Notes on Greub-Rheinboldt inequalities

Abstract

In this paper, we focus on matrix Greub-Rheinboldt inequalities for commutative positive definite Hermitian matrix pairs. Some improvements, which yield sharpened bounds compared with existing results, are presented.

1 Introduction and preliminaries

Let $M_{m,n}$ denote the space of $m\times n$ complex matrices and write $M_n\equiv M_{n,n}$. The identity matrix in $M_n$ is denoted by $I_n$. As usual, $A^{\ast }={(\overline{A})}^T$ denotes the conjugate transpose of the matrix A. A matrix $A\in M_n$ is an Hermite matrix if $A^{\ast }=A$. An Hermitian matrix A is said to be positive semi-definite or nonnegative definite, written as $A\ge 0$, if $x^{\ast }Ax\ge 0$, $\mathrm{\forall }x\in {\mathbb{C}}^n$. A is further called positive definite, symbolized $A>0$, if $x^{\ast }Ax>0$ for all nonzero $x\in {\mathbb{C}}^n$. An equivalent condition for $A\in M_n$ to be positive definite is that A is an Hermitian matrix and all eigenvalues of A are positive.

Denote by ${\lambda }_1\le {\lambda }_2\le \cdots \le {\lambda }_n$ the eigenvalues of an Hermitian matrix A. The matrix version of the well-known Kantorovich inequality for a positive definite matrix A is stated as follows (see, e.g., [1, 2]):

$$1\le \frac{x^{\ast }Ax{x}^{\ast }{A}^{-1}x}{{(x^{\ast }x)}^2}\le \frac{{({\lambda }_1+{\lambda }_n)}^2}{4{\lambda }_1{\lambda }_n}$$

(1.1)

for any nonzero vector $x\in {\mathbb{C}}^n$.

An equivalent form of this result is the inequality

$$0\le \frac{x^{\ast }Ax{x}^{\ast }{A}^{-1}x}{{(x^{\ast }x)}^2}-1\le \frac{{({\lambda }_1-{\lambda }_n)}^2}{4{\lambda }_1{\lambda }_n}$$

(1.2)

valid for any nonzero vector $x\in {\mathbb{C}}^n$.

This famous inequality plays an important role in statistics (see [3, 4]; for the latest work on applications in statistics, we refer to Seddighin’s work [3]) and numerical analysis, for example, studying the rates of convergence and error bounds of solving systems of equations (see in [5, 6]).

In 2008, Dragomir gave a refinement of the additive version of the operator Kantorovich inequality [7],

$$0\le K(A;x)-1\le \frac{1}{4}\frac{{(M-m)}^2}{mM}-{[Re〈C_{m,M}(A)x,x〉Re〈C_{\frac{1}{m},\frac{1}{M}}\left(A^{-1}\right)x,x〉]}^{1/2},$$

(1.3)

where A is a self-adjoint bounded linear operator on a complex Hilbert space, $0<m<M$, such that $mI\le A\le MI$ in the partial operator order, $K(A;x):=〈Ax,x〉〈A^{-1}x,x〉$, and $C_{\alpha ,\beta }(A):=(A-\overline{\alpha }I)(\beta I-A)$.

A further improvement of the matrix version of (1.3) is proposed in [8], where the classical Kantorovich inequality (1.1) is modified to apply not only to positive definite, but also to all invertible Hermitian matrices.

We adopt the following transform for a positive definite Hermitian matrix $A\in M_n$ with eigenvalues $0<{\lambda }_1\le {\lambda }_2\le \cdots \le {\lambda }_n$:

$$C(A,x)=x^{\ast }({\lambda }_{n}I-A)(A-{\lambda }_{1}I)x,$$

(1.4)

and

$$C(A^{-1},x)=x^{\ast }(\frac{1}{{\lambda }_1}I-A^{-1})(A^{-1}-\frac{1}{{\lambda }_n}I)x.$$

(1.5)

Then the following inequality holds [8]:

$$0\le x^{\ast }Ax\cdot x^{\ast }{A}^{-1}x-1\le \frac{{({\lambda }_1-{\lambda }_n)}^2}{4{\lambda }_1{\lambda }_n}-\sqrt{C(A,x)\cdot C(A^{-1},x)}\le \frac{{({\lambda }_1-{\lambda }_n)}^2}{4{\lambda }_1{\lambda }_n}.$$

(1.6)

The result above is an improvement of the Kantorovich inequality (1.1).

