Determinant and eigenvalue inequalities involving nonnegative matrices

Abstract

We discuss some bounds for the determinant of a matrix when all its eigenvalues are positive as in case of totally positive matrices. We further show how positive unital linear maps can be used to obtain some bounds for the eigenvalues of Hermitian and nonnegative matrices.

1 Introduction

Let ${\mathbb {M}}\left( n\right)$ denote the algebra of all $n\times n$ complex matrices. An element $A=\left( a_{ij}\right) \in {\mathbb {M}}(n)$ is called nonnegative (positive) if $a_{ij}\ge 0$ ($a_{ij}>0)$ for all i and j. An element $A\in {\mathbb {M}}(n)$ is called totally nonnegative (totally positive) if every minor of A is nonnegative (positive). A totally nonnegative matrix is called oscillatory if $A^{k}$ is totally positive for some positive integer k. The eigenvalues of an oscillatory matrix are all distinct positive numbers. The eigenvalues of a totally nonnegative square matrix are all nonnegative. See [1, 12].

The Hadamard inequality [14] for positive definite Hermitian matrices says that the determinant of a positive definite matrix is less than or equal to the product of its diagonal entries, that is

$$\begin{aligned} \det A\le \overset{n}{\underset{i=1}{\prod }}a_{ii}. \end{aligned}$$

(1.1)

The refinements, extensions, generalizations, and several other related bounds for the determinant of a positive definite Hermitian matrix have been investigated in the literature; for instance, see [13, 20]. The inequality (1.1) need not hold good if A is not Hermitian. For example, on using the Gersgorin disk theorem, one can see that the eigenvalues of

$$\begin{aligned} A= \begin{bmatrix} 1 &{} 0 &{} \frac{1}{2} \\ \frac{1}{2} &{} 3 &{} 0 \\ 0 &{} 2 &{} 6 \end{bmatrix} \end{aligned}$$

are all positive. However, $\det A=18.5,$ while the product of its diagonal entries is 18. It is then natural to investigate some other such quick estimates for the determinant when the eigenvalues of the matrix (not necessarily Hermitian) are all positive as in case of the totally positive matrices. One such inequality

$$\begin{aligned} \left( \det A\right) ^{\frac{1}{n}}\le \frac{\text {tr} A}{n} \end{aligned}$$

(1.2)

is an immediate consequence of the arithmetic mean-geometric mean inequality (AM-GM inequality) for n positive numbers. We here obtain an upper bound for detA in terms of $\text {tr} A$ and $\text {tr} A^{2}.$ This also provides a refinement of the inequality (1.2) (Theorem 1 and Corollary 1, below). The upper bound for the determinant in terms of the expressions involving entries of the matrix is given (Corollary 2). We further show that the Hadamard inequality (1.1) is valid for nonnegative matrices in the special case when all its diagonal entries are equal and eigenvalues are all positive and also obtain a refinement of (1.1) in this case (Corollary 3). This also suggests a refinement of (1.1) for a positive definite Hermitian matrix in the particular case when the diagonal entries of the matrix are all equal (Corollary 4).

Some estimates for the eigenvalues of nonnegative matrices are also obtained here as the particular cases of some inequalities involving positive linear maps. This is in analogy and motivated by the work of Bhatia and Sharma [7,8,9]. They have obtained various bounds for the spread (diameter of spectrum $\sigma (A)$) of a matrix using positive linear maps. In a similar fashion, we obtain here bounds for the spectral radius

$$\begin{aligned} \rho (A)=\max \left\{ \left| \lambda \right| :\lambda \in \sigma (A)\right\} \end{aligned}$$

(1.3)

of nonnegative matrices (Theorem 2).

In case the eigenvalues and diagonal entries of A are all real, we assume for simplicity that they are, respectively, arranged as

$$\begin{aligned} \lambda _{1}(A)\le \lambda _{2}(A)\le \cdots \le \lambda _{n}(A) \end{aligned}$$

(1.4)

and

$$\begin{aligned} a_{1}\le a_{2}\le \cdots \le a_{n}. \end{aligned}$$

(1.5)

Bhatia and Sharma [10] have recently shown that for an Hermitian element $A\in {\mathbb {M}}(n)$ whose off-diagonal entries are either all nonnegative or purely imaginary, $\lambda _{2}(A)\le a_{3}.$ In case the off-diagonal entries are all purely imaginary, then for $1\le k\le \lceil \frac{n}{2}\rceil ,$ $\lambda _{k}(A)\le a_{2k-1}\text { and }\lambda _{n-k+1}(A)\ge a_{n-2k+2}.$ Also, see Berman and Farber [3] and Charles et al. [11]. We extend these arguments further and discuss some lower bounds for $\rho (A)$ (Theorem 3 and Corollary 5). An upper bound for the largest eigenvalue of a symmetric Z-matrix is also given (Theorem 4).

2 Main results

We need the following lemma in the proof of Theorem 1.

