1 Introduction

Let \( \mathbb {M} _{n}\) be the algebra of all \(n\times n\) complex matrices. For \(A\in \mathbb {M} _{n}\), we denote the eigenvalues of \(\left| A\right| =\left( A^{*}A\right) ^{1/2}\) by \(s_{1}(A)\ge s_{2}(A)\ge \cdots \ge s_{n}(A)\), they are called the singular values of A. Note that \(s_{j}(A)=s_{j}(A^{ *})=s_{j}(\left| A\right| )\) for \(j=1,2,\ldots ,n\). Note that the spectral (usual operator) norm \(\left\| .\right\| \) is the largest singular value, i.e. \(\left\| A\right\| =s_{1}(A)\), and the Schatten p-norms \(\left\| .\right\| _{p}\) are defined interms of the singular values, where \(\left\| A\right\| _{p}=\left( {{\sum _{j=1}^{n} s_{j}^{p}}}(A)\right) ^{1/p}\) for \(1\le p\le \infty \). Apart from the spectral (usual operator) norm and the Schatten p-norms, we have the wider class of unitarily invariant norms \(\left| \left| \left| .\right| \right| \right| \). Unitarily invariant norms are characterized by the invariance property which states that \( \left| \left| \left| UAV\right| \right| \right| =\left| \left| \left| A\right| \right| \right| \) for all \(A\in \mathbb {M} _{n}\) and for all unitary matrices U and V. Unitarily invariant norms are increasing functions of singular values (see, e.g., [4] or [9]).

For \(A,B,X\in \mathbb {M} _{n}\), a matrix of the form \(AX-XA\) is called a commutator, a matrix of the form \(AX-XB\) is called a generalized commutator, a matrix of the form \(AX+XA\) is called anticommutator, and a matrix of the form \(AX+XB\) is called a generalized anticommutator. In this paper, we present singular value inequalities for these types of matrices.

Kittaneh in [11] has proved that if \(A,B\in \mathbb {M} _{n}\) are positive semidefinite, then

$$\begin{aligned} s_{j}(A+B)\le s_{j}\left( \left( A+\left| B^{1/2}A^{1/2}\right| \right) \oplus \left( B+\left| A^{1/2}B^{1/2}\right| \right) \right) \end{aligned}$$
(1.1)

for \(j=1,2,\ldots ,2n\). Inequality (1.1) can be extended to unitarily invariant norms. For \(j=1\), this inequality is the spectral norm inequality,

$$\begin{aligned} \left| \left| A+B\right| \right| \le \max \left\{ \left\| A+\left| B^{1/2}A^{1/2}\right| \right\| ,\left\| B+\left| A^{1/2}B^{1/2}\right| \right\| \right\} . \end{aligned}$$
(1.2)

Specifying inequality (1.1) to the Schatten p-norms, we get

$$\begin{aligned} \left\| A+B\right\| _{p}\le \left( \left\| A+\left| B^{1/2}A^{1/2}\right| \right\| _{p}^{p}+\left\| B+\left| A^{1/2}B^{1/2}\right| \right\| _{p}^{p}\right) ^{1/p} \end{aligned}$$
(1.3)

for \(1\le p\le \infty .\) Kittaneh in [10] has been proved that if \(A,B\in \mathbb {M} _{n}\) are positive semidefinite, then

$$\begin{aligned} \left| \left| \left| A+B\right| \right| \right| \le \left| \left| \left| A\oplus B\right| \right| \right| +\left| \left| \left| A^{1/2}B^{1/2}\oplus A^{1/2}B^{1/2}\right| \right| \right| . \end{aligned}$$
(1.4)

It should be mentioned here that inequality (1.4) is trivial consequence of inequality (1.1) by application of triangular inequality. Specifying inequality (1.4) to the spectral norm \( \left\| .\right\| \), leads to

$$\begin{aligned} \left| \left| A+B\right| \right| \le \max \left\{ \left\| A\right\| ,\left\| B\right\| \right\} +\left\| A^{1/2}B^{1/2}\right\| . \end{aligned}$$
(1.5)

