Reducible matrices

Since all the terms of the sum are positive, we must have Xn

i=1

v_iN_i

u_i(t) Ni −C₁

−

2

= 0, and hence,

u_i(t) Ni −C₁

−

2

= 0 ⇒ u_i(t)

Ni −C₁ ≥0 for eachi∈ {1, . . . , n}. Therefore, u_i(t)≥C₁N_i.

To be able to extend the results from the previous section to reducible matrices, we need to be able to divide by N_i. But this, in general, is not possible. However, further developments (as will be seen below) allow for the vectorNto have zeros.

As seen in Example 3.3.1, Lemma 3.2.3 does not hold for reducible matrices, even when a positive right eigenvector exists. This is because the eigenvalue0may have algebraic multiplicity greater than one and v may have zeros. Indeed ifN>0, then the right eigenspace corresponding to 0 are spanned by the vectors

N_i = (0,Nⁱ,0,y^g+1_i ,· · · ,y^s_i);i= 1, . . . , g; (3.19) whereNⁱ >0 is a normalised right eigenvector forA_i corresponding to0and

y^g+1_i = (−A_g+1)⁻¹A_g+1,iNⁱ y^j_i = (−Aj)⁻¹



Aj,iNⁱ+

j−1X

h=g+1

A_j,hy^h_i



.

The vectorsy_i^h are well defined since s(A_h)<0for allh=g+ 1, . . . , s. The matrices A_j and A_h,j described here are the matrices in the normal form (2.8). SinceA_h, h=g+ 1, . . . , sare all irreducible, Theorem 2.1 of [51] guarantees that the matrices(−A_h)⁻¹ are all strictly positive, hencey^h_i ≥0. The left eigenspace is then spanned by the vectors

v_i= (0,vⁱ,0) ∀i= 1, . . . , g vⁱA_i=0. (3.20) In order to state analogous results (to Lemma 3.2.3 and Theorem 3.2.5) for reducible matrices, regardless of whether a positive eigenvector exists or not, we need the following definitions.

We partition the set {1, . . . , n} into sets I_k = {n1 +· · ·+n_k−1 + 1, . . . , n1 +· · ·+n_k} for 1≤k≤scorresponding to matrices A_k that haver as an eigenvalue. Define

I^Θ=I_k1 ∪ · · · ∪I_kl, (3.21) whereΘ ={k₁, . . . , k_l} ⊂ {1, . . . , s}is arbitrary. Further, we define

X^Θ =Span{e_i}i∈I^Θ, (3.22)

where ei = (δ_ik)_1≤k≤n and δ_ik is the Kronecker delta. Let X_k = X^{k} so that for general Θ defined above, X^Θ = X_k1 ⊕ · · · ⊕X_kl. Let N= (N¹,· · · ,N^s) be an eigenvector of A. For

a non-negative left eigenvector of A, we shall write v= (v¹,· · · ,v^s). Then using the normal form ofA in (2.8), we see that vⁱ =0 ifs(A_i)<0.

To make this indexing clearer, suppose in the normal form of A in (2.8), and the matrices A₁, A_g and A_s−1 all have r = 0 as an eigenvalue (and s(A_i) < 0 for i 6= 1, g, s−1). Then I1 = {1, . . . , n1}, Ig = {n1 +· · ·+ng−1 + 1, n1 +· · ·+ng−1 + 2, . . . , n1 +· · ·+ng} and I_s−1 ={n₁+· · ·+n_s−2+ 1, n₁+· · ·+n_s−2+ 2, . . . , n₁+· · ·+n_s−1}. I^Θ=I₁∪I_g∪I_s−1, whereΘ ={r1, r2, r3}={1, g, s−1} ⊂ {1, . . . , s}.

Consider an arbitrary reducible ML matrixAwith dominant eigenvalues(A) = 0. The following result holds for this matrix.

Lemma 3.3.2. Let Θ be an arbitrary set of indices from the set {1, . . . , s} such that X^Θ is invariant underA. Let A^Θ=A|X^Θ :X^Θ→X^Θ,N^Θ= (N^k)_k∈Θ and v^Θ= (v^k)_k∈Θ. Then

1. v^ΘA^Θ=0.

