A note on variance bounds and location of eigenvalues

TL;DR

This paper explores advanced variance bounds for real and complex numbers and extends these to eigenvalues and matrix spread, providing refined mathematical tools for spectral analysis.

Contribution

It introduces new extensions and refinements of variance bounds and applies them to eigenvalues and matrix spread analysis.

Findings

Derived tighter variance bounds for real and complex numbers

Extended bounds to eigenvalues and matrix spread

Provided mathematical tools for spectral analysis

Abstract

We discuss some extensions and refinements of the variance bounds for both real and complex numbers. The related bounds for the eigenvalues and spread of a matrix are also derived here.

Equations

$$\frac{1}{n}\stackrel{n}{\sum _{i=1}}{z}_i=\tilde{z}.$$

$$\frac{1}{n}\stackrel{n}{\sum _{i=1}}{\left|z_i-\tilde{z}\right|}^2=S_z^2$$

$$\frac{1}{n}\stackrel{n}{\sum _{i=1}}{\left(z_i-\tilde{z}\right)}^2=S^2$$

$${\sigma }_z^2=\frac{\left|S^2\right|+S_z^2}{2}$$

$$s^2\le \frac{{\left(b-a\right)}^2}{4}$$

$${\sigma }_z^2\le \underset{i,j}{\max }\frac{{\left|z_i-z_j\right|}^2}{4}.$$

$$S_z^2\le r_z^2\le \underset{i,j}{\max }\frac{{\left|z_i-z_j\right|}^2}{3},$$

$$\underset{i,j}{\max }\frac{\left|z_i-z_j\right|}{2}\le r_z\le \underset{i,j}{\max }\frac{\left|z_i-z_j\right|}{\sqrt{3}}.$$

$$s^2\ge \frac{r}{n-r}{\left({\alpha }_r-\overline{x}\right)}^2,$$

$$S^2\ge \frac{{\left(b-a\right)}^2}{2n}$$

$$\frac{4}{n}\text{tr}{B}^2\le \text{Spd}{\left(A\right)}^2\le 2\text{tr}{B}^2,$$

$$\frac{{\left(b-a\right)}^2}{2n}+\frac{2}{n-2}{\left(\overline{x}-\frac{a+b}{2}\right)}^2\le s^2\le \left(b-\overline{x}\right)\left(\overline{x}-a\right),$$

$$S_z^2=\frac{1}{n}\stackrel{n}{\sum _{i=1}}{\left|z_i-c+c-\tilde{z}\right|}^2=\frac{1}{n}\stackrel{n}{\sum _{i=1}}{\left|z_i-c\right|}^2-\frac{1}{n}\stackrel{n}{\sum _{i=1}}{\left|\tilde{z}-c\right|}^2.$$

$$\frac{1}{n}\stackrel{n}{\sum _{i=1}}{\left|z_i-c\right|}^2=r_z^2.$$

$$S_z^2+{\left|\tilde{z}-c\right|}^2=r_z^2.$$

$$\left|\stackrel{n}{\sum _{i=1}}{\left(z_i-\tilde{z}\right)}^2\right|=\stackrel{n}{\sum _{i=1}}{\left|z_i-\tilde{z}\right|}^2.$$

$$\left|\stackrel{n}{\sum _{i=1}}{a}_i\right|\le \stackrel{n}{\sum _{i=1}}\left|a_i\right|$$

$${\left(\frac{z_i-\tilde{z}}{z_j-\tilde{z}}\right)}^2>0.$$

$$S_{Z_1\cup Z_2}^2=\frac{n_1}{n_1+n_2}{S}_{Z_1}^2+\frac{n_2}{n_1+n_2}{S}_{Z_2}^2+\frac{n_1{n}_2}{{\left(n_1+n_2\right)}^2}{\left|{\tilde{Z}}_1-{\tilde{Z}}_2\right|}^2.$$

$$S_{Z_1\cup Z_2}^2=\frac{1}{n_1+n_2}\left(\underset{j=1}{\sum ^{n_1}}{\left|z_j-\tilde{a}\right|}^2+\underset{j=n_1+1}{\sum ^{n_1+n_2}}{\left|z_j-\tilde{a}\right|}^2\right),$$

