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Unformatted text preview: r for T corresponding to the eigenvalue λ, the components of x must satisfy the vector equation T x = λx. In terms of components,
the vector equation becomes
x2 = λx1 , x3 = λx2 , . . . , xn = λxn−1 , x1 = λxn . (2.19) Thus x3 = λ2 x1 , x4 = λ3 x1 , and so forth, yielding ﬁnally the equation
x1 = λn x1 .
x2 = λn x2 , ... , xn = λn xn . Since at least one of the xi ’s is nonzero, we must have
λn = 1. (2.20) This equation is easily solved via Euler’s formula:
1 = e2πi ⇒ (e2πi/n )n = 1, and similarly [(e2πi/n )j ]n = 1, for 0 ≤ j ≤ n − 1. Thus the solutions to (2.20) are
λ = ηj , for 0 ≤ j ≤ n − 1, where
η = e2πi/n . (2.21) For each choice of j , we can try to ﬁnd eigenvectors corresponding to η j . If
we set x1 = 1, we can conclude from (2.19) that
x2 = η j , x3 = η 2j , ... , xn = η (n−1)j , thereby obtaining a nonzero solution to the eigenvector equation. Thus for each
j , 0 ≤ j ≤ n − 1, we do indeed have an eigenvector 1 ηj ej = η 2 j , ·
for the eigenvalu...
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This document was uploaded on 01/12/2014.
- Winter '14