Lemma 1

Let $x_{1},x_{2},\ldots ,x_{n}$ denote n positive numbers and let

$$\begin{aligned} A_{n}=\frac{1}{n}\sum _{i=1}^{n}x_{i},\ G_{n}=\left( \overset{n}{\underset{i=1}{\prod }}x_{i}\right) ^{\frac{1}{n}}\ \text {and}\ S_{n}^{2}=\frac{1}{n}\sum _{i=1}^{n}\left( x_{i}-A_{n}\right) ^{2}. \end{aligned}$$

Then

$$\begin{aligned} A_{n}^{2}-G_{n}^{2}\le \left( n-1\right) S_{n}^{2}. \end{aligned}$$

(2.1)

Proof

The inequality (2.1) becomes equality for $n=1,2.$ We prove the general case by induction. For $n\ge 2,$ (2.1) is equivalent to

$$\begin{aligned} m_{2_{n}}^{\prime }\ge \frac{n}{n-1}A_{n}^{2}-\frac{1}{n-1}G_{n}^{2}, \end{aligned}$$

(2.2)

where $m_{2_{n}}^{\prime }=A_{n}^{2}+S_{n}^{2}.$ Suppose (2.2) holds for $n=k.$ We then show that

$$\begin{aligned} m_{2_{k+1}}^{\prime }\ge \frac{k+1}{k}A_{k+1}^{2}-\frac{1}{k}G_{k+1}^{2}. \end{aligned}$$

(2.3)

Write $x_{k+1}=x$ and assume without restricting generality that $x\le x_{i}$ for $i=1,2,\ldots ,k.$ We insert $A_{k+1}= \frac{kA_{k}+x}{k+1},$ $G_{k+1}=G_{k}^{\frac{k}{k+1}}x^{\frac{1}{k+1}}$ and $m_{2_{k+1}}^{\prime }=\frac{km_{2_{k}}^{\prime }+x^{2}}{k+1}$ in (2.3) and write it as

$$\begin{aligned} k^{2}\left( m_{2_{k}}^{\prime }-A_{k}^{2}\right) +\left( k-1\right) x^{2}-2kxA_{k}+\left( k+1\right) G_{k}^{\frac{2k}{k+1}}x^{\frac{2}{k+1}}\ge 0. \end{aligned}$$

(2.4)

On using the AM–GM inequality $\alpha a+\beta b\ge \left( \alpha +\beta \right) \left( a^{\alpha }b^{\beta }\right) ^{\frac{1}{ \alpha +\beta }}$ for positive numbers $a,b,\alpha$ and $\beta ,$ we find that

$$\begin{aligned} \left( k-1\right) x^{2}+\left( k+1\right) G_{k}^{\frac{2k}{k+1}}x^{\frac{2}{ k+1}}\ge 2kxG_{k}. \end{aligned}$$

(2.5)

The inequality (2.5) implies that the inequality (2.4), and hence, (2.3) holds if and only if

$$\begin{aligned} k^{2}\left( m_{2_{k}}^{\prime }-A_{k}^{2}\right) -2kx\left( A_{k}-G_{k}\right) \ge 0. \end{aligned}$$

(2.6)

It is enough to prove (2.6) for the greatest lower bound of $m_{2_{k}}^{\prime }$ which can be obtained from (2.2). Thus, (2.6) holds if

$$\begin{aligned} \frac{k^{2}}{k-1}\left( A_{k}^{2}-G_{k}^{2}\right) -2kx\left( A_{k}-G_{k}\right) \ge 0. \end{aligned}$$

(2.7)

We have, by AM–GM inequality, $A_{k}\ge G_{k}.$ For $A_{k}=G_{k},$ (2.7) is obviously true. For $A_{k}>G_{k},$ (2.7) holds if and only if

$$\begin{aligned} \frac{A_{k}+G_{k}}{2}\ge \frac{k-1}{k}x. \end{aligned}$$

This is true as $\frac{A_{k}+G_{k}}{2}\ge \underset{i}{\min }\ x_{i}\ge x\ge \frac{k-1}{k}x.$ $\square$

Theorem 1

If the eigenvalues of an element $A\in {\mathbb {M}}(n)$ are all positive,  then

$$\begin{aligned} \det A\le \left( \frac{\left( {\textrm{tr}}A\right) ^{2}-{\textrm{tr}}A^{2}}{n(n-1)} \right) ^{\frac{n}{2}}\le \det \left( \frac{{\textrm{tr}}AI-A}{n-1}\right) , \quad n\ge 2. \end{aligned}$$

(2.8)

For $n=2,$ (2.8) becomes equality.

