Eigenvalues and Eigenvectors 7.1 Eigenvalues and Eigenvecto

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1 7.1 November Eigenvalues and Eigenvecto

2 Goals Suppose A is square matrix of order n. Eigenvalues of A will be defined. Eigenvectors of A, corresponding to each eigenvalue, will be defined. Eigenspaces of A, corresponding to each eigenvalue, will be defined. 7.1 Eigenvalues and Eigenvecto

3 Main Definition Definition Suppose A is square matrix of order n A scalar λ is said to be a eigenvalue of A, if Equivalently, if Ax = λx for some vector x 0. (λi A)x = 0 for some vector x 0. The vector x is called an eigenvector corresponding to λ (or simply of λ). The zero vector 0 is not considered an eigenvectors. Remark. eigen means characteristic, in German. 7.1 Eigenvalues and Eigenvecto

4 Eigenspaces Suppose λ is an eigenvalue of A. Definition Set of all solutions E(λ) of this system of linear equation(s) (λi A)x = 0 is called the eigenspace of A corresponding to λ. Theorem 7.1 The eigenspace E(λ) is a subspace of R n, because it is the null space of the homogeneous system (λi A)x = 0. Note, the eigenspace E(λ) consisits of all the eigen vectors of λ and Eigenvalues and Eigenvecto

5 Theorem 7.2 Let A be a square matrix of order n. Then A scalar λ is an eigenvalue of A if and only if λi A = 0. A vector x is an eigenvector, of λ if and only x 0 and (λi A)x = 0. Proof. λ is an eigenvalue of A if and only if, for some x 0, (λi A)x = 0 λi A = 0. This establishes the first statement. The later is restatement of the definition. The proof is complete. 7.1 Eigenvalues and Eigenvecto

6 The Characteristic Equation Let A be a square matrix of order n. Then the equation λi A = 0 is called the characteristic equation of A. Expanding the determinant λi A, it follows λi A = λ n +c n 1 λ n 1 + +c 1 λ+c 0, which is a polynomial in λ, of degree n. This polynomial is called the characteristic polynomial of A. 7.1 Eigenvalues and Eigenvecto

7 Computing To compute the eigenvalues of A solve the characteristic equation λi A = 0. So, an eigen value would be a real or a complex number. Given an eigenvalue λ i to compute the eigenspace E(λ i ), solve the linear system (λ i I A)x = 0. As usual, to solve this, use reduction to row echelon form. Since λ i is an eigenvalue, at least one row of the echlon form will be zero. 7.1 Eigenvalues and Eigenvecto

8 Theorem 7.3 Theorem 7.3 If A is a diagonal matrix, then its eigenvalues are the diagonal entries. Proof. Let A = a a a nn be a diagonal matrix. Then, characteristic equation: λ a λi A = 0 λ a 22 0 = λ a nn 7.1 Eigenvalues and Eigenvecto

9 Continued Which is So, the eigenvalues are (λ a 11 )(λ a 22 ) (λ a nn ) = 0 The proof is completel λ = a 11,a 22,,a nn Theorem. If A is a triangular matrix, then its eigenvalues are the diagonal entries. Proof. Similar to the above. Reading Assignment: Read 7.1 Examples Eigenvalues and Eigenvecto

10 Exercise 6 Let A = Verify that λ 1 = 5 is a eigenvalue of A and x 1 = (1,2, 1) T is a corresponding eigenvector. Solution: We need to check Ax 1 = 5x 1, We have Ax 1 = So, assertion is verified = = = 5x Eigenvalues and Eigenvecto

11 Continued Verify that λ 2 = 3 is a eigenvalue of A and x 2 = ( 2,1,0) T is a corresponding eigenvector. Solution: We need to check Ax 2 = 3x 2, We have Ax 2 = = 3 So, assertion is verified = 3x 2. = Eigenvalues and Eigenvecto

12 Continued Verify that λ 3 = 3 is a eigenvalue of A and x 3 = (3,0,1) T is a corresponding eigenvector. Solution: We need to check Ax 3 = 3x 3, We have Ax 3 = So, assertion is verified = = = 3x Eigenvalues and Eigenvecto

13 Exercise 14 Let A = Determine whether x = (1,1,0) T is an eigenvector of A. Solution: We have Ax = = 2 λ for all λ. So, x is not an eigenvector of A. 7.1 Eigenvalues and Eigenvecto

14 Exercise 14b Determine whether x = ( 5,2,1) T is an eigenvector of A. Solution: We have Ax = = 0 = 0 2 = 0x So, x is an eigenvector and corresponding eigenvalue is λ = Eigenvalues and Eigenvecto

15 Exercise 14 c Determine whether x = (0,0,0) T is an eigenvector of A. Solution: No, 0 is, by definition, never an eigenvector. 7.1 Eigenvalues and Eigenvecto

16 Exercise 14 d Determine whether x = (2 6 3, 2 6+6,3) T is an eigenvector of A. Solution: We have = Ax = = (2 6+4) So, x is an eigenvector of A, for the eigenvalue λ = Eigenvalues and Eigenvecto

17 Exercise 20 Let A = (a) Find the characteristic equation of A. Solution: The characteristic polynomial is λ det(λi A) = 3 λ λ 3 = (λ+5)(λ 7)(λ 3). So, the characteristic equation is. (λ+5)(λ 7)(λ 3) = Eigenvalues and Eigenvecto

18 Exercise 20: Continued Find eigenvalues (and corresponding eigenvectors) of A. Solution: Solving the characteristic equation, the eigenvalues are λ = 5,7,3. To find an eigenvector corresponding to λ = 5, solve ( 5I A)x = 0 or x y = z 0 Solving, we get x = 16 9 t y = 4 9 t z = t. 7.1 Eigenvalues and Eigenvecto

19 Exercise 20: Continued So, the eigenspace of λ = 5 is {( 16 9 t, 4 ) } 9 t,t : t R. In particular, with t = 1, an eigenvector, for eigenvalue λ = 5, is ( 16 9, 4 9,1) T. 7.1 Eigenvalues and Eigenvecto

20 Exercise 20: Continued To find an eigenvector corresponding to λ = 7, wehave to solve (7I A)x = 0 or x y = z 0 Solving, we get So, that eigenspace of λ = 7 is x = 0 y = 2t z = t. {(0, 2t,t) : t R}. In particular, with t = 1, an eigenvector, for eigenvalue λ = 7, is (0, 2,1) T. 7.1 Eigenvalues and Eigenvecto

21 Exercise 20: Continued To find an eigenvector corresponding to λ = 3, wehave to solve (3I A)x = 0 or x y = z 0 Solving, we get So, that eigenspace of λ = 3 is x = 0 y = 0 z = t. {(0,0,t) : t R}. In particular, with t = 1, an eigenvector, for eigenvalue λ = 3, is (0,0,1) T. 7.1 Eigenvalues and Eigenvecto

22 Exercise 66 Let A = Find the dimension of the eigenspace corresponding to the eigenvalue λ = 3. Solution: The eigenspace E(3) is the solution space of the system (3I A)x = x, or x y z = Eigenvalues and Eigenvecto

23 Exercise 66: Continued or x y z = The coefficient matrix C = has rank Since rank(c)+nullity(c) = 3, nullity(c) = 1. Therefore, dime(3) = nullity(c) = Eigenvalues and Eigenvecto

24 : 7.1 See site. 7.1 Eigenvalues and Eigenvecto