$$\tilde{a}=\frac{1}{n_1+n_2}\stackrel{n_1+n_2}{\sum _{j=1}}{z}_j.$$

$${\left|z_j-\tilde{a}\right|}^2$$

$$\underset{j=1}{\sum ^{n_1}}\left(z_j-{\tilde{Z}}_1\right)=0\text{ and }\left|{\tilde{Z}}_1-\tilde{a}\right|=\frac{n_2}{n_1+n_2}\left|{\tilde{Z}}_1-{\tilde{Z}}_2\right|.$$

$$\underset{j=1}{\sum ^{n_1}}{\left|z_j-\tilde{a}\right|}^2=\underset{j=1}{\sum ^{n_1}}{\left|z_j-{\tilde{Z}}_1\right|}^2+\frac{n_1{n}_2^2}{{\left(n_1+n_2\right)}^2}{\left|{\tilde{Z}}_1-{\tilde{Z}}_2\right|}^2.$$

$$\underset{j=n_1+1}{\sum ^{n_1+n_2}}{\left|z_j-\tilde{a}\right|}^2=\underset{j=n_1+1}{\sum ^{n_1+n_2}}{\left|z_j-{\tilde{Z}}_2\right|}^2+\frac{n_1^2{n}_2}{{\left(n_1+n_2\right)}^2}{\left|{\tilde{Z}}_1-{\tilde{Z}}_2\right|}^2.$$

$${\left|{\tilde{\gamma }}_r-\tilde{z}\right|}^2\le \frac{n-r}{r}{\sigma }_z^2$$

$$S_z^2=\frac{r}{n}{S}_{z\left(r\right)}^2+\frac{n-r}{n}{S}_{z\left(n-r\right)}^2+\frac{r\left(n-r\right)}{n^2}{\left|{\gamma }_r-{\gamma }_{n-r}\right|}^2.$$

$$\left|{\gamma }_r-{\gamma }_{n-r}\right|=\left|{\gamma }_r-\frac{1}{n-r}\left(\underset{i=1}{\sum ^n}{z}_i-\underset{i=1}{\sum ^r{z}_i}\right)\right|=\left|\frac{n}{n-r}\left({\gamma }_r-\tilde{z}\right)\right|.$$

$$S_z^2=\frac{r}{n}{S}_{z\left(r\right)}^2+\frac{n-r}{n}{S}_{z\left(n-r\right)}^2+\frac{r}{n-r}{\left|{\gamma }_r-\tilde{z}\right|}^2.$$

$$S^2=\frac{r}{n}{S}_r^2+\frac{n-r}{n}{S}_{n-r}^2+\frac{r}{n-r}{\left({\gamma }_r-\tilde{z}\right)}^2.$$

Full text

**A note on variance bounds and location of eigenvalues **

R. Sharma, A. Sharma and R. Saini

Department of Mathematics & Statistics

Himachal Pradesh University

Shimla - 171005,

India

email: rajesh_hpu_math@yahoo.co.in

**Abstract. **We discuss some extensions and refinements of the variance bounds for both real and complex numbers. The related bounds for the eigenvalues and spread of a matrix are also derived here.

**AMS classification.  ** 15A42, 26C10, 60E15

Key words and phrases.  Variance, complex numbers, leptokurtic distributions, eigenvalues, trace, spread, polynomial, span, Jung’s theorem in the plane.

1 Introduction

Let $z_{1},z_{2},...,z_{n}$ denote $n$ complex numbers. Their arithmetic mean is the number

[TABLE]

In literature, the number

[TABLE]

or its equivalent expressions have been studied in several different contexts and notations and is termed as the variance of complex numbers at many places. For example, see Audenaert $\left[2\right]$, Bhatia and Sharma $\left[4,5\right],$ Merikoski and Kumar $\left[13\right]$, and Park $\left[17\right]$.

