A note on the Frobenius conditional number with positive definite matrices

Abstract

In this article, we focus on the lower bounds of the Frobenius condition number. Using the generalized Schwarz inequality, we present some lower bounds for the Frobenius condition number of a positive definite matrix depending on its trace, determinant, and Frobenius norm. Also, we give some results on a kind of matrices with special structure, the positive definite matrices with centrosymmetric structure.

1 Introduction and preliminaries

In this article, we use the following notations. Let ℂ^n×nand ℝ^n×nbe the space of n×n complex and real matrices, respectively. The identity matrix in ℂ^n×nis denoted by I = I _n . Let A^T, Ā, A^H, and tr(A) denote the transpose, the conjugate, the conjugate transpose, and the trace of a matrix A, respectively. Re(a) stands for the real part of a number a. The Frobenius inner product < ·, · > _F in ℂ^m×nis defined as < A,B > _F = Re(tr(B^HA)), for A,B ∈ ℂ^m×n, i.e., < A,B > is the real part of the trace of B^HA. The induced matrix norm is $\left|\right|A|{|}_F=\sqrt{<A,A{>}_F}=\sqrt{\mathsf{\text{Re}}\left(\mathsf{\text{tr}}\left(A^{H}A\right)\right)}=\sqrt{\mathsf{\text{tr}}\left(A^{H}A\right)}$, which is called the Frobenius (Euclidean) norm. The Frobenius inner product allows us to define the consine of the angle between two given real n × n matrices as

$$cos\left(\mathsf{\text{A}},\mathsf{\text{B}}\right)=\frac{<A,B{>}_F}{\left|\right|A\left|{|}_F\right|\left|B\right|{|}_F}.$$

(1.1)

The cosine of the angle between two real n × n depends on the Frobenius inner product and the Frobenius norms of given matrices. A matrix A ∈ ℂ^n×nis Hermitian if A^H= A. An Hermitian matrix A is said to be positive semidefinite or nonnegative definite, written as A ≥ 0, if (see, e.g., [[1], p. 159])

$$x^{H}Ax\ge 0,\forall x\in {ℂ}^n,$$

(1.2)

A is further called positive definite, symbolized A > 0, if the strict inequality in (1.2) holds for all nonzero x ∈ ℂⁿ. An equivalent condition for A ∈ ℂⁿto be positive definite is that A is Hermitian and all eigenvalues of A are positive real numbers.

The quanity

$$\mu \left(A\right)=\left\{\begin{array}{cc}\hfill \left|\right|A\left|\right|\cdot \left|\right|A^{-1}\left|\right|,\hfill & \hfill \mathsf{\text{if}}A\mathsf{\text{is}}\mathsf{\text{nonsingular}};\hfill \\ \hfill \infty \hfill & \hfill \mathsf{\text{if}}A\mathsf{\text{is}}\mathsf{\text{singular}}.\hfill \end{array}\right.$$

(1.3)

is called the condition number of matrix μ(A) with respect to the matrix norm || · ||. Notice that μ(A) = ||A^-1|| · ||A|| ≥ ||A^-1A|| = ||I|| ≥ 1 for any matrix norm (see, e.g., [[2], p. 336]). The condition number μ(A) of a nonsingular matrix A plays an important role in the numerical solution of linear systems since it measures the sensitivity of the solution of linear system Ax = b to the perturbations on A and b. There are several methods that allow to find good approximations of the condition number of a general square matrix.

We first introduce some inequalities. Buzano [3] obtained the following extension of the celebrated Schwarz inequality in a real or complex inner product space (H, < ·, ·>).

Lemma 1.1 ([3]). For any a, b, x ∈ H, there is

$$|<a,x><x,b>|\le \frac{1}{2}\left(\left|\right|a\left|\right|\cdot \left|\right|b\left|\right|+|<a.b>|\right)\left|\right|x|{|}^2.$$

(1.4)

It is clear that for a = b, (1.4) becomes the standard Schwarz inequality

$$|<a,x{>}^2|\le \left|\right|a\left|{|}^2\right|\left|x\right|{|}^2,a,x\in H$$

(1.5)

with equality if and only if there exists a scalar λ such that x = λa.

Also Dragomir [4] has stated the following inequality.

