Notes_on_ERO_operations_and_Eigenvalues_and_Eigenvectors

# 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 Denition. 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 dierential 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 coecient 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 innitely 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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