some inequalities connecting the singular values of a complex matrix with the perron roots of...

Journal of Mathematical Sciences, Vol. 199, No. 4, June, 2014

SOME INEQUALITIES CONNECTING THE SINGULAR VALUES OF A COMPLEXMATRIX WITH THE PERRON ROOTS OF RELATED NONNEGATIVE MATRICES

L. Yu. Kolotilina∗ UDC 512.643

In the paper, some inequalities interrelating the singular values of a square matrix A with complex entries with thePerron roots of the associated nonnegative matrices PA = |A|◦|A| and (PA+PT

A )/2 are derived. The results obtainedare applied to (0, 1)-matrices. Bibliography: 8 titles.

1. Introduction

Let A = (aij) ∈ Cn×n, n ≥ 1, be a square matrix with complex entries and let a nonnegative matrix P ∈ C

n×n

satisfy the conditionP ≥ |A|,

where |A| = (|aij |). Then, by the well-known Wielandt lemma (see, e.g., [5, Chapter XIII, §2, Lemma 2] or [3,Chapter 2, Theorem 2.14]), the eigenvalues of A, λi(A), i = 1, . . . , n, satisfy the inequality

|λi(A)| ≤ ρ(P ), i = 1, . . . , n. (1.1)

Here and below, for a nonnegative matrix P , by ρ(P ) we denote its Perron root, which is the largest positiveeigenvalue of P and coincides with its spectral radius.

Let σi(A) = λ1/2i (AA∗), i = 1, . . . , n, be the singular values of A ordered nonincreasingly, i.e.,

σ1(A) ≥ · · · ≥ σn(A).

Then from (1.1) we immediately obtain that

σ21(A) ≤ ρ(|A||A|T ) = σ2

1(|A|), (1.2)

which is an upper bound for the largest singular value of A in terms of the Perron root of the related nonnegativematrix |A||A|T .

In this paper, we associate with A yet another nonnegative matrix, which is defined as follows:

PA = (|aij |2) = |A| ◦ |A|. (1.3)

Here, ◦ means the Hadamard (entrywise) product of matrices.Observe that the matrix PA possesses the following obvious properties:

(i) For α ∈ C and A ∈ Cn×n, n ≥ 1,

PαA = |α|2PA.

(ii) If U is a unitary matrix, then PUe = e and PTU e = e. In particular, by the Frobenius theorem (e.g., see [8,

Chapter II, Section 2.1, Theorem 1.1]), this implies that

ρ(PU ) = ρ

(PU + PT

U

2

)= 1.

(iii) For an arbitrary nonsingular diagonal matrix D, it holds that

PD−1AD = |D|−2PA|D|2. (1.4)

In particular, from (1.4) it follows thatρ(PD−1AD) = ρ(PA), (1.5)

i.e., the Perron root of PA is invariant with respect to diagonal conjugation of A.(iv) If S is a permutation matrix, then, obviously,

PS−1AS = S−1PAS. (1.6)

(v) If E = (εij) ∈ Cn×n, n ≥ 1, and |εij | = 1, i, j = 1, . . . , n, then

PA◦E = PA.

∗St.Petersburg Department of the Steklov Mathematical Institute, St.Petersburg, Russia, e-mail: [email protected].

Translated from Zapiski Nauchnykh Seminarov POMI, Vol. 419, 2013, pp. 121–138. Original article submitted July 1,2013.438 1072-3374/14/1994-0438 ©2014 Springer Science+Business Media New York

(vi) The row sums of PA, denoted by ri(PA), i = 1, . . . , n, satisfy the obvious but useful relations

ri(PA) = (AA∗)ii, i = 1, . . . , n. (1.7)

(vii) If A is a (0, 1)-matrix, thenPA = A.

The paper is organized as follows. In Sec. 2, we prove a number of inequalities interrelating the singularvalues of A with ρ(PA) and ρ

(PA+PT

A

2

). Section 3 considers applications to (0, 1)-matrices.