The number

[TABLE]

is also important in this context. If $z_{i}$’s are all real we denote them by $x_{i}$’s with $a=\min x_{i}$ and $b=\max x_{i}.$ The arithmetic mean by $\overline{x}$ and variance by the lower case letter $s^{2}.$ In this case $S_{z}=\left|S\right|=S=s$ but in general $S_{z}$ rather than $\left|S\right|$ is more consistent with $s.$ For instance, $s=0$ $\left(S_{z}=0\right)$ if and only if all the $x_{i}$’s $\left(z_{i}\text{'s}\right)$ are equal. This is not the case with $\left|S\right|;$ for example, for three distinct complex numbers $0,$ $\pm\frac{1}{2}+i\frac{\sqrt{3}}{2}$ we have $S=0.$ It however turns out that for some purposes $s^{2}$ is more consistent with

[TABLE]

than $S_{z}^{2}.$ Note that the analogue of the Popoviciu inequality [18]

[TABLE]

for the complex numbers says that

[TABLE]

But it is not always true that $S_{z}^{2}\leq\underset{i,j}{\max}\frac{\left|z_{i}-z_{j}\right|^{2}}{4}.$ For example, for $z_{1}=-\frac{1}{2}+i\frac{\sqrt{3}}{2},$ $z_{2}=0$ and $z_{3}=\frac{1}{2}+i\frac{\sqrt{3}}{2},$ $S_{z}^{2}=\frac{1}{3}$ and $\underset{i,j}{\max}\left|z_{i}-z_{j}\right|=1.$

The corresponding inequality for $S_{z}^{2}$ is

[TABLE]

where $r_{z}$ is the radius of the smallest disk containing all the numbers $z_{i}$’s, see $\left[\text{4, 5}\right].$

A classical theorem of Jung [9] says that the complex numbers $z_{i}$’s in a plane can be contained in a closed  disk of radius $\underset{i,j}{\max}\frac{\left|z_{i}-z_{j}\right|}{\sqrt{3}}.$ We thus have

[TABLE]

In this context it is in interesting to note a case when the given complex numbers lie on the boundary of the smallest disk containing them. We here show that if the complex numbers lie on a circle with centre at their arithmetic mean then this circle is the smallest circle enclosing these points, (see Theorem 2.1 & 3.1 below). Further, if the complex numbers are all collinear then $\left|S\right|=\sigma_{z}=$ $S_{z}$, and conversely, ( Theorem 2.2). A necessary and sufficient condition is given for which the numbers $\sigma_{z}$, $S_{z}$ and $\left|S\right|$ are all equal, ( Theorem 2.2 ). We obtain a complex analogue of the inequality, Mallows and  Richter [11],

[TABLE]

where $\alpha_{r}$ is the arithmetic mean of any subset of $r$ numbers chosen from the real numbers $x_{1},x_{2},\ldots,x_{n},$ (Theorem 2.3).

On the other hand we find in literature that the inequality (1.5) and its complementary Nagy’s inequality [13],

[TABLE]

also provide bounds for the spread of a complex $n\times n$ matrix $A$ when the eigenvalues $\lambda_{i}\left(A\right)$ of $A$ are all real. The spread of a matrix $A$ is the maximum distance between two eigenvalues of a matrix, Spd$\left(A\right)=\lambda_{\max}\left(A\right)-\lambda_{\min}\left(A\right).$ We have,

[TABLE]

where $B=A-\frac{\text{tr}A}{n}I$ and tr$A$ denotes the trace of $A,$ see [6, 23].

We show that the inequalities, [3, 21],

[TABLE]

provide some further refinements of the inequalities (1.5) and (1.9) and consequently we get better bounds for the spread of a matrix for some special cases, ( Theorem 2.4, 2.5, 3.2). A refinement of the inequality (1.5) is obtained for Leptokurtic and Mesokurtic distributions, (Theorem 2.6). We obtain refinements of the eigenvalue bounds in some special cases, (Theorem 3.3, 3.4). Likewise, the bounds for the span of a polynomial are given, (Theorem 3.5).

2 Main Results

**Theorem 2.1. **If the complex numbers $z_{i}$’s lie on a circle in the complex plane with centre $\widetilde{z}$ and radius $r_{z},$ then $r_{z}$ is the radius of the smallest disk containing all the points $z_{i}$’s.

