Linear Algebra, Theory And Applications, 2012a

384 NUMERICAL METHODS FOR FINDING EIGENVALUES
Now let the eigenvalues of A be λ 1 ≤ λ 2 ≤ ··· ≤ λ n and Ax k = λ k x k where {x k } n k=1 is
the above orthonormal basis of eigenvectors mentioned in the corollary. Then if x is an
arbitrary vector, there exist constants, a i such that
n∑
x = a i x i .
Also,
Therefore,
|x| 2 =
n∑
a i x ∗ i
i=1
= ∑ ij
x ∗ Ax
|x| 2 =
=
i=1
n∑
a j x j
j=1
a i a j x ∗ i x j = ∑ ij
a i a j δ ij =
n∑
|a i | 2 .
i=1
( ∑ (
n
i=1 a ix ∗ i ) ∑n
)
j=1 a jλ j x j
∑ n
i=1 |a i| 2
∑
ij a ia j λ j x ∗ i x ∑
j ij
∑ n
i=1 |a i| 2 =
a ia j λ j δ ij
∑ n
i=1 |a i| 2
=
∑ n
i=1 |a i| 2 λ i
∑ n
i=1 |a i| 2 ∈ [λ 1 ,λ n ] .
In other words, the Rayleigh quotient is always between the largest and the smallest eigenvalues
of A. When x = x n , the Rayleigh quotient equals the largest eigenvalue and when x = x 1
the Rayleigh quotient equals the smallest eigenvalue. Suppose you calculate a Rayleigh quotient.
How close is it to some eigenvalue?
Theorem 15.1.9 Let x ≠ 0 and form the Rayleigh quotient,
x ∗ Ax
|x| 2 ≡ q.
Then there exists an eigenvalue of A, denoted here by λ q such that
|Ax − qx|
|λ q − q| ≤ . (15.6)
|x|
Proof: Let x = ∑ n
k=1 a kx k where {x k } n k=1
is the orthonormal basis of eigenvectors.
|Ax − qx| 2 = (Ax − qx) ∗ (Ax − qx)
( n
) ∗ (
∑
n
)
∑
= a k λ k x k − qa k x k a k λ k x k − qa k x k
k=1
k=1
⎛
⎞ (
n∑
n
)
∑
= ⎝ (λ j − q) a j x ∗ ⎠
j (λ k − q) a k x k
j=1
k=1
= ∑ j,k
(λ j − q) a j (λ k − q) a k x ∗ j x k
=
n∑
|a k | 2 (λ k − q) 2
k=1