Throughout the paper, the following notation is used:• In is the identity matrix of order n ≥ 1;• e = [1, . . . , 1]T is the unit vector;• for A ∈ C

n×n, ρ(A) = max1≤i≤n

|λi(A)| is the spectral radius of A;

• the eigenvalues of a Hermitian matrix A = A∗ ∈ Cn×n, n ≥ 1, are ordered nonincreasingly, i.e.,

λ1(A) ≥ · · · ≥ λn(A);

• Dn is the group of (nonsingular) diagonal matrices of order n with positive diagonal entries.

2. Main results

First we establish the following elementary result.

Theorem 2.1. Let A ∈ Cn×n, n ≥ 1. Then

σ2n(A) ≤ ρ(PA) ≤ σ2

1(A). (2.1)

Furthermore, if A is irreducible, then equality occurs on either side of (2.1) if and only if

σn(A) = · · · = σ1(A). (2.2)

Proof. By the Frobenius theorem and (1.7), we have

ρ(PA) ≤ max1≤i≤n

ri(PA) = max1≤i≤n

(AA∗)ii ≤ λ1(AA∗) = σ21(A). (2.3)

Here, the last inequality stems from the classical Cauchy theorem (e.g., see [7, Theorem 4.3.15]).Similarly,

ρ(PA) ≥ min1≤i≤n

ri(PA) = min1≤i≤n

(AA∗)ii ≥ σ2n(A).

This proves inequalities (2.1)Consider the equality cases.If condition (2.2) is fulfilled, then, obviously, both relations in (2.1) are equalities.Conversely, assume that A is irreducible and that

ρ(PA) = σ21(A). (2.4)

In this case, (2.3) is a string of equalities and, by the Frobenius theorem, we have

(AA∗)ii = ρ(PA), i = 1, . . . , n.

Therefore,n∑

i=1

σ2i (A)

n=

tr (AA∗)n

= ρ(PA) = σ21(A),

and (2.2) follows.The case of equality on the left-hand side of (2.1) is treated similarly. �By applying Theorem 2.1 to the matrix D−1AD and using (1.5), we obtain the following generalization of

Theorem 2.1.

Corollary 2.1. Let A ∈ Cn×n, n ≥ 1. Then, for any nonsingular diagonal matrix D of order n,

σ2n(D

−1AD) ≤ ρ(PA) ≤ σ21(D

−1AD). (2.5)


σn(D−1AD) = · · · = σ1(D−1AD).

439

Corollary 2.1 immediately implies the following two-sided bounds for ρ(PA).

Corollary 2.2. Let A ∈ Cn×n, n ≥ 1. Then

maxD

σ2n(D

−1AD) ≤ ρ(PA) ≤ minD

σ2n(D

−1AD),

where the minimum and maximum are taken over all nonsingular diagonal matrices D of order n.Furthermore, if A is irreducible, then each of the relations is an equality if and only if there is a nonsingular

diagonal matrix Δ such that all the singular values of Δ−1AΔ are equal.

The left-hand-side inequality of Theorem 2.1 can be strengthened by changing σ2n(A) for the geometric mean

of the squared singular values of A. In this way, an upper bound for |detA| in terms of the Perron root of PA

is obtained.


|detA|2/n = [det(AA∗)]1/n =

[n∏

i=1

σ2i (A)

]1/n

≤ ρ(PA). (2.6)

Furthermore, if A is irreducible, then equality in (2.6) holds if and only if there exists a diagonal matrixΔ ∈ Dn such that

σ1(Δ−1AΔ) = · · · = σn(Δ−1AΔ). (2.7)

Proof. By properties of the determinant and by the arithmetic-mean-geometric-mean inequality, for any nonsin-gular diagonal matrix D = diag (d1, . . . , dn) we have

[det(AA∗)]1/n =[det(D−1/2AD1/2)(D−1/2AD1/2)∗

]1/n

=

[n∏

i=1

σ2i (D

−1/2AD1/2)

]1/n

≤

n∑i=1

σ2i (D

−1/2AD1/2)

n

=tr

[(D−1/2AD1/2)(D−1/2AD1/2)∗

]n

=

n∑i,j=1

d−1i |aij |2djn

=eTD−1PADe

n.