**Proof. **For any complex number $c,$ we can write (1.2) in the form

[TABLE]

Under the condition of the theorem, $\left|z_{i}-c\right|=r_{z}$ for all $i=1,2,\ldots,n$ and therefore

[TABLE]

Combining (2.1) and (2.2), we get that

[TABLE]

From the first inequality (1.7), $r_{z}\geq S_{z}.$ So the minimum value of the $r_{z}$ is $S_{z}.$ This implies that if $r_{z}=S_{z}$ then $r_{z}$ is the radius of the smallest disk containing the points $z_{i}$’s. For $\widetilde{z}=c,$ (2.3) gives $r_{z}=S_{z}.$ This proves the theorem. $\blacksquare$

**Theorem 2.2. **Let $z_{1},z_{2},...,z_{n}$ be the points in the finite complex plane and let $S_{z},$ $S$ and $\sigma_{z}$ be defined as in (1.2), (1.3) and (1.4), respectively. Then, $S_{z}=\left|S\right|=\sigma_{z}$ if and only if all the points $z_{1},z_{2},...,z_{n}$ lie on a straight line.

**Proof. **In the complex plane the convex combination of complex numbers lie in the convex hull of these numbers. It follows that if the points $z_{i}$’s are collinear then $\widetilde{z}$ also lies on the straight line passing through $z_{i}$’s.

From (1.2) - (1.4), we see that $S_{z}=\left|S\right|=\sigma_{z}$ if and only if

[TABLE]

The equality occurs in triangle inequality

[TABLE]

if and only if the ratio of any two non-zero terms is positive that is $\frac{a_{i}}{a_{j}}>0,$ $\ i,j=1,2,...,n,$ see Ahlfors [1]. This means (2.4) holds true if and only if the ratio of any two non zero terms in (2.4) is positive, that is

[TABLE]

The square of a complex number $z$ is positive if and only  if $z$ is real  and therefore (2.5) implies that $\frac{z_{i}-\widetilde{z}}{z_{j}-\widetilde{z}}$ is real. Also, $\frac{z_{i}-\widetilde{z}}{z_{j}-\widetilde{z}}$ is real if and only if $z_{i}$ lies on the straight line passing through $z_{j}$ and $\widetilde{z}.$ If $z_{k}-\widetilde{z}=0$ for some $k$ then $z_{k}=\widetilde{z}$ and so $z_{k}$ lies on the straight line passing through $z_{j}$ and $\widetilde{z}.$ $\blacksquare$

We need following lemma to extend the inequality (1.8) for complex numbers.

Lemma 2.1 Let $Z_{1}=\left\{z_{1},z_{2},...,z_{n_{1}}\right\}$ and $Z_{2}=\left\{z_{n_{1}+1},z_{n_{1}+2},...,z_{n_{1}+n_{2}}\right\}$ be two sets of complex numbers. Denote by $\widetilde{Z}_{i}$ and $S_{Z_{i}}^{2}$ the arithmetic mean and variance of $Z_{i}$’s, $i=1,2,$ respectively. Then the combined variance $S_{Z_{1}\cup Z_{2}}^{2}$ of the set $Z_{1}\cup Z_{2}$ is given by

[TABLE]

Proof. The combined variance of the set $Z_{1}\cup Z_{2}$ of $n_{1}+n_{2}$ numbers can be written as

[TABLE]

where

[TABLE]

We note that

[TABLE]

[TABLE]

Therefore,

[TABLE]

On using similar arguments, we have

[TABLE]

The assertions of the theorem now follow on using (2.8) and (2.9) in (2.7). $\blacksquare$

**Theorem 2.3. **Let $\widetilde{\gamma}_{r}$ be the arithmetic mean of any subset of $r$ numbers chosen from the set of $n$ complex numbers $z_{1},z_{2},\ldots,z_{n}$ and let $\sigma_{z}^{2}$ be defined as in (1.4). Then the inequality

[TABLE]

holds true for $1\leq r\leq n.$

**Proof. **Let $Z_{1}$ and $Z_{2}$ be the disjoint sets of $r$ and $n-r$ numbers chosen from the numbers $z_{1},z_{2},\ldots,z_{n},$ respectively. Denote by $S_{z\left(r\right)}^{2}$ and $S_{z\left(n-r\right)}^{2}$ the variance of $Z_{1}$ and $Z_{2},$ respectively. We now apply Lemma 2.1 and find that