Davidson and Power in [8] has been shown a weaker version of inequality (1.5). Bourin in [7] provides an equivalent formulation of inequality (1.5). Specifying inequality (1.4) to the Schatten p-norms, we have

$$\begin{aligned} \left\| A+B\right\| _{p}\le \left( \left\| A\right\| _{p}^{p}+\left\| B\right\| _{p}^{p}\right) ^{1/p}+2^{1/p}\left\| A^{1/2}B^{1/2}\right\| _{p}. \end{aligned}$$
(1.6)

For recent studies and details for generalizations of singular value inequalities, we refer to [1, 2] and [3]. In this paper, we give a remarkable generalizations of the inequalities (1.1), (1.2), and (1.3). Several applications are also given.

2 Main results

To reach our findings, we need the following lemmas. The first lemma has been shown by Bhatia and Kittaneh in [5]. The second lemma has been proved by Bourin in [6]. The third lemma has been given by Bhatia in [4].

Lemma 2.1

Let \(A,B\in \mathbb {M} _{n}\). Then

$$\begin{aligned} s_{j}(AB^{*})\le \frac{1}{2}s_{j}(A^{*}A+B^{*}B) \end{aligned}$$

for \(j=1,2,\ldots ,n\)

Lemma 2.2

Let \(A,B\in \mathbb {M} _{n}\) be normal and let f be a nonnegative concave function on \(\left[ 0,\infty \right) \). Then

$$\begin{aligned} \left| \left| \left| f(\left| A+B\right| )\right| \right| \right| \le \left| \left| \left| f(\left| A\right| )+f(\left| B\right| )\right| \right| \right| \end{aligned}$$

for every unitarily invariant norm.

Lemma 2.3

Let \(A,B\in \mathbb {M} _{n}\) such that AB is Hermitian. Then

$$\begin{aligned} \left| \left| \left| AB\right| \right| \right| \le \left| \left| \left| \text {Re}(BA)\right| \right| \right| . \end{aligned}$$

From now until the end of the paper, we will assume that all functions considered are continuous and all matrices denoted by the symbol A or B are positive semidefinite. Our first result is the following singular value inequality for generalized commutator.

Theorem 2.4

Let \(A,B,X\in \mathbb {M} _{n}\). Then

$$\begin{aligned} s_{j}(AX-XB)\le s_{j}\left( K\oplus L\right) \end{aligned}$$
(2.1)

for \(j=1,2,\ldots ,2n\), where

$$\begin{aligned} K=\frac{1}{2}A+\frac{1}{2}A^{1/2}\left| X^{*}\right| ^{2}A^{1/2}\text { } \end{aligned}$$

and

$$\begin{aligned} L=\frac{1}{2}B+\frac{1}{2}B^{1/2}\left| X\right| ^{2}B^{1/2}. \end{aligned}$$

Proof

Let

$$\begin{aligned} S= & {} \left[ \begin{array}{c@{\quad }c} A^{1/2} &{}\quad XB^{1/2} \\ 0 &{}\quad 0 \end{array} \right] ,\\ R^{*}= & {} \left[ \begin{array}{c@{\quad }c} A^{1/2}X &{}\quad 0 \\ -B^{1/2} &{}\quad 0 \end{array} \right] ,\\ M= & {} \left[ \begin{array}{c@{\quad }c} A &{}\quad A^{1/2}XB^{1/2} \\ B^{1/2}X^{*}A^{1/2} &{}\quad B^{1/2}\left| X\right| ^{2}B^{1/2} \end{array} \right] , \end{aligned}$$

and

$$\begin{aligned} N=\left[ \begin{array}{c@{\quad }c} A^{1/2}\left| X^{*}\right| ^{2}A^{1/2} &{}\quad -A^{1/2}XB^{1/2} \\ -B^{1/2}X^{*}A^{1/2} &{}\quad B \end{array} \right] . \end{aligned}$$

