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First year Algebra: eigenvalues and eigenvectors Let A be a n n matrix. is an eigenvalue for A if there exists a vector u 0 such that Au u. If such a vector u exists, it is said to be an eigenvector associated with the eigenvalue . (I) How to find the eigenvalues and eigenvectors Suppose that is an eigenvalue for A . x1 Then there exists a non-zero vector u ... such that Au u or, x n equivalently, such that ( A I n )u 0 , where I n is the n n identity matrix. Since ( A I n )u 0 has a non-trivial solution u , the matrix ( A I n ) is not invertible, i.e. A I n 0. Note that A I n is a polynomial in , so you get the eigenvalues of A by finding the roots of A I n . Say 0 is such an eigenvalue. In order to find the eigenvectors associated with 0 , you have to solve the system Au 0 u for x1 ,..., xn . (II) Example: two by two matrices 3 2 . Let A 2 1 To find the eigenvalues of A , form the matrix A I 2 , find its determinant A I 2 and solve the equation A I 2 0 : 2 3 2 1 0 3 . A I 2 1 2 1 0 1 2 Now, A I 2 (3 )(1 ) (2)(2) 2 2 1 , so that the eigenvalues of A are the roots of 2 2 1 ( 1) 2 , i.e. A has a repeated eigenvalue: 0 1 . Now we have to solve the system Au 0 u . Here 0 1 , so that Au 0 u Au u , which yields x 3 2 x1 1 2 1 x2 x2 or, equivalently, 2 2 x1 0 . 2 2 x2 0 x A vector u 1 is therefore an eigenvector associated with the eigenvalue x2 0 1 if and only if its coordinates satisfy 2 x1 2 x1 2 x2 0 2 x2 0 , i.e. if and only if x1 x2 . Hence the eigenvectors associated with the eigenvalue 0 1 are of the 1 form u , where is a real number. 1 (III) Example: three by three matrices 1 3 1 1 3 Let A 7 5 1 . Then A I 3 7 5 6 6 2 6 6 A I 3 (3 ) 5 6 1 1 and 2 1 7 1 7 5 3 12 16 . 2 6 2 6 6 In general, it is not easy to find the roots of a polynomial of degree 3. We try some simple values and hope we will be lucky. For 2 , you get A I 3 (2) 3 12(2) 16 0 , so that 2 is a root of A I 3 . Hence ( 2) is a factor of A I 3 and A I 3 can be written in the form A I 3 ( 2)(a2 b c) a3 2 (b 2a) (c 2b) 2c. We now have A I 3 3 12 16 a3 2 (b 2a) (c 2b) 2c, so that a, b and c must satisfy a 1 b 2a 0 , c 2b 12 2c 16 which yields a 1 b2. c8 We are now in position to find all the factors of A I 3 : A I 3 ( 2)(2 2 8) ( 2) 2 ( 4). A therefore has two eigenvalues: 0 2 and 1 4 . To get the eigenvectors associated with 0 2 , we solve ( A 2I 3 )u 0 : 1 1 1 x1 0 x1 ( A 2 I 3 )u 7 7 1 x2 0 7 x1 6 6 0 x 0 6 x 1 3 x2 7 x2 6 x2 x3 0 x x2 x3 0 1 . x3 0 0 Hence the eigenvectors associated with the eigenvalue 0 2 are of the 1 form 1 . 0 To get the eigenvectors associated with 1 4 , we solve ( A 4I 3 )u 0 : 7 1 1 x1 0 7 x1 ( A 2 I 3 )u 7 1 1 x2 0 7 x1 6 6 6 x 0 6 x 1 3 x2 x2 6 x2 x3 0 x 0 x3 0 1 . x 2 x3 6 x3 0 Hence the eigenvectors associated with the eigenvalue 1 4 are of the form 0 1 . 1 (III) Exercises Find the eigenvalues and the eigenvectors of the matrix A , where (i) 0 2 A 3 5 (ii) 0 5 A 0 1 (iii) 1 3 3 A 3 5 3 6 6 4