eigenstuff

# Example find the eigenvalues and associated

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Example : Find the eigenvalues and associated eigenvectors of the matrix A = 7 0 - 3 - 9 - 2 3 18 0 - 8 . First we compute det( A - λ I ) via a cofactor expansion along the second column: 7 - λ 0 - 3 - 9 - 2 - λ 3 18 0 - 8 - λ = ( - 2 - λ )( - 1) 4 7 - λ - 3 18 - 8 - λ = - (2 + λ )[(7 - λ )( - 8 - λ ) + 54] = - ( λ + 2)( λ 2 + λ - 2) = - ( λ + 2) 2 ( λ - 1) . Thus A has two distinct eigenvalues, λ 1 = - 2 and λ 3 = 1. (Note that we might say λ 2 = - 2, since, as a root, - 2 has multiplicity two. This is why we labelled the eigenvalue 1 as λ 3 .) Now, to find the associated eigenvectors, we solve the equation ( A - λ j I ) x = 0 for j = 1 , 2 , 3. Using the eigenvalue λ 3 = 1, we have ( A - I ) x = 6 x 1 - 3 x 3 - 9 x 1 - 3 x 2 + 3 x 3 18 x 1 - 9 x 3 = 0 0 0 x 3 = 2 x 1 and x 2 = x 3 - 3 x 1 x 3 = 2 x 1 and x 2 = - x 1 . 3

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So the eigenvectors associated with λ 3 = 1 are all scalar multiples of u 3 = 1 - 1 2 . Now, to find eigenvectors associated with λ 1 = - 2 we solve ( A + 2 I ) x = 0 . We have ( A + 2 I ) x = 9 x 1 - 3 x 3 - 9 x 1 + 3 x 3 18 x 1 - 6 x 3 = 0 0 0 x 3 = 3 x 1 . Something different happened here in that we acquired no information about x 2 . In fact, we have found that x 2 can be chosen arbitrarily, and independently of x 1 and x 3 (whereas x 3 cannot be chosen independently of x 1 ). This allows us to choose two linearly independent eigenvectors associated with the eigenvalue λ = - 2, such as u 1 = (1 , 0 , 3) and u 2 = (1 , 1 , 3). It is a fact that all other eigenvectors associated with λ 2 = - 2 are in the span of these two; that is, all others can be written as linear combinations c 1 u 1 + c 2 u 2 using an appropriate choices of the constants c 1 and c 2 . Example : Find the eigenvalues and associated eigenvectors of the matrix A = - 1 2 0 - 1 . We compute det( A - λ I ) = - 1 - λ 2 0 - 1 - λ = ( λ + 1) 2 . Setting this equal to zero we get that λ = - 1 is a (repeated) eigenvalue. To find any associated eigenvectors we must solve for x = ( x 1 , x 2 ) so that ( A + I ) x = 0 ; that is, 0 2 0 0 x 1 x 2 = 2 x 2 0 = 0 0 x 2 = 0 . Thus, the eigenvectors corresponding to the eigenvalue λ = - 1 are the vectors whose second component is zero, which means that we are talking about all scalar multiples of u = (1 , 0). 4
Notice that our work above shows that there are no eigenvectors associated with λ = - 1 which are linearly independent of u . This may go against your intuition based upon the results of the example before this one, where an eigenvalue of multiplicity two had two linearly independent associated eigenvectors. Nevertheless, it is a (somewhat disparaging)
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