Proof

The eigenvalues $\lambda _{1}(A),\lambda _{2}(A),\ldots ,\lambda _{n}(A)$ of A are all positive. We apply the AM–GM inequality to $\left( {\begin{array}{c}n\\ 2\end{array}}\right)$ numbers $\lambda _{i}(A)\lambda _{j}(A),$ $i< j$ and get

$$\begin{aligned} \frac{1}{\left( {\begin{array}{c}n\\ 2\end{array}}\right) }\underset{i<j}{\overset{n}{\sum }}\lambda _{i}(A)\lambda _{j}(A)\ge \left( \overset{n}{\underset{i<j}{\prod }}\lambda _{i}(A)\lambda _{j}(A)\right) ^{\frac{1}{\left( {\begin{array}{c}n\\ 2\end{array}}\right) }}=\left( \overset{n}{\underset{i=1}{\prod }}\lambda _{i}(A)\right) ^{\frac{2}{n}}=\left( \det A\right) ^{\frac{2}{n}}. \end{aligned}$$

(2.9)

Further, by a simple calculation, we have

$$\begin{aligned} \frac{1}{ \left( {\begin{array}{c}n\\ 2\end{array}}\right) }\underset{i<j}{\overset{n}{\sum }}\lambda _{i}(A)\lambda _{j}(A)=\frac{\left( \text {tr}A\right) ^{2}-\text {tr}A^{2}}{n(n-1)}. \end{aligned}$$

(2.10)

Combining (2.9) and (2.10), we immediately get the first inequality (2.8).

Denote $\frac{\text {tr}AI-A}{n-1}$ by H. The eigenvalues of H are $\lambda _{i}(H)=\frac{\text {tr}A-\lambda _{i}(A) }{n-1}>0,$ $i=1,2,\ldots ,n.$ On using the Lemma 1, we find

$$\begin{aligned} \left( \overset{n}{\underset{i=1}{\prod }}\lambda _{i}(H)\right) ^{\frac{2}{n }}\ge n\left( \frac{1}{n}\overset{n}{\underset{i=1}{\sum }}\lambda _{i}(H)\right) ^{2}-\left( n-1\right) \left( \frac{1}{n}\overset{n}{\underset{i=1}{\sum }}\lambda _{i}^{2}(H)\right) . \end{aligned}$$

(2.11)

Also, it is easily seen that

$$\begin{aligned} \overset{n}{\underset{i=1}{\sum }}\lambda _{i}(H)=\text {tr}A,\quad \overset{n}{\underset{i=1}{\prod }}\lambda _{i}(H)= \det \left( \frac{\text {tr}AI-A}{n-1}\right) , \end{aligned}$$

(2.12)

and

$$\begin{aligned} \frac{1}{n}\overset{n}{\underset{i=1}{\sum }}\lambda _{i}^{2}(H)=\frac{ \left( n-2\right) \left( \text {tr}A\right) ^{2}+\text {tr}A^{2}}{n(n-1)^{2}}. \end{aligned}$$

(2.13)

Inserting (2.12) and (2.13) in (2.11) and simplifying the resulting expression, we immediately get the second inequality (2.8). $\square$

Corollary 1

With notations and conditions as in Theorem 1, we have

$$\begin{aligned} \left( \det A\right) ^{\frac{1}{n}}\le \left( \det \left( \frac{\text {tr} AI-A}{n-1}\right) \right) ^{\frac{1}{n}}\le \frac{\text {tr}A}{n},\quad n\ge 2. \end{aligned}$$

(2.14)

Proof

The second inequality (2.14) follows immediately on applying (1.2) to the matrix $X=\frac{\text {tr}AI-A}{n-1}.$ The first inequality (2.14) follows from (2.8).

Alternatively,let $\alpha _{i}$ and $\beta _{i}$ be the arithmetic mean and geometric mean, respectively, of $n-1$ numbers obtained by excluding the number $x_{n-i+1}$ from the numbers $x_{1},x_{2},\ldots ,x_{n},$ that is

$$\begin{aligned} \alpha _{i}=\frac{1}{n-1}\sum _{\begin{array}{c} j=1\\ j\ne n-i+1 \end{array}}^{n}x_{j}\quad \text {and}\quad \beta _{i}=\left( \prod _{\begin{array}{c} j=1\\ j\ne n-i+1 \end{array}}^{n}x_{j}\right) ^{\frac{1}{n-1}}. \end{aligned}$$

On using the AM–GM inequality, we have $\alpha _{i}\ge \beta _{i}$ for each $i=1,2,\ldots ,n.$ We then have

$$\begin{aligned} \prod _{i=1}^{n}\alpha _{i}\ge \prod _{i=1}^{n}\beta _{i}=\prod _{i=1}^{n}x_{i}. \end{aligned}$$

(2.15)

On applying (2.15) to the n positive eigenvalues $\lambda _{i}(A),$ we get

$$\begin{aligned} \prod _{i=1}^{n}\alpha _{i}(A)\ge \prod _{i=1}^{n}\beta _{i}(A)=\prod _{i=1}^{n}\lambda _{i}(A), \end{aligned}$$

(2.16)

where

$$\begin{aligned} \alpha _{i}(A)=\frac{1}{n-1}\sum _{\begin{array}{c} j=1\\ j\ne n-i+1 \end{array}}^{n}\lambda _{j}(A) \quad \text {and}\quad \beta _{i}(A)=\left( \prod _{\begin{array}{c} j=1\\ j\ne n-i+1 \end{array}}^{n}\lambda _{j}(A)\right) ^{\frac{1}{n-1}}. \end{aligned}$$

(2.17)

Further, $(n-1)\alpha _{i}(A)=\text {tr}A-\lambda _{n-i+1}(A).$ Therefore, from (2.16) and (2.17), we get

$$\begin{aligned} \prod _{i=1}^{n}\frac{\text {tr}A-\lambda _{n-i+1}(A)}{n-1}\ge \prod _{i=1}^{n}\lambda _{i}(A). \end{aligned}$$