2. If A_ij =0 wheneveri∈Θandj /∈Θ, thenA^ΘN^Θ=0.

3. If A^ΘN^Θ=0 andA_kN^k=0 for somek∈Θ, then A_k,l =0 for all k > l∈Θ. Hence if A_kN^k=0 for all k∈Θ, thenv^lA_l =0 for all l∈Θ.

Proof. ForX^Θ to be invariant, A_ij =0 fori /∈Θandj ∈Θ. SincevA=0, then if k∈Θ, v^kA_k+

Xs

l=k+1

v^lA_l,k =0.

This together with the invariance ofX^Θ under A implies that v^kA_k+

Xs

l=k+1,l∈Θ

v^lA_l,k =0 ∀k∈Θ, hencev^ΘA^Θ=0.

To prove item (2), pickk∈Θ. Then 0=A_kN^k+

k−1X

l=1

A_k,lN^l=A_kN^k+

k−1X

l=1,l∈Θ

A_k,lN^l=A^ΘN^Θ.

Finally, fork∈Θ,

(A^ΘN^Θ)_k=

k−1X

l=1

AN^l+AN^k=

k−1X

l=1

AN^l=0.

Ifl∈Θ,thenN^l>0 andAk,l ≥0 henceAk,l =0. Therefore, 0=v^lA_l+

Xs

k=l+1

v^kA_k,l=v^lA_l. This gives the required result.

Lemma 3.3.3. Let I^Θ be an arbitrary set of indices defined by (3.21) and N= (N₁, . . . , N_n) be any right eigenvector ofAwithN_i >0fori∈I^ΘandA_rN^r =0for anyr ∈Θ. Then there is a positive constantα such that for any non zero vector m= (m₁, . . . , m_n)∈Rⁿ satisfying

X

i∈Ik

vimi = 0 (3.23)

for eachk∈Θ, the following inequality holds:

X

i∈I^Θ

X

j∈I^Θ

viaijNj

m_j N_j −mi

N_i 2

> αX

i∈I^Θ

vi

N_im²_i. (3.24)

Proof. By Lemma 3.3.2,v^k>0for k∈Θ. Thus kmk=

sX

i∈Θ

v_i N_im²_i

is a norm on the space Y ={e_k}k∈I^Θ. If m=0, the result holds trivially, so we assume that m6=0. We now divide both sides of Equation 3.24 bykmk² to get

X

i∈I^Θ

X

j∈I^Θ

viaijNj

m˜_j N_j −m˜i

N_i 2

> α; (3.25)

where m˜ = m/km. Note that the vector m˜ satisfies the assumptions of the lemma. Now suppose that (3.3.3) is false. Then there exists a sequence ( ˜m^l)_l≥0 on the unit sphere in Y satisfying the relation

X

i∈I^Θ

X

j∈I^Θ

v_ia_ijN_j m˜^l_j N_j −m˜^l_i

N_i

!2

≤ 1

l. (3.26)

Since the sequence is on the unit sphere in Y, it contains a convergent subsequence, whose limit ism˜˜ ∈Y. This limit also exists on the unit sphere (Bolzano-Weierstrass) and it satisfies the assumptions of the lemma. Taking limits on (3.26), we get

X

i∈I^Θ

X

j∈I^Θ

viaijNj

˜˜ m_j N_j −m˜˜i

N_i

!2

= 0. (3.27)

For ri, . . . , rk∈I^Θ, equation (3.27) is equivalent to the equation below:

X

i∈Ir1

X

j∈Ir1

v_ia_ijN_j m˜˜_j N_j −m˜˜_i

N_i

!2

+· · ·+ X

i∈I_rk

X

j∈I_rk

v_ia_ijN_j m˜˜_j N_j −m˜˜_i

N_i

!2

= 0. (3.28) For i=j the terms of the term are all equal to zero. For i6=j,aij ≥0 and v_k >0, N_k >0 provided k ∈ I^Θ. So equation (3.27) holds if and only if each term on the left hand side of the equation is zero. Consider the term corresponding to Ari, for 1 ≤ i ≤ k. Since Ari is irreducible, for every pairk, j ∈I_ri, there exists a sequence of indicesj, k_r, k_r−1,· · · , k₁, k such that a_k,k1a_k1,k2· · ·a_kr−1,kra_kr,j>0. Therefore, Equation 3.28 holds if and only if

˜˜ m_j

N_j = m˜˜_kr

N_kr =· · ·= m˜˜_k1 N_k1 = m˜˜_k

N_k.