Lemma 1.2[4]. For any a,b,x ∈ H, and x ≠ 0, there is the following

$$\left|\frac{<a,x><x,b>}{\left|\right|x|{|}^2}-\frac{<a,b>}{2}\right|\le \frac{\left|\right|a\left|\right|\cdot \left|\right|b\left|\right|}{2}$$

(1.6)

Dannan [5] showed the following inequality by using arithmetic-geometric inequality.

Lemma 1.3[5]. For n-square positive definite matrices A and B,

$$n{\left(\mathsf{\text{detA}}\cdot det\mathsf{\text{B}}\right)}^{m∕n}\le \mathsf{\text{tr}}\left({\mathsf{\text{A}}}^{\mathsf{\text{m}}}{\mathsf{\text{B}}}^{\mathsf{\text{m}}}\right),$$

(1.7)

where m is a positive integer.

By taking A = I, B = A^-1, and m = 1 in (1.7), we obtain

$$n{\left(\mathsf{\text{detI}}\cdot det{\mathsf{\text{A}}}^{-1}\right)}^{1∕\mathsf{\text{n}}}\le \mathsf{\text{tr}}\left(\mathsf{\text{I}}\cdot {\mathsf{\text{A}}}^{-1}\right),$$

(1.8)

$$n{\left(\frac{1}{\mathsf{\text{detA}}}\right)}^{1∕n}\le \mathsf{\text{tr}}\left({\mathsf{\text{A}}}^{-1}\right)$$

(1.9)

In [6], Türkmen and Ulukök proposed the following,

Lemma 1.4[6]. Let both A and B be n-square positive definite matrices, then

$$\cos (A,I)\cos (B,I)\le \frac{1}{2}\cos (A,B)+1],$$

(1.10)

$$cos\left(A,A^{-1}\right)\le cos\left(A,I\right)cos\left(A^{-1},I\right)\le \frac{1}{2}\left[cos\left(A,A^{-1}\right)+1\right]\le 1,$$

(1.11)

$$cos\left(A,I\right)\le 1,cos\left(A^{-1},I\right)\le 1,$$

(1.12)

As a consequence, in the following section, we give some bounds for the Frobenius condition numbers by considering inequalities given in this section.

2 Main results

Theorem 2.1. Let A be a positive definite real matrix, α be any real number. Then,

$$2n\frac{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{1+\alpha }}{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{2\alpha }{\left(\mathsf{\text{detA}}\right)}^{\left(1-\alpha \right)∕n}}-n\le {\mu }_F\left(A\right);$$

(2.1)

$$2\frac{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{1+\alpha }\mathsf{\text{tr}}{\mathsf{\text{A}}}^{\alpha -1}}{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{2\alpha }}-n\le {\mu }_F\left(A\right).$$

(2.2)

where μ _F (A) is the Frobenius conditional number of A.

Proof. Let X, A, B be positive definite real matrices. From Lemma 1.2, we have the following

$$\left|\frac{<A,X{>}_F<X,B{>}_F}{\left|\right|X|{|}_F^2}\frac{<A,B{>}_F}{2}\right|\le \frac{\left|\right|A\left|{|}_F\right|\left|B\right|{|}_F}{2},$$

(2.3)

i.e.,

$$\left|\frac{\mathsf{\text{tr}}\left({\mathsf{\text{A}}}^{\mathsf{\text{T}}}\mathsf{\text{X}}\right)\mathsf{\text{tr}}\left({\mathsf{\text{X}}}^{\mathsf{\text{T}}}\mathsf{\text{B}}\right)}{\left|\right|X|{|}_F^2}-\frac{\mathsf{\text{tr}}\left({\mathsf{\text{A}}}^{\mathsf{\text{T}}}\mathsf{\text{B}}\right)}{2}\right|\le \frac{\left|\right|A\left|{|}_F\right|\left|B\right|{|}_F}{2}.$$

(2.4)

Let B = A^-1, then (2.4) turns into

$$\left|\frac{\mathsf{\text{tr}}\left({\mathsf{\text{A}}}^{\mathsf{\text{T}}}\mathsf{\text{X}}\right)\mathsf{\text{tr}}\left({\mathsf{\text{X}}}^{\mathsf{\text{T}}}{\mathsf{\text{A}}}^{-1}\right)}{\left|\right|X|{|}_F^2}-\frac{\mathsf{\text{tr}}\left({\mathsf{\text{A}}}^{\mathsf{\text{T}}}{\mathsf{\text{A}}}^{-1}\right)}{2}\right|\le \frac{\left|\right|A\left|{|}_F\right|\left|A^{-1}\right|{|}_F}{2}.$$

(2.5)

Since both X and A are positive definite, we have

$$\left|\frac{\mathsf{\text{tr}}\left(\mathsf{\text{AX}}\right)\mathsf{\text{tr}}\left(\mathsf{\text{X}}{\mathsf{\text{A}}}^{-1}\right)}{\left|\right|X|{|}_F^2}-\frac{n}{2}\right|\le \frac{\left|\right|A\left|{|}_F\right|\left|A^{-1}\right|{|}_F}{2}=\frac{{\mu }_F\left(A\right)}{2},$$

(2.6)

where μ _F (A) is the Frobenius condition number of A.