[TABLE]

Further,

[TABLE]

and therefore (2.1) can be written as

[TABLE]

On using similar arguments, we have

[TABLE]

On applying triangle inequality we find from (2.13) that

[TABLE]

From (2.12) and (2.14), we get that

[TABLE]

Again by triangle inequality, $S_{z\left(r\right)}^{2}\geq\left|S_{r}^{2}\right|,$ $S_{z\left(n-r\right)}^{2}\geq\left|S_{n-r}^{2}\right|$ and therefore

[TABLE]

The inequality (2.10) now follows from (2.15) and (2.16). $\blacksquare$

The inequality (2.10) is an extension of Mallows and Richter inequality [11]. For $r=1,$ we obtain the generalisation of the well known Samuelson’s inequality [20],

[TABLE]

Likewise, we can prove the following extension of Nagy’s inequality [13],

[TABLE]

Note that** **for $r=1$, $S_{1}=0$ and therefore from (2.13) on using triangle inequality we get that

[TABLE]

Similarly, from (2.12), we have

[TABLE]

and by addition we obtain the inequality

[TABLE]

It then follows inductively that the inequality

[TABLE]

holds true for $m=1,2,\ldots,n$ and therefore for $m=2,$ we have

[TABLE]

for all $i,j=1,2,...,n,$ $i\neq j.$ The inequality (2.18) implies (2.17). Also, see [24].

Theorem 2.4. For $0\leq a<\overline{x}\leq s,$ we have

[TABLE]

and

with $n\geq 3$

[TABLE]

Proof. The second inequality (1.11) implies  that

[TABLE]

and  therefore for $0\leq a<\overline{x},$ we can write

[TABLE]

It is clear that $\alpha\leq$ $\frac{a+b}{2}$ and since $f\left(x\right)=\left(x-a\right)\left(b-x\right)$ increases in the interval $\left[a,\frac{a+b}{2}\right],a<b,$ we find that

[TABLE]

Combining (2.22) and the second inequality (1.11); we immediately get (2.19).

Further, it follows from (2.21) that for $0<\overline{x}\leq s,$

[TABLE]

Combining (2.23) with the first inequality (1.11); a little computation leads to (2.20). $\blacksquare$

It may be noted here that the inequality (2.20) can equivalently be written as

[TABLE]

where $m_{2}^{\prime}=s^{2}+\overline{x}^{2}.$

We mention an alternative proof of (2.24). From the second inequality (1.11),

[TABLE]

Also, for $0\leq a<\overline{x}\leq s,$ from the inequality (1.5), we have $\overline{x}\leq s\leq\frac{b-a}{2}\leq\frac{b}{2}$ and for $\overline{x}\leq\frac{b}{2},$

[TABLE]

The inequality (2.24) follows from (2.25) and (2.26).

**Theorem 2.5. **For $a<0$ and $2\overline{x}\geq ns,$ we have

[TABLE]

and with $n\geq 3,$

[TABLE]

Proof. We write (1.8) in the form

[TABLE]

Using arithmetic geometric mean inequality,

[TABLE]

Thus, from (2.29) and (2.30),

[TABLE]

It follows from (2.31) that

[TABLE]

and consequently, for $a<0$ and $\overline{x}>0,$ we have

[TABLE]

It is clear that $\beta\geq\frac{a+b}{2}$ for $2\overline{x}\geq ns$ and since $f\left(x\right)=\left(\overline{x}-a\right)\left(b-\overline{x}\right)$ decreases in the interval $\left[\frac{a+b}{2},b\right],a<b,$ we find that

[TABLE]

Combining  (2.33) with the second inequality (1.11); we immediately get (2.27).