A generalized form of the Kantorovich inequality presented by Greub and Rheinboldt [1] in 1959 is known as the Greub-Rheinboldt inequality in operator theoretic terms, which is also an important and early example of the so-called complementary inequality referred to in [9],

$$〈Ax,Ax〉〈Bx,Bx〉\le \frac{{(M_1{M}_2+m_1{m}_2)}^2}{4m_1{m}_2{M}_1{M}_2}{〈Ax,Bx〉}^2,$$

(1.7)

where A and B are commuting positive definite self-adjoint operators on a Hilbert space, with upper and lower bounds $M_i$ and $m_i$, $i=1,2$, respectively.

In 1997, Fujii et al. [10] generalized the Greub-Rheinboldt inequality to pairs of invertible operators that may not even commute,

$$〈A^2\mathrm{♯}{B}^{2}x,x〉\le {〈A^2,x〉}^{1/2}{〈B^2,x〉}^{1/2}\le \frac{m_1{m}_2+M_2{M}_2}{2\sqrt{m_1{m}_2{M}_1}{M}_2}〈A^2\mathrm{♯}{B}^{2}x,x〉{〈Ax,Bx〉}^2,$$

(1.8)

where A, B are invertible positive operators satisfying $0<m_1\le A\le M_1$ and $0<m_2\le B\le M_2$, and $A\mathrm{♯}B=A^{1/2}{(A^{-1/2}B{A}^{-1/2})}^{1/2}{A}^{1/2}$. By using the viewpoint of interaction antieigenvalue, Gustafson [9] sharpened the Greub-Rheinboldt inequality (1.7) to obtain the following result:

$$〈Ax,Ax〉〈Bx,Bx〉\le \frac{{(m(A{B}^{-1})+M(A{B}^{-1}))}^2}{4m(A{B}^{-1})M(A{B}^{-1})}{〈Ax,Bx〉}^2,$$

(1.9)

where A and B are commuting positive definite self-adjoint operators on a Hilbert space.

Let A and B be two positive definite Hermite matrices and $AB=BA$ with real eigenvalues ${\lambda }_1\le {\lambda }_2\le \cdots \le {\lambda }_n$ and ${\mu }_1\le {\mu }_2\le \cdots \le {\mu }_n$, respectively. Moreover, let $〈Ax,Bx〉:={(Ax)}^{\ast }Bx=x^{\ast }{A}^{\ast }Bx$. Then a matrix version of (1.9) is

$$\frac{x^{\ast }{A}^{2}x\cdot x^{\ast }{B}^{2}x}{{(x^{\ast }ABx)}^2}\le \frac{{({\lambda }_1{\mu }_1+{\lambda }_n{\mu }_n)}^2}{4{\lambda }_1{\lambda }_n{\mu }_1{\mu }_n}$$

(1.10)

for any nonzero vector $x\in {\mathbb{C}}^n$.

In 2005, Seddighin [11] extended the Greub-Rheinboldt inequality (1.9) to pairs of normal operators and established for what vectors the Greub-Rheinboldt inequality becomes equality.

Let V be an $n\times r$ matrix such that $V^{\ast }V=I_r$, i.e., V is suborthogonal. Another well-known matrix version of the Kantorovich inequality asserts that

$$V^{\ast }{A}^{2}V\le \frac{{(m+M)}^2}{4mM}{\left(V^{\ast }AV\right)}^2$$

(1.11)

for any $A>0$, $V^{\ast }V=I$, and $0<mI<A<MI$.