Then for \(j=1,2,\ldots ,2n\), we have

$$\begin{aligned} s_{j}(AX-XB)= & {} s_{j}(SR^{*}) \\\le & {} \frac{1}{2}s_{j}(S^{*}S+R^{*}R) (\text {by Lemma}~2.1)\\= & {} s_{j}\left( \frac{1}{2}M+\frac{1}{2}N\right) \\= & {} s_{j}\left( \left[ \begin{array}{c@{\quad }c} K &{}\quad 0 \\ 0 &{}\quad L \end{array} \right] \right) \\= & {} s_{j}\left( K\oplus L\right) . \end{aligned}$$

Our inequality has thus been proved. \(\square \)

Remark 2.5

Letting \(X=I\) in inequality (2.1), we give

$$\begin{aligned} s_{j}(A-B)\le s_{j}(A\oplus B) \end{aligned}$$
(2.2)

for \(j=1,2,\ldots ,2n\). Inequality (2.2) has been proved by Zhan in [12].

Remark 2.6

Letting \(B=A\) in inequality (2.1), we give the following singular value inequality for commutator.

$$\begin{aligned} s_{j}(AX-XA)\le s_{j}\left( Y\oplus Z\right) \end{aligned}$$

for \(j=1,2,\ldots ,2n\), where

$$\begin{aligned} Y=\frac{1}{2}A+\frac{1}{2}A^{1/2}\left| X^{*}\right| ^{2}A^{1/2} \end{aligned}$$

and

$$\begin{aligned} Z=\frac{1}{2}A+\frac{1}{2}A^{1/2}\left| X\right| ^{2}A^{1/2}. \end{aligned}$$

We present the following generalization of inequality (1.1), which is singular value inequality for generalized anticommutator.

Theorem 2.7

Let \(A,B,X\in \mathbb {M} _{n}\). Then

$$\begin{aligned} s_{j}(AX+XB)\le s_{j}(C\oplus D) \end{aligned}$$
(2.3)

for \(j=1,2,\ldots ,2n\), where

$$\begin{aligned} C= & {} K+\left| B^{1/2}X^{*}A^{1/2}\right| ,\\ D= & {} L+\left| A^{1/2}XB^{1/2}\right| , \\ K= & {} \frac{1}{2}A+\frac{1}{2}A^{1/2}\left| X^{*}\right| ^{2}A^{1/2}, \end{aligned}$$

and

$$\begin{aligned} L=\frac{1}{2}B+\frac{1}{2}B^{1/2}\left| X\right| ^{2}B^{1/2}. \end{aligned}$$

Proof

Let

$$\begin{aligned} S= & {} \left[ \begin{array}{c@{\quad }c} A^{1/2} &{} XB^{1/2} \\ 0 &{} 0 \end{array} \right] ,\\ T= & {} \left[ \begin{array}{c@{\quad }c} X^{*}A^{1/2} &{} B^{1/2} \\ 0 &{} 0 \end{array} \right] ,\\ U= & {} \left[ \begin{array}{c@{\quad }c} A &{} A^{1/2}XB^{1/2} \\ B^{1/2}X^{*}A^{1/2} &{} B^{1/2}\left| X\right| ^{2}B^{1/2} \end{array} \right] , \end{aligned}$$

and

$$\begin{aligned} V=\left[ \begin{array}{c@{\quad }c} A^{1/2}\left| X^{*}\right| ^{2}A^{1/2} &{} A^{1/2}XB^{1/2} \\ B^{1/2}X^{*}A^{1/2} &{} B \end{array} \right] . \end{aligned}$$

