11.4. POSITIVE MATRICES 299

and so the complex numbers,Ai jx j, Aikxk

must have the same argument for every k, j because equality holds in the triangle in-equality. Therefore, there exists a complex number, µ i such that

Ai jx j = µ iAi j∣∣x j∣∣ (11.13)

and so, letting r ∈ N,Ai jx jµ

rj = µ iAi j

∣∣x j∣∣µr

j.

Summing on j yields∑

jAi jx jµ

rj = µ i ∑

jAi j∣∣x j∣∣µr

j. (11.14)

Also, summing 11.13 on j and using that λ is an eigenvalue for x, it follows from 11.12that

λxi = ∑j

Ai jx j = µ i ∑j

Ai j∣∣x j∣∣= µ iλ 0 |xi| . (11.15)

From 11.14 and 11.15,

∑j

Ai jx jµrj = µ i ∑

jAi j∣∣x j∣∣µr

j = µ i ∑j

Ai j

see 11.15︷ ︸︸ ︷µ j

∣∣x j∣∣µr−1

j

= µ i ∑j

Ai j

λ 0

)x jµ

r−1j = µ i

λ 0

)∑

jAi jx jµ

r−1j

Now from 11.14 with r replaced by r−1, this equals

µ2i

λ 0

)∑

jAi j∣∣x j∣∣µr−1

j = µ2i

λ 0

)∑

jAi jµ j

∣∣x j∣∣µr−2

j = µ2i

λ 0

)2

∑j

Ai jx jµr−2j .

Continuing this way,

∑j

Ai jx jµrj = µ

ki

λ 0

)k

∑j

Ai jx jµr−kj

and eventually, this shows

∑j

Ai jx jµrj = µ

ri

λ 0

)r

∑j

Ai jx j =

λ 0

)r

λ (xiµri )

and this says(

λ

λ 0

)r+1is an eigenvalue for

(A

λ 0

)with the eigenvector being

(x1µr1, · · · ,xnµ

rn)

T .

Now recall that r ∈ N was arbitrary and so this has shown that(λ

λ 0

)2

,

λ 0

)3

,

λ 0

)4

, · · ·

11.4. POSITIVE MATRICES 299and so the complex numbers,Aijxj, AikXkmust have the same argument for every k, j because equality holds in the triangle in-equality. Therefore, there exists a complex number, 1; such thatAijXj = MAij |x;| (11.13)and so, letting r € N,Aijx iH), = BiAiy |xj| Hj.Summing on j yieldsVAixju = 4; VAiy |xj| U5 (11.14)Jj jAlso, summing 11.13 on j and using that A is an eigenvalue for x, it follows from 11.12thatAx; =) Aijxj = MY Aij |x;| = MjAo |x|. (11.15)Jj JFrom 11.14 and 11.15,see 11.15r r r-1DeAix jee = we VAG bey] = es Ae |}J J JA i=H Lay, -) pt = = bi (5) Dave!jNow from 11.14 with r replaced by r — 1, this equalsAr r rew2 (FE) Savfolay! 02 (Ge) Sava ay ?—a2 (2) BasajContinuing this way,a\kLAist = Lj ‘(Z) YAizxju;Xo 7and eventually, this showsr r A " A " rAix jHi = Hi (Z-) VAijx) = (Z-) A (xiLt;)J Jr+l Aand this says (+) is an eigenvalue for (4 Io ) with the eigenvector being(MiMi y+ Xn) «Now recall that r € N was arbitrary and so this has shown thatAy (a)