By taking X = A^α(α is an arbitrary real number) into (2.6), there exists

$$\left|\frac{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{1+\alpha }\mathsf{\text{tr}}{\mathsf{\text{A}}}^{-\left(1-\alpha \right)}}{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{2\alpha }}-\frac{n}{2}\right|\le \frac{{\mu }_F\left(A\right)}{2}.$$

(2.7)

Thus, it follows that

$$\frac{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{1+\alpha }\mathsf{\text{tr}}{\mathsf{\text{A}}}^{-\left(1-\alpha \right)}}{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{2\alpha }}-\frac{n}{2}\le \frac{{\mu }_F\left(A\right)}{2},$$

(2.8)

i.e.,

$$2\frac{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{1+\alpha }\mathsf{\text{tr}}{\mathsf{\text{A}}}^{\alpha -1}}{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{2\alpha }}-\frac{n}{\le {\mu }_F\left(A\right)}.$$

(2.9)

From (1.9), by replacing A with A^1-α, we get

$$0<n\cdot \frac{1}{{\left(detA\right)}^{\left(1-\alpha \right)∕n}}\le \mathsf{\text{tr}}\left({\mathsf{\text{A}}}^{-\left(1-\alpha \right)}\right).$$

(2.10)

Taking (2.10) into (2.8), we can write

$$n\cdot \frac{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{1+\alpha }}{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{2\alpha }}\cdot \frac{1}{{\left(detA\right)}^{\left(1-\alpha \right)∕n}}-\frac{n}{2}\le \frac{{\mu }_F\left(A\right)}{2},$$

(2.11)

i.e.,

$$2n\cdot \frac{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{1+\alpha }}{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{2\alpha }{\left(\mathsf{\text{detA}}\right)}^{\left(1-\alpha \right)∕n}}-n\le {\mu }_F\left(A\right).$$

(2.12)

In particular, let α = 1, and by taking it into (2.7), we have the following

$$\left|\frac{\mathsf{\text{tr}}{\mathsf{\text{A}}}^2\mathsf{\text{trI}}}{\mathsf{\text{tr}}{\mathsf{\text{A}}}^2}-\frac{n}{2}\right|\le \frac{{\mu }_F\left(A\right)}{2},$$

(2.13)

i.e.,

$$n\le {\mu }_F\left(A\right).$$

(2.14)

Note, when $\alpha =\frac{1}{2}$, (2.12) becomes

$$2n\cdot \frac{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{3∕2}}{trA\cdot {\left(detA\right)}^{1∕2n}}-n\le {\mu }_F\left(A\right),$$

(2.15)

Taking α = 1 into (2.12), we obtain that

$$2\frac{\mathsf{\text{tr}}A}{{\left(detA\right)}^{1∕n}}-n\le {\mu }_F\left(A\right).$$

(2.16)

(2.15), (2.16) can be found in [6].

Example 2.2.

$$A=\left[\begin{array}{cc}\hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2^{-1}\hfill \end{array}\right].$$

Here trA = 2.5, detA = 1 and n = 2. Then, from (2.15) and (2.16), we obtain two lower bounds of μ _F (A):

$${\mu }_F\left(A\right)\ge 2n\cdot \frac{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{3∕2}}{\mathsf{\text{trA}}\cdot {\left(\mathsf{\text{detA}}\right)}^{1∕2\mathsf{\text{n}}}}=3.091168,\mathsf{\text{and}}{\mu }_F\left(A\right)\ge 2\cdot \frac{\mathsf{\text{tr}}A}{{\left(detA\right)}^{1∕n}}=3.$$

Taking α = 1/4 into (2.1) and (2.2) from Theorem 2.1, another two lower bounds are obtained as follows:

$${\mu }_F\left(A\right)\ge 2n\frac{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{1+\alpha }}{\mathsf{\text{tr}}{A}^{2\alpha }{\left(detA\right)}^{\left(1-\alpha \right)∕n}}-n=3.277585,\mathsf{\text{and}}{\mu }_F\left(A\right)\ge 2\frac{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{1+\alpha }\mathsf{\text{tr}}{\mathsf{\text{A}}}^{\alpha -1}}{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{2\alpha }}-n=4.006938.$$

In fact, μ _F (A) = 4.25. Thus, Theorem 2.1 is indeed a generalization of (2.15) and (2.16) given in [6].

