7.1 November 6 7.1 Eigenvalues and Eigenvecto
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
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
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 0. 7.1 Eigenvalues and Eigenvecto
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
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
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
Theorem 7.3 Theorem 7.3 If A is a diagonal matrix, then its eigenvalues are the diagonal entries. Proof. Let A = a 11 0 0 0 a 22 0 0 0 a nn be a diagonal matrix. Then, characteristic equation: λ a 11 0 0 λi A = 0 λ a 22 0 = 0 0 0 λ a nn 7.1 Eigenvalues and Eigenvecto
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 4-7. 7.1 Eigenvalues and Eigenvecto
Exercise 6 Let 2 2 3 A = 2 1 6. 1 2 0 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 2 2 3 Ax 1 = 2 1 6 1 2 0 So, assertion is verified. 1 2 1 = 5 10 5 = 5 1 2 1 = 5x 1. 7.1 Eigenvalues and Eigenvecto
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 2 2 3 Ax 2 = 2 1 6 1 2 0 = 3 So, assertion is verified. 2 1 0 2 1 0 = 3x 2. = 6 3 0 7.1 Eigenvalues and Eigenvecto
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 2 2 3 Ax 3 = 2 1 6 1 2 0 So, assertion is verified. 3 0 1 = 9 0 3 = 3 3 0 1 = 3x 3. 7.1 Eigenvalues and Eigenvecto
Exercise 14 Let A = 1 0 5 0 2 4 1 2 9 Determine whether x = (1,1,0) T is an eigenvector of A. Solution: We have 1 0 5 1 1 1 Ax = 0 2 4 1 = 2 λ 1 1 2 9 0 1 0. for all λ. So, x is not an eigenvector of A. 7.1 Eigenvalues and Eigenvecto
Exercise 14b Determine whether x = ( 5,2,1) T is an eigenvector of A. Solution: We have 1 0 5 5 0 5 Ax = 0 2 4 2 = 0 = 0 2 = 0x. 1 2 9 1 0 1 So, x is an eigenvector and corresponding eigenvalue is λ = 0. 7.1 Eigenvalues and Eigenvecto
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
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+12 4 6 6 6+12 1 0 5 0 2 4 1 2 9 2 6 3 2 6+6 3 = (2 6+4) 2 6 3 2 6+6 3 So, x is an eigenvector of A, for the eigenvalue λ = 2 6+4.. 7.1 Eigenvalues and Eigenvecto
Exercise 20 Let A = 5 0 0 3 7 0 4 2 3 (a) Find the characteristic equation of A. Solution: The characteristic polynomial is λ+5 0 0 det(λi A) = 3 λ 7 0 4 2 λ 3 = (λ+5)(λ 7)(λ 3). So, the characteristic equation is. (λ+5)(λ 7)(λ 3) = 0. 7.1 Eigenvalues and Eigenvecto
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 0 0 0 x 0 3 12 0 y = 0. 4 2 8 z 0 Solving, we get x = 16 9 t y = 4 9 t z = t. 7.1 Eigenvalues and Eigenvecto
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
Exercise 20: Continued To find an eigenvector corresponding to λ = 7, wehave to solve (7I A)x = 0 or 12 0 0 x 0 3 0 0 y = 0. 4 2 4 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
Exercise 20: Continued To find an eigenvector corresponding to λ = 3, wehave to solve (3I A)x = 0 or 8 0 0 x 0 3 4 0 y = 0. 4 2 0 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
Exercise 66 Let A = 3 1 1 0 3 1 0 0 3 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 3 3 1 1 0 3 3 1 0 0 3 3. x y z = 0 0 0 7.1 Eigenvalues and Eigenvecto
Exercise 66: Continued or 0 1 1 0 0 1 0 0 0 x y z = The coefficient matrix 0 1 1 C = 0 0 1 has rank 2. 0 0 0 Since rank(c)+nullity(c) = 3, nullity(c) = 1. Therefore, dime(3) = nullity(c) = 1. 0 0 0 7.1 Eigenvalues and Eigenvecto
: 7.1 See site. 7.1 Eigenvalues and Eigenvecto