxn invertible matrix Not all matrices are diagonalizable A 2 = 0 (?) If A = PDP 1 for some invertible P and diagonal D A 2 = PD 2 P 1 = 0 D 2 = 0 D = 0 A= 0 D is diagonal
P = p 1 p n Diagonalizable If A is diagonalizable A = PDP 1 AP = PD D = d 1 0 0 d n AP = Ap 1 Ap n PD = P d 1 e 1 d n e n = Pd 1 e 1 Pd n e n = d 1 Pe 1 d n Pe n = d 1 p 1 d n p n Ap i = d i p i p i is an eigenvector of A corresponding to eigenvalue d i
Diagonalizable If A is diagonalizable A = PDP 1 = = = P = p 1 p n D = d 1 0 0 d n p i is an eigenvector of A corresponding to eigenvalue d i There are n eigenvectors that form an invertible matrix There are n independent eigenvectors The eigenvectors of A can form a basis for R n.
Diagonalizable If A is diagonalizable A = PDP 1 P = p 1 p n D = d 1 0 0 d n p i is an eigenvector of A corresponding to eigenvalue d i How to diagonalize a matrix A? Step 1: Step 2: Find n L.I. eigenvectors corresponding if possible, and form an invertible P The eigenvalues corresponding to the eigenvectors in P form the diagonal matrix D.
Diagonalizable A set of eigenvectors that correspond to distinct eigenvalues is linear independent. det A ti n Factorization = t λ 1 m 1 t λ 2 m 2 t λ k m k Eigenvalue: Eigenspace: (dimension) λ 1 λ 2 λ k d 1 d 2 d k m 1 m 2 m k Independent
Diagonalizable A set of eigenvectors that correspond to distinct eigenvalues is linear independent. Eigenvalue: λ 1 λ 2 Eigenvector: v 1 v 2 v k = c 1 v 1 +c 2 v 2 + +c k 1 v k 1 λ m v m Assume dependent a contradiction - Av k = c 1 Av 1 +c 2 Av 2 + +c k 1 Av k 1 k v k = c 1 1 v 1 +c 2 2 v 2 + +c k 1 k 1 v ( k ) k 1 k v k = c 1 k v 1 +c 2 k v 2 + +c k 1 k v k 1 0 = c 1 ( 1 k ) v 1 +c 2 ( 2 k )v 2 + +c k 1 ( k 1 k )v k 1 Not c 1 = c 2 = = c k 1 = 0 Same eigenvalue a contradiction
Diagonalizable P = p 1 p n d 1 0 D = If A is diagonalizable 0 d n A = PDP 1 p i is an eigenvector of A corresponding to eigenvalue d i det A ti n = t λ m 1 1 t λ m 2 2 t λ m k k Eigenvalue: Eigenspace: λ 1 λ 2 λ k You can t find more! Basis for λ 1 Basis for λ 2 Basis for λ 3 Independent Eigenvectors
Diagonalizable - Example Diagonalize a given matrix characteristic polynomial is (t + 1) 2 (t 3) eigenvalue 3 0 B 1 = 1 1 eigenvalue 1 1 0 B 2 = 0, 1 0 1 1 0 0 A = 0 1 2 0 2 1 A = PDP 1, where eigenvalues: 3, 1 0 1 0 P = 1 0 1 1 0 1 3 0 0 D = 0 1 0 0 0 1
Application of Diagonalization If A is diagonalizable, Example: Study.727 FB.273 A = PDP 1 A m = PD m P 1.85 Study FB.97.03.85.97.15.03 Study.85 FB.15.15.85.15 Study
To Study FB From Study FB.15.85 Study FB.97.03 = A Study.727 FB.273.85.97.15.03 Study.85 FB.15.85.15 Study.727.273.85.15 1 0 A = PDP 1 A m = PD m P 1
Diagonalizable Diagonalize a given matrix RREF p 1 1 = 1 RREF p 2 = 1 5 (invertible)
Application of Diagonalization When m, A m = 1/6 1/6 5/6 5/6 The beginning condition does not influence.
Test for a Diagonalizable Matrix An n x n matrix A is diagonalizable if and only if both the following conditions are met. The characteristic polynomial of A factors into a product of linear factors. det A ti n Factorization = t λ m 1 1 t λ m 2 2 t λ m k k For each eigenvalue of A, the multiplicity of equals the dimension of the corresponding eigenspace.
Independent Eigenvectors An n x n matrix A is diagonalizable = The eigenvectors of A can form a basis for R n. = det A ti n m = t λ 1 m 1 t λ 2 m 2 t λ k k Eigenvalue: Eigenspace: (dimension) λ 1 λ 2 λ k d 1 d 2 d k = m 1 = m 2 = m k
This lecture Reference: Chapter 5.4 T B v B T v B simple B 1 Properly selected v T T v B Properly selected
Review P = p 1 p n eigenvector D = eigenvalue d 1 0 0 d n An n x n matrix A is diagonalizable (A = PDP 1 ) = The eigenvectors of A can form a basis for R n. det A ti n m = t λ 1 m 1 t λ 2 m 2 t λ k k Eigenvalue: Eigenspace: λ 1 λ 2 λ k Basis for λ 1 Basis for λ 2 Basis for λ 3 Independent Eigenvectors
Diagonalization of Linear Operator Example 1: T x 1 x 2 x 3 = 8x 1 + 9x 2 6x 1 7x 2 3x 1 + 3x 2 x 3 det The standard matrix is A = 8 -t 9 0 6 7 -t 0 3 3 1 the characteristic polynomial is (t + 1) 2 (t 2) eigenvalues: 1, 2 -t Eigenvalue -1: Eigenvalue 2: B 1 = 1 1 0, 0 0 1 B 2 = 3 2 1 B 1 B 2 is a basis of R 3 T is diagonalizable.
Diagonalization of Linear Operator Example 2: T x 1 x 2 x 3 = x 1 + x 2 + 2x 3 x 1 x 2 0 det The standard matrix is A = the characteristic polynomial is t 2 (t + 2) eigenvalues: 0, 2 the reduced row echelon form of A 0I 3 = A is 1 -t 1 2 1 1 -t0 0 0 0 1 1 0 0 0 1 0 0 0 -t the eigenspaces corresponding to the eigenvalue 0 has the dimension 1 < 2 = algebraic multiplicity of the eigenvalue 0 T is not diagonalizable.
Diagonalization of Linear Operator If a linear operator T is diagonalizable PB 1 1 Properly selected T B v B T v B v D simple Eigenvectors form the good system A = PDP 1 T v B P Properly selected
Diagonalization of Linear Operator T x 1 x 2 x 3 = 1 0 3 1 0 2 0 1 1 1 8x 1 + 9x 2 6x 1 7x 2 3x 1 + 3x 2 x 3 B 1 = v -1: 2: 1 1 0, 0 0 1 T v B 2 = T B v B T v B 8 9 0 6 7 0 3 3 1 3 2 1 1 0 0 0 1 0 0 0 2 1 0 3 1 0 2 0 1 1