Pde

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Unformatted text preview: e η j . Moreover, each eigenspace is one-dimensional. 56 In the dynamical system that we are considering, of course, we need to solve the eigenvalue-eigenvector problem for A, not for T . Fortunately, however, since A and T commute, the eigenvectors for T are also eigenvectors for A. Indeed, since AT = T A, T (Aej ) = A(T ej ) = A(η j ej ) = η j (Aej ), and this equation states that Aej is an eigenvector for T with the same eigenvalue as ej . Since the eigenspaces of T are all one-dimensional, Aej must be a multiple of ej ; in other words, Aej = λj ej , for some number λj . To find the eigenvalues λj for A, we simply act on ej by A: we find that the first component of the vector Aej = λj ej is −2 + η j + η (n−1)j = −2 + (e2πij/n + e−2πij/n ) = −2 + 2 cos(2πj/n), where we have used Euler’s formula once again. On the other hand, the first component of ej is 1, so we immediately conclude that λj = −2 + 2 cos(2πj/n). It follows from the familiar formula cos(2α) = cos2 α − sin2 α that λj = −2 + 2[cos(πj/n)]2 − 2[sin(πj/n)]2 = −4[sin(πj/n)]2 . Note that λj = λn−j , and hence the ei...
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This document was uploaded on 01/12/2014.

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