Understanding Diagonalization of Matrices
Learn about diagonalization of matrices through eigenvectors and eigenvalues, the concept of eigenspace, characteristic polynomials, and how to determine if a matrix is diagonalizable. Discover the process of diagonalizing a matrix and finding the invertible matrix P and diagonal matrix D. Explore the significance of eigenvectors forming a basis for R^n.
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Diagonalization Hung-yi Lee
Review If ?? = ?? (? is a vector, ? is a scalar) ? is an eigenvector of A ? is an eigenvalue of A that corresponds to ? Eigenvectors corresponding to ? are nonzero solution of (A In)v = 0 Eigenvectors corresponding to ? = ????(A In) ? eigenspace excluding zero vector Eigenspace of ?: Eigenvectors corresponding to ? + ? A scalar ? is an eigenvalue of A ??? ? ??? = 0
Review Characteristic polynomial of A is ??? ? ??? Factorization multiplicity ?1? ?2 ?2 ? ?? ?? = ? ?1 ?1 ?2 ?? Eigenvalue: ?1 ?? Eigenspace: (dimension) ?2 ?? ?1 ?2
Outline An nxn matrix A is called diagonalizable if ? = ??? 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 ? = ??? 1 D: nxn diagonal matrix P: nxn invertible matrix Not all matrices are diagonalizable (?) A2= 0 If A = PDP 1for some invertible P and diagonal D A2= PD2P 1= 0 D2= 0 D = 0 D is diagonal A= 0
? = ?1 ?? Diagonalizable ?1 0 0 ? = ?? If A is diagonalizable ? = ??? 1 ?? = ?? ?? = ??1 ??? ?? = ? ?1?1 = ??1?1 ???? ????? = ?1??1 = ?1?1 ????? ???? ???= ???? ?? is an eigenvector of A corresponding to eigenvalue ??
? = ?1 ?? Diagonalizable ?1 0 0 ? = ?? If A is diagonalizable ?? is an eigenvector of A corresponding to eigenvalue ?? ? = ??? 1 = There are n eigenvectors that form an invertible matrix = There are n independent eigenvectors = The eigenvectors of A can form a basis for Rn.
? = ?1 ?? Diagonalizable ?1 0 0 ? = ?? If A is diagonalizable ?? is an eigenvector of A corresponding to eigenvalue ?? ? = ??? 1 How to diagonalize a matrix A? Find n L.I. eigenvectors corresponding if possible, and form an invertible P Step 1: The eigenvalues corresponding to the eigenvectors in P form the diagonal matrix D. Step 2:
Diagonalizable A set of eigenvectors that correspond to distinct eigenvalues is linear independent. ??? ? ??? Factorization ?1? ?2 ?2 ? ?? ?? = ? ?1 ?1 ?1 ?2 ?2 ?? Eigenvalue: ?? ?? ?2 ?1 Eigenspace: (dimension) Independent
Diagonalizable A set of eigenvectors that correspond to distinct eigenvalues is linear independent. ?2 ?1 ?1 ?? ?? Eigenvalue: Assume dependent ?2 Eigenvector: a contradiction vk= c1v1+c2v2+ +ck 1vk 1 Avk= c1Av1+c2Av2+ +ck 1Avk 1 kvk= c1 1v1+c2 2v2+ +ck 1 k 1vk 1 kvk= c1 kv1+c2 kv2+ +ck 1 kvk 1 0 = c1( 1 k) v1+c2( 2 k)v2+ +ck 1( k 1 k)vk 1 Not c1 = c2 = = ck 1= 0 ( k) - a contradiction Same eigenvalue
? = ?1 ?? Diagonalizable ?1 0 0 ? = ?? If A is diagonalizable ?? is an eigenvector of A corresponding to eigenvalue ?? ? = ??? 1 ??? ? ??? = ? ?1 ?1? ?2 ?2 ? ?? ?? ?1 ?2 ?? Eigenvalue: Eigenspace: Basis for ?3 Basis for ?1 Basis for ?2 Independent Eigenvectors You can t find more!
Diagonalizable - Example 1 0 0 0 0 2 1 = Diagonalize a given matrix 1 2 A characteristic polynomial is (t + 1)2(t 3) eigenvalues: 3, 1 A = PDP 1, where eigenvalue 3 0 1 1 0 1 1 1 0 0 0 1 = P B1= 1 eigenvalue 1 3 0 0 0 0 0 1 0 , 1 0 0 = 1 D B2= 0 1 1
Application of Diagonalization If A is diagonalizable, ? = ??? 1 ??= ???? 1 Example: .15 Study FB .85 .97 .03 .85 .85 Study Study Study .15 .03 .85 .727 .15 FB FB .97 .273 .15
From Study FB .15 Study FB .85 .97 To Study .03 FB = ? .85 .85 Study Study Study .15 .03 .85 .15 .727 FB FB .97 .273 .15 .727 .273 1 0 .85 .15 ? = ??? 1 ??= ???? 1
Diagonalizable Diagonalize a given matrix 1 = 1 p 1 RREF 1 (invertible) = p 2 5 RREF
Application of Diagonalization When ? , The beginning condition does not influence. 1/6 5/6 1/6 5/6 ??=
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. ??? ? ??? = ? ?1 Factorization ?1? ?2 ?2 ? ?? ?? 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 Rn. = ??? ? ??? ?1? ?2 ?2 ? ?? ?? = ? ?1 ?1 ?1 ?2 ?2 ?? Eigenvalue: ?? = ?? = ?2 = ?1 Eigenspace: (dimension)
This lecture Reference: Chapter 5.4 ?B ?B ? ?B simple ? 1 ? Properly selected Properly selected ? ? ? ?
? = ?1 ?? ?1 0 0 eigenvector Review ? = ?? eigenvalue An n x n matrix A is diagonalizable (? = ??? 1) = The eigenvectors of A can form a basis for Rn. ??? ? ??? = ? ?1 ?1? ?2 ?2 ? ?? ?? ?1 ?2 ?? Eigenvalue: Eigenspace: Basis for ?3 Basis for ?1 Basis for ?2 Independent Eigenvectors
Diagonalization of Linear Operator ?1 ?2 ?3 8?1+ 9?2 6?1 7?2 3?1+ 3?2 ?3 Example 1: ? = det -t 8 9 0 0 The standard matrix is ? = -t 6 3 7 3 -t 1 the characteristic polynomial is (t + 1)2(t 2) eigenvalues: 1, 2 Eigenvalue -1: Eigenvalue 2: B1 B2is a basis of R3 T is diagonalizable. 1 1 0 0 0 1 3 B1= B2= , 2 1
Diagonalization of Linear Operator ?1 ?2 ?3 ?1+ ?2+ 2?3 ?1 ?2 0 ? = Example 2: -t 1 1 0 1 2 0 0 det The standard matrix is ? = 1 0 -t -t the characteristic polynomial is t2(t + 2) eigenvalues: 0, 2 the reduced row echelon form of A 0I3= A is 1 0 0 1 0 0 0 1 0 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 ?B ? ?B ? ?B simple Eigenvectors form the good system ? 1 ? 1 ? ? Properly selected Properly selected ? = ??? 1 ? ? ?
Diagonalization of Linear Operator -1: 2: ?1 ?2 ?3 8?1+ 9?2 6?1 7?2 3?1+ 3?2 ?3 3 1 1 0 0 0 1 B2= ? = B1= 2 1 , ?B ?B ? ?B 1 0 0 0 0 0 2 1 0 1 1 1 0 0 0 1 3 1 1 0 0 0 1 3 2 1 2 1 8 9 0 0 6 3 7 3 1 ? ? ?