Therefore,m˜˜_k=νⁱN_k for some constantνⁱ, hence from (3.23), we obtain 0 = X

k∈I_ri

v_km˜˜_k=νⁱ X

k∈I_ri

N_kv_k

and since N^ri,v^ri >0, we have νⁱ = 0. But this also means that m˜˜_k= 0 for all k∈I_ri. If we repeat this procedure for allk∈Θ, we obtain m˜˜ =0 in Y, a contradiction.

Remark 3.3.4. If A has a positive right eigenvector, then any vector m∈ Rⁿ which satisfies (3.23) also satisfies

Xn

i=1

v_im_i = 0 trivially.

Example 3.3.5. Let

A=













2 1 0 0 0

1 2 0 0 0

0 0 0 0.5 0 0 0 18 0 0 1 1 0 1 0.5











 .

For this matrix,r = 3, and if we pickN= (5/61)(1,1,1,6,16/5)^T andv= (61/70)(1,1,6,1,0), we see thatAN= 3N andvA= 3v. We select a vectorm such that

X5

i=1

v_im_i = 0.

Let m = (−k, k, k1,−6k1, m)^T, where m, k, k1 ∈ R are arbitrary. This vector satisfies the requirements of Lemma 3.2.3.

X5

i=1

X5

j=1

v_ia_ijN_j m_j

N_j − m_i N_i

2

= X5

i=1

v_i

"

5 61a_i1

(−k)61 5 −m_i

N_i 2

+a_i2 5 61

k61

5 −m_i N_i

2

+a_i3 5 61

(k₁)61

5 −m_i Ni

2

+a_i430 61

(−6k₁)61 30− m_i

Ni

2

+a_i516 61

(m)61

16− m_i Ni

2#

= 122²

25(7)(k²+ 9k₁²) = 122²

175(k²+ 9k₁²).

and the right hand side of the inequality is X5

i=1

v_i Ni

m²_i = 61 70× 61

5

(−k)²+k²+ 6k₁²+1

6.(−6k₁)²

= 61²

175 k²+ 6k²₁

<

X5

i=1

X5

j=1

v_ia_ijN_j m_j

N_j −m_i N_i

2

.

Therefore, for someα >0, the result holds.

From Example 3.3.5 above, we observe that I₁ = {1,2};I₂ = {3,4}, I^Θ = {1,2,3,4} and Θ ={1,2}. In the first case whenm= (−k, k, k1,−6k1, m)^T, we observe that

X2

i=1

vimi= 0 = X4

i=3

vimi,

hence the assumption in (3.23) is satisfied, and hence the result holds. In general, any vector m of the form m = (−k, k, k₁,−6k₁, m)^T, m ∈ R yields a positive result for this matrix in Example 3.3.5, regardless of the vectorv used.

Remark 3.3.6. IfA is a 3×3 block triangular matrix in normal form withg = 2,A₁ and A₂ being1×1 blocks (i.e scalars) andN>0,v≥0, then the only vectorsmfor which equation 3.23 holds are vectors of the form (0,0,m³)^T, wherem³ is a vector whose dimension depends on the dimension of A3. This is because ifm= (m1, m2,m³)^T, then (3.23) holds if and only if m₁ =m₂ = 0. But for such a vector, the left hand side of the inequalities (3.7) and (3.24) are always 0. Therefore, there is no vector m for which the lemma holds. This explains why Example 3.3.1 gives a negative answer.

In general, ifN>0and2≤g≤sandAiare one dimensional, withr(Ai) =r(A), i= 1, . . . , g, then there is no non-trivial vector mthat satisfies the result.