Lemma 2.3. Let a₁, a₂,..., a _n be positive numbers, and

$$f\left(x\right)=a_1^x+a_2^x+\cdot \cdot \cdot +a_n^x+a_1^{-x}+\cdot \cdot \cdot +a_n^{-x}.$$

Then, f(x) is monotonously increasing for x ∈ [0,+ ∞).

Proof. It is obvious that

$$\begin{array}{ll}\hfill f^{\prime }\left(x\right)& =a_1^{x}ln{a}_1+a_2^{x}ln{a}_2+\cdots +a_n^{x}ln{a}_n-a_1^{-x}ln{a}_1-\cdots -a_n^{x}ln{a}_n\\ =ln{a}_1\left(a_1^x-a_1^{-x}\right)+ln{a}_2\left(a_2^x-a_1^{-x}\right)+ln{a}_n\left(a_n^x-a_n^{-x}\right).\end{array}$$

When a _i ≥ 1, and x ∈ [0, + ∞), $ln{a}_i\ge 0,a_i^x\ge 1$, and $a_i^{-x}\le 1$. Thus, $ln{a}_i\left(a_i^x-a_i^{-x}\right)\ge 0$.

When 0 < a _i < 1, and x ∈ [0, + ∞), we have $ln{a}_i<0,0<a_i^x\le 1$, and $a_i^{-x}\ge 1$. Thus, $ln{a}_i\left(a_i^x-a_i^{-x}\right)\ge 0$. Therefore, f'(x) ≥ 0, and f(x) is increasing for x ∈ [0, ∞).

Theorem 2.4. Let A be a positive definite real matrix. Then

$$\frac{n\sqrt{n}\cdot {∥A^{1∕2}∥}_F}{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{1∕2}}\le {\mu }_F\left(A\right).$$

(2.17)

Proof. Since A is positive definite, from Lemma 1.4, we have

$$\cos (A^{1/2},A^{-1/2}\le \cos (A^{1/2},I).$$

(2.18)

That is

$$\frac{n}{{\mu }_F\left(A^{1∕2}\right)}=\frac{\mathsf{\text{tr}}\left({\mathsf{\text{A}}}^{1∕2}{\mathsf{\text{A}}}^{-1∕2}\right)}{\left|\right|A^{1∕2}|{|}_F\cdot |\left|{\left(A^{1∕2}\right)}^{-1}\right|{|}_F}\le \frac{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{1∕2}}{\sqrt{n}\cdot \left|\right|A^{1∕2}|{|}_F}.$$

(2.19)

Let all the positive real eigenvalues of A be λ₁, λ₂,..., λ _n > 0. Then, the eigenvalues of A^-1 are 1/λ₁, 1/λ₂,..., 1/λ _n > 0, the eigenvalues of A² are ${\lambda }_1^2,{\lambda }_2^2,...,{\lambda }_n^2>0$, and the eigenvalue of A^-2 are ${\lambda }_1^{-2},{\lambda }_2^{-2},...,{\lambda }_n^{-2}>0$. Next, we will prove the following by induction.

$$\left(\sum _{i=1}^n{\lambda }_i^2\right)\left(\sum _{i=1}^{n}1∕{\lambda }_i^2\right)\ge \left(\sum _{i=1}^n{\lambda }_i\right)\left(\sum _{i=1}^{n}1∕{\lambda }_i\right).$$

(2.20)

In case n = 1, it is obvious that (2.20) holds.

