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Chapter 7 Eigenvalues and Eigenvectors 7.1 Eigenvalues and Eigenvectors 7.2 Diagonalization 7.3 Symmetric Matrices and Orthogonal Diagonalization 7.4 Application of Eigenvalues and Eigenvectors 7.5 Principal Component Analysis 7.1 7.1 Eigenvalues and Eigenvectors n Eigenvalue problem (特徵值問題) (one of the most important problems in the linear algebra): If A is an n´n matrix, do there exist nonzero vectors x in Rn such that Ax is a scalar multiple of x? (The term eigenvalue is from the German word Eigenwert, meaning “proper value”) n Eigenvalue (特徵值) and Eigenvector (特徵向量): A: an n´n matrix l: a scalar (could be zero) ※ Geometric Interpretation n y x: a nonzero vector in R Ax = l x Eigenvalue Ax = l x x Eigenvector x 7.2 n Ex 1: Verifying eigenvalues and eigenvectors é2 0 ù é1 ù é0ù A=ê ú x1 = ê ú x 2 = ê ú ë0 - 1û ë0 û ë1 û Eigenvalue ※ In fact, for each eigenvalue, it é 2 0 ù é1 ù é 2 ù é1 ù has infinitely many eigenvectors. Ax1 = ê ú ê ú = ê ú = 2 ê ú = 2x1 For l = 2, [3 0]T or [5 0]T are ë0 -1û ë0û ë 0 û ë0û both corresponding eigenvectors. Moreover, ([3 0] + [5 0])T is still Eigenvector an eigenvector. The proof is in Thm. 7.1. Eigenvalue é 2 0 ù é0ù é 0 ù é0ù Ax2 = ê ú ê ú = ê ú = -1 ê ú = (-1)x2 ë0 -1û ë1 û ë -1û ë1 û Eigenvector 7.3 n Thm. 7.1: The eigenspace corresponding to l of matrix A If A is an n´n matrix with an eigenvalue l, then the set of all eigenvectors of l together with the zero vector is a subspace of Rn. This subspace is called the eigenspace (特徵空間) of l Pf: x1 and x2 are eigenvectors corresponding to l (i.e., Ax1 = l x1 , Ax2 = l x2 ) (1) A(x1 + x 2 ) = Ax1 + Ax 2 = l x1 + l x 2 = l (x1 + x 2 ) (i.e., x1 + x 2 is also an eigenvector corresponding to λ) (2) A(cx1 ) = c( Ax1 ) = c(l x1 ) = l (cx1 ) (i.e., cx1 is also an eigenvector corresponding to l ) Since this set is closed under vector addition and scalar multiplication, this set is a subspace of Rn according to Theorem 4.5 7.4 nEx 3: Examples of eigenspaces on the xy-plane For the matrix A as follows, the corresponding eigenvalues are l1 = –1 and l2 = 1: é - 1 0ù A=ê ú ë 0 1 û Sol: For the eigenvalue l1 = –1, corresponding vectors are any vectors on the x-axis é x ù é -1 0ù é x ù é - x ù é x ù ※ Thus, the eigenspace Aê ú = ê ú ê ú = ê ú = -1 ê ú corresponding to l = –1 is the x- ë0û ë 0 1û ë0û ë 0 û ë0û axis, which is a subspace of R2 For the eigenvalue l2 = 1, corresponding vectors are any vectors on the y-axis é 0 ù é -1 0ù é 0 ù é 0 ù é 0 ù ※ Thus, the eigenspace Aê ú = ê ú ê ú = ê ú = 1ê ú corresponding to l = 1 is the y- ë y û ë 0 1û ë y û ë y û ë y û axis, which is a subspace of R2 7.5 ※ Geometrically speaking, multiplying a vector (x, y) in R2 by the matrix A corresponds to a reflection to the y-axis, i.e., left multiplying A to v can transform v to another vector in the same vector space é xù æ é xù é 0 ù ö é xù é0ù Av = A ê ú = A