Singular value inequalities of matrices via increasing functions

Abstract

Let A, B, X, and Y be $n\times n$ complex matrices such that A is self-adjoint, $B\geq 0$, $\pm A\leq B$, $\max ( \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} ) \leq 1$, and let f be a nonnegative increasing convex function on $[ 0,\infty )$ satisfying $f(0)=0$. Then

$$ 2s_{j}\bigl(f \bigl( \bigl\vert XAY^{\ast } \bigr\vert \bigr) \bigr)\leq \max \bigl\{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \bigr\} s_{j}\bigl(f(B+A)\oplus f(B-A)\bigr) $$

for $j=1,2,\ldots,n$. This singular value inequality extends an inequality of Audeh and Kittaneh. Several generalizations for singular value and norm inequalities of matrices are also given.

1 Introduction

Let $\mathbb{M}_{n}(\mathbb{C})$ denote the algebra of all $n\times n$ complex matrices. For $A\in \mathbb{M}_{n}(\mathbb{C})$, the singular values of A are denoted by $s_{{1}}(A)\geq s_{2}(A)\geq \cdots\geq s_{{n}}(A)$, they are precisely the eigenvalues of the positive operator $\vert A \vert = ( A^{\ast }A ) ^{1/2}$. Singular values have several properties: Let $A,B\in \mathbb{M}_{n}(\mathbb{C})$. Then

(a) $s_{j}(A)=s_{j}(A^{\ast })=s_{j}( \vert A \vert )$ for $j=1,2,\ldots,n$.

(b) $s_{j}(AA^{\ast })=s_{j}(A^{\ast }A)$ for $j=1,2,\ldots,n$.

(c) If $A,B\in \mathbb{M}_{n}(\mathbb{C})$, then $s_{j}(A)\leq s_{j}(B)$ if and only if $s_{j}(A\oplus A)\leq s_{j}(B\oplus B)$ for $j=1,2,\ldots,n$.

Bhatia and Kittaneh proved in [15] the following inequalities:

• (i) If $A,B\in \mathbb{M}_{n}(\mathbb{C})$ such that A is self-adjoint, $B\geq 0$, and $\pm A\leq B$, then

  $$ s_{j}(A)\leq s_{j}(B\oplus B) $$

  (1)

  for $j=1,2,\ldots,n$.

• (ii) If $A,B\in \mathbb{M}_{n}(\mathbb{C})$, then

  $$ s_{j}\bigl(AB^{\ast }+BA^{\ast }\bigr)\leq s_{j}\bigl(\bigl(AA^{\ast }+BB^{\ast }\bigr)\oplus \bigl(AA^{ \ast }+BB^{\ast }\bigr)\bigr) $$

  (2)

  for $j=1,2,\ldots n$. Audeh and Kittaneh pointed out in [8] that:

• (i) If $A,B\in \mathbb{M}_{n}(\mathbb{C})$ such that A is self-adjoint, $B\geq 0$, and $\pm A\leq B$, then

  $$ 2s_{j}(A)\leq s_{j}\bigl((B+A)\oplus (B-A)\bigr) $$

  (3)

  $for$ $j=1,2,\ldots,n$.

• (ii) If $A,B,C\in \mathbb{M}_{n}(\mathbb{C})$ such that $\left[\begin{array}{cc}A& B\\ B^{\ast }& C\end{array}\right]\ge 0$, then

  $$ s_{j}(B)\leq s_{j}(A\oplus C) $$

  (4)

  for $j=1,2,\ldots,n$.

• (iii) If $A,B\in \mathbb{M}_{n}(\mathbb{C})$, then

  $$ s_{j}(A+B)\leq s_{j}\bigl(\bigl( \vert A \vert + \vert B \vert \bigr)\oplus \bigl( \bigl\vert A^{\ast } \bigr\vert + \bigl\vert B^{ \ast } \bigr\vert \bigr)\bigr) $$

  (5)

  for $j=1,2,\ldots,n$.

Tao proved in [24] that if $A,B,C\in \mathbb{M}_{n}(\mathbb{C})$ such that $[\begin{array}{cc}A& B\\ B^{\ast }& C\end{array}]\ge 0$, then

$$2s_j(B)\le s_j\left[\begin{array}{cc}A& B\\ B^{\ast }& C\end{array}\right]$$

(6)

for $j=1,2,\ldots,n$. In addition, Bhatia and Kittaneh showed in [14] that if $A,B\in \mathbb{M}_{n}(\mathbb{C})$, then

$$ 2s_{j}\bigl(AB^{\ast }\bigr)\leq s_{j} \bigl(A^{\ast }A+B^{\ast }B\bigr) $$

(7)

for $j=1,2,\ldots,n$. For more details and comprehensive results related to this topic, we refer to [1–7, 9, 10] and [17]. In this paper, we provide considerable generalizations of inequalities (1)–(6).