Mond and Pečarić proved the following matrix version inequality (see (7) in [12]):

$${\left(V^{\ast }{A}^{2}V\right)}^{1/2}-V^{\ast }AV\le \frac{{(M-m)}^2}{4(M-m)}I$$

(1.12)

for $A>0$ and $V^{\ast }V=I$. For more related properties and applications, see, e.g., [13–15].

In the next section, we propose some refinements about the matrix Kantorovich-type inequalities (1.2), the Greub-Rheinboldt inequality for commutative positive definite Hermitian matrix pairs, and (1.10) for positive definite matrices, yielding sharpened upper bounds compared with original results, together with an improvement to (1.12).

2 Main results

In this section, we first introduce some lemmas.

Lemma 2.1 (in [8], Lemma 2.2)

Let $A\in M_n$ be a positive definite Hermitian matrix. The following inequalities hold:

$${\lambda }_1{\parallel x\parallel }^2\le x^{\ast }Ax\le {\lambda }_n{\parallel x\parallel }^2,0\le ({\lambda }_n{\parallel x\parallel }^2-x^{\ast }Ax)(x^{\ast }Ax-{\lambda }_1{\parallel x\parallel }^2)\le \frac{1}{4}{({\lambda }_n-{\lambda }_1)}^2{\parallel x\parallel }^4,$$

and

$$\begin{array}{r}\frac{1}{{\lambda }_n}{\parallel x\parallel }^2\le x^{\ast }{A}^{-1}x\le \frac{1}{{\lambda }_1}{\parallel x\parallel }^2,\\ 0\le (\frac{1}{{\lambda }_1}{\parallel x\parallel }^2-x^{\ast }{A}^{-1}x)(x^{\ast }{A}^{-1}x-\frac{1}{{\lambda }_n}{\parallel x\parallel }^2)\le \frac{{({\lambda }_n-{\lambda }_1)}^2}{4{({\lambda }_1{\lambda }_n)}^2}{\parallel x\parallel }^4\end{array}$$

(2.1)

for any $x\in {\mathbb{C}}^n$.

Let A, B be two invertible commuting Hermite matrices. Denote by ${\lambda }_1\le {\lambda }_2\le \cdots \le {\lambda }_n$ and ${\mu }_1\le {\mu }_2\le \cdots \le {\mu }_n$ the eigenvalues of A and B, respectively. Then there exists a unitary matrix $U\in M_n$ such that $A=U\mathrm{\Lambda }{U}^{\ast }$, $B=UM{U}^{\ast }$, where $\mathrm{\Lambda }=diag({\lambda }_1,\dots ,{\lambda }_n)$, $M=diag({\hat{\mu }}_1,\dots ,{\hat{\mu }}_n)$. Note that ${\hat{\mu }}_1,{\hat{\mu }}_2,\dots ,{\hat{\mu }}_n$ is a permutation of ${\mu }_1,{\mu }_2,\dots ,{\mu }_n$. Let ${\sigma }_k=\frac{{\lambda }_k}{{\hat{\mu }}_k}$ ($k=1,\dots ,n$), then it is easy to see that all eigenvalues of $A{B}^{-1}$ are ${\sigma }_1,{\sigma }_2,\dots ,{\sigma }_n$. Without loss of generality, we may assume that ${\sigma }_1={min}_k\{\frac{{\lambda }_k}{{\hat{\mu }}_k}\}$, ${\sigma }_n={max}_k\{\frac{{\lambda }_k}{{\hat{\mu }}_k}\}$ and ${\sigma }_1\le \cdots \le {\sigma }_n$. For convenience, we introduce the notation

$$D(AB,x)=x^{\ast }A({\sigma }_{n}I-A{B}^{-1})(A{B}^{-1}-{\sigma }_{1}I)Bx.$$

(2.2)