ç ê ú + ê ú ÷ = A ê ú + A ê ú ë yû è ë0û ë y û ø ë0û ë yû é xù é 0 ù é- xù = -1 ê ú + 1 ê ú = ê ú ë0û ë y û ë y û 7.6 n Thm. 7.2: Finding eigenvalues and eigenvectors of a matrix AÎMn´n Let A be an n´n matrix. (1) An eigenvalue of A is a scalar l such that det(l I - A) = 0 (2) The eigenvectors of A corresponding to l are the nonzero solutions of (l I - A)x = 0 n Note: follwing the definition of the eigenvalue problem Ax = l x Þ Ax = l Ix Þ (l I - A)x = 0 (homogeneous system) (l I - A)x = 0 has nonzero solutions for x iff det(l I - A) = 0 (The above iff results comes from the equivalent conditions on Slide 4.101) n Characteristic equation (特徵方程式) of A: det(l I - A) = 0 n Characteristic polynomial (特徵多項式) of AÎMn´n: det(l I - A) = (l I - A) = l n + cn-1l n-1 + ! + c1l + c0 7.7 n Ex 4: Finding eigenvalues and eigenvectors é2 - 12 ù A=ê ú ë 1 - 5 û Sol: Characteristic equation: l -2 12 det(l I - A) = -1 l +5 = l 2 + 3l + 2 = (l + 1)(l + 2) = 0 Þ l = -1, - 2 Eigenvalue: l1 = -1, l2 = -2 7.8 é -3 12 ù é x1 ù é0 ù (1) l1 = -1 Þ (l1 I - A)x = ê ú ê ú =ê ú ë -1 4 û ë x2 û ë0 û é -3 12 ù G.-J. E. é1 -4 ù Þê ú ¾¾¾® ê ú ë - 1 4 û ë 0 0 û é x1 ù é 4t ù é 4 ù Þ ê ú = ê ú = t ê ú, t ¹ 0 ë x2 û ë t û ë 1 û é -4 12 ù é x1 ù é0 ù (2) l2 = -2 Þ (l2 I - A)x = ê ú ê ú =ê ú ë -1 3 û ë x2 û ë0 û é -4 12 ù G.-J. E. é1 -3ù Þê ú ¾¾¾® ê ú ë -1 3 û ë 0 0 û é x1 ù é3s ù é 3ù Þ ê ú = ê ú = sê ú, s ¹ 0 ë x2 û ë s û ë1û 7.9 n Ex 5: Finding eigenvalues and eigenvectors Find the eigenvalues and corresponding eigenvectors for the matrix A. What is the dimension of the eigenspace of each eigenvalue? é2 1 0ù A = ê0 2 0 ú ê ú ë 0 0 2 û Sol: Characteristic equation: l - 2 -1 0 lI - A = 0 l -2 0 = (l - 2)3 = 0 0 0 l -2 Eigenvalue: l = 2 7.10 The eigenspace of λ = 2: é0 -1 0ù é x1 ù é0ù (l I - A)x = êê0 0 0úú êê x2 úú = êê0úú êë0 0 0úû êë x3 úû êë0úû é x1 ù é s ù é1ù é0ù ê x2 ú = ê 0 ú = s ê0 ú + t ê 0 ú , s , t ¹ 0 ê ú ê ú ê ú ê ú x ë 3û ë û t ë 0 û ë1 û ì é1 ù é 0 ù ü ï ê ú ê ú ï s 0 + t í ê ú ê ú 0 s , t Î R ý : the eigenspace of A corresponding to l = 2 ï ê0 ú ê1 ú ï î ë û ë û þ Thus, the dimension of its eigenspace is 2 7.11 n Notes: (1) If an eigenvalue l1 occurs as a multiple root (k times) for the characteristic polynominal, then l1 has multiplicity (重數) k (2) The multiplicity of an eigenvalue is greater than or equal to the dimension of its eigenspace. (In Ex. 5, k is 3 and the dimension of its eigenspace is 2) 7.12 n Ex 6：Find the eigenvalues of the matrix A and find a basis for each of the corresponding eigenspaces é1 0 0 0 ù ê0 1 5 - 10 ú A=ê ú ê1 0 2 0 ú êë1 0 0 3 úû Sol: Characteristic equation: l -1 0 0 0 ※ According to the note on the previous slide, the 0 l - 1 -5 10 dimension of the lI - A = -1 0 l -2 0 eigenspace of λ1 = 1 is at most to be 2 -1 0 0 l -3 ※ For λ2 = 2 and λ3 = 3, the = (l - 1) 2 (l - 2)(l - 3) = 0 demensions of their eigenspaces are at most to Eigenvalues: l1 = 1, l2 = 2, l3 = 3 be 1 7.13 é0 0 0 0 ù é x1 ù é0 ù ê0 0 -5 10 úú êê x2 úú êê0 úú (1) l1 = 1 Þ (l1 I - A)x = ê = ê -1 0 -1 0 ú x3 ê ú ê0 ú ê úê ú ê ú ë -1 0 0 -2 û ë x4 û ë0 û é x1 ù é -2t ù é0 ù é -2 ù ê ú ê s ú G.