From (2.33), we also have

[TABLE]

The inequality (2.28) follows from (2.34) and the first inequality (1.11). $\blacksquare$

Sharma et al. [22] have proved that

[TABLE]

where $m_{2}=s^{2}$ and $m_{4}=\frac{1}{n}\overset{n}{\underset{i=1}{\sum}}\left(x_{i}-\overline{x}\right)^{4}.$

If the distribution is Leptokurtic or Mesokurtic, we have, see [10],

[TABLE]

We prove a refinement of the inequality (1.5) in the following theorem.

**Theorem 2.6. **For a Leptokurtic or Mesokurtic distribution, we have

[TABLE]

**Proof. **Under the  assumptions of the theorem the inequalities (2.35) and (2.36) hold true. By (2.36), $3S^{4}\leq m_{4}$ and we obtain from (2.18) that

[TABLE]

This gives the first inequality (2.37). The second inequality (2.37) follows from (2.38) on using arithmetic - geometric mean inequality, $\left(\overline{x}-a\right)\left(b-\overline{x}\right)\leq\frac{\left(b-a\right)^{2}}{4}.$ $\blacksquare$

We remark that the inequality (2.37) also holds true for both discrete and continuous distributions.

3 Bounds for eigenvalues

Let $\mathbb{M}\left(n\right)$ denote the algebra of all complex $n\times n$ matrices. We assume that the eigenvalues $\lambda_{i}\left(A\right)$ of $A=\left(a_{ij}\right)\in\mathbb{M}\left(n\right)$ are all real, and may define respectively their arithmetic mean and variance to be

[TABLE]

and

[TABLE]

where $B=A-\frac{\text{tr}A}{n}I.$

The spread of a matrix is the greatest distance between its eigenvalues. The notion of the spread was introduced by Mirsky [14,15] and several authors have studied bounds for the spread of a matrix, see [6,8,13,24].

**Theorem 3.1. **If trace of a unitary matrix $U\in\mathbb{M}\left(n\right)$ is zero then the unit circle is the smallest circle enclosing the eigenvalues of $U,$ and greatest lower bound on the Spd$\left(U\right)$ is $\sqrt{3}.$

**Proof. **The eigenvalues of a unitary matrix $U$ all lie on the unit circle and by assumption of the theorem tr$U=0.$ So, the eigenvalues $\lambda_{i}\left(U\right)$’s satisfy the conditions of the Theorem 2.1 and hence  theunit circle is the smallest circle containing $\lambda_{i}\left(U\right)$’s. It also follows from the second inequality (1.7) that Spd$\left(U\right)\geq\sqrt{3}.$ $\blacksquare$

**Example 1. **The basic circulant matrix $C$ with first row $\left(0,1,0\ldots,0\right)$ is a unitary matrix and its trace is zero. By Theorem 3.1 the unit disk is the smallest disk containing eigenvalues of $C$ and Spd$C\geq\sqrt{3}.$ Also, for $n=3$ we have Spd$C=\sqrt{3}.$

The following theorem is a consequence of Theorem 2.4 and provides refinements of the inequalities (1.10). $\blacksquare$

**Theorem 3.2. **Let the eigenvalues of an element $A\in\mathbb{M}\left(n\right)$ be all non negative and let $0<$tr$A\leq\left(n\text{tr}B\right)^{\frac{1}{2}}.$ Then

[TABLE]

and

[TABLE]

**Proof. **Under the condition tr$A\leq\left(n\text{tr}B\right)^{\frac{1}{2}}$ , we have

[TABLE]

Further, the eigenvalues of $A$ are all non-negative, therefore $0<\lambda_{\min}\leq\overline{\lambda}\leq s_{\lambda}$ and Spd$\left(A\right)=\lambda_{\max}\left(A\right)-\lambda_{\min}\left(A\right).$ So we can apply Theorem 2.1, the inequalities (3.3) and (3.4) follow on using (3.1) and (3.2) in (2.10) and (2.20), respectively. $\blacksquare$

Example 2. Let

[TABLE]

From (1.9), $81.393\leq$Spd$\left(A\right)\leq 115.11.$The matrix $A$ is positive definite and tr$A\leq\left(n\text{tr}B^{2}\right)^{\frac{1}{2}}.$ So, from our bounds (3.3) and (3.4) we have better estimate $85\leq$Spd$\left(A\right)\leq 109.77.$ $\blacksquare$

