Kenneth Kuttler

420 CHAPTER 15. NUMERICAL METHODS, EIGENVALUES

For x ∈ Fn, x ̸= 0, the Rayleigh quotient is defined by x∗Ax|x|2

. Now let the eigenvalues

of A be λ 1 ≤ λ 2 ≤ ·· · ≤ λ n and Axk = λ kxk where {xk}nk=1 is the above orthonormal

basis of eigenvectors mentioned in the corollary. Then if x is an arbitrary vector, thereexist constants, ai such that x= ∑

ni=1 aixi. Also,

|x|2 =n

∑i=1

aix∗i

∑j=1

a jx j = ∑i j

aia jx∗i x j = ∑

i jaia jδ i j =

∑i=1|ai|2 .

Therefore,

x∗Ax

|x|2=

(∑ni=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 smallest eigen-values of A. When x= xn, the Rayleigh quotient equals the largest eigenvalue and whenx= x1 the Rayleigh quotient equals the smallest eigenvalue. Suppose you calculate aRayleigh quotient. How close is it to some eigenvalue?

Theorem 15.2.6 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|. (15.4)

Proof: Let x= ∑nk=1 akxk where {xk}n

k=1 is the orthonormal basis of eigenvectors.

|Ax−qx|2 = (Ax−qx)∗ (Ax−qx)

∑k=1

akλ kxk−qakxk

)∗( n

∑k=1

akλ kxk−qakxk

)

∑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 =

∑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

∑k=1|ak|2 = (λ q−q)2 |x|2

which implies 15.4. ■

420 CHAPTER 15. NUMERICAL METHODS, EIGENVALUESFor « € F”, x £0, the Rayleigh quotient is defined by —. Now let the eigenvaluesof A be Ay < Ad < +++ < Ay and Ax, = Apay where {x,};_, is the above orthonormalbasis of eigenvectors mentioned in the corollary. Then if x is an arbitrary vector, thereexist constants, a; such that a = )"_, aja;. Also,n n n||? = aia; y ajxvj= \Vajaja; x; = aa j6ij = y la;|”.i=l j=l ij ij i=lTherefore,xv Ax (vey ix; ) (rn ajA ja) Lij GiajA jxj7 x;ie Eh lal ro lal- 2_ Lij aia jA jOij _ Le lal Ai E[A1,An]-n 2 n 2i=1 lai| i=1 laiIn other words, the Rayleigh quotient is always between the largest and the smallest eigen-values of A. When x = 2», the Rayleigh quotient equals the largest eigenvalue and whenx =a, the Rayleigh quotient equals the smallest eigenvalue. Suppose you calculate aRayleigh quotient. How close is it to some eigenvalue?Theorem 15.2.6 Let x 4 0 and form the Rayleigh quotient,xv Axja?Ill8Then there exists an eigenvalue of A, denoted here by Aq such that\Aax — ga||Ag—a| < \ar(15.4)Proof: Let « = Y7_, a,x, where {a, };_, is the orthonormal basis of eigenvectors.|Aa — gar|” = (Aw — qax)* (Aw — qa)=k=1= i (Aj aa) e (Ax —q) ass)j=! k=1= LA. q) a; ( (Ap — q) AX; j= Yo lal (Ap = qyik= e AA KLE — sae) (x AA KLE — wn)Now pick the eigenvalue A, which is closest to g. Thenn n|Ax —gax|? =) |ag|? (Ax — a) Yi la a—@) |ek=1 k=1which implies 15.4.