-J.E. x ê1 ú ê 0 ú Þ ê 2 ú = ê ú = s ê ú + t ê ú , s, t ¹ 0 ê x3 ú ê 2t ú ê0ú ê 2 ú ê ú ê ú ê ú ê ú ë x4 û ë t û ë0û ë 1 û ì é0 ù é - 2 ù ü ïê ú ê ú ï ïê1ú ê 0 ú ï Þí , ý is a basis for the eigenspace ïê0ú ê 2 ú ï corresponding to l1 = 1 ïêë0úû êë 1 úû ï î þ ※The dimension of the eigenspace of λ1 = 1 is 2 7.14 é1 0 0 0 ù é x1 ù é0 ù ê0 1 -5 10 úú êê x2 úú êê0úú (2) l2 = 2 Þ (l2 I - A)x = ê = ê -1 0 0 0 ú x3 ê ú ê0 ú ê úê ú ê ú ë -1 0 0 -1û ë x4 û ë0û é x1 ù é 0 ù é0 ù ê ú ê5t ú ê5 ú G.-J.E. x Þ ê 2ú = ê ú = t ê ú, t ¹ 0 ê x3 ú ê t ú ê1 ú ê ú ê ú ê ú ë x4 û ë 0 û ë0 û ì é0 ù ü ïê ú ï ïê5ú ï is a basis for the eigenspace Þí ý ïê1ú ï corresponding to l2 = 2 ïêë0úû ï î þ ※The dimension of the eigenspace of λ2 = 2 is 1 7.15 é2 0 0 0 ù é x1 ù é0 ù ê0 2 -5 10 úú êê x2 úú êê0 úú (3) l3 = 3 Þ (l3 I - A)x = ê = ê -1 0 1 0 ú ê x3 ú ê0 ú ê úê ú ê ú ë -1 0 0 0 û ë x4 û ë0 û é x1 ù é 0 ù é 0 ù ê ú ê -5t ú ê -5ú G.-J.E. x Þ ê 2ú = ê ú = t ê ú, t ¹ 0 ê x3 ú ê 0 ú ê 0 ú ê ú ê ú ê ú ë x4 û ë t û ë 1 û ìé 0 ù ü ïê ú ï ïê- 5ú ï is a basis for the eigenspace Þí ý ï ê 0 ú ï corresponding to l3 = 3 ïîêë 1 úû ïþ ※The dimension of the eigenspace of λ3 = 3 is 1 7.16 n Thm. 7.3: Eigenvalues for triangular matrices If A is an n´n triangular matrix, then its eigenvalues are the entries on its main diagonal n Ex 7: Finding eigenvalues for triangular and diagonal matrices é -1 0 0 0ù 0 ê0 2 0 0 úú 0 é2 0 0ù ê (a) A = êê -1 1 0 úú (b) A = ê 0 0 0 0 0ú ê ú êë 5 3 -3úû ê0 0 0 -4 0 ú êë 0 0 0 0 3 úû Sol: l -2 0 0 (a) l I - A = 1 l -1 0 = (l - 2)(l - 1)(l + 3) = 0 -5 -3 l + 3 ※According to Thm. 3.2, the determinant of a triangular Þ l1 = 2, l2 = 1, l3 = -3 matrix is the product of the entries on the main diagonal (b) l1 = -1, l2 = 2, l3 = 0, l4 = -4, l5 = 3 7.17 n Eigenvalues and eigenvectors of linear transformations: A number l is called an eigenvalue of a linear transformation T : V ® V if there is a nonzero vector x such that T (x) = l x. The vector x is called an eigenvector of T corresponding to l , and the set of all eigenvectors of l (together with the zero vector) is called the eigenspace of l ※ The definition of linear transformation functions should be introduced in Ch 6 ※ Here I briefly introduce the linear transformation and its some basic properties ※ The typical example of a linear transformation function is that each component of the resulting vector is the linear combination of the components in the input vector x n An example for a linear transformation T: R3→R3 T ( x1 , x2 , x3 ) = ( x1 + 3x2 ,3x1 + x2 , -2 x3 ) 7.18 n Theorem: Standard matrix for a linear transformation Let T : R n ® R n be a linear trtansformation such that é a11 ù é a12 ù é a1n ù êa ú êa ú êa ú T (e1 ) = ê 21 ú , T (e2 ) = ê 22 ú , !, T (en ) = ê 2 n ú , ê " ú ê " ú ê " ú ê ú ê ú ê ú a ë n1 û a ë n2 û ë ann û where {e1 , e 2 , !, e n } is a standard basis for R n . Then an n ´ n matrix A, whose i -th column correspond to T (ei ), é a11 a12 ! a1n ù êa a22 ! a2 n úú A = [T (e1 ) T (e2 ) ! T (en ) ] = ê 21 , ê " " # " ú ê ú ë an1 an 2 ! ann û satisfies that T (x) = Ax for every x in R n . A is called the standard matrix for T (T的標準矩陣) 7.19 n Consider the same linear transformation T(x1, x2, x3) = (x1 + 3x2, 3x1 + x2, –2x3) é1 ù é1 ù é0 ù é 3ù é0ù é 0 ù Þ T (e1 ) = T ( êê0 úú) = êê3úú , T (e2 ) = T ( êê1 úú) = êê1 úú , T (e3 ) = T ( êê0 úú) = êê 0 úú êë0úû êë0úû êë0úû êë0úû êë1 úû êë -2úû n Thus, the above linear transformation T is with the following corresponding standard matrix A such that T(x) = Ax é1 3 0 ù é1 3 0 ù é x1 ù é x1 + 3x2 ù A = êê3 1 0 úú Þ Ax = êê3 1 0 úú êê x2 úú = êê3x1 + x2 úú êë0 0 -2 úû êë0 0 -2 úû êë x3 úû êë -2 x3 úû ※ The statement on Slide 7.18 is valid because for any linear transformation T: V →V, there is a corresponding square matrix such that T(x) = Ax. Consequently, the eignvalues and eigenvectors of a linear transformation T are in essence the eigenvalues and eigenvectors of the corresponding square matrix A 7.20 n Ex 8: Finding eigenvalues and eigenvectors for standard matrices Find the eigenvalues and corresponding eigenvectors for é1 3 0 ù ※ A is the standard matrix for T(x1, x2, A = êê 3 1 0 úú x3) = (x1 + 3x2, 3x1 + x2, –2x3) (see Slides 7.19 and 7.20) êë0 0 -2 úû Sol: él - 1 - 3 0 ù lI - A = ê - 3 l - 1 0 ú = (l + 2) (l - 4) = 0 2 ê ú êë 0 0 l + 2úû Þ eigenvalues l1 = 4, l2 = -2 For l1 = 4, the corresponding eigenvector is (1, 1, 0). For l2 = -2, the corresponding eigenvectors are (1, - 1, 0) and (0, 0, 1). 7.21 n Transformation matrix A ' for nonstandard bases Suppose B is the standard basis of R n . Since the coordinate matrix of a vector relative to the standard basis consists of the components of that vector, i.e., for any x in R n , x = [x]B , the theorem on Slide 7.19 can be restated as follows. T (x) = Ax Þ [T (x) ]B = A [ x ]B , where A = éë[T (e1 ) ]B [T (e 2 ) ]B ![T (e n ) ]B ùû is the standard matrix for T or the matrix of T relative to the standard basis B The above theorem can be extended to consider a nonstandard basis B ', which consists of {v1 , v 2 ,! , v n } [T (x)]B ' = A '[ x]B ' , where A ' = éë[T ( v1 )]B ' [T ( v 2 )]B ' ![T ( v n ) ]B ' ùû is the transformation matrix for T relative to the basis B ' ※ On the next two slides, an example is provided to verify numerically that this extension is valid 7.22 n EX. Consider an arbitrary nonstandard basis B ' to be {v1, v2, v3}= {(1, 1, 0), (1, –1, 0), (0, 0, 1)}, and find the transformation matrix A ' such that [T (x)]B ' = A ' [ x]B 'corresponding to the same linear transformation T(x1, x2, x3) = (x1 + 3x2, 3x1 + x2, –2x3) é é1 ù ù é 4ù é 4ù é é1ù ù é -2ù é0ù ê ú ê ú [T ( v1 )]B ' = êT ( êê1 úú ) ú = êê 4 úú = êê 0 úú , [T ( v 2 )]B ' = êT ( êê -1úú ) ú = êê 2 úú = êê -2 úú , êë êë0 úû úû êë 0 úû B ' êë 0 úû êë êë 0 úû úû êë 0 úû B ' êë 0 úû B' B' é é0ù ù é0ù é0ù ê ú [T ( v3 )]B ' = êT ( êê0 úú ) ú = êê 0 úú = êê 0 úú êë êë1 úû úû êë -2úû B ' ëê -2 úû B' é4 0 0 ù Þ A ' = êê0 -2 0 úú êë0 0 -2 úû 7.23 n Consider x = (5, –1, 4), and check that [T (x)]B ' = A ' [ x]B ' corresponding to the linear transformation T(x1, x2, x3) = (x1 + 3x2, 3x1 + x2, –2x3) é é5ù ù é2ù é8ù é5ù é 2ù ê ê ú ú [T (x)]B ' = êT ( ê -1ú) ú = êê14 úú = êê -6úú , [ x ]B ' = êê -1úú = êê 3úú , êë êë 4 úû úû êë -8úû B ' êë -8úû êë 4 úû B ' êë 4úû B' é4 0 0 ù é2ù é 8 ù Þ A ' [ x ]B ' = êê 0 -2 0 úú êê 3 úú = êê -6 úú = [T (x)]B ' êë0 0 -2 úû êë 4 úû êë -8úû 7.24 n For a special basis 𝐵! = 𝐯", 𝐯#, … , 𝐯$ , where 𝐯% ’s are eigenvectors of the standard matrix 𝐴, 𝐴! is obtained immediately to be diagonal due to 𝑇 𝐯% = 𝐴𝐯% = 𝜆% 𝐯% and 𝜆% 𝐯% &! = 0𝐯" + 0𝐯# + ⋯ + 𝜆% 𝐯% + ⋯ + 0𝐯$ &! = 0 ⋯ 0 𝜆% 0 ⋯ 0 ' Let B ' be a basis of R 3 made up of three linearly independent eigenvectors of A, e.g., B ' = {v1 , v 2 , v 3} = {(1, 1, 0), (1, - 1, 0), (0, 0, 1)} in Ex. 8 Then A ', the transformation matrix for T relative to the basis B ', defined as [[T ( v1 )]B ' [T ( v 2 )]B ' [T ( v 3 )]B ' ] (see Slide 7.22), is diagonal, and the main diagonal entries are corresponding eigenvalues (see Slides 7.23) for l1 = 4 !"# for l2 =-2 !"" " #""" $ é4 0 0ù B ' = {(1, 1, 0), (1, - 1, 0), (0, 0, 1)} A' = ê0 - 2 0 ú ê ú Eigenvectors of A ë 0 0 - 2 û Eigenvalues of A 7.25 Keywords in Section 7.1: n eigenvalue problem: 特徵值問題 n eigenvalue: 特徵值 n eigenvector: 特徵向量 n characteristic equation: 特徵方程式 n characteristic polynomial: 特徵多項式 n eigenspace: 特徵空間 n multiplicity: 重數 n linear transformation: 線性轉換 n diagonalization: 對角化 7.26 7.2 Diagonalization n Diagonalization problem (對角化問題): For a square matrix A, does there exist an invertible matrix P such that P–1AP is diagonal? n Diagonalizable matrix (可對角化矩陣): Definition 1: A square matrix A is called diagonalizable if there exists an invertible matrix P such that P–1AP is a diagonal matrix (i.e., P diagonalizes A) Definition 2: A square matrix A is called diagonalizable if A is similar to a diagonal matrix ※ In Sec. 6.4, two square matrices A and B are similar if there exists an invertible matrix P such that B = P–1AP. n Notes: In this section, I will show that the eigenvalue and eigenvector problem is closely related to the diagonalization problem 7.27 n Thm. 7.4: Similar matrices have the same eigenvalues If A and B are similar n´n matrices, then they have the same eigenvalues Pf: -1 For any diagonal matrix in the A and B are similar Þ B = P AP form of D = λI, P–1DP = D Consider the characteristic equation of B: l I - B = l I - P -1 AP = P -1l IP - P -1 AP = P -1 (l I - A) P = P -1 l I - A P = P -1 P l I - A = P -1P l I - A = lI - A Since A and B have the same