In case $n=2,\left({\sum }_{i=1}^n{\lambda }_i^2\right)\left({\sum }_{i=1}^{n}1∕{\lambda }_i^2\right)=2+{\left({\lambda }_1∕{\lambda }_2\right)}^2+{\left({\lambda }_2∕{\lambda }_1\right)}^2$. From Lemma 2.3, 2 + (λ₁/λ₂)² + (λ₂/λ₁)² ≥ 2 + λ₁/λ₂ + λ₂/λ₁ = (λ₁ + λ₂)(1/λ₁ + 1/λ₂). Thus, (2.20) holds. Suppose that (2.20) holds, when n = k, i.e.,

$$\left(\sum _{i=1}^k{\lambda }_i^2\right)\left(\sum _{i=1}^{k}1∕{\lambda }_i^2\right)\ge \left(\sum _{i=1}^k{\lambda }_i\right)\left(\sum _{i=1}^{k}1∕{\lambda }_i\right).$$

(2.21)

In case n = k + 1,

$$\begin{array}{c}\left({\displaystyle {\sum }_{i=1}^{k+1}{\lambda }_i^2}\right)\left({\displaystyle {\sum }_{i=1}^{k+1}1/{\lambda }_i^2}\right)=\left({\displaystyle {\sum }_{i=1}^k{\lambda }_i^2+{\lambda }_{k+1}^2}\right)\left({\displaystyle {\sum }_{i=1}^{k}1/{\lambda }_i^2+1/{\lambda }_{k+1}^2}\right)\\ =\left({\displaystyle {\sum }_{i=1}^k{\lambda }_i^2}\right)\left({\displaystyle {\sum }_{i=1}^{k}1/{\lambda }_i^2}\right)+{{\displaystyle {\sum }_{i=1}^k({\lambda }_i/{\lambda }_{k+1})}}^2+{\displaystyle {\sum }_{i=1}^k{({\lambda }_{k+1}/{\lambda }_i)}^2+1.}\end{array}$$

(2.22)

By Lemma 2.3, we get

$$\begin{array}{ll}\hfill \left(\sum _{i=1}^{k+1}{\lambda }_i^2\right)\left(\sum _{i=1}^{k+1}1∕{\lambda }_i^2\right)& \ge \left(\sum _{i=1}^k{\lambda }_i\right)\left(\sum _{i=1}^{k}1∕{\lambda }_i\right)+\sum _{i=1}^k\left({\lambda }_i∕{\lambda }_{k+1}\right)+\sum _{i=1}^k\left({\lambda }_{k+1}∕{\lambda }_i\right)+1\\ =\left(\sum _{i=1}^{k+1}{\lambda }_i\right)\left(\sum _{i=1}^{k+1}1∕{\lambda }_i\right).\end{array}$$

(2.23)

Thus, when n = k + 1, (2.20) holds. On the other hand, $\left|\right|A|{|}_F=\sqrt{\mathsf{\text{tr}}\left({\mathsf{\text{A}}}^2\right)}=\sqrt{{\sum }_{i=1}^n{\lambda }_i^2},|\left|A^{-1}\right|{|}_F=\sqrt{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{-2}}=\sqrt{{\sum }_{i=1}^n{\lambda }_i^{-2}},\left|\right|A^{1∕2}|{|}_F=\sqrt{\mathsf{\text{trA}}}=\sqrt{{\sum }_{i=1}^n{\lambda }_i}$ and $\left|\right|A^{-1∕2}|{|}_F=\sqrt{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{-1}}=\sqrt{{\sum }_{i=1}^n{\lambda }_i^{-1}}$. Therefore,

$$\left|\right|A\left|{|}_F^2\right|\left|A^{-1}\right|{|}_F^2\ge \left|\right|A^{1∕2}\left|{|}_F^2\right|\left|A^{-1∕2}\right|{|}_F^2,$$

(35)

i.e.,

$${\mu }_F\left(A\right)\ge {\mu }_F\left(A^{1∕2}\right).$$

(2.25)

Taking (2.25) into (2.19), we obtain

$$\frac{n}{{\mu }_F\left(A\right)}\le \frac{n}{{\mu }_F\left(A^{1∕2}\right)}\le \frac{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{1∕2}}{\sqrt{n}\left|\right|A^{1∕2}|{|}_F},$$

(2.26)

i.e.,

$$\frac{n\sqrt{n}\left|\right|A^{1∕2}|{|}_F}{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{1∕2}}\le {\mu }_F\left(A\right).$$

(2.27)

Remark 2.5. (2.17) can be extended to any 0 < α ≤ 1, i.e., $\frac{n\sqrt{n}\left|\right|A^{\alpha }|{|}_F}{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{\alpha }}\le {\mu }_F\left(A\right)$.

