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.