characteristic equation, they are with the same eigenvalues ※ Note that the eigenvectors of A and B are not identical 7.28 n Ex 1: Eigenvalue problems and diagonalization programs é1 3 0 ù A = ê3 1 0 ú ê ú ë 0 0 - 2 û Sol: Characteristic equation: l - 1 -3 0 l I - A = -3 l -1 0 = (l - 4)(l + 2) 2 = 0 0 0 l+2 The eigenvalues : l1 = 4, l2 = -2, l3 = -2 é1 ù ê1 ú (1) l = 4 Þ the eigenvector 1 ê ú p = êë0 úû 7.29 é1ù é0ù ê -1ú , p = ê0 ú (2) l = -2 Þ the eigenvector 2 ê ú p = 3 ê ú êë 0 úû êë1 úû é1 1 0 ù é4 0 0 ù P = [p1 p 2 p3 ] = êê1 -1 0 úú , and P -1 AP = êê0 -2 0 úú êë0 0 1 úû êë0 0 -2 úû n Note: If P = [p 2 p1 p3 ] é 1 1 0ù é -2 0 0 ù = êê -1 1 0 úú Þ P -1 AP = êê 0 4 0 úú êë 0 0 1 úû êë 0 0 -2 úû ※ The above example can verify Thm. 7.4 since the eigenvalues for both A and P–1AP are the same to be 4, –2, and –2 ※ The reason why the matrix P is constructed with the eigenvectors of A is demonstrated in Thm. 7.5 on the next slide 7.30 n Thm. 7.5: Condition for diagonalization An n´n matrix A is diagonalizable if and only if it has n linearly independent eigenvectors ※ If there are n linearly independent eigenvectors, it does not imply that there are n distinct eigenvalues. In an extreme case, it is possible to have only one eigenvalue with the multiplicity n, and there are n linearly independent eigenvectors for this eigenvalue ※ On the other hand, if there are n distinct eigenvalues, then there are n linearly independent eigenvectors (see Thm. 7.6), and thus A must be diagonalizable Pf: (Þ) Since A is diagonalizable, there exists an invertible P s.t. D = P -1 AP is diagonal. Let P = [p1 p 2 ! p n ] and D = diag (l1 , l2 ,!, ln ), then él1 0 ! 0ù ê0 l ! 0 úú PD = [p1 p 2 ! p n ] ê 2 ê" " # "ú ê ú ë0 0 ! ln û = [l1p1 l2p 2 ! lnp n ] 7.31 ! AP = PD (since D = P -1 AP ) \[ Ap1 Ap 2 " Ap n ] = [l1p1 l2p 2 " lnp n ] Þ Api = li pi , i = 1, 2,!, n (The above equations imply the column vectors pi of P are eigenvectors of A, and the diagonal entries li in D are eigenvalues of A) Because A is diagonalizable Þ P is invertible Þ Columns in P, i.e., p1 , p 2 ,! , p n , are linearly independent (see Slide 4.101 in the lecture note) Thus, A has n linearly independent eigenvectors (Ü) Since A has n linearly independent eigenvectors p1 , p 2 ,! p n with corresponding eigenvalues l1 , l2 ,! ln (could be the same), then Þ Api = li pi , i = 1, 2,!, n Let P = [p1 p2 ! pn ] 7.32 AP = A[p1 p 2 ! p n ] = [ Ap1 Ap 2 ! Ap n ] = [l1p1 l2p 2 ! lnp n ] él1 0 ! 0ù ê0 l ! 