Likewise, we can obtain another refinement of the inequality (1.10) on applying Theorem 2.5. if $\lambda_{\min}\left(A\right)<0$ and $0<2$tr$A\leq\left(n^{3}\text{tr}B^{2}\right)^{\frac{1}{2}},$ then

[TABLE]

and

[TABLE]

Further, Wolkowicz and Styan [23] have shown that if the eigenvalues of $A\in\mathbb{M}\left(n\right)$ are all real and $\lambda_{1}\left(A\right)\leq\lambda_{i}\left(A\right)\leq\lambda_{n}\left(A\right),$ $i=1,2,\ldots,n,$ then

[TABLE]

and

[TABLE]

The inequalities (3.8) and (3.9) follow respectively from the inequalities, [7,20],

[TABLE]

and

[TABLE]

We now discuss extensions of these inequalities for the case when any one eigenvalue of $A$ is known as in case of stochastic and singular matrices.

It is clear from Lemma 2.1 that if $s_{n-1}^{2}$ is the variance of $n-1$ numbers obtained by excluding a number $x_{j}$ from the real numbers $x_{1},x_{2},\ldots,x_{n},$ then

[TABLE]

**Theorem 3.3. **Let the eigenvalues of $A\in\mathbb{M}\left(n\right)$ be all real. Let $\nu\left(A\right)$ be an eigenvalue of $A$ and denote the remaining eigenvalues by $\nu_{i}\left(A\right),$ $\nu_{1}\left(A\right)\leq\nu_{i}\left(A\right)\leq\nu_{n-1}\left(A\right),$ $i=1,2,\ldots,n-1.$ Then, for $n\geq 3,$

[TABLE]

and

[TABLE]

**Proof. **The arithmetic mean $\overline{\nu}\left(A\right)$ of $n-1$ eigenvalues $\nu_{i}\left(A\right)$ can be written as

[TABLE]

By the use of (3.13) the variance of these eigenvalues is

[TABLE]

On applying (3.10) to $n-1$ numbers $\nu_{1}\left(A\right),\nu_{2}\left(A\right),\ldots,\nu_{n-1}\left(A\right)$ and  using (3.16) and (3.17); we immediately get (3.12). Likewise, (3.15) follows  from (3.11). $\blacksquare$

**Theorem 3.4. **Under the conditions of the Theorem 3.3, we have

[TABLE]

and

[TABLE]

**Proof. **On using the inequalities (1.5) and (1.9), for $n-1$ numbers $\nu_{1}\left(A\right),\nu_{2}\left(A\right),\ldots,\nu_{n-1}\left(A\right),$ we have

[TABLE]

Inserting (3.17) in (3.20), we immediately get (3.18) and (3.19) on simplifications. $\blacksquare$

Example 3. Let

[TABLE]

From the inequalities (3.8) , we have$-9.0688\leq\lambda_{1}\left(A\right)\leq.644.$ The largest eigenvalue of $A$ is $16$ as all row sum of the positive definite matrix $A$ is $16$. From (3.14) we have better estimate for the smallest root, $-7.521\leq\lambda_{1}\left(A\right)\leq-2.7610.\blacksquare$

We now consider polynomials with real zeros. Let $f$ be a monic polynomial

[TABLE]

with only real zeros. Then the length $b-a$ of the smallest interval $\left[a,b\right]$ containing all the zeros of $f$ is called Span of $f,$ see $\left[\text{12, 19}\right].$ Denote by $D_{n}$ the span of $f$ then

[TABLE]

See Corollary 6.1.4 and Theorem 6.1.6 in [19].

We prove a refinement of (3.22) in the following theorem.

**Theorem 3.5. **Let the zeros of the polynomial (3.21) be all non-negative and let $na_{2}\leq\left(n-2\right)a_{1}^{2}.$ Then

[TABLE]

and with $n\geq 3,$

[TABLE]

**Proof. **Let $x_{1},x_{2},\ldots,x_{n}$ be the roots of the polynomial (3.21). Then, on using relation between roots and coefficient of polynomial, we have

[TABLE]

and

[TABLE]

The assertions of the theorem now follow on applying Theorem 2.4.