298 CHAPTER 11. LIMITS OF VECTORS AND MATRICES

Lemma 11.4.8 Let M be a matrix of the form

M =

(A 0B C

)or

M =

(A B0 C

)where A is an r× r matrix and C is an (n− r)× (n− r) matrix. Then it follows thatdet(M) = det(A)det(B) and σ (M) = σ (A)∪σ (C) .

Proof: To verify the claim about the determinants, note(A 0B C

)=

(A 00 I

)(I 0B C

)Therefore,

det

(A 0B C

)= det

(A 00 I

)det

(I 0B C

).

But it is clear from the method of Laplace expansion that

det

(A 00 I

)= detA

and from the multilinear properties of the determinant and row operations that

det

(I 0B C

)= det

(I 00 C

)= detC.

The case where M is upper block triangular is similar.This immediately implies σ (M) = σ (A)∪σ (C) .

Theorem 11.4.9 Let A > 0 and let λ 0 be given in 11.11. If λ is an eigenvalue for A suchthat |λ |= λ 0, then λ/λ 0 is a root of unity. Thus (λ/λ 0)

m = 1 for some m ∈ N.

Proof: Applying Theorem 11.4.7 to AT , there exists v > 0 such that ATv = λ 0v. Inthe first part of the argument it is assumed v >> 0. Now suppose Ax = λx,x ̸= 0 andthat |λ |= λ 0. Then

A |x| ≥ |λ | |x|= λ 0 |x|

and it follows that if A |x|> |λ | |x| , then since v >> 0,

λ 0 (v, |x|)< (v,A |x|) =(ATv, |x|

)= λ 0 (v, |x|) ,

a contradiction. Therefore,A |x|= λ 0 |x| . (11.12)

It follows that ∣∣∣∣∣∑jAi jx j

∣∣∣∣∣= λ 0 |xi|= ∑j

Ai j∣∣x j∣∣

298 CHAPTER 11. LIMITS OF VECTORS AND MATRICESLemma 11.4.8 Let M be a matrix of the form“(52)w-(5)where A is an r X r matrix and C is an (n—r) X (n—r) matrix. Then it follows thatdet (M) = det (A) det(B) and o (M) =o (A)Uo(C).orProof: To verify the claim about the determinants, note(2)-(2)(2 8)w(2 t)oa(2)a(4 2)But it is clear from the method of Laplace expansion thatdet A 0 =detA07and from the multilinear properties of the determinant and row operations thatIdet 0 = det r 0 = detC.BC 0 CThe case where M is upper block triangular is similar.This immediately implies o (M) = 0 (A)Uo(C).Therefore,Theorem 11.4.9 Let A > 0 and let Xo be given in 11.11. If A is an eigenvalue for A suchthat |A| = Ao, then A/Ao is a root of unity. Thus (A/Ao)"”" = 1 for somem EN.Proof: Applying Theorem 11.4.7 to A’, there exists v > 0 such that A?v = Agv. Inthe first part of the argument it is assumed v >> 0. Now suppose Ax = Ax,a #0 andthat |A| = Ao. ThenAla > |A||a] = Ao|a|and it follows that if A |a| > |A||a|, then since v >> 0,Ao (v, |x|) < (v,A |x|) = (ATv, |x|) = Ao (v, |x|),a contradiction. Therefore,Ala| =Ao|a|. (11.12)It follows thatAix;J= Ao|ai| = YU Ai; |x|J