Then for \(j=1,2,\ldots ,2n\), we have

$$\begin{aligned} s_{j}(AX+XB)= & {} s_{j}(ST^{*}) \\\le & {} \frac{1}{2}s_{j}(S^{*}S+T^{*}T), (\text {by Lemma} 2.1) \\= & {} \frac{1}{2}s_{j}\left( U+V\right) \\= & {} s_{j}\left( \left[ \begin{array}{c@{\quad }c} K &{} A^{1/2}XB^{1/2} \\ B^{1/2}X^{*}A^{1/2} &{} L \end{array} \right] \right) \\= & {} s_{j}\left( \left[ \begin{array}{c@{\quad }c} K &{} 0 \\ 0 &{} L \end{array} \right] +\left[ \begin{array}{c@{\quad }c} 0 &{} A^{1/2}XB^{1/2} \\ B^{1/2}X^{*}A^{1/2} &{} 0 \end{array} \right] \right) \\\le & {} s_{j}\left( \left[ \begin{array}{c@{\quad }c} K &{} 0 \\ 0 &{} L \end{array} \right] +\left[ \begin{array}{c@{\quad }c} \left| B^{1/2}X^{*}A^{1/2}\right| &{} 0 \\ 0 &{} \left| A^{1/2}XB^{1/2}\right| \end{array} \right] \right) \\= & {} s_{j}\left( \left[ \begin{array}{c@{\quad }c} K+\left| B^{1/2}X^{*}A^{1/2}\right| &{} 0 \\ 0 &{} L+\left| A^{1/2}XB^{1/2}\right| \end{array} \right] \right) \\= & {} s_{j}\left( \left[ \begin{array}{c@{\quad }c} C &{} 0 \\ 0 &{} D \end{array} \right] \right) . \end{aligned}$$

Inequality (2.3) has thus been substantiated. \(\square \)

Remark 2.8

Letting \(X=I\) in inequality (2.3), we give inequality (1.1). In that sense inequality (2.3) is certainly a generalization of inequality (1.1).

We are now in a position to present our next norm inequality, which is a generalization of the generalized anticommutator.

Theorem 2.9

Let \(A,B,X\in \mathbb {M} _{n}\) and let f be a nonnegative increasing concave function on \(\left[ 0,\infty \right) \). Then

$$\begin{aligned} \left| \left| \left| f(\left| (AX+XB)\oplus 0\right| )\right| \right| \right| \le \left| \left| \left| I\oplus J\right| \right| \right| \end{aligned}$$
(2.4)

for every unitarily invariant norm, where

$$\begin{aligned} I= & {} \left( f\left( K\right) +f\left( \left| B^{1/2}X^{*}A^{1/2}\right| \right) \right) ,\\ J= & {} \left( f\left( L\right) +f\left( \left| A^{1/2}XB^{1/2}\right| \right) \right) , \\ K= & {} \frac{1}{2}A+\frac{1}{2}A^{1/2}\left| X^{*}\right| ^{2}A^{1/2}, \end{aligned}$$

and

$$\begin{aligned} L=\frac{1}{2}B+\frac{1}{2}B^{1/2}\left| X\right| ^{2}B^{1/2}. \end{aligned}$$

Proof

Let

$$\begin{aligned} S= & {} \left[ \begin{array}{c@{\quad }c} A^{1/2} &{} XB^{1/2} \\ 0 &{} 0 \end{array} \right] ,\\ T= & {} \left[ \begin{array}{c@{\quad }c} X^{*}A^{1/2} &{} B^{1/2} \\ 0 &{} 0 \end{array} \right] ,\\ E= & {} \left[ \begin{array}{c@{\quad }c} A &{} A^{1/2}XB^{1/2} \\ B^{1/2}X^{*}A^{1/2} &{} B^{1/2}\left| X\right| ^{2}B^{1/2} \end{array} \right] , \end{aligned}$$

and

$$\begin{aligned} F=\left[ \begin{array}{c@{\quad }c} A^{1/2}\left| X^{*}\right| ^{2}A^{1/2} &{} A^{1/2}XB^{1/2} \\ B^{1/2}X^{*}A^{1/2} &{} B \end{array} \right] . \end{aligned}$$