If ${\sigma }_1{\sigma }_n>0$, then we can define

$$D({(AB)}^{-1},x)=x^{\ast }A(\frac{1}{{\sigma }_1}I-A^{-1}B)(A^{-1}B-\frac{1}{{\sigma }_n}I)Bx.$$

(2.3)

Lemma 2.2 Let A and B be two positive definite commuting matrices with eigenvalues $0<{\lambda }_1\le {\lambda }_2\le \cdots \le {\lambda }_n$, $0<{\mu }_1\le {\mu }_2\le \cdots \le {\mu }_n$, respectively. ${\sigma }_1\le {\sigma }_2\le \cdots \le {\sigma }_n$, $D(AB,x)$ and $D({(AB)}^{-1},x)$ are as before. Then for any $x\in {\mathbb{C}}^n$,

$$\begin{array}{r}0\le D(AB,x)\le \frac{1}{4}{({\sigma }_n-{\sigma }_1)}^2\left|x^{\ast }ABx\right|,\\ 0\le D({(AB)}^{-1},x)\le \frac{{({\sigma }_n-{\sigma }_1)}^2}{4{({\sigma }_1{\sigma }_n)}^2}\left|x^{\ast }ABx\right|\end{array}$$

(2.4)

for any $x\in {\mathbb{C}}^n$.

Proof From (2.2),

$$\begin{array}{rl}D(AB,x)& =x^{\ast }A({\sigma }_{n}I-A{B}^{-1})(A{B}^{-1}-{\sigma }_{1}I)Bx\\ =x^{\ast }U\mathrm{\Lambda }{U}^{\ast }({\sigma }_{n}I-U\mathrm{\Lambda }{U}^{\ast }U{M}^{-1}{U}^{\ast })(U\mathrm{\Lambda }{U}^{\ast }U{M}^{-1}{U}^{\ast }-{\sigma }_{1}I)UM{U}^{\ast }x\\ =x^{\ast }U\mathrm{\Lambda }({\sigma }_{n}I-\mathrm{\Lambda }{M}^{-1})(\mathrm{\Lambda }{M}^{-1}-{\sigma }_{1}I)M{U}^{\ast }x.\end{array}$$

(2.5)

Let $z={(z_1,\dots ,z_n)}^T={(\mathrm{\Lambda }M)}^{1/2}{U}^{\ast }x$. Thus, ${\parallel z\parallel }^2=z^{\ast }z=x^{\ast }U(\mathrm{\Lambda }M)U^{\ast }x=x^{\ast }ABx$. Then

$$D(AB,x)=z^{\ast }({\sigma }_{n}I-\mathrm{\Lambda }{M}^{-1})(\mathrm{\Lambda }{M}^{-1}-{\sigma }_{1}I)z=\sum _{i=1}^n({\sigma }_n-{\sigma }_i)({\sigma }_i-{\sigma }_1)z_i^2\ge 0.$$

(2.6)

On the other hand,

$$\sum _{i=1}^n({\sigma }_n-{\sigma }_i)({\sigma }_i-{\sigma }_1)z_i^2\le \frac{{({\sigma }_n-{\sigma }_1)}^2}{4}{\parallel z\parallel }^2.$$

(2.7)

Thus,

$$D(AB,x)\le \frac{{({\sigma }_n-{\sigma }_1)}^2}{4}{\parallel z\parallel }^2=\frac{{({\sigma }_n-{\sigma }_1)}^2}{4}\left|x^{\ast }ABx\right|.$$

(2.8)

The proof of $D({(AB)}^{-1},x)$ is similar. □

Theorem 2.3 With the assumptions of Lemma  2.2,

$$0\le \frac{x^{\ast }{A}^{2}x\cdot x^{\ast }{B}^{2}x}{{(x^{\ast }ABx)}^2}-1\le \frac{{({\sigma }_n-{\sigma }_1)}^2}{4{\sigma }_1{\sigma }_n}-\frac{1}{|x^{\ast }ABx|}\sqrt{D(AB,x)\cdot D({(AB)}^{-1},x)}.$$

