Eigenvalues & Eigenvectors Eigenvalues and Eigenvectors Eigenvalue problem (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”) Eigenvalue Geometric Interpretation A: an n×n matrix λ: a scalar (could be zero ) x: a nonzero vector in Rn Eigenvalue Ax= λ x Eigenvector 2 Eigenvalues and Eigenvectors Example 1: Verifying eigenvalues and eigenvectors. 2 0 1 0 A = x = x = 0 −1 1 0 2 1 Eigenvalue 201 2 1 In fact, for each eigenvalue, it has infinitely many eigenvectors. Ax1 = ===2 2 x 1 010− 0 0 For λ = 2, [3 0]T or [5 0]T are both corresponding eigenvectors. Eigenvector Moreover, ([3 0] + [5 0]) T is still an eigenvector. Eigenvalue The proof is in Theorem. 7.1. 200 0 0 Ax= = =−1 =− ( 1) x 2 011− − 1 1 2 Eigenvector 3 Eigenspace Theorem 7.1: (The eigenspace corresponding to λ of matrix A) If A is an n×n matrix with an eigenvalue λ, then the set of all eigenvectors of λ together with the zero vector is a subspace of Rn. This subspace is called the eigenspace of λ. Proof: x1 and x2 are eigenvectors corresponding to λ (i.e., Ax1=λ x 1, A x 2 = λ x 2 ) (1) A (xx12+=+ ) A xx 1 A 2 =+=λ xx 12 λ λ ( xx 12 + ) (i.e., x1+ x 2 is also an eigenvector corr esponding to λ) (2)( Acx1 )= cA ( x 1 ) = c (λ x 1 ) = λ ( c x 1 ) (i.e., cx1 is also an eigenvector corre sponding to λ) Since this set is closed under vector addition and scalar multiplication, this set is a subspace of Rn according to Theorem 4.5. 4 Eigenspace Eamplex 3: Examples of eigenspaces on the xy -plane For the matrix A as follows, the corresponding eigenvalues are λ1 = –1 and λ2 = 1: −1 0 A = 0 1 Solution For the eigenvalue λ1 = –1, corresponding vectors are any vectors on the x-axis Thus, the eigenspace corresponding to λ = –1 x−1 0 xxx − 2 A= = =− 1 is the x-axis, which is a subspace of R . 0 010 0 0 For the eigenvalue λ2 = 1, corresponding vectors are any vectors on the y-axis 0− 100 0 0 Thus, the eigenspace corresponding to λ = 1 A= = = 1 is the y-axis, which is a subspace of R2. y0 1 yyy 5 Eigenspace Geometrically speaking, multiplying a vector (x, y) in R2 by the matrix A in the example, corresponds to a reflection to the y-axis. x x0 x 0 AAAv = = += AA + y 0 y 0 y x 0 − x =−1 + 1 = 0 y y 6 Finding eigenvalues and eigenvectors Theorem 7.2: ( Finding eigenvalues and eigenvectors of a matrix A∈Mn×n ) Let A be an n×n matrix. (1) An eigenvalue of A is a scalar λ such that det(λI− A ) = 0 . (2) The eigenvectors of A corresponding to λ are the nonzero solutions of (λI− A ) x = Θ . 7 Finding eigenvalues and eigenvectors Note: some definitions of the eigenvalue problem: homogeneous system Axxxx= λ ⇒ AI= λ ⇒ ( λ IA− ) x = Θ (λI− A ) x = Θ has nonzero solutions for x iff det(λI− A ) = 0 Characteristic equation of A: det(λI− A ) = 0 Characteristic polynomial of A∈Mn×n: n n −1 det(λIAIA−= ) ( λ −=+ ) λλ cn−1 +++ cc 1 λ 0 8 Finding eigenvalues and eigenvectors Example 4: Finding eigenvalues and eigenvectors 2 −12 A = 1 − 5 Solution: Characteristic equation: λ − 2 12 det(λI− A ) = −1λ + 5 =λλ2 +3 +=+ 2 ( λ 1)( λ += 2) 0 ⇒ λ = − ,1 − 2 Eigenvalue: λ1 = − ,1 λ2 = −2 9 Finding eigenvalues and eigenvectors −312 x 0 λ = − ⇒ 1 (1) 1 1 (λ1I− A ) x = = −1 4 x2 0 −312 G.-J. E. 14 − ⇒ → −14 00 x1 4t 4 ⇒ = =t , t ≠ 0 x2 t 1 −4 12 x 0 ⇒ (λ I− A ) x =1 = (2) λ2 = − 2 2 −1 3 x2 0 −412 G.-J. E. 13 − ⇒ → −13 00 x1 3s 3 ⇒ = =s , s ≠ 0 x2 s 1 10 Finding eigenvalues and eigenvectors Example 5: Find the eigenvalues and corresponding eigenvectors for the matrix A. What is the dimension of the eigenspace of each eigenvalue? 