Then for \(j=1,2,\ldots ,2n\), we have

$$\begin{aligned} s_{j}(f(\left| (AX+XB)\oplus 0\right| ))= & {} s_{j}(f(\left| ST^{*}\right| )) \\= & {} f(s_{j}(ST^{*})) \\\le & {} f\left( \frac{1}{2}s_{j}(S^{*}S+T^{*}T)\right) (\text {by Lemma} 2.1) \\= & {} f\left( \frac{1}{2}s_{j}\left( E+F\right) \right) \\= & {} s_{j}\left( f\left( \left| \frac{1}{2}E+\frac{1}{2}F\right| \right) \right) \\= & {} s_{j}\left( f\left( \left| \left[ \begin{array}{c@{\quad }c} K &{} A^{1/2}XB^{1/2} \\ B^{1/2}X^{*}A^{1/2} &{} L \end{array} \right] \right| \right) \right) . \end{aligned}$$

This implies that,

$$\begin{aligned} \left| \left| \left| f(\left| (AX+XB)\oplus 0\right| )\right| \right| \right|\le & {} \left| \left| \left| f\left( \left| \left[ \begin{array}{c@{\quad }c} K &{} A^{1/2}XB^{1/2} \\ B^{1/2}X^{*}A^{1/2} &{} L \end{array} \right] \right| \right) \right| \right| \right| \\= & {} \left| \left| \left| f\left( \left| \begin{array}{c} \left[ \begin{array}{c@{\quad }c} K &{} 0 \\ 0 &{} L \end{array} \right] + \\ \left[ \begin{array}{c@{\quad }c} 0 &{} A^{1/2}XB^{1/2} \\ B^{1/2}X^{*}A^{1/2} &{} 0 \end{array} \right] \end{array} \right| \right) \right| \right| \right| \\\le & {} \left| \left| \left| \begin{array}{c} f\left( \left| \left[ \begin{array}{c@{\quad }c} K &{} 0 \\ 0 &{} L \end{array} \right] \right| \right) + \\ f\left( \left| \begin{array}{c@{\quad }c} 0 &{} A^{1/2}XB^{1/2} \\ B^{1/2}X^{*}A^{1/2} &{} 0 \end{array} \right| \right) \end{array} \right| \right| \right| , \\&(\text {by Lemma} 2.2), \\\le & {} \left| \left| \left| \begin{array}{c} f\left( \left[ \begin{array}{c@{\quad }c} K &{} 0 \\ 0 &{} L \end{array} \right] \right) + \\ f\left( \left[ \begin{array}{c@{\quad }c} \left| B^{1/2}X^{*}A^{1/2}\right| &{} 0 \\ 0 &{} \left| A^{1/2}XB^{1/2}\right| \end{array} \right] \right) \end{array} \right| \right| \right| \\= & {} \left| \left| \left| \begin{array}{c} \left[ \begin{array}{c@{\quad }c} f\left( K\right) &{} 0 \\ 0 &{} f\left( L\right) \end{array} \right] + \\ \left[ \begin{array}{c@{\quad }c} f\left( \left| B^{1/2}X^{*}A^{1/2}\right| \right) &{} 0 \\ 0 &{} f\left( \left| A^{1/2}XB^{1/2}\right| \right) \end{array} \right] \end{array} \right| \right| \right| \\= & {} \ \left| \left| \left| \text {\ }\left[ \begin{array}{c@{\quad }c} I &{} 0 \\ 0 &{} J \end{array} \right] \right| \right| \right| \\= & {} \left| \left| \left| I\oplus J\right| \right| \right| , \end{aligned}$$

which is precisely inequality (2.4). \(\square \)

Remark 2.10

Letting \(f(t)=t\) in inequality (2.4), we give norm inequality for generalized anticommutator. In that sense, inequality (2.4) is certainly a generalization of generalized anticommutator norm inequalities.

Specifying inequality (2.4) to the spectral norm and the Schatten p-norms, we give the following norm inequalities for generalized anticommutator which are generalizations of the inequalities (1.2) and (1.3), respectively.