(2.18)

The eigenvalues of the matrix tr$AI-A$ are tr$A-\lambda _{j}(A),$ $j=1,2,\ldots ,n.$ Therefore, (2.18) gives the first inequality (2.14). $\square$

It may be noted that the first inequality (2.14) yields the inequality

$$\begin{aligned} \det \left( \text {tr}AI-A\right) \ge \left( n-1\right) ^{n}\det A \end{aligned}$$

which clearly becomes equality for $A=I.$

Corollary 2

Let the eigenvalues of $A=\left( a_{ij}\right) \in {\mathbb {M}}(n)$ be all positive. Then

$$\begin{aligned} \det A\le \left( \frac{1}{\left( {\begin{array}{c}n\\ 2\end{array}}\right) }\underset{i<j}{\overset{n}{\sum }} (a_{ii}a_{jj}-a_{ij}a_{ji})\right) ^{\frac{n}{2}}\le \det \left( \frac{{\textrm{tr}}AI-A}{n-1}\right) \le \left( \frac{1}{n}\sum _{i=1}^{n}a_{ii}\right) ^{n}.\nonumber \\ \end{aligned}$$

(2.19)

Proof

Let $\lambda _{i}(A),$ $i=1,2,\ldots ,n,$ be the eigenvalues of $A\in {\mathbb {M}}(n).$ The sum of the product of the eigenvalues taken two at a time is equal to the sum of all the $2\times 2$ principal minors of A (there are $\left( {\begin{array}{c}n\\ 2\end{array}}\right)$ of them), see [15]. We therefore have

$$\begin{aligned} \frac{\left( \text {tr}A\right) ^{2}-\text {tr}A^{2}}{n(n-1)}=\frac{1}{ \left( {\begin{array}{c}n\\ 2\end{array}}\right) }\underset{i<j}{\overset{n}{\sum }}\lambda _{i}(A)\lambda _{j}(A)=\frac{1}{ \left( {\begin{array}{c}n\\ 2\end{array}}\right) }\underset{i<j}{\overset{n}{\sum }}\left( a_{ii}a_{jj}-a_{ij}a_{ji}\right) . \end{aligned}$$

(2.20)

Inserting (2.20) in the inequalities (2.8), we immediately get the first and second inequalities (2.19). The third inequality (2.19) follows from the second inequality (2.14) on using tr$A$ $=\sum _{i=1}^{n}a_{ii}.$ $\square$

Corollary 3

Let the diagonal entries of a nonnegative matrix $A=\left( a_{ij}\right) \in {\mathbb {M}}(n)$ be all equal and positive. If the eigenvalues of A are all positive,  then

$$\begin{aligned} \det A\le \underset{i=1}{\overset{n}{\prod }}a_{ii}= a_{11}^{n}. \end{aligned}$$

(2.21)

Proof

The matrix $A=\left( a_{ij}\right) \in {\mathbb {M}}(n)$ is nonnegative and its diagonal entries are all equal and positive; therefore, $a_{ii}=a_{11},$ $a_{ij}a_{ji}\ge 0$ and $a_{ii}>0$ for all $i,j=1,2,\ldots ,n.$ It then follows that:

$$\begin{aligned} \frac{1}{ \left( {\begin{array}{c}n\\ 2\end{array}}\right) }\underset{i<j}{\overset{n}{\sum }}\left( a_{ii}a_{jj}-a_{ij}a_{ji}\right) \le \frac{1}{ \left( {\begin{array}{c}n\\ 2\end{array}}\right) }\underset{i<j}{\overset{n}{\sum }}a_{ii}a_{jj}=a_{11}^{2}. \end{aligned}$$

(2.22)

Combining the first inequality (2.19) and (2.22), we immediately get (2.21). $\square$

Example 1

The matrix

$$\begin{aligned} A= \begin{bmatrix} 1 &{} \frac{1}{2} &{} \frac{1}{5} \\ 1 &{} 1 &{} \frac{1}{2} \\ \frac{1}{3} &{} \frac{1}{2} &{} \frac{1}{2} \end{bmatrix} \end{aligned}$$

is totally positive. Its eigenvalues are all positive, and therefore, by the Corollary 3, we have $\det A\le 0.5.$ Also, $\det A=\frac{7}{60}=0.11667.$

We now prove a refinement of the Hadamard inequality (1.1) in the special case when the diagonal entries of A are all equal.