2 1 0 A = 0 2 0 0 0 2 Solution: Characteristic equation: λ −2 − 1 0 λI−= A 0 λ − 2 0 =−= (2)0 λ 3 0 0λ − 2 Eigenvalue: λ = 2 11 Finding eigenvalues and eigenvectors The eigenspace of λ = 2: 010− x1 0 λ − = = (I A )000x x 2 0 000 x3 0 x1 s 1 0 x = 0 = s0 + t0, s,t ≠ 0 2 x3 t 0 1 1 0 ststR00,+ ∈ :the eigenspace of A corresponding to λ = 2 0 1 Thus, the dimension of its eigenspace is 2. 12 Finding eigenvalues and eigenvectors Notes: (1) If an eigenvalue λ1 occurs as a multiple root ( k times) for the characteristic polynominal, then λ1 has multiplicity k. (2) The multiplicity of an eigenvalue is greater than or equal to the dimension of its eigenspace. In Example. 5, k is 3 and the dimension of its eigenspace is 2. 13 Finding eigenvalues and eigenvectors Example 6: Find the eigenvalues of the matrix A and find a basis for each of the corresponding eigenspaces 1 0 0 0 0 1 5 −10 A = 1 0 2 0 1 0 0 3 Solution: Characteristic equation: According to the note on the previous slide, the dimension of λ −10 0 0 λ the eigenspace of 1 = 1 is at most 0λ − 1 − 5 10 to be 2. λI− A = λ λ −1 0λ − 20 For 2 = 2 and 3 = 3, the −10 0λ − 3 demensions of their eigenspaces are at most to be 1. =−(λ 1)2 ( λ − 2)( λ −= 3) 0 Eigenvalues: λ1 = ,1 λ2 = ,2 λ3 = 3 14 Finding eigenvalues and eigenvectors 0000 x1 0 00− 510x 0 (1) λ =1 ⇒ (λ I− A ) x = 2 = 1 1 −1010 − x 0 3 −1002 − x4 0 x1 −2t 0 − 2 G.-J.E. x s 1 0 ⇒ 2 ==+ s t , st, ≠ 0 x 2t 0 2 3 x4 t 0 1 0 − 2 1 0 ⇒ , is a basis for the eigenspace corresponding to λ1 = 1. 0 2 0 1 λ The dimension of the eigenspace of 1 = 1 is 2. 15 Finding eigenvalues and eigenvectors 1000 x1 0 01− 510x 0 (2) λ = 2 ⇒ (λ I− A ) x = 2 = 2 2 −1000 x 0 3 −100 − 1 x4 0 x1 0 0 G.-J.E. x 5t 5 ⇒ 2 = =t , t ≠ 0 x t 1 3 x4 0 0 0 5 λ ⇒ is a basis for the eigenspace corresponding to 2 = 2 1 0 λ The dimension of the eigenspace of 2 = 2 is 1. 16 Finding eigenvalues and eigenvectors 2000 x1 0 0 2− 5 10x 0 (3) λ = 3 ⇒ (λ I− A ) x = 2 = 3 3 −1010 x 0 3 −1000 x4 0 x1 0 0 G.-J.E. x −5t − 5 ⇒ 2 = =t , t ≠ 0 x 0 0 3 x4 t 1 0 − 5 ⇒ λ is a basis for the eigenspace corresponding to 3 = 3. 0 1 λ The dimension of the eigenspace of 3 = 3 is 1. 17 Eigenvalues for triangular matrices Theorem 7.3: If A is an n×n triangular matrix, then its eigenvalues are the entries on its main diagonal. 18 Eigenvalues for triangular matrices Example 7: Finding eigenvalues for triangular and diagonal matrices −100 0 0 2 0 0 0 20 0 0 (a) A = − 1 1 0 (b) A = 0 00 0 0 5 3− 3 0 00− 40 0 00 0 3 Solution: λ − 2 0 0 (a) λI−= A 1 λ − 1 0 =−−+= ( λλλ 2)( 1)( 3) 0 −5 − 3λ + 3 According to Theorem 3.2, the ⇒ λ=2, λ =1, λ =− 3 determinant of a triangular 1 2 3 matrix is the product of the entries on the main diagonal . (b) λ1=−1, λ 234 =2, λλ =0, =−4, λ 5 = 3 19 Eigenvalues and eigenvectors of L.T. A number λ is called an eigenvalue of a li near transformation TVV:→ if there is a nonzero vector x such that T ( x )= λ x . The vector x is called an eigenvector of T c orresponding to λ, and the set of all eigenvectors of λ (together with the zero vector) is called the eigenspace of λ. 20 Eigenvalues and eigenvectors of L.T. The definition of linear transformation functions was introduced in Chapter 6. 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. An example for a linear transformation T: R3→R3 Txxx(,,)(123=+ x 1 3,3 xxx 212 +− ,2) x 3 21 Eigenvalues and eigenvectors of L.T. Example 8: Find eigenvalues and eigenvectors for standard matrices Find the eigenvalues and corresponding e igenvectors for 1 3 0 A is the standard matrix for A = 3 1 0 T(x1, x2, x3) = ( x1 + 3x2, 3x1 + x2, –2x3) 0 0− 2 Solution λ −1 − 3 0 2 λI − A = − 3 λ −1 0 =+(λ 2) ( λ −= 4) 0 0 0 λ + 2 ⇒ eigenvalues λ1=4, λ 2 = − 2 For λ1 = 4, the corresponding eigenvector is (1, 1, 0). For λ2 = −2, the corresponding eigenvectors are (1, − 1 , 0) and (0, 0, 1). 22 Eigenvalues and eigenvectors of L.T.