(2.8)

Now, in order to prove (2.6), it is sufficient to provide a matrix D ∈ Dn such that

eTD−1PADe

n= ρ(PA). (2.9)

To this end, first let A (and, consequently, also PA) be irreducible. In this case, by the Perron–Frobeniustheorem (e.g., see [3, Chapter 2, Theorem 1.4]), there is a positive vector x = (xi) such that

PAx = ρ(PA)x. (2.10)

Define the matrix D by the relationDe = x. (2.11)

Then D ∈ Dn and (2.10) can be written as

D−1PADe = ρ(PA)e, (2.12)

which implies (2.9).If A is reducible we may assume that it is a block triangular matrix of the form

A =

⎡⎢⎣

A11 ∗. . .

0 Arr

⎤⎥⎦ , r ≥ 2,

440

where all the diagonal blocks Aii, i = 1, . . . , r, are irreducible matrices of orders ni, i = 1, . . . , r, and n =r∑

i=1

ni.

Then, obviously,

det(AA∗) =r∏

i=1

det(AiiA∗ii) (2.13)

and

PA =

⎡⎢⎣

PA11 ∗. . .

0 PArr

⎤⎥⎦ , r ≥ 2.

As we have already demonstrated,

det(AiiA∗ii) ≤ ρ(PAii

)ni , i = 1, . . . , r. (2.14)

By (2.14) and the monotonicity property of the Perron root with respect to principal submatrices (e.g., see [3,Chapter 2, Corollary 1.6]), we have

ρ(PAii) ≤ ρ(PA), i = 1, . . . , r. (2.15)

Finally, using (2.13), (2.14), and (2.15), we derive

det(AA∗) ≤r∏

i=1

ρ(PAii)ni ≤ ρ(PA)

r∑i=1

ni

= ρ(PA)n.

This completes the proof of (2.6) in the reducible case.Consider the case of equality in (2.6). First assume that

[det(AA∗)]1/n = ρ(PA) (2.16)

and let A be irreducible. In this case, the matrix D defined in (2.11) has positive diagonal entries, and from(2.8), (2.12), and (2.16) it follows that

[n∏

i=1

σ2i (D

−1/2AD1/2)

]1/n

=

n∑i=1

σ2i (D

−1/2AD1/2)

n= ρ(PA).

By virtue of the arithmetic-mean-geometric-mean inequality, this means that

σ2i (D

−1/2AD1/2) = ρ(PA), i = 1, . . . , n,

and (2.7) with Δ = D1/2 follows.The sufficiency of (2.7) for (2.16) to hold stems from (2.5).This completes the proof of Theorem 2.2. �

Remark 2.1. For a real matrix A free of zero entries, inequality (2.6) was first proved in [6].

Since both det(AA∗) and ρ(PA) are invariant with respect to changing A for D−1AD, where D is an arbitrarynonsingular diagonal matrix, from Theorem 2.2 we immediately obtain the following more general result.

Corollary 2.3. Let A ∈ Cn×n, n ≥ 1, and let D ∈ C

n×n be an arbitrary nonsingular diagonal matrix. Then[

n∏i=1

σ2i (D

−1AD)

]1/n

≤ ρ(PA). (2.17)

Furthermore, if A is irreducible, then equality in (2.17) occurs if and only if (2.7) holds with a matrix Δ ∈ Dn.

Remark 2.2. From the proof of Theorem 2.2 it follows that in the irreducible case, there exists a diagonalmatrix D ∈ Dn such that

n∑i=1

σ2i (D

−1/2AD1/2)

n≤ ρ(PA).

441

However, in the general case, for D = In the inequalityn∑

i=1

σ2i (A)

n≤ ρ(PA) (2.18)

is not necessarily valid, unless PA = PTA .

This is illustrated by the following example.

Example 2.1. For the matrix A =[

0 21 0

]we have

PA =[

0 41 0

]and ρ(PA) = 2 =

√det(AAT ).

However,n∑

i=1

σ2i (A)

n=

4 + 12

> 2 = ρ(PA).

