Notes_on_ERO_operations_and_Eigenvalues_and_Eigenvectors

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Unformatted text preview: Notes on ERO operations and Eigenvalues and Eigenvectors Elementary Row Operations Denition. An elementary row operation (ERO) on an augmented matrix pro- duces a new augmented matrix corresponding to a new (but equivalent) system of linear equations. Two matrices are said to be row-equivalent if one can be obtained from the other by a nite sequence of elementary row operations. Theorem. There are matrix. These are: 3 elementary row operations possible on an augmented 1. Interchanging two rows. 2. Multiplying a row by a non-zero constant. 3. Adding a multiple of a row to another row. Example. An example of each of the ERO's are given below. 1. Interchanging the 1st and 2nd rows. 0 −1 2 1 2 −3 3 4 0 3 =⇒ 4 1 −1 0 2 2 1 −3 0 3 3 4 4 1 R1 ↔ R2 2. Multiply the 1st row by 2 1 5 −4 3 −2 1 2 6 −2 −3 0 =⇒ 1 2 1 r1 → R1 2 to produce a new 1st row. 1 1 5 −2 3 −2 3 −3 1 −1 0 2 3. Add −2 times the 1st row to the 3rd row to produce a new 3rd row. 1 1 2 0 3 2 1 −4 −2 5 3 −1 =⇒ −2 1 0 0 2 3 −3 −4 −2 13 3 −1 −8 −2r1 + r3 → R3 Eigenvalues and Eigenvectors Eigenvalues is one of the most important topics in Linear Algebra. Some applications of eigenvalues are studying populations growth, solving systems of dierential equations, in quadratic forms, and in engineering and science. The concept of eigenvalues comes from the eigenvalue problem. Eigenvalue Problem If A is a n × n matrix, do there exist non-zero matrices x of size n × 1 such that Ax is a scalar multiple of x? The fundamental equation to the eigenvalue problem is Ax = λx. where λ is the eigenvalue of A and x is the eigenvector of A.       1 4 1 2 Example. Let A = 2 3 , and x1 = 1 and x2 = −1 . Verify that λ1 = 5 is an eigenvalue of A corresponding to x1 and λ2 = −1 is an eigenvalue of A corresponding to x2 . Solution.      1 4 1 5 = Ax1 = 2 3 1 5      1 4 2 −2 Ax2 = = 2 3 −1 1   1 = λ 1 x1 1   2 = (−1) = λ 2 x2 . −1 = (5) Finding Eigenvalues and Eigenvectors The procedure to nd the eigenvalues and its corresponding eigenvectors is given below. Given the fundamental equations, we have Ax = λx Ax = λIx O = λIx − Ax O = (λI − A)x which is a homogenous system. The homogenous system has non-zero solutions if and only if the coecient matrix (λI − A) is singular; that is, if and only if, the determinant of |λI − A| = 0. The equation |λI − A| = 0 is called the characteristic equation of A. 2 Example. Find the eigenvalues and eigenvectors of the matrix A =  1 2  4 . 3 Solution. The characteristic equation of A is     1 0 1 4 0 = |λI − A| = λ − 0 1 2 3     λ 0 1 4 − 0 = 0 λ 2 3 λ − 1 −4 0 = −2 λ − 3 (λ − 1)(λ − 3) − (−2)(−4) 0 = 0 = λ2 − 4λ − 5 0 = (λ − 5)(λ + 1). This yields two eigenvalues, λ1 = 5 and λ2 = −1. We now go on to nd the eigenvectors of each of the eigenvectors. For λ1 = 5, substitute into the homogenous system, we have  λ−1 −2 −4 λ−3  0 0 =⇒ =   (5) − 1 −4 0 −2 (5) − 3 0   4 −4 0 . −2 2 0 Then row-reducing the system, we have   4 −4 0 −2 2 0   1 −1 0 −2 2 0   1 −1 0 . 0 0 0 1 r1 → R1 4 2r1 + r2 → R2 The row of zeros tells us that the system has innitely many solutions and thus has to be represented by parametric representation. The system in equation form is x1 − x2 = 0. Let x2 = t, t ∈ R. Then x1 = x2 = t as well. Then the eigenvectors x is given by   x x= 1 x2    t 1 = =t , t ∈ R. t 1 Thus, all eigenvectors corresponding to λ1 = 5 are non-zero scalar multiples of   1 . 1 3 For λ1 = −1, substitute into the homogenous system, we have  λ−1 −2 −4 λ−3  0 0   (−1) − 1 −4 0 −2 (−1) − 3 0   −2 −4 0 . −2 −4 0 =⇒ = Then row-reducing the system, we have   −2 −4 0 −2 −4 0   1 2 0 −2 −4 0   1 2 0 . 0 0 0 1 − r1 → R1 2 2r1 + r2 → R2 The system in equation form is x1 + 2x2 = 0. Let x2 = t, t ∈ R. Then x1 = −2x2 = −2t as well. Then the eigenvectors x is given by x=   x1 = x2     −2t −2 =t , t ∈ R. t 1 Thus, all eigenvectors corresponding to λ1 = −1 are non-zero scalar multiples of   2 . 1 Exercise. Find the eigenvalues and corresponding eigenvectors of the matrix 1 A= 1 −1 2 2 −1 −2 1 . 0 Solution. The eigenvalues and corresponding eigenvectors are −2 x1 = 1 1 2 x2 = −1 1 −2 x3 = −1 . 1 λ1 = 1 λ2 = −1 λ3 = 3 4 ...
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