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Unformatted text preview: e η j . Moreover, each eigenspace is onedimensional.
56 In the dynamical system that we are considering, of course, we need to solve
the eigenvalueeigenvector 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 onedimensional, 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.
 Winter '14
 Equations

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