The next theorem provides a correct modification of (2.18), which is an upper bound for the arithmetic mean ofthe squared singular values of A, and, since ρ(PA) ≤ ρ

(PA+PT

A

2

)(see, e.g., [2, Theorem 12.6]), it also strengthens

the lower bound of Theorem 2.1 for σ21(A).


n∑i=1

σ2i (A)

n≤ ρ

(PA + PT

A

2

)≤ σ2

1(A). (2.19)

Furthermore, equality occurs on the left-hand side of (2.19) if and only if

PA + PTA

2e = ρ

(PA + PT

A

2

)e, (2.20)

and if PA + PTA is irreducible, then equality occurs on the right-hand side of (2.19) if and only if

σ1(A) = · · · = σn(A). (2.21)

Proof. The left-hand-side inequality in (2.19) is derived using the obvious relation eTPAe = eTPTA e and the

Rayleigh–Ritz theorem (e.g., see [7, Theorem 4.2.2]) as follows:n∑

i=1

σ2i (A)

n=

tr (AA∗)n

=eTPAe

n=

eT(

PA+PTA

2

)e

n≤ ρ

(PA + PT

A

2

).

The right-hand-side inequality in (2.19) is derived as follows. For i = 1, . . . , n we have

σ21(A) =

σ21(A) + σ2

1(A∗)

2≥ (AA∗)ii + (A∗A)ii

2=

ri(PA) + ri(PTA )

2= ri

(PA + PT

A

2

),

implying that, by the Frobenius theorem,

σ21(A) ≥ max

1≤i≤nri

(PA + PT

A

2

)≥ ρ

(PA + PT

A

2

). (2.22)

Consider the equality cases.First consider the right-hand-side inequality in (2.19).Since from (2.21) it readily follows that both bounds in (2.19) coincide, the sufficiency of (2.21) for the

right-hand-side inequality in (2.19) to be an equality is trivial.In order to prove the necessity of (2.21), assume that PA + PT

A is irreducible and that

ρ

(PA + PT

A

2

)= σ2

1(A).

442

In this case, by the Frobenius theorem, from (2.22) it follows that

σ21(A) = ri

(PA + PT

A

2

)= ρ

(PA + PT

A

2

), i = 1, . . . , n. (2.23)

Since, as has been mentioned above,

eTPAe

n=

eTPA+PT

A

2 e

nand since, by (2.23),

eTPA+PT

A

2 e

n= σ2

1(A),

we haven∑

i=1

σ2i (A)

n=

tr (AA∗)n

=eTPAe

n= σ2

1(A).

This implies that A has equal singular values, and the necessity of (2.21) is established.Now let the left-hand-side inequality in (2.19) hold with equality, i.e.,

ρ

(PA + PT

A

2

)=

n∑i=1

σ2i (A)

n. (2.24)

Since, as has been indicated above,n∑

i=1

σ2i (A)

n=

tr (AA∗)n

=eTPAe

n=

eTPA+PT

A

2 e

n, (2.25)

from (2.24) it follows that

eTPA+PT

A

2 e

n= ρ

(PA + PT

A

2

),

which implies that e is an eigenvector of PA+PTA

2 corresponding to its largest eigenvalue, i.e.,

PA + PTA

2e = ρ

(PA + PT

A

2

)e.

This proves (2.20).Conversely, if condition (2.20) is fulfilled, then, in view of (2.25), we have

n∑i=1

σ2i (A)

n=

eTPA+PT

A

2 e

n= ρ

(PA + PT

A

2

).

This proves the sufficiency of (2.20) for the left-hand-side inequality in (2.19) to be an equality. �

Remark 2.3. If a matrix A ∈ Cn×n, n ≥ 1, has equal singular values, then from Theorems 2.1 and 2.3 it follows

that

ρ(PA) = ρ

(PA + PT

A

2

).

The latter relation also follows from properties (i) and (ii) in Sec. 1 because, as is readily seen from the singularvalue decomposition, any matrix with equal singular values is a scalar multiple of a unitary matrix.

Remark 2.4. Sincen∑

i=1

σ2i (A) = tr (AA∗) =

n∑i,j=1

|aij |2 = eTPAe,

the left-hand side of (2.19) is readily computable, and the left-hand-side inequality in (2.19) can be regarded asa lower bound for ρ

(PA+PT

A

2

).