Corollary 2.11

Let \(A,B,X\in \mathbb {M} _{n}\). Then

$$\begin{aligned} \left\| AX+XB\right\| \le \max \left\{ \left\| K+\left| B^{1/2}X^{*}A^{1/2}\right| \right\| ,\left\| L+\left| A^{1/2}XB^{1/2}\right| \right\| \right\} \end{aligned}$$
(2.5)

where

$$\begin{aligned} K=\frac{1}{2}A+\frac{1}{2}A^{1/2}\left| X^{*}\right| ^{2}A^{1/2} \end{aligned}$$

and

$$\begin{aligned} L=\frac{1}{2}B+\frac{1}{2}B^{1/2}\left| X\right| ^{2}B^{1/2}. \end{aligned}$$

Proof

Inequality (2.3) follows by substituting \(f(t)=t\) and by considering the spectral norm in Theorem 2.9. \(\square \)

Remark 2.12

Letting \(X=I\) in Corollary 2.11, we give inequality (1.2).

Corollary 2.13

Let \(A,B,X\in \mathbb {M} _{n}\). Then for \(1\le p\le \infty \), we have

$$\begin{aligned} \left\| AX+XB\right\| _{p}\le \left( \left\| K+\left| B^{1/2}X^{*}A^{1/2}\right| \right\| _{p}^{p}+\left\| L+\left| A^{1/2}XB^{1/2}\right| \right\| _{p}^{p}\right) ^{1/p} \end{aligned}$$
(2.6)

where

$$\begin{aligned} K=\frac{1}{2}A+\frac{1}{2}A^{1/2}\left| X^{*}\right| ^{2}A^{1/2} \end{aligned}$$

and

$$\begin{aligned} L=\frac{1}{2}B+\frac{1}{2}B^{1/2}\left| X\right| ^{2}B^{1/2}. \end{aligned}$$

Proof

Inequality (2.6) follows by substituting \(f(t)=t\) and by considering the Schatten p-norms in Theorem 2.9. \(\square \)

Remark 2.14

Letting \(X=I\) in Corollary 2.13, we give inequality (1.3).

The following two corollaries are applications of Theorem 2.9.

Corollary 2.15

Let \(A,B,X\in \mathbb {M} _{n}\). Then

$$\begin{aligned} \left| \left| \left| \log \left( \left| (AX+XB)\right| +I\right) \right| \right| \right| \le \left| \left| \left| M\oplus N\right| \right| \right| \end{aligned}$$

for every unitarily invariant norm, where

$$\begin{aligned} M= & {} \left( \log \left( K+I\right) +\log \left( \left| B^{1/2}X^{*}A^{1/2}\right| +I\right) \right) , \\ N= & {} \left( \log \left( L+I\right) +\log \left( \left| A^{1/2}XB^{1/2}\right| +I\right) \right) ,\\ K= & {} \frac{1}{2}A+\frac{1}{2}A^{1/2}\left| X^{*}\right| ^{2}A^{1/2}, \end{aligned}$$

and

$$\begin{aligned} L=\frac{1}{2}B+\frac{1}{2}B^{1/2}\left| X\right| ^{2}B^{1/2}. \end{aligned}$$

Proof

The inequality is an immediate consequence of Theorem 2.9 by letting \(f(t)=\log (t+1)\). \(\square \)

Corollary 2.16

Let \(A,B,X\in \mathbb {M} _{n}\). Then, for \(r\in \left( 0,1\right] \), we have

$$\begin{aligned} \left| \left| \left| \left| (AX+XB)\right| ^{r}\right| \right| \right| \le \left| \left| \left| P\oplus Q\right| \right| \right| \end{aligned}$$

for every unitarily invariant norm, where

$$\begin{aligned} P= & {} \left( K^{r}+\left| B^{1/2}X^{*}A^{1/2}\right| ^{r}\right) , \\ Q= & {} \left( L^{r}+\left| A^{1/2}XB^{1/2}\right| ^{r}\right) , \\ K= & {} \frac{1}{2}A+\frac{1}{2}A^{1/2}\left| X^{*}\right| ^{2}A^{1/2}, \end{aligned}$$

and

$$\begin{aligned} L=\frac{1}{2}B+\frac{1}{2}B^{1/2}\left| X\right| ^{2}B^{1/2}. \end{aligned}$$

Proof

The inequality is an immediate consequence of Theorem 2.9 by letting \(f(t)=t^{r}\) and \(r\in \left( 0,1\right] \). \(\square \)