(2.9)

Proof Let $z={(\mathrm{\Lambda }M)}^{1/2}{U}^{\ast }x$, $E=\mathrm{\Lambda }{M}^{-1}=diag(\frac{{\lambda }_n}{{\hat{\mu }}_n},\dots ,\frac{{\lambda }_1}{{\hat{\mu }}_1})=diag({\sigma }_n,\dots ,{\sigma }_1)$. Then

$$\frac{x^{\ast }{A}^{2}x\cdot x^{\ast }{B}^{2}x}{{(x^{\ast }ABx)}^2}=\frac{z^{\ast }Ez\cdot z^{\ast }{E}^{-1}z}{{(z^{\ast }z)}^2}.$$

(2.10)

From (1.2) and (1.6),

$$\begin{array}{rl}0& \le \frac{z^{\ast }Ez\cdot z^{\ast }{E}^{-1}z}{{(z^{\ast }z)}^2}-1\le \frac{{({\sigma }_n-{\sigma }_1)}^2}{4{\sigma }_1{\sigma }_n}-\sqrt{C(E,\frac{z}{\parallel z\parallel })\cdot C(E^{-1},\frac{z}{\parallel z\parallel })}\\ =\frac{{({\sigma }_n-{\sigma }_1)}^2}{4{\sigma }_1{\sigma }_n}-\frac{1}{{\parallel z\parallel }^2}\sqrt{C(E,z)\cdot C(E^{-1},z)}.\end{array}$$

(2.11)

From (2.5) and (2.10), we have

$$z^{\ast }z=x^{\ast }ABx,C(E,z)=D(AB,x),C(E^{-1},z)=D({(AB)}^{-1},x).$$

(2.12)

By substituting (2.12) and (2.10) into (2.11), the inequality becomes

$$0\le \frac{x^{\ast }{A}^{2}x\cdot x^{\ast }{B}^{2}x}{{(x^{\ast }ABx)}^2}-1\le \frac{{({\sigma }_n-{\sigma }_1)}^2}{4{\sigma }_1{\sigma }_n}-\frac{1}{|x^{\ast }ABx|}\sqrt{D(AB,x)\cdot D({(AB)}^{-1},x)}.$$

 □

Corollary 2.4 Let A and B be two positive definite commuting matrices with eigenvalues $0<{\lambda }_1\le \cdots \le {\lambda }_n$, $0<{\mu }_1\le \cdots \le {\mu }_n$, respectively. Then

$$\frac{x^{\ast }{A}^{2}x\cdot x^{\ast }{B}^{2}x}{{(x^{\ast }ABx)}^2}\le \frac{{({\lambda }_1{\mu }_1+{\lambda }_n{\mu }_n)}^2}{4{\lambda }_1{\mu }_1{\lambda }_n{\mu }_n}-\frac{1}{|x^{\ast }ABx|}\sqrt{D(AB,x)\cdot D({(AB)}^{-1},x)}$$

(2.13)

holds for any nonzero vector $x\in {\mathbb{C}}^n$.

Proof

By Theorem 2.3, we have the following:

$$0\le \frac{x^{\ast }{A}^{2}x\cdot x^{\ast }{B}^{2}x}{{(x^{\ast }ABx)}^2}\le \frac{{({\sigma }_1+{\sigma }_n)}^2}{4{\sigma }_1{\sigma }_n}-\frac{1}{|x^{\ast }ABx|}\sqrt{D(AB,x)\cdot D({(AB)}^{-1},x)}.$$

(2.14)