443

Remark 2.5. For the matrix A =[

0 21 0

](see Example 2.1) we have ρ

(PA+PT

A

2

)= 5

2 =

2∑i=1

σ2i (A)

2 , i.e., the

left-hand-side inequality in (2.19) is an equality, whereas the right-hand-side inequality is strict. This exampledemonstrates that the two inequalities in (2.19) are not necessarily strict simultaneously.

We conclude this section by presenting an improvement of the right-hand-side inequalities in Theorems 2.1and 2.3. To this end, we will need some additional notation, which is recalled below.

For a matrix A = (aij) ∈ Cn×n, n ≥ 1, by GA = (〈n〉, EA) we denote the directed graph of A with vertex set

〈n〉 = {1, . . . , n} and arc setEA = {(i, j) : i, j ∈ 〈n〉 and aij = 0};

C(A) is the set of all (simple) circuits in GA. Recall that a circuit of length k ≥ 1 in GA is an ordered setγ = (i1, . . . , ik, ik+1), where all the vertices i1, . . . , ik ∈ 〈n〉 are distinct, ik+1 = i1, and for every j = 1, . . . , k thearc (ij , ij+1) belongs to EA. The set {i1, . . . , ik} is called the support of γ and is denoted by γ. The length ofthe circuit γ is denoted by |γ|; |γ| = k. Also recall that if C(A) = ∅, then

g(A) = g(GA) = minγ∈C(A)

|γ|

is called the girth of GA.Recall the following theorem due to Al’pin, which reduces to the Frobenius theorem if all the diagonal matrix

entries are positive.

Theorem 2.4 ([1]). Let A be a nonnegative matrix of order n ≥ 1 free of zero rows. Then

minγ∈C(A)

⎡⎣∏i∈γ

ri(A)

⎤⎦1/|γ|

≤ ρ(A) ≤ maxγ∈C(A)

⎡⎣∏i∈γ

ri(A)

⎤⎦1/|γ|

. (2.26)


minγ∈C(A)

⎡⎣∏i∈γ

ri(A)

⎤⎦1/|γ|

= maxγ∈C(A)

⎡⎣∏i∈γ

ri(A)

⎤⎦1/|γ|

. (2.27)

Al’pin’s theorem readily implies the following corollary, generalizing an earlier Brualdi’s result [4, Corol-lary 4.6], originally established for nonnegative irreducible matrices with zero diagonal entries.

Corollary 2.4. Let A be a nonnegative matrix of order n ≥ 1 such that

r1(A) ≥ · · · ≥ rn(A). (2.28)

Then

ρ(A) ≤⎡⎣g(A)∏

i=1

ri(A)

⎤⎦1/g(A)

. (2.29)

Corollary 2.4 makes it possible to derive the following improvement of the right-hand-side inequality in The-orem 2.1 mentioned above.

Theorem 2.5. Let a matrix A ∈ Cn×n, n ≥ 1, be free of zero rows. Then

ρ(PA) ≤

g(A)∑i=1

σ2i (A)

g(A). (2.30)

Furthermore, if A is irreducible, then equality occurs in (2.30) if and only if

σ1(A) = · · · = σn(A). (2.31)

Proof. In view of relation (1.6), we may assume, without loss of generality, that

r1(PA) ≥ · · · ≥ rn(PA).

444

Under this assumption, using Corollary 2.4, the arithmetic-mean-geometric-mean inequality, and Cauchy’s the-orem, we derive (2.30) as follows:

ρ(PA) ≤⎡⎣g(A)∏

i=1

ri(PA)

⎤⎦1/g(A)

≤

g(A)∑i=1

ri(PA)

g(A)=

g(A)∑i=1

(AA∗)ii

g(A)

=tr Bg(A)

=

g(A)∑i=1

λi(B)

g(A)≤

g(A)∑i=1

λi(AA∗)

g(A)=

g(A)∑i=1

σ2i (A)

g(A).

