Nevertheless it is a somewhat disparaging fact that

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linearly independent associated eigenvectors. Nevertheless, it is a (somewhat disparaging) fact that eigenvalues can have fewer linearly independent eigenvectors than their multiplicity suggests. Example : Find the eigenvalues and associated eigenvectors of the matrix A = 2 - 1 1 2 . We compute det( A - λ I ) = 2 - λ - 1 1 2 - λ = ( λ - 2) 2 + 1 = λ 2 - 4 λ + 5 . The roots of this polynomial are λ 1 = 2+ i and λ 2 = 2 - i ; that is, the eigenvalues are not real numbers. This is a common occurrence, and we can press on to find the eigenvectors just as we have in the past with real eigenvalues. To find eigenvectors associated with λ 1 = 2 + i , we look for x satisfying ( A - (2 + i ) I ) x = 0 - i - 1 1 - i x 1 x 2 = 0 0 - ix 1 - x 2 x 1 - ix 2 = 0 0 x 1 = ix 2 . Thus all eigenvectors associated with λ 1 = 2+ i are scalar multiples of u 1 = ( i, 1). Proceeding with λ 2 = 2 - i , we have ( A - (2 - i ) I ) x = 0 i - 1 1 i x 1 x 2 = 0 0 ix 1 - x 2 x 1 + ix 2 = 0 0 x 1 = - ix 2 , which shows all eigenvectors associated with λ 2 = 2 - i to be scalar multiples of u 2 = ( - i, 1). Notice that u 2 , the eigenvector associated with the eigenvalue λ 2 = 2 - i in the last example, is the complex conjugate of u 1 , the eigenvector associated with the eigenvalue λ 1 = 2 + i . It is indeed a fact that, if A ∈ M n × n ( R ) has a nonreal eigenvalue λ 1 = λ + with corresponding eigenvector ξ 1 , then it also has eigenvalue λ 2 = λ - with corresponding eigenvector ξ 2 = ¯ ξ 1 . 5
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