Let $f(x)=\frac{{(1+x)}^2}{4x}$. It can be easily deduced that $f(x)$ is monotone increasing on $[1,+\mathrm{\infty })$. Let ${\alpha }_1=\frac{{\mu }_1}{{\lambda }_n}$, ${\alpha }_n=\frac{{\mu }_n}{{\lambda }_1}$. From the definition of ${\sigma }_1$ and ${\sigma }_n$, we know that $\frac{{\alpha }_n}{{\alpha }_1}\ge \frac{{\sigma }_n}{{\sigma }_1}\ge 1$. Thus,

$$\frac{{({\sigma }_1+{\sigma }_n)}^2}{4{\sigma }_1{\sigma }_n}=f\left(\frac{{\sigma }_n}{{\sigma }_1}\right)\le f\left(\frac{{\alpha }_n}{{\alpha }_1}\right)=\frac{{({\lambda }_1{\mu }_1+{\lambda }_1{\mu }_1)}^2}{4{\lambda }_1{\mu }_1{\lambda }_1{\mu }_1}.$$

That is,

$$0\le \frac{x^{\ast }{A}^{2}x\cdot x^{\ast }{B}^{2}x}{{(x^{\ast }ABx)}^2}\le \frac{{({\lambda }_1{\mu }_1+{\lambda }_1{\mu }_1)}^2}{4{\lambda }_1{\mu }_1{\lambda }_1{\mu }_1}-\frac{1}{|x^{\ast }ABx|}\sqrt{D(AB,x)\cdot D({(AB)}^{-1},x)}.$$

(2.15)

 □

Remark From Lemma 2.2 and (2.15), we can obtain a sharpened bound for the classical Kantorovich-type inequality, i.e., the Greub-Rheinboldt inequality.

Besides the discussion on the Greub-Rheinboldt inequality (1.9), we are also interested in another form of Kantorovich-type inequality aforementioned. We turn our attention to the inequalities (1.11) and (1.12) in the remainder of this paper.

Let A be an $n\times n$ positive (semi-) definite Hermitian matrix with (nonzero) eigenvalues contained in the interval $[m,M]$, where $0<m<M$. Let V be $n\times r$ matrices.

As is declared in (1.11), for $A>0$, $V^{\ast }V=I$, and m, M mentioned above, the following inequality holds:

$$V^{\ast }{A}^{2}V\le \frac{{(m+M)}^2}{4mM}{\left(V^{\ast }AV\right)}^2.$$

It is not difficult to see that as $V^{\ast }V=I$, then $V{V}^{\ast }=V{V}^{+}\le I$, where + indicates the Moore-Penrose inverse. Multiplying from the right and from the left by $V^{\ast }A$ and AV respectively, we have $V^{\ast }{A}^{2}V\ge {(V^{\ast }AV)}^2$ for $A>0$. From the well-known Löwner-Heinz inequality, we have ${(V^{\ast }{A}^{2}V)}^{1/2}\ge V^{\ast }AV$ and the following inequality (see in [16]):

$${\left(V^{\ast }{A}^{2}V\right)}^{1/2}\le \frac{m+M}{2\sqrt{mM}}{V}^{\ast }AV.$$

For $z\in [m,M]$, $m>0$, the convexity of $(z^{-1}+z/mM)$ implies that

$$z^{-1}\le \frac{m+M}{mM}-\frac{z}{mM}.$$

(2.16)

If A has the representation $A=\mathrm{\Gamma }{D}_{\alpha }{\mathrm{\Gamma }}^{\ast }$, where Γ is unitary and $D_{\alpha }=diag({\alpha }_1,\dots ,{\alpha }_n)$, and if $0<m\le {\alpha }_i\le M$, $i=1,\dots ,n$, then from (2.16) it follows that

$$D_{\alpha }^{-1}\le \frac{m+M}{mM}I-\frac{D_{\alpha }}{mM}.$$

(2.17)

After multiplying from the right and from the left by Γ and ${\mathrm{\Gamma }}^{\ast }$, it is not difficult to see that (2.17) yields the following [17]:

$$A^{-1}\le \frac{m+M}{mM}I-\frac{A}{mM}.$$

(2.18)

Based on (2.18), we derive several results on the inequality (1.12).

