**Unformatted text preview: **n. Since these λj ’s are all distinct (cos θ is a strictly decreasing function of θ on
(0, π ), and a = 0 = c), A must be diagonalizable—recall (7.2.6). Finally, the
k
k
k th component of any eigenvector associated with λj satisﬁes xk = αr1 + βr2
with α + β = 0, so
xk = α Copyright c 2000 SIAM c
a k /2 eiπjk/(n+1) − e−iπjk/(n+1) = 2iα c
a k /2 sin j kπ
n+1 . Buy online from SIAM
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516
Chapter 7
Eigenvalues and Eigenvectors
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Setting α = 1/2i yields a particular eigenvector associated with λj as It is illegal to print, duplicate, or distribute this material
Please report violations to [email protected] ⎛ (c/a)1/2 sin (1jπ/(n + 1)) ⎞
⎜ (c/a)2/2 sin (2jπ/(n + 1)) ⎟
⎜
⎟
3/2
xj = ⎜ (c/a) sin (3jπ/(n + 1)) ⎟ .
⎜
⎟
.
⎝
⎠
.
.
n/2
(c/a)
sin (njπ/(n + 1))
Because the λj ’s are distinct, {x1 , x2 , . . . , xn } is a complete linearly independent set—recall (7.2.3)—so P = x1 | x2 | · · · | xn diagonalizes A. D
E It’s often the case that a right-hand and left-hand eigenvector for some
eigenvalue is known. Rather than starting from scratch to ﬁnd additional eigenpairs, the known information can be used to reduce or “deﬂate” the problem to
a smaller one as described in the following example. T
H Example 7.2.6 IG
R Deﬂation. Suppose that right-hand and left-hand eigenvectors x and y∗ for an
eigenvalue λ of A ∈ n×n are already known, so Ax = λx and y∗ A = λy∗ .
Furthermore, suppose y∗ x = 0 —such eigenvectors are guaranteed to exist if λ
is simple or if A is diagonalizable (Exercises 7.2.23 and 7.2.22).
Problem: Use x and y∗ to deﬂate the size of the remaining eigenvalue problem. Y
P Solution: Scale x and y∗ so that y∗ x = 1, and construct Xn×n−1 so that its
columns are an orthonormal basis for y⊥ . An easy way of doing this is to build
˜
˜
a reﬂector R = y | X having y = y/ y 2 as its ﬁrst column as described...

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