(2.32)

Here, B = (AA∗)[1, . . . , g(A)] is the left-upper-corner principal submatrix of order g(A) of the matrix AA∗.Consider the equality case, assuming that A is irreducible.If

ρ(PA) =

g(A)∑i=1

σ2i (A)

g(A), (2.33)

then from (2.32) and the arithmetic-mean-geometric-mean inequality it follows that

ρ(PA) = r1(PA) = · · · = rg(A)(PA).

In this case, by the Frobenius theorem, we have

ρ(PA) = r1(PA) = · · · = rn(PA),

or, equivalently,ρ(PA) = (AA∗)11 = · · · = (AA∗)nn. (2.34)

From (2.34) and (2.33) we infer

ρ(PA) =tr (AA∗)

n=

n∑i=1

σ2i (A)

n=

g(A)∑i=1

σ2i (A)

g(A).

As is readily seen, the latter relation for the nonincreasingly ordered singular values of A implies (2.31).Conversely, if condition (2.31) is fulfilled, then (2.33) stems from Theorem 2.1. �

Remark 2.6. For matrices with nonzero diagonal entries, for which, obviously, g(A) = 1, the bound (2.30)reduces to the respective bound of Theorem 2.1.

By applying Theorem 2.5 to the matrix D−1AD and using property (1.5), we obtain the following generaliza-tion of Theorem 2.5.

Corollary 2.5. Under the assumptions of Theorem 2.5, for an arbitrary nonsingular diagonal matrix D of ordern it holds that

ρ(PA) ≤

g(A)∑i=1

σ2i (D

−1AD)

g(A). (2.35)

Furthermore, if the matrix A is irreducible, then (2.35) is an equality if and only if all the singular values ofD−1AD are equal.

From Corollary 2.5 we immediately obtain the result below.

Corollary 2.6. Under the assumptions of Theorem 2.5,

ρ(PA) ≤ minD

g(A)∑i=1

σ2i (D

−1AD)

g(A), (2.36)

where the minimum is taken over all nonsingular diagonal matrices D of order n.Furthermore, if the matrix A is irreducible, then (2.36) is an equality if and only if there is a nonsingular

diagonal matrix Δ such that all the singular values of Δ−1AΔ are equal.

445

In order to establish the counterpart of (2.30) for the Perron root of the symmetric part of PA, first we notethat g

(A+AT

2

)≤ 2, and if A has a nonzero diagonal entry, then g

(A+AT

2

)= 1. For this reason, in order to

improve the right-hand-side bound of Theorem 2.3, we must assume that A has zero principal diagonal. Thedesired counterpart of Theorem 2.5 is as follows.

Theorem 2.6. Let A ∈ Cn×n, n ≥ 1, and let all the diagonal entries of A be zero. Then

ρ

(PA + PT

A

2

)≤ σ2

1(A) + σ22(A)

2. (2.37)

Furthermore, if PA+PTA

2 is irreducible, then equality occurs in (2.37) if and only if

σ1(A) = · · · = σn(A). (2.38)

Proof. Assume, without loss of generality, that

r1

(PA + PT

A

2

)≥ · · · ≥ rn

(PA + PT

A

2

).

Under this assumption, by Corollary 2.4 and the arithmetic-mean-geometric-mean inequality, we have

ρ

(PA + PT

A

2

)≤

[r1

(PA + PT

A

2

)r2

(PA + PT

A

2

)]1/2

≤r1

(PA+PT

A

2

)+ r2

(PA+PT

A

2

)2

=r1(PA) + r2(PA)

4+

r1(PTA ) + r2(PT

A )4

=(AA∗)11 + (AA∗)22 + (A∗A)11 + (A∗A)22

4=

tr B + tr C4

,

(2.39)

where by B and C we denote the left-upper-corner principal submatrices of AA∗ and A∗A, respectively, of order2. As in the proof of Theorem 2.5, we conclude that

ρ

(PA + PT

A

2

)≤ λ1(AA∗) + λ2(AA∗) + λ1(A∗A) + λ2(A∗A)

4=

σ21(A) + σ2

2(A)2

,

which completes the proof of (2.37).It remains to consider the equality cases. If PA+PT