Corollary 4

Let $A\in {\mathbb {M}}(n)$ be positive definite. If the diagonal entries of A are all equal,  then

$$\begin{aligned} \det A\le \det \left( \frac{\text {tr}AI-A}{n-1}\right) \le \underset{i=1}{\overset{n}{\prod }}a_{ii}= a_{11}^{n}. \end{aligned}$$

(2.23)

Proof

The diagonal entries of $X= \frac{\text {tr}AI-A}{n-1}$ are $x_{ii}=a_{ii}$ for all $i=1,2,\ldots ,n.$ Then, the second inequality (2.23) follows on applying (1.1) to the matrix X. The first inequality (2.23) follows from (2.8). $\square$

Example 2

Let

$$\begin{aligned} A= \begin{bmatrix} 2 &{} 1 &{} 1 \\ 1 &{} 2 &{} 1 \\ 1 &{} 1 &{} 2 \end{bmatrix},\quad X=\frac{\text {tr}AI-A}{n-1}= \begin{bmatrix} 2 &{} -\frac{1}{2} &{} -\frac{1}{2} \\ -\frac{1}{2} &{} 2 &{} -\frac{1}{2} \\ -\frac{1}{2} &{} -\frac{1}{2} &{} 2 \end{bmatrix}. \end{aligned}$$

Then, corresponds to (2.23), $\det A=4\le \det X=6.25\le \underset{i=1}{\overset{n}{\prod }}a_{ii}=8.$

Likewise, for the totally positive matrix $A=(a_{ij})\in {\mathbb {M}}(5),$ $a_{ij}=e^{-\left( i-j\right) ^{2}},$ we have $\det A=0.52602\le \det X=0.96609\le \overset{n}{\underset{i=1}{\prod }}a_{ii}=1.$

3 Bounds for the spectral radius

Our aim here is to study the inequalities involving eigenvalues of nonnegative matrices in connection with the positive linear maps. A linear map $\Phi :{\mathbb {M}}(n)\rightarrow {\mathbb {M}}(k)$ is said to be positive if $\Phi (A)$ is positive semidefinite $\left( \Phi (A)\ge O\right)$ whenever $A\ge O.$ It is called unital if $\Phi (I_{n})=I_{k}.$ In the special case when $k=1$ the map from ${\mathbb {M}}(n)$ to ${\mathbb {C}}$ is called linear functional and it is customary to represent it by the lower case letter $\varphi ,$ see Bhatia [5]. For some related inequalities, see Kian et al. [17].

Our proofs of theorems in this section rely upon the following basic facts of matrix analysis.

For any Hermitian element $A\in {\mathbb {M}}(n),$ we have

$$\begin{aligned} \lambda _{\text {min}}(A)\le y^{*}Ay\le \lambda _{\text {max}}(A) \end{aligned}$$

(3.1)

for all unit vectors $y\in {\mathbb {C}}^{n},$ see [4].

Let $A,B\in {\mathbb {M}}(n)$ and let $\left| A\right| =\left( \left| a_{ij}\right| \right) .$ Then, by [15, Theorem 8.1.18], if $B-\left| A\right|$ is nonnegative

$$\begin{aligned} \rho (A)\le \rho (\left| A\right| )\le \rho (B), \end{aligned}$$

(3.2)

where $\rho (A)$ is defined as in (1.3).

The Kadison theorem [16] says that for any positive unital linear map $\Phi$ and for any Hermitian element $A\in {\mathbb {M}}(n),$ we have

$$\begin{aligned} \Phi (A^{2})\ge \Phi (A)^{2}. \end{aligned}$$

(3.3)

Bhatia and Davis [6] proved that for a positive unital linear map $\Phi :{\mathbb {M}}(n)\rightarrow {\mathbb {M}}(k)$ and for any Hermitian element $A\in {\mathbb {M}}(n)$ whose spectrum is contained in the interval $\left[ m,M \right] ,$ we have

$$\begin{aligned} \Phi (A^{2})-\Phi (A)^{2}\le \left( MI_{k}-\Phi (A)\right) \left( \Phi (A)-mI_{k}\right) \le \frac{\left( M-m\right) ^{2}}{4}I_{k}. \end{aligned}$$

(3.4)

Theorem 2

Let $A=\left( a_{ij}\right) \in {\mathbb {M}}(n)$ be nonnegative. Let $x_{ij}= {\min }\left\{ a_{ij},a_{ji}\right\}$ for all i and j and let $X=\left( x_{ij}\right) \in {\mathbb {M}}(n).$ Then, for any positive unital linear map $\Phi :{\mathbb {M}}(n)\rightarrow {\mathbb {M}}(k),$ we have

$$\begin{aligned} \rho (A)I_{k}\ge \Phi (X). \end{aligned}$$

(3.5)

Proof

It is clear that X is a real symmetric matrix and $\rho (X)$ is its largest eigenvalue. Then, on using (3.1), we have $y^{*} \left( \rho (X)I_{n}-X\right) y\ge 0$ for all $y\in {\mathbb {C}}^{n}.$ The matrix $\rho (X)I_{n}-X$ is therefore positive semidefinite. The map $\Phi :{\mathbb {M}}(n)\rightarrow {\mathbb {M}}(k)$ is positive; therefore, the matrix $\Phi \left( \rho (X)I_{n}-X\right)$ is also positive semidefinite. The map $\Phi$ is unital, $\Phi (I_{n})=I_{k}.$ Therefore, by the linearity of $\Phi$

$$\begin{aligned} \rho (X)I_{k}\ge \Phi (X). \end{aligned}$$

(3.6)

Further, A,  X and $A-X$ are all nonnegative. Therefore, by (3.2)

$$\begin{aligned} \rho (A)\ge \rho (X). \end{aligned}$$

(3.7)

