To find the eigenvectors v v 1 v 2 corresponding to ?

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To find the eigenvectors v = v 1 v 2 corresponding to λ 1 = 3 , we need to solve ( A - λ 2 I ) v = 0 that is, 4 2 - 1 1 - 3 1 0 0 1 v 1 v 2 = 0 0 . 1 2 - 1 - 2 v 1 v 2 = 0 0 v 1 + 2 v 2 = 0 - v 1 - 2 v 2 = 0
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Since the second equation is just a scalar multiple of the first, this system has infinitely many solutions of the form v 1 = - 2 v 2 . Then the eigenvectors corresponding to λ 2 = 3 are all of the form v = - 2 c c = c - 2 1 where c is arbitrary. Why Eigenvectors? We have seen that the solutions to the homogeneous equation ˙ y = A y form a vector space of dimension n . Then every solution can be written in terms of n linearly- independent solutions. Eigenvectors give us a way of producing such solutions. We will show that every eigenvector v of the matrix A gives a solution of the homogeneous equation of the form y = e λx v . Suppose that v is an eigenvector of A with eigenvalue λ . Let y be as above. Then we compute ˙ y = λe λx v . Plugging y and ˙ y into the homogeneous equation we get λe λx v = A ( e λx v ) . Since e λx is a scalar we can move it past A to get λe λx v = e λx ( A v ) . But A v = λ v since v is an eigenvector, so equivalently we have λe λx v = e λx λ v . Thus y = λe λx v is a solution as desired. We summarize these results as follows: The Eigenvector/Eigenvalue Method Given the homogeneous equation ˙ y = A y . (1) Find the eigenvalues of A by solving det( A - λI ) = 0 . (2) For each eigenvalue λ , find a corresponding eigenvector v by solving the system ( A - λI ) v = 0 for v . (3) For each pair ( λ, v ) from steps (1) and (2), y = e λx v is a solution to the homogeneous equation.
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Case 1: Distinct real eigenvalues In this case the eigenvector/eigenvalue method allows us to find n linearly independent solutions, as a result of the following theorem: Theorem If λ 1 , λ 2 , . . . , λ k are distinct eigenvalues of A with corresponding eigenvectors v 1 , v 2 , . . . , v k , respectively, then these eigenvectors are linearly independent.
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