Unitarily invariant norms on $\mathbb{M}_{n}$ are denoted by $|\!|\!|.|\!|\!|$, recall that these norms satisfying $|\!|\!|UAV|\!|\!|=|\!|\!|A|\!|\!|$ for all $U,V,A\in \mathbb{M}_{n}$ such that U and V are unitary. Important classes of such norms are the Schatten p-norms defined by $\Vert A \Vert _{p}= ( \sum_{j=1}^{n}s_{j}^{p}(A) ) ^{1/p}$ where $p\geq 1$ and the spectral norm defined by $\Vert A \Vert =s_{1}(A)$. For the general theory of unitarily invariant norms, we refer the reader to [13], [16], and [23]. It follows easily from the basic properties of unitarily invariant norms that

$$ \big|\!\big|\!\big|A^{\ast }A\big|\!\big|\!\big|=\big|\!\big|\!\big|AA^{\ast } \big|\!\big|\!\big|. $$

(8)

Bhatia and Davis proved in [13] that if $A,X,B\in \mathbb{M}_{n}$, then

$$ 2\big|\!\big|\!\big|AXB^{\ast }\big|\!\big|\!\big|\leq \big|\!\big|\!\big|A^{\ast }AX+XB^{ \ast }B \big|\!\big|\!\big|. $$

(9)

This is a generalization of the arithmetic–geometric mean inequality for unitarily invariant norms. In this paper, we provide a considerable generalization of inequality (9). Hou and Du proved in [19] that if $A\in \mathbb{M}_{n}$, then

$$ \Vert A \Vert \leq \bigl\Vert \bigl( \Vert A_{ij} \Vert \bigr) _{1\leq i,j\leq n} \bigr\Vert . $$

(10)

Popovici and Sebestyen showed in [22] the following inequalities:

• 1. If $A_{1},A_{2},\ldots,A_{n}\in \mathbb{M}_{n}$ are positive, then

  $$ \Biggl\Vert \sum_{k=1}^{n}A_{k} \Biggr\Vert \leq \bigl\Vert \bigl( \bigl\Vert A_{i}^{1/2}A_{j}^{1/2} \bigr\Vert \bigr) _{1 \leq i,j\leq n} \bigr\Vert . $$

  (11)

• 2. If $A_{1},A_{2},\ldots,A_{n}\in \mathbb{M}_{n}$ are positive, then

  $$ \Biggl\Vert \sum_{k=1}^{n}A_{k}A_{k}^{\ast } \Biggr\Vert \leq \bigl\Vert \bigl( \bigl\Vert A_{i}^{\ast }A_{j} \bigr\Vert \bigr) _{1\leq i,j\leq n} \bigr\Vert . $$

  (12)

We provide inequalities that are more general and sharper than inequalities (11) and (12).

Zou in [26] demonstrated the following generalization of arithmetic–geometric mean inequality: Let $A,X,B\in \mathbb{M}_{n}$ such that X is positive semidefinite Then

$$ 2\big|\!\big|\!\big|AXB^{\ast }\big|\!\big|\!\big|\leq \big|\!\big|\!\big|\bigl(A^{\ast }A+B^{ \ast }B \bigr)^{1/2}X\bigl(A^{\ast }A+B^{\ast }B\bigr)^{1/2} \big|\!\big|\!\big|. $$

(13)

Among our results, we obtain a generalization of inequality (13).

2 Singular value inequalities

The following lemmas are essential for supporting our conclusions. The first lemma is an immediate consequence of the min-max principle (see, e.g., [2, p. 75]). The second and third lemmas were shown in [11].

Lemma 1

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

$$ s_{j}(AXB)\leq \Vert A \Vert \Vert B \Vert s_{j}(X) $$

(14)

for $j=1,2,\ldots,n$.

Lemma 2

Let $A\in \mathbb{M}_{n}(\mathbb{C})$ and let f be a nonnegative increasing function on an interval I. Then

$$ f\bigl(s_{j}(A)\bigr)=s_{j}\bigl(f\bigl( \vert A \vert \bigr)\bigr) $$

for $j=1,2,\ldots,n$. If A is Hermitian and f is increasing on an interval I, then

$$ f\bigl(\lambda _{j}(A)\bigr)=\lambda _{j}\bigl(f(A)\bigr) $$

for $j=1,2,\ldots,n$.