3 The Frobenius condition number of a centrosymmetric positive definite matrix

Definition 3.1 (see [7]). Let A = (a _ij )_{p xq}∈ ℝ^p×q. A is a centrosymmetric matrix, if

$$a_{ij}=a_{p-i+1,q-j+1},1\le i\le p,1\le j\le q,or{J}_{p}A{J}_q=A,$$

where J _n = (e _n , e_n-1,..., e₁), e _i denotes the unit vector with the i-th entry 1.

Using the partition of matrix, the central symmetric character of a square centrosym-metric matrix can be described as follows (see [7]):

Lemma 3.2. Let A = (a _ij )_n×n(n = 2m) be centrosymmetric. Then, A has the form,

$$A=\left[\begin{array}{cc}\hfill B\hfill & \hfill J_{m}C{J}_m\hfill \\ \hfill C\hfill & \hfill J_{m}B{J}_m\hfill \end{array}\right],P^{T}AP=\left[\begin{array}{cc}\hfill B-J_{m}C\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill B+J_{m}C\hfill \end{array}\right],$$

(3.1)

Where B,C ∈ ℂ^m×m, $P=\frac{1}{2}\left[\begin{array}{cc}\hfill I_m\hfill & \hfill I_m\hfill \\ \hfill -J_m\hfill & \hfill J_m\hfill \end{array}\right]$.

Lemma 3.3. Let Abeann × n(n = 2m) centrosymmetric positive definite matrix with the following form

$$A=\left[\begin{array}{cc}\hfill B\hfill & \hfill J_{m}C{J}_m\hfill \\ \hfill C\hfill & \hfill J_{m}B{J}_m\hfill \end{array}\right],$$

where B,C ∈ ℝ^m×m. Then there exists an orthogonal matrix P such that

$$P^{T}AP=\left[\begin{array}{c}\hfill M\hfill \\ \hfill N\hfill \end{array}\right],$$

where M = B - J _m C, N = B + J _m C and M^-1, N^-1are positive definite matrices.

Proof. From Lemma 3.2, there exists an orthogonal matrix P such that

$$H=P^{T}AP=\left[\begin{array}{c}\hfill M\hfill \\ \hfill N\hfill \end{array}\right].$$

Since A is positive definite, then any eigenvalue of A is positive real number. Thus, the eigenvalues of H are all positive real numbers. That is to say, all eigenvalues of M and N are positive real numbers. Thus, M and N are both positive definite. It is obvious that M^-1 and N^-1 are positive definite matrices.

Lemma 3.4. Let A, B are positive definite real matrices. Then,

$$\frac{\mathsf{\text{tr}}A}{{\left(\mathsf{\text{detB}}\right)}^{1∕n}}\le \left|\right|A\left|{|}_F\right|\left|B^{-1}\right|{|}_F.$$

(3.2)

Proof. Let X be positive definite. By Lemma 1.2, we have

$$\left|\frac{<A,X{>}_F<X,B{>}_F}{\left|\right|X|{|}_F^2}-\frac{<A,B{>}_F}{2}\right|\le \frac{\left|\right|A\left|{|}_{F}B\right|{|}_F}{2}.$$

(3.3)

Thus,

$$\left|\frac{\mathsf{\text{tr}}\left({\mathsf{\text{A}}}^{\mathsf{\text{T}}}\mathsf{\text{X}}\right)\mathsf{\text{tr}}\left({\mathsf{\text{X}}}^{\mathsf{\text{T}}}\mathsf{\text{B}}\right)}{\left|\right|X|{|}_F^2}-\frac{\mathsf{\text{tr}}\left({\mathsf{\text{A}}}^{\mathsf{\text{T}}}\mathsf{\text{B}}\right)}{2}\right|\le \frac{\left|\right|A\left|{|}_{F}B\right|{|}_F}{2}.$$

(3.4)

By replacing X, B with I _n and B^-1, respectively, in inequality (3.4), we can obtain

$$\left|\frac{\mathsf{\text{tr}}\left(\mathsf{\text{A}}\right)\mathsf{\text{tr}}\left({\mathsf{\text{B}}}^{-1}\right)}{n}-\frac{\mathsf{\text{tr}}\left({\mathsf{\text{A}}}^{\mathsf{\text{T}}}{\mathsf{\text{B}}}^{-1}\right)}{2}\right|\le \frac{\left|\right|A\left|{|}_F\right|\left|B^{-1}\right|{|}_F}{2}.$$

(3.5)