Combining (3.6) and (3.7), we immediately get (3.5). $\square$

On using Theorem 2, we can derive various lower bounds for $\rho (A)$ for different choices of $\Phi .$ For instance, for the positive unital linear functional $\varphi (A)=\frac{ 1}{n}\underset{i,j}{\overset{n}{\sum }}a_{ij},$ we have

$$\begin{aligned} \rho (A)\ge \frac{1}{n}\underset{i,j}{\overset{n}{\sum }}x_{ij}. \end{aligned}$$

(3.8)

For a nonnegative symmetric matrix, (3.8) improves the lower bound in the classical inequalities, see [15]

$$\begin{aligned} \max \left\{ \underset{i}{\min }\ R_{i},\ \underset{j}{\min }\ C_{j}\right\} \le \rho (A)\le \min \left\{ \underset{i}{\max }\ R_{i}, \ \underset{j}{\max }\ C_{j}\right\} , \end{aligned}$$

(3.9)

where $R_{i}=\sum _{j=1}^{n}a_{ij}$ and $C_{j}=\sum _{i=1}^{n}a_{ij}$ are the row sums and column sums, respectively. This is so as

$$\begin{aligned} \frac{1}{n}\underset{i,j}{\overset{n}{\sum }}a_{ij}=\frac{1}{n}\underset{i=1}{\overset{n}{\sum }}R_{i}\ge \underset{i}{\min }\ R_{i}. \end{aligned}$$

For $\varphi (A)=\frac{a_{ii}+a_{jj}+a_{ij}+a_{ji}}{2},$ $i\ne j,$ (3.5) yields

$$\begin{aligned} \rho (A)\ge \underset{i,j}{\max }\left\{ \frac{x_{ii}+x_{jj}}{2} +x_{ij}\right\} . \end{aligned}$$

(3.10)

The inequalities (3.8) and (3.10) are independent.

Example 3

Marcus and Minc [18] compare the various bounds of $\rho (A)$ for the matrix

$$\begin{aligned} A= \begin{bmatrix} 1 &{} 1 &{} 2 \\ 2 &{} 1 &{} 3 \\ 2 &{} 3 &{} 5 \end{bmatrix}. \end{aligned}$$

Their best estimate gives $\rho (A)\ge 5.162.$ The estimates of Wolkowicz and Styan [21] give $\rho (A)\ge 2.33.$ From (3.9), $\rho (A)\ge 5,$ while from (3.8) and (3.10), we, respectively, have $\rho (A)\ge 6.33$ and $\rho (A)\ge 6.$ The actual value of $\rho (A)$ to three decimal places is 7.531.

For the matrix

$$\begin{aligned} B= \begin{bmatrix} 1 &{} 1 &{} 0 \\ 2 &{} 1 &{} 3 \\ 0 &{} 3 &{} 5 \end{bmatrix}, \end{aligned}$$

(3.8), (3.9), and (3.10) give $\rho (B)\ge 5,$ $\rho (B)\ge 3$ and $\rho (B)\ge 6,$ respectively. We also conclude from these estimates of $\rho (A)$ and $\rho (B)$ that (3.8) and (3.10) are independent.

We need the following lemma in the proof of subsequent theorem.

Lemma 2

Let $A=\left( a_{ij}\right) \in {\mathbb {M}}(n)$ be Hermitian and let its eigenvalues and diagonal entries be arranged as in (1.4) and (1.5), respectively. Let $A=D+N,$ $D={\textrm{diag}} (a_{11},a_{22},\ldots ,a_{nn}).$ If $\lambda _{n}\left( N\right) \ge \left| \lambda _{1}\left( N\right) \right| ,$ then for $n\ge 3$

$$\begin{aligned} \lambda _{1}(A)+\lambda _{n}(A)\ge a_{1}+a_{2}, \end{aligned}$$

(3.11)

and if $\lambda _{n}\left( N\right) \le \left| \lambda _{1}\left( N\right) \right|$

$$\begin{aligned} \lambda _{1}(A)+\lambda _{n}(A)\le a_{n-1}+a_{n}. \end{aligned}$$

(3.12)

Proof

Let B be the matrix obtained by replacing the smallest diagonal entry of A by $a_{2}$ and keeping all other entries of A and B same. Then, $B-A$ is a diagonal matrix and its only non-zero diagonal entry is $a_{2}-a_{1}\ge 0.$ Therefore, $B-A$ is positive semidefinite.

