358 CHAPTER 14. NUMERICAL METHODS, EIGENVALUES

which clearly converges to 0 as n→ ∞ since |λ1| > |λk|. An application of the ratio test orroot test for each term in the sum will show this. When k = 1, this block is

λ−n1 Jn

1 = λ−n1 Jn

k =

r1∑i=0

(n

i

)λ−n1 λn−i

1 N i1 =

(n

r1

)[λ−r11 Nr1

1 + en]

where limn→∞ en = 0 because it is a sum of bounded matrices which are multiplied by(ni

)/(nr1

). This quotient converges to 0 as n → ∞ because i < r1. It follows that 14.2 is of

the form

yn =λn1

snsn−1 · · · s1

(n

r1

)P

(λ−r11 Nr1

1 + en 0

0 En

)P−1x ≡ λn1

snsn−1 · · · s1

(n

r1

)wn

where En → 0, en → 0. Let(P−1x

)m1

denote the first m1 entries of the vector P−1x.

Unless a very unlucky choice for x was picked, it will follow that(P−1x

)m1

/∈ ker (Nr11 ) .

Then for large n, yn is close to the vector

λn1snsn−1 · · · s1

(n

r1

)P

(λ−r11 Nr1

1 0

0 0

)P−1x ≡ λn1

snsn−1 · · · s1

(n

r1

)w ≡ z ̸= 0

However, this is an eigenvector because

(A− λ1I)w =

A−λ1I︷ ︸︸ ︷P (J − λ1I)P

−1P

(λ−r11 Nr1

1 0

0 0

)P−1x =

P

N1

. . .

Jm − λ1I

P−1P

λ−r11 Nr1

1

. . .

0

P−1x

= P

(N1λ

−r11 Nr1

1 0

0 0

)P−1x = 0

Recall Nr1+11 = 0. Now you could recover an approximation to the eigenvalue as follows.

(Ayn,yn)

(yn,yn)≊

(Az, z)

(z, z)= λ1

Here ≊ means “approximately equal”. However, there is a more convenient way to identifythe eigenvalue in terms of the scaling factors sk.∥∥∥∥ λn1

sn · · · s1

(n

r1

)(wn −w)

∥∥∥∥∞

≊ 0

Pick the largest nonzero entry of w, wl. Then for large n, it is also likely the case thatthe largest entry of wn will be in the lth position because wm is close to w. From theconstruction,

λn1sn · · · s1

(n

r1

)wnl = 1 ≊

λn1sn · · · s1

(n

r1

)wl

In other words, for large nλn1

sn · · · s1

(n

r1

)≊ 1/wl

308 CHAPTER 14. NUMERICAL METHODS, EIGENVALUESwhich clearly converges to 0 as n — oo since |A;| > |Ax|- An application of the ratio test orroot test for each term in the sum will show this. When k = 1, this block isry—n yn —n Jn n —n\n—-i nzi n —-r 1AVI = APPT => ("Jai MONE = (") [Ay NT? + en|i=0where limy+o0 €n = 0 because it is a sum of bounded matrices which are multiplied by(7)/()- This quotient converges to 0 as n — oo because i < r,. It follows that 14.2 is ofthe formy= ("p(n ee 0 ) Ps at ("ws8n8n—1°°° 81 \T1 0 En SnSn—1°°° 81 \11where FE, > 0,e, — 0. Let (P'x),,, denote the first m entries of the vector P~!x.Unless a very unlucky choice for x was picked, it will follow that (Px), & ker (Ny').Then for large n, yy, is close to the vectorAT ("J( AVTNT! 0 ) Pixs AM (")w=z40Sn$n—1°°°S1 \T1 0 0 Sn$n—1°°° S81 \T1However, this is an eigenvector becauseA-AiIea==—x_—ee~ —Ti NTUa atw = PU PP ( Ai 0 t ; ) PoxNy APN]?P ms PP me Po'xJim — aT 0—Pri ry— p( MAM 9 / py0 0Recall Ny 1+! _ 9, Now you could recover an approximation to the eigenvalue as follows.(Ayns¥n) ~ (Azz)Gn¥n) ~~ nyHere © means “approximately equal”. However, there is a more convenient way to identifythe eigenvalue in terms of the scaling factors s,.rn”—— (") (mn — | = 0Sn-°° $1 \T1 ooPick the largest nonzero entry of w,w;. Then for large n, it is also likely the case thatthe largest entry of w,, will be in the I” position because w,, is close to w. From theconstruction,Mt n AL n— Wn! = 1 & —— WiSn * $1 T1 Sn Sy, TLIn other words, for large n