A

2 is irreducible and

ρ

(PA + PT

A

2

)=

σ21(A) + σ2

2(A)2

, (2.40)

then from (2.39) it follows that

ρ

(PA + PT

A

2

)= r1

(PA + PT

A

2

)= r2

(PA + PT

A

2

),

whence, by the Frobenius theorem, we conclude that

ρ

(PA + PT

A

2

)= r1

(PA + PT

A

2

)= · · · = rn

(PA + PT

A

2

),

or, equivalently,

ρ

(PA + PT

A

2

)=

(AA∗)11 + (A∗A)112

= · · · = (AA∗)nn + (A∗A)nn2

,

implying that

ρ

(PA + PT

A

2

)=

tr AA∗ + tr A∗A2n

=

n∑i=1

σ2i (A)

n. (2.41)

Now, comparing (2.40) with (2.41), we arrive at (2.38).Conversely, if all the singular values of A are equal, then (2.40) stems from Theorem 2.3. �

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3. Applications to (0, 1)-matrices

As was mentioned in the Introduction, if A is a (0, 1)-matrix, then, obviously,

PA = |A| ◦ |A| = A. (3.1)

Thus, in this case, in all the results of Sec. 2 one can change ρ(PA) for ρ(A). In this way, one obtains interrelationsbetween the singular values of A and the Perron roots of A and its symmetric part (A+AT )/2.

In particular, from Theorems 2.2 and 2.5 we immediately obtain the following result for (0, 1)-matrices.

Corollary 3.1. Let A be a (0, 1)-matrix of order n, n ≥ 1, and let g(A) be the girth of the associated digraph GA.Then

[n∏

i=1

σ2i (A)

]1/n

≤ ρ(A) ≤

g(A)∑i=1

σ2i (A)

g(A). (3.2)

Furthermore, if A is irreducible, then equality occurs on the left-hand side of (3.2) if and only if there existsa diagonal matrix Δ ∈ Dn such that

σ1(Δ−1AΔ) = · · · = σn(Δ−1AΔ), (3.3)

and equality occurs on the right-hand side of (3.2) if and only if

σ1(A) = · · · = σn(A), (3.4)

i.e., (3.3) holds with Δ = In.

For the Perron root of the symmetric part of a (0, 1)-matrix, Theorems 2.3 and 2.6 yield the following two-sidedbounds in terms of its singular values.

Corollary 3.2. Let A = (aij) be a (0, 1)-matrix of order n, n ≥ 1. Thenn∑

i=1

σ2i (A)

n≤ ρ

(A+AT

2

)≤ σ2

1(A), (3.5)

and ifa11 = · · · = ann = 0, (3.6)

then

ρ

(A+AT

2

)≤ σ2

1(A) + σ22(A)

2. (3.7)

Furthermore, equality occurs on the left-hand side of (3.5) if and only if A+AT

2 has equal row sums, and ifA+AT

2 is irreducible, then the right-hand side of (3.5) is an equality if and only if A has equal singular values.Finally, if A+AT

2 is an irreducible matrix satisfying (3.6), then (3.7) holds with equality if and only if A hasequal singular values.

Translated by L. Yu. Kolotilina.

REFERENCES

1. Yu. A. Al’pin, “Bounds for the Perron root of a nonnegative matrix taking into consideration the propertiesof its graph,” Mat. Zametki, 58, 635–637 (1982).

2. A. R. Amir-Moez and A. L. Fass, Elements of Linear Spaces, Pergamon Press (1962).3. A. Berman and R. J. Plemmons, Nonnegative Matrices in the Mathematical Sciences, Academic Press, New

York etc. (1979).4. R. A. Brualdi, “Matrices, eigenvalues, and directed graphs,” Linear Multilinear Algebra, 11, 143–165 (1982).5. F. R. Gantmakher, Matrix Theory [in Russian], Nauka, Moscow (1967).6. G. Gu, “Spectral radius bounds for positive matrices with applications,” unpublished.7. R. A. Horn and C. R. Johnson, Matrix Analysis, Cambridge University Press (1986).8. H. Minc, Nonnegative Matrices, John Wiley and Sons, New York etc. (1988).

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some inequalities connecting the singular values of a complex matrix with the perron roots of...

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