Under the condition of the lemma, we have

$$\begin{aligned} \lambda _{1}\left( N\right) +\lambda _{n}\left( N\right) \ge 0. \end{aligned}$$

(3.13)

Let $C=a_{2}I_{n}+N.$ Therefore, $\lambda _{i}\left( N\right) =\lambda _{i}(C)-a_{2}.$ Then, from (3.13), we get that

$$\begin{aligned} \lambda _{1}(C)+\lambda _{n}(C)\ge 2a_{2}. \end{aligned}$$

(3.14)

It is clear that $B-C$ is a diagonal matrix with diagonal entries $0,0,a_{3}-a_{2},\ldots ,a_{n}-a_{2}$ in some order. Also, $a_{j}\ge a_{2}$ for $j=3,\ldots ,n.$ Therefore, $B-C$ is positive semidefinite. Therefore, by Weyl’s inequality, $\lambda _{i}(B)\ge \lambda _{i}(C)$ for all $i=1,2,\ldots ,n.$ Then, by (3.14)

$$\begin{aligned} \lambda _{1}(B)+\lambda _{n}(B)\ge \lambda _{1}(C)+\lambda _{n}(C)\ge 2a_{2}. \end{aligned}$$

(3.15)

We can write

$$\begin{aligned} 2a_{2}=\text {tr}B-\underset{i=3}{\overset{n}{\sum }}a_{i}\quad \text {and}\quad \lambda _{1}(B)+\lambda _{n}(B)=\text {tr}B-\underset{i=2}{\overset{n-1}{\sum }}\lambda _{i}(B). \end{aligned}$$

(3.16)

From (3.15) and (3.16), we obtain

$$\begin{aligned} \underset{i=2}{\overset{n-1}{\sum }}\lambda _{i}(B)\le \underset{i=3}{ \overset{n}{\sum }}a_{i}. \end{aligned}$$

(3.17)

Further, $B-A$ is positive semidefinite; therefore, by Weyl’s inequality $\lambda _{i}(B)\ge \lambda _{i}(A)$ for all $i=1,2,\ldots ,n$ and it follows from (3.17) that $\sum _{i=2}^{n-1}\lambda _{i}(A)\le \sum _{i=3}^{n}a_{i}.$ This yields (3.11) on using the fact that tr$A=\sum _{i=1}^{n}\lambda _{i}(A)=\sum _{i=1}^{n}a_{i}.$ The arguments for (3.12) are same. $\square$

Theorem 3

Let the diagonal entries and eigenvalues of a nonnegative symmetric matrix $A=\left( a_{ij}\right) \in {\mathbb {M}}(n)$ be arranged as in (1.4) and (1.5), respectively. Then,  for any positive unital linear map $\Phi :{\mathbb {M}}(n)\rightarrow {\mathbb {M}}(k),$ we have

$$\begin{aligned} \lambda _{n}(A)I_{k}\ge \frac{a_{1}+a_{2}}{2}I_{k}+\sqrt{\Phi (A^{2})-\Phi (A)^{2}}. \end{aligned}$$

(3.18)

Proof

By Kadison’s inequality (3.3), the matrix $\Phi (A^{2})-\Phi (A)^{2}$ is positive semidefinite. Therefore, it has a unique positive square root $\sqrt{\Phi (A^{2})-\Phi (A)^{2}},$ [5]. It then follows from (3.4) that:

$$\begin{aligned} \left( \lambda _{n}(A)-\lambda _{1 }(A)\right) I_{k}\ge 2\sqrt{\Phi (A^{2})-\Phi (A)^{2}}. \end{aligned}$$

(3.19)

For a nonnegative symmetric matrix $S\in {\mathbb {M}}(n),$ we have $\rho (S)\ge \left| \lambda \right|$ where $\lambda$ is any eigenvalue of S. We write $A=D+N$ where $D=\text {diag}(a_{11},a_{22},\ldots ,a_{nn}).$ The matrix A is nonnegative and, therefore, $N=A-D$ is also nonnegative. Therefore, $\lambda _{n}\left( N\right) \ge \left| \lambda _{1}\left( N\right) \right| .$ Then, by Lemma 2, we have

$$\begin{aligned} \left( \lambda _{n }(A)+\lambda _{1}(A)\right) I_{k}\ge \left( a_{1}+a_{2}\right) I_{k}. \end{aligned}$$

(3.20)

The inequalities (3.19) and (3.20) yield the inequality (3.18). $\square$

Corollary 5

With notations and conditions as in Theorem 3, we have for any nonnegative matrix $A=\left( a_{ij}\right) \in {\mathbb {M}}(n)$

$$\begin{aligned} \rho (A)I_{k}\ge \frac{a_{1}+a_{2}}{2}I_{k}+\sqrt{\Phi (X^{2})-\Phi (X)^{2}}, \end{aligned}$$

(3.21)

where $X=(x_{ij})\in {\mathbb {M}}(n)$ with $x_{ij}={\min }\left\{ a_{ij},a_{ji}\right\} .$

Proof

For a nonnegative symmetric matrix X,  $\lambda _{n }(X)=\rho (X).$ On applying (3.18) to the matrix X,  and then using (3.7), we immediately get (3.21). $\square$

We mention some particular cases of Corollary 5.