That is,

$$\frac{\mathsf{\text{tr}}\left(\mathsf{\text{A}}\right)\mathsf{\text{tr}}\left({\mathsf{\text{B}}}^{-1}\right)}{n}-\frac{\mathsf{\text{tr}}\left({\mathsf{\text{A}}}^{\mathsf{\text{T}}}{\mathsf{\text{B}}}^{-1}\right)}{2}\le \frac{\left|\right|A\left|{|}_F\right|\left|B^{-1}\right|{|}_F}{2}.$$

(3.6)

From Schwarz inequality (1.5),

$$\mathsf{\text{tr}}\left(A^T{B}^{-1}\right)=<A,B^{-1}{>}_F\le \left|\right|A\left|{|}_F\right|\left|B^{-1}\right|{|}_F.$$

(3.7)

By taking (3.7) into (3.6), we have

$$\frac{\mathsf{\text{tr}}\left(\mathsf{\text{A}}\right)\mathsf{\text{tr}}\left(B^{-1}\right)}{n}\le \left|\right|A\left|{|}_F\right|\left|B^{-1}\right|{|}_F\le \frac{\left|\right|A\left|{|}_F\right|\left|B^{-1}\right|{|}_F}{2}+\frac{\mathsf{\text{tr}}\left({\mathsf{\text{A}}}^{\mathsf{\text{T}}}{\mathsf{\text{B}}}^{-1}\right)}{2}.$$

(3.8)

From (1.9),

$$n{\left(1∕\mathsf{\text{detB}}\right)}^{1∕n}\le \mathsf{\text{tr}}{\mathsf{\text{B}}}^{-1}.$$

(3.9)

Therefore,

$$\frac{\mathsf{\text{tr}}A}{{\left(detB\right)}^{1∕n}}\le \left|\right|A\left|{|}_F\right|\left|B^{-1}\right|{|}_F.$$

(3.10)

Theorem 3.5. Let A be a centrosymmetric positive definite matrix with the form

$$A=\left[\begin{array}{cc}\hfill B\hfill & \hfill J_{m}C{J}_m\hfill \\ \hfill C\hfill & \hfill J_{m}B{J}_m\hfill \end{array}\right],B,C\in {ℝ}^{m\times m}.$$

Let M = B - J _m C, N = B + J _m C. Then,

$${\mu }_F\left(A\right)\ge \sqrt{{\left(2\frac{\mathsf{\text{tr}}M}{{\left(detM\right)}^{1∕m}}-m\right)}^2+{\left(2\frac{\mathsf{\text{tr}}N}{{\left(detN\right)}^{1∕m}}-m\right)}^2+{\left(\frac{\mathsf{\text{tr}}M}{{\left(detN\right)}^{1∕m}}\right)}^2+{\left(\frac{\mathsf{\text{tr}}N}{{\left(detM\right)}^{1∕m}}\right)}^2}.$$

(3.11)

Proof. By Lemma 3.2, there exists an orthogonal matrix P such that

$$H=P^{T}AP=\left[\begin{array}{c}\hfill B-J_{m}C\hfill \\ \hfill B+J_{m}C\hfill \end{array}\right]=\left[\begin{array}{c}\hfill M\hfill \\ \hfill N\hfill \end{array}\right].$$

Thus,

$${\mu }_F\left(A\right)=\left|\right|A\left|{|}_F\right|\left|A^{-1}\right|{|}_F={\mu }_F\left(H\right)=\left|\right|H\left|{|}_F\right|\left|H^{-1}\right|{|}_F=\sqrt{\left|\right|M|{|}_F^2+|\left|N\right|{|}_F^2}\sqrt{\left|\right|M^{-1}|{|}_F^2+|\left|N^{-1}\right|{|}_F^2}.$$

(3.12)

Therefore,

$$\begin{array}{ll}\hfill {\mu }_F^2\left(A\right)& =\left(\left|\right|M|{|}_F^2+|\left|N\right|{|}_F^2\right)\left(\left|\right|M^{-1}|{|}_F^2+|\left|N^{-1}\right|{|}_F^2\right)\\ =\left|\right|M\left|{|}_F^2\right|\left|M^{-1}\right|{|}_F^2+\left|\right|N\left|{|}_F^2\right|\left|N^{-1}\right|{|}_F^2+\left|\right|M\left|{|}_F^2\right|\left|N^{-1}\right|{|}_F^2+\left|\right|N\left|{|}_F^2\right|\left|M^{-1}\right|{|}_F^2\\ ={\mu }_F^2\left(M\right)+{\mu }_F^2\left(N\right)+{\left(\left|\right|M\left|{|}_F\right|\left|N^{-1}\right|{|}_F\right)}^2+{\left(\left|\right|N\left|{|}_F\right|\left|M^{-1}\right|{|}_F\right)}^2.\end{array}$$

