358 CHAPTER 15. NUMERICAL METHODS, EIGENVALUES
Therefore,
x∗Ax|x|2
=(∑n
i=1 aix∗i )(
∑nj=1 a jλ jx j
)∑
ni=1 |ai|2
=∑i j aia jλ jx∗i x j
∑ni=1 |ai|2
=∑i j aia jλ jδ i j
∑ni=1 |ai|2
=∑
ni=1 |ai|2 λ i
∑ni=1 |ai|2
∈ [λ 1,λ n] .
In other words, the Rayleigh quotient is always between the largest and the smallesteigenvalues of A. When x = xn, the Rayleigh quotient equals the largest eigenvalue andwhen x = x1 the Rayleigh quotient equals the smallest eigenvalue. Suppose you calculatea Rayleigh quotient. How close is it to some eigenvalue?
Theorem 15.4.2 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∣∣λ q−q∣∣≤ |Ax−qx|
|x|. (15.2)
Proof: Let x = ∑nk=1 akxk where {xk}n
k=1 is the orthonormal basis of eigenvectors.
|Ax−qx|2 = (Ax−qx)∗ (Ax−qx)
=
(n
∑k=1
akλ kxk−qakxk
)∗( n
∑k=1
akλ kxk−qakxk
)
=
(n
∑j=1
(λ j−q)a jx∗j
)(n
∑k=1
(λ k−q)akxk
)= ∑
j,k(λ j−q)a j (λ k−q)akx∗jxk
=n
∑k=1|ak|2 (λ k−q)2
Now pick the eigenvalue, λ q which is closest to q. Then
|Ax−qx|2 =n
∑k=1|ak|2 (λ k−q)2 ≥ (λ q−q)2
n
∑k=1|ak|2 = (λ q−q)2 |x|2
which implies 15.2. ■
Example 15.4.3 Consider the symmetric matrix A =
1 2 32 2 13 1 4
. Let
x =(1,1,1)T .
How close is the Rayleigh quotient to some eigenvalue of A? Find the eigenvector andeigenvalue to several decimal places.