For $\varphi (A)=a_{ii}$ for any fixed $i=1,2,\ldots ,n,$ we have from (3.21)

$$\begin{aligned} \rho (A)\ge \frac{a_{1}+a_{2}}{2}+\underset{j}{\max }\sqrt{\underset{k\ne j }{\underset{k=1}{\overset{n}{\sum }}}x_{jk}^{2}}, \end{aligned}$$

and for $\varphi (A)=\frac{a_{ii}+a_{jj}}{2},$ $i\ne j,$ we have

$$\begin{aligned} \rho (A)\ge \frac{a_{1}+a_{2}}{2}+\frac{1}{2}\underset{i,j}{\max }\sqrt{2\left( \underset{k\ne i }{\underset{k=1}{\overset{n}{\sum }}} x_{ik} ^{2}+\underset{k\ne j }{\underset{k=1}{\overset{n}{\sum }}} x_{jk} ^{2}\right) +\left( a_{ii}-a_{jj}\right) ^{2} }. \end{aligned}$$

For $\varphi (A)=\frac{\text {tr}A}{n},$ we have

$$\begin{aligned} \rho (A)\ge \frac{a_{1}+a_{2}}{2}+\sqrt{\frac{\text {tr}X^{2}}{n}-\left( \frac{\text {tr}X}{n}\right) ^{2}}. \end{aligned}$$

(3.22)

The inequality (3.22) is independent of the Wolkowicz and Styan [21] bound

$$\begin{aligned} \rho (A)\ge \frac{\text {tr}A}{n}+\frac{1}{\sqrt{n-1}}\sqrt{\frac{\text {tr} A^{2}}{n}-\left( \frac{\text {tr}A}{n}\right) ^{2}}. \end{aligned}$$

(3.23)

The inequality (3.23) is also valid for any $A\in {\mathbb {M}}(n)$ with all its eigenvalues real. In case if A is symmetric and the diagonal entries of A are all equal, then (3.22) is always better than (3.23).

Example 4

Let

$$\begin{aligned} A= \begin{bmatrix} 4 &{} 0 &{} 2 &{} 3 \\ 0 &{} 5 &{} 0 &{} 1 \\ 2 &{} 0 &{} 6 &{} 0 \\ 3 &{} 1 &{} 0 &{} 7 \end{bmatrix},\quad B= \begin{bmatrix} 4 &{} 1 &{} 1 &{} 2 &{} 2 \\ 1 &{} 5 &{} 1 &{} 1 &{} 1 \\ 1 &{} 1 &{} 6 &{} 1 &{} 1 \\ 2 &{} 1 &{} 1 &{} 7 &{} 1 \\ 2 &{} 1 &{} 1 &{} 1 &{} 8 \end{bmatrix}. \end{aligned}$$

The estimates of Wolkowicz and Styan [21] give $\lambda _{4}(A)\ge 7.158$ and $\lambda _{5}(B)\ge 7.449,$ while from our estimate (3.22), we have $\lambda _{4}(A)\ge 7.3723$ and $\lambda _{5}(B)\ge 7.3983.$ This also shows that for nonnegative symmetric matrices, the inequalities (3.22) and (3.23) are independent. The actual values of $\lambda _{4}(A)$ and $\lambda _{5}(B)$ to three decimal places are 9.376 and 11.171,  respectively.

We now consider Z-matrices. An element $A=(a_{ij})\in {\mathbb {M}}(n)$ is a Z-matrix if $a_{ij}\le 0$ for all $i\ne j;$ see [2]. We derive an upper bound for the largest eigenvalue of a symmetric Z-matrix on using the Nagy’s inequality [19] that says that for n real numbers $x_{i}$’s

$$\begin{aligned} \frac{1}{n}\underset{i=1}{\overset{n}{\sum }}x_{i}^{2}-\left( \frac{1}{n} \underset{i=1}{\overset{n}{\sum }}x_{i}\right) ^{2}\ge \underset{i,j}{\max }\frac{\left( x_{i}-x_{j}\right) ^{2}}{2n}. \end{aligned}$$

(3.24)

Theorem 4

Let $A=\left( a_{ij}\right) \in {\mathbb {M}}(n)$ be a symmetric Z-matrix. Then

$$\begin{aligned} \lambda _{\max } (A)\le \frac{a_{n-1}+a_{n}}{2}+\sqrt{\frac{1}{2}\left( {\textrm{tr}}A^{2}- \frac{\left( {\textrm{tr}}A\right) ^{2}}{n}\right) }. \end{aligned}$$

(3.25)

Proof

Let the eigenvalues of A be arranged as in (1.4). Then, on applying (3.24) to the n eigenvalues of A,  we get

$$\begin{aligned} \lambda _{n}(A)-\lambda _{1}(A)\le \sqrt{2}\sqrt{\left( \text {tr} A^{2}-\frac{\left( \text {tr}A\right) ^{2}}{n}\right) }. \end{aligned}$$

(3.26)

It is clear that if the diagonal entries of a Z-matrix Z are all zero, then $\lambda _{n} (Z)\le \left| \lambda _{1}(Z)\right| .$ The matrix A is Z-matrix; therefore, $N=A-D$ is also a Z-matrix where D is the diagonal part of A. Therefore, $\lambda _{n}\left( N\right) \le \left| \lambda _{1}\left( N\right) \right| ,$ and therefore, by Lemma 2, we have

$$\begin{aligned} \lambda _{n}(A)+\lambda _{1}(A)\le a_{n-1}+a_{n}. \end{aligned}$$

(3.27)

We also have $\lambda _{n}(A)=\lambda _{\max }(A).$ Therefore, the inequality (3.25) follows immediately on adding the inequalities (3.26) and (3.27). $\square$