04-25-03

# 04-25-03 - Lecture 30 Andrei Antonenko 1 Eigenvectors...

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Lecture 30 Andrei Antonenko April 25, 2003 1 Eigenvectors. Eigenvalues Last lecture we saw, that in order to ﬁnd vectors, “stretched” by the operator with matrix A , we need to solve the characteristic equation det( A - λI ) = 0 , (1) which will give us diﬀerent λ i ’s — coeﬃcients, showing, how the vectors are changed after applying the operator. Now we will give the following deﬁnition. Deﬁnition 1.1. Let V be a vector space, and let A be a linear operator in vector space V . Then the vector x is called eigenvector of the operator A is there exist a number λ , which is called eigenvalue such that A ( x ) = λx. So, our goal is to ﬁnd eigenvectors, since the following proposition holds: Proposition 1.2. Let V be an n -dimensional vector space, and A be a linear operator. Then if there are n linearly independent eigenvectors, then the matrix of A is diagonal in the basis, consisting of eigenvectors. So far we know how to ﬁnd λ i ’s — eigenvalues of the operator. In order to ﬁnd eigenvectors, we need to solve the system ( A - λ i I ) x = 0 (2) for every found eigenvalue λ i . We will give an example of computing eigenvalues and eigenvectors. Example 1.3. Let A = ˆ 1 - 3 1 5 ! . Let’s compute its eigenvalues and eigenvectors. A - λI = ˆ 1 - λ - 3 1 5 - λ ! 1

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det( A - λI ) = (1 - λ )(5 - λ ) + 3 = 5 - 6 λ + λ 2 + 3 = λ 2 - 6 λ + 8 . The roots of this equation are λ 1 = 2 and λ 2 = 4 . Now we’ll ﬁnd eigenvectors corresponding to these eigenvalues. λ = 2 . Let’s subtract λ ’s from the diagonal. We’ll get the following matrix, and the system: ˆ - 1 - 3 1 3 ! , ( - x - 3 y = 0 x + 3 y = 0 From this system it follows that x = - 3 y , so each vector of the form ( - 3 c,c ) , i.e. ( - 3 , 1) is an eigenvector corresponding to the eigenvalue λ = 2 . λ = 4 . Let’s subtract λ ’s from the diagonal. We’ll get the following matrix, and the system: ˆ - 3 - 3 1 1 ! , ( -
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04-25-03 - Lecture 30 Andrei Antonenko 1 Eigenvectors...

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