From Lemma 3.4,

$$\frac{\mathsf{\text{tr}}M}{{\left(detN\right)}^{1∕m}}\le \left|\right|M\left|{|}_F\right|\left|N^{-1}\right|{|}_F,\frac{\mathsf{\text{tr}}N}{{\left(detM\right)}^{1∕m}}\le \left|\right|N\left|{|}_F\right|\left|M^{-1}\right|{|}_F.$$

(3.14)

From (2.18),

$$2\frac{\mathsf{\text{tr}}M}{{\left(detM\right)}^{1∕m}}-m\le {\mu }_F\left(M\right),\mathsf{\text{and}}2\frac{\mathsf{\text{tr}}N}{{\left(detN\right)}^{1∕m}}-m\le {\mu }_F\left(N\right).$$

(3.15)

Thus,

$${\mu }_F\left(A\right)\ge \sqrt{{\left(2\frac{\mathsf{\text{tr}}M}{{\left(detM\right)}^{1∕m}}-m\right)}^2+{\left(2\frac{\mathsf{\text{tr}}N}{{\left(detN\right)}^{1∕m}}-m\right)}^2+{\left(\frac{\mathsf{\text{tr}}M}{{\left(detN\right)}^{1∕m}}\right)}^2+{\left(\frac{\mathsf{\text{tr}}N}{{\left(detM\right)}^{1∕m}}\right)}^2.}$$

Example 3.6.

$$A=\left[\begin{array}{cccc}\hfill 5\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 5\hfill & \hfill 3\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill 5\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 5\hfill \end{array}\right],\mathsf{\text{and}}{P}^{T}AP=\left[\begin{array}{c}\hfill M\hfill \\ \hfill N\hfill \end{array}\right]=\left[\begin{array}{c}\hfill \left[\begin{array}{cc}\hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill \end{array}\right]\hfill \\ \hfill \left[\begin{array}{cc}\hfill 8\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 8\hfill \end{array}\right]\hfill \end{array}\right].$$

Here n = 4, m = 2, det (A) = 256, tr A = 20, tr M = 4, tr N = 16, det(M) = 4, det(N) = 64. From Theorem 3.5, a lower bound of μ _F (A) is as follows:

$${\mu }_F\left(A\right)\ge \sqrt{{\left(2\frac{\mathsf{\text{trM}}}{{\left(\mathsf{\text{detM}}\right)}^{1∕m}}-m\right)}^2+{\left(2\frac{\mathsf{\text{trN}}}{{\left(\mathsf{\text{detN}}\right)}^{1∕m}}-m\right)}^2+{\left(\frac{\mathsf{\text{trM}}}{{\left(\mathsf{\text{detN}}\right)}^{1∕\mathsf{\text{m}}}}\right)}^2+{\left(\frac{\mathsf{\text{trN}}}{{\left(\mathsf{\text{detM}}\right)}^{1∕\mathsf{\text{m}}}}\right)}^2}=8.5.$$

In fact

$${\mu }_F\left(A\right)=\sqrt{{\mu }_F^2\left(M\right)+{\mu }_F^2\left(N\right)+{\left(\left|\right|M\left|{|}_F\right|\left|N^{-1}\right|{|}_F\right)}^2+{\left(\left|\right|N\left|{|}_F\right|\left|M^{-1}\right|{|}_F\right)}^2}=8.5.$$

On the other hand, the lower bounds of μ _F (A) in (2.15) and (2.16) provided by [6] are

$${\mu }_F\left(A\right)\ge 2n\cdot \frac{\mathsf{\text{tr}}{\mathsf{\text{A}}}^{3∕2}}{\mathsf{\text{tr}}A\cdot {\left(\mathsf{\text{detA}}\right)}^{1∕2\mathsf{\text{n}}}}-n=6.1823376,\mathsf{\text{and}}{\mu }_F\left(A\right)\ge 2\frac{\mathsf{\text{tr}}A}{{\left(detA\right)}^{1∕n}}-n=6.$$

It can easily be seen that, in this example, the best lower bound is the first one given by Theorem 3.5.