Theorem 2.5 For any $A>0$ and $V^{\ast }V=I$,

$${\left(V^{\ast }{A}^{2}V\right)}^{1/2}-V^{\ast }AV\le \frac{{(M-m)}^2}{4(M+m)}I-D^2(A,V),$$

(2.19)

where $D(A,V)={(\frac{1}{m+M}{V}^{\ast }{A}^{2}V)}^{1/2}-\frac{{(M+m)}^{1/2}}{2}I$.

Proof From (2.18) and $A>0$, we can get

$$-A\le -\frac{mM}{(M+m)}I-\frac{1}{(M+m)}{A}^2.$$

(2.20)

Since $V^{\ast }V=I$, (2.20) can be turned into

$$-V^{\ast }AV\le -\frac{mM}{(M+m)}I-\frac{1}{(M+m)}{V}^{\ast }{A}^{2}V.$$

(2.21)

By adding ${(V^{\ast }{A}^{2}V)}^{1/2}\ge 0$ to both sides of the inequality (2.21), we obtain that

$${\left(V^{\ast }{A}^{2}V\right)}^{1/2}-V^{\ast }AV\le {\left(V^{\ast }{A}^{2}V\right)}^{1/2}-\frac{mM}{(M+m)}I-\frac{1}{(M+m)}{V}^{\ast }{A}^{2}V,$$

(2.22)

i.e.,

$$\begin{array}{rcl}{\left(V^{\ast }{A}^{2}V\right)}^{1/2}-V^{\ast }AV& \le & \frac{{(M-m)}^2}{4(M+m)}I-\frac{1}{(M+m)}{V}^{\ast }{A}^{2}V+{\left(V^{\ast }{A}^{2}V\right)}^{1/2}-\frac{(M+m)}{4}I\\ =& \frac{{(M-m)}^2}{4(M+m)}I-{[{\left(\frac{1}{M+m}{V}^{\ast }{A}^{2}V\right)}^{1/2}-\frac{{(M+m)}^{1/2}}{2}I]}^2.\end{array}$$

(2.23)

Thus, we finally have

$${\left(V^{\ast }{A}^{2}V\right)}^{1/2}-V^{\ast }AV\le \frac{{(m-M)}^2}{4(M+m)}I-D^2(A,V),$$

where $D(A,V)={(\frac{1}{(m+M)}{V}^{\ast }{A}^{2}V)}^{1/2}-\frac{{(M+m)}^{1/2}}{2}I$. □

Remark It is obvious that $D^2(A,V)\ge 0$. Thus, Theorem 2.5 indeed presents an improvement of the Kantorovich-type inequality (1.12) in [12].

For an application to the Hadamard product, we have the following corollary.

Corollary 2.6 Let $A_1$ and $A_2$ be $n\times n$ positive definite matrices with eigenvalues of $A_1\otimes A_2$ contained in the interval $[m,M]$. Then

$${(A_1^2\circ A_2^2)}^{1/2}-A_1\circ A_2\le \frac{{(M-m)}^2}{4(m+M)}I-D^2(A_1\otimes A_2,V),$$

where V is the selection matrix of order $n^2\times n$ with the property $V^{\ast }(A_1\otimes A_2)V=A_1\circ A_2$ (⊗ and ∘ indicate the tensor and the Hadamard product, respectively).

3 Conclusion

In this paper, we introduce some new bounds for several Kantorovich-type inequalities for commutative positive definite Hermitian matrix pairs. As a particular situation, in Corollary 2.4, when A and B are both positive definite, the result provides a sharpened upper bound for the matrix version of the well-known Greub-Rheinboldt inequality. Moreover, it holds for negative definite Hermite matrices. Also, a refinement of Kantorovich-type inequalities concerning positive definite matrices is presented together with an application to the Hadamard product.