0 úú = [p1 p 2 ! p n ] ê 2 = PD ê" " # "ú ê ú ë0 0 ! ln û Since p1 , p 2 ,! , p n are linearly independent Þ P is invertible (see Slide 4.101 in the lecture note) " AP = PD \ P -1 AP = D Þ A is diagonalizable (according to the definition of the diagonalizable matrix on Slide 7.27) ※ Note that pi 's are linearly independent eigenvectors and the diagonal entries li in the resulting diagonalized D are eigenvalues of A 7.33 n Ex 4: A matrix that is not diagonalizable Show that the following matrix is not diagonalizable é1 2 ù A=ê ú ë 0 1 û Sol: Characteristic equation: l - 1 -2 lI - A = = (l - 1) 2 = 0 0 l -1 The eigenvalue l1 = 1, and then solve (l1I - A)x = 0 for eigenvectors é0 -2 ù é1 ù l1 I - A = I - A = ê ú Þ eigenvector p1 = ê ú ë0 0 û ë0û Since A does not have two linearly independent eigenvectors, A is not diagonalizable 7.34 n Steps for diagonalizing an n´n square matrix: Step 1: Find n linearly independent eigenvectors p1 , p2 ,!p n for A with corresponding eigenvalues l1 , l2 ,!, ln Step 2: Let P = [p1 p 2 ! p n ] Step 3: él1 0 ! 0ù -1 ê 0 l2 ! 0ú P AP = D = ê ú ê" " # "ú êë 0 0 ! ln úû where Api = li pi , i = 1, 2,!, n 7.35 n Ex 5: Diagonalizing a matrix é 1 - 1 - 1ù A=ê 1 3 1ú ê ú êë- 3 1 - 1úû Find a matrix P such that P -1 AP is diagonal. Sol: Characteristic equation: l -1 1 1 l I - A = -1 l -3 -1 = (l - 2)(l + 2)(l - 3) = 0 3 -1 l +1 The eigenvalues : l1 = 2, l2 = -2, l3 = 3 7.36 é1 1 1ù é1 0 1 ù é x1 ù é0 ù l ê -1 -1 -1ú ¾¾¾ ê0 1 0 ú ê x ú = ê0 ú l1 = 2 Þ 1 I - A = ê ú G.-J. E. ® ê úê 2ú ê ú êë 3 -1 3 úû êë0 0 0 úû êë x3 úû êë0 úû é x1 ù é -t ù é -1ù ê x ú = ê 0 ú Þ eigenvector p = ê 0 ú ê 2ú ê ú 1 ê ú êë x3 úû êë t úû êë 1 úû é -3 1 1 ù é1 0 - 14 ù é x1 ù é0 ù ê ú G.-J. E.® ê0 1 1 ú ê x ú = ê0 ú l2 = -2 Þ l2 I - A = ê -1 -5 -1ú ¾¾¾ ê 4 úê 2ú ê ú êë 3 -1 -1úû êë0 0 0 úû êë x3 úû êë0 úû é x1 ù é 14 t ù é1ù ê x ú = ê - 1 t ú Þ eigenvector p = ê -1ú ê 2ú ê 4 ú 2 ê ú êë x3 úû êë t úû êë 4 úû 7.37 é2 1 1ù é1 0 1 ù é x1 ù é0 ù ê ú G.-J. E. ê úê ú ê ú l3 = 3 Þ l3 I - A = ê -1 0 -1ú ¾¾¾® ê0 1 -1ú ê x2 ú = ê0 ú êë 3 -1 4 úû êë0 0 0 úû êë x3 úû êë0 úû é x1 ù é -t ù é -1ù ê x ú = ê t ú Þ eigenvector p = ê 1 ú ê 2ú ê ú 3 ê ú êë x3 úû êë t úû êë 1 úû é -1 1 -1ù P = [p1 p 2 p3 ] = êê 0 -1 1 úú and it follows that êë 1 4 1 úû é2 0 0ù P -1 AP = êê 0 -2 0 úú êë 0 0 3úû 7.38 n Note: a quick way to calculate Ak based on the diagonalization technique él1 0 ! 0ù él1k 0 ! 0ù ê0 l ê ú ! 0 úú ê 0 l2k ! 0ú (1) D = ê 2 Þ D = k ê" " # "ú ê" " # " ú ê ú ê kú ë0 0 ! ln û êë 0 0 ! ln úû (2) D = P -1 AP Þ D k = " P -# 1 AP P -1 AP $ "#$ "#$ ! P -1 AP = P -1 k AP repeat k times él1k 0 ! 0ù ê ú k -1 ê 0 l2k ! 0ú A = PD P , where D = k k ê% % & % ú ê kú êë 0 0 ! ln úû 7.39 nThm. 7.6: Sufficient conditions for diagonalization If an n´n matrix A has n distinct eigenvalues, then the corresponding eigenvectors are linearly independent and thus A is diagonalizable according to Thm. 7.5. Pf: Let λ1, λ2, …, λn be distinct eigenvalues and corresponding eigenvectors be x1, x2, …, xn. In addition, consider that the first m eigenvectors are linearly independent, but the first m+1 eigenvectors are linearly dependent, i.e., xm+1 = c1x1 + c2 x2 + ! + cm xm , (1) where ci’s are not all zero. Multiplying both sides of Eq. (1) by A yields Ax m+1 = Ac1x1 + Ac2 x 2 + ! + Acm x m lm+1xm+1 = c1l1x1 + c2l2 x 2 + ! + cm lm x m (2) 7.40