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[ 0 => 'Eigenvalues and Eigenvectors', 1 => 'We review here the basics of computing eigenvalues and eigenvectors. Eigenvalues and eigenvectors play a prominent role in the study of ordinary differential equations and in many applications in the physical sciences. Expect to see them come up in a variety of contexts!', 2 => 'Definitions', 3 => false, 4 => false, 5 => 'Let A be an nn matrix. The number is an eigenvalue of A if there exists a non-zero vector v such that', 6 => 'Av=v ', 7 => 'In this case, vector v is called an eigenvector of A corresponding to . ', 8 => false, 9 => false, 10 => false, 11 => false, 12 => 'Computing Eigenvalues and Eigenvectors', 13 => false, 14 => 'We can rewrite the condition Av=v as', 15 => '(A−I)v=0 ', 16 => 'where I is the nn identity matrix. Now, in order for a non-zero vector v to satisfy this equation, A−I must not be invertible. ', 17 => 'Otherwise, if A−I has an inverse,', 18 => '(A−I)−1(A−I)v v = = (A−I)−10 0 ', 19 => 'But we are looking for a non-zero vector v .', 20 => 'That is, the determinant of A−I must equal 0. We call p()=det(A−I) the characteristic polynomial of A. The eigenvalues of A are simply the roots of the characteristic polynomial of A.', 21 => false, 22 => 'Example', 23 => false, 24 => 'Let A=2 −1 −4 −1 . Then', 25 => 'p() = = = = det2− −1 −4 −1− (2−)(−1−)−(−4)(−1) 2−−6 (−3)(+2) ', 26 => 'Thus, 1=3 and 2=−2 are the eigenvalues of A.', 27 => false, 28 => 'To find eigenvectors v= v1 v2 vn corresponding to an eigenvalue , we simply solve the system of linear equations given by', 29 => '(A−I)v=0 ', 30 => false, 31 => 'Example', 32 => false, 33 => 'The matrix A=2 −1 −4 −1 of the previous example has eigenvalues 1=3 and 2=−2. Let's find the eigenvectors corresponding to 1=3. Let v=v2v1 . Then (A−3I)v=0 gives us', 34 => '2−3 −1 −4 −1−3 v1 v2 =0 0 ', 35 => 'from which we obtain the duplicate equations', 36 => '−v1−4v2 −v1−4v2 = = 0 0 ', 37 => 'If we let v2=t, then v1=−4t. All eigenvectors corresponding to 1=3 are multiples of 1−4 and thus the eigenspace corresponding to 1=3 is given by the span of 1−4 . That is, 1−4 is a basis of the eigenspace corresponding to 1=3.', 38 => false, 39 => 'Repeating this process with 2=−2, we find that', 40 => '4v1−4V2 −v1+v2 = = 0 0 ', 41 => 'If we let v2=t then v1=t as well. Thus, an eigenvector corresponding to 2=−2 is 11 and the eigenspace corresponding to 2=−2 is given by the span of 11 . 11 is a basis for the eigenspace corresponding to 2=−2.', 42 => false, 43 => 'In the following example, we see a two-dimensional eigenspace.', 44 => false, 45 => 'Example', 46 => false, 47 => 'Let A= 5 4 −4 8 1 −4 16 8 −11 . Then p()=det 5− 4 −4 8 1− −4 16 8 −11− =(−1)(+3)2 after some algebra! Thus, 1=1 and 2=−3 are the eigenvalues of A. Eigenvectors v= v1 v2 v3 corresponding to 1=1 must satisfy', 48 => false, 49 => '4v1 4v1 −4v1 + − 8v2 4v2 + + − 16v3 8v3 12v3 = = = 0 0 0 ', 50 => 'Letting v3=t, we find from the second equation that v1=−2t, and then v2=−t. All eigenvectors corresponding to 1=1 are multiples of −2 −1 1 , and so the eigenspace corresponding to 1=1 is given by the span of −2 −1 1 . −2 −1 1 is a basis for the eigenspace corresponding to 1=1.', 51 => false, 52 => 'Eigenvectors corresponding to 2=−3 must satisfy', 53 => false, 54 => '8v1 4v1 −4v1 + + − 8v2 4v2 4v2 + + − 16v3 8v3 8v3 = = = 0 0 0 ', 55 => 'The equations here are just multiples of each other! If we let v3=t and v2=s, then v1=−s−2t. Eigenvectors corresponding to 2=−3 have the form', 56 => ' −1 1 0 s+ −2 0 1 t', 57 => 'Thus, the eigenspace corresponding to 2=−3 is two-dimensional and is spanned by −1 1 0 and −2 0 1 . −1 1 0 −2 0 1 is a basis for the eigenspace corresponding to 2=−3.', 58 => false, 59 => 'Notes', 60 => 'Eigenvalues and eigenvectors can be complex-valued as well as real-valued. ', 61 => false, 62 => 'The dimension of the eigenspace corresponding to an eigenvalue is less than or equal to the multiplicity of that eigenvalue. ', 63 => false, 64 => 'The techniques used here are practical for 22 and 33 matrices. Eigenvalues and eigenvectors of larger matrices are often found using other techniques, such as iterative methods.', 65 => 'Key Concepts', 66 => false, 67 => 'Let A be an nn matrix. The eigenvalues of A are the roots of the characteristic polynomial', 68 => 'p()=det(A−I)', 69 => 'For each eigenvalue , we find eigenvectors v= v1 v2 vn by solving the linear system', 70 => '(A−I)v=0 ', 71 => 'The set of all vectors v satisfying Av=v is called the eigenspace of A corresponding to .' ]