Lemma 3

Let f be a monotone convex function on an interval I such that $0\in I$ and $f(0)\leq 0$, and let $A,X\in \mathbb{M}_{n}(\mathbb{C})$ such that A is Hermitian and X is a contraction. Then

$$ \lambda _{j}\bigl(f\bigl(X^{\ast }AX\bigr)\bigr)\leq \lambda _{j}\bigl(X^{\ast }f(A)X\bigr) $$

for $j=1,2,\ldots,n$.

The first result in this paper is now ready to be presented.

Theorem 1

Let $A,B,X,Y\in \mathbb{M}_{n}(\mathbb{C})$ such that $\max \{ \Vert X \Vert , \Vert Y \Vert \}\leq 1$, and let f be a nonnegative increasing convex function on $[ 0,\infty )$ satisfying $f(0)=0$. Then

$$ s_{j} \bigl( f \bigl( \bigl\vert XAB^{\ast }Y^{\ast } \bigr\vert \bigr) \bigr) \leq \max \bigl\{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \bigr\} s_{j} \bigl( f \bigl(AA^{\ast }\bigr) \oplus f\bigl(BB^{\ast }\bigr) \bigr) $$

(15)

for $j=1,2,\ldots,n$.

Proof

Consider the operator matrix $Q= [ ( XA ) ^{\ast }\ (YB)^{\ast } ] \in M_{n,2n}(\mathbb{C} )$.

$$P=Q^{\ast }Q=\left[\begin{array}{cc}XA{A}^{\ast }{X}^{\ast }& XA{B}^{\ast }{Y}^{\ast }\\ YB{A}^{\ast }{X}^{\ast }& YB{B}^{\ast }{Y}^{\ast }\end{array}\right]\ge 0$$

for any $A,B,X,Y\in M_{2n,2n}(\mathbb{C} )$.

By making use of inequality (4) and by letting

$$R=\left[\begin{array}{cc}A{A}^{\ast }& 0\\ 0& B{B}^{\ast }\end{array}\right]$$

gives

$$\begin{array}{rcl}{s}_j\left(f\left(\right|XA{B}^{\ast }{Y}^{\ast }\left|\right)\right)& =& f\left(s_j\left(XA{B}^{\ast }{Y}^{\ast }\right)\right)\\ & \le & f\left(s_j(XA{A}^{\ast }{X}^{\ast }\oplus YB{B}^{\ast }{Y}^{\ast })\right)\\ & =& f\left(s_j\left(\left[\begin{array}{cc}X& 0\\ 0& Y\end{array}\right]R\left[\begin{array}{cc}{X}^{\ast }& 0\\ 0& Y^{\ast }\end{array}\right]\right)\right)\\ & =& f\left({\lambda }_j\left(\left[\begin{array}{cc}X& 0\\ 0& Y\end{array}\right]R\left[\begin{array}{cc}{X}^{\ast }& 0\\ 0& Y^{\ast }\end{array}\right]\right)\right)\\ & =& {\lambda }_j\left(f\left(\left[\begin{array}{cc}X& 0\\ 0& Y\end{array}\right]R\left[\begin{array}{cc}{X}^{\ast }& 0\\ 0& Y^{\ast }\end{array}\right]\right)\right),\\ & & \text{(by Lemma 2)}\\ & \le & {\lambda }_j\left((\left[\begin{array}{cc}X& 0\\ 0& Y\end{array}\right]f(R)\left[\begin{array}{cc}{X}^{\ast }& 0\\ 0& Y^{\ast }\end{array}\right])\right),\\ & & \text{(by Lemma 3)}\\ & =& s_j\left((\left[\begin{array}{cc}X& 0\\ 0& Y\end{array}\right]f(R)\left[\begin{array}{cc}{X}^{\ast }& 0\\ 0& Y^{\ast }\end{array}\right])\right)\\ & \le & \parallel \left[\begin{array}{cc}X& 0\\ 0& Y\end{array}\right]\parallel \parallel \left[\begin{array}{cc}{X}^{\ast }& 0\\ 0& Y^{\ast }\end{array}\right]\parallel s_j(f(R)),\\ & & \text{(by Lemma 1)}\\ & =& {\parallel \left[\begin{array}{cc}X& 0\\ 0& Y\end{array}\right]\parallel }^2{s}_j(f(R))\\ & =& max\{{\parallel X\parallel }^2,{\parallel Y\parallel }^2\}{s}_j[f(R)]\\ & =& max\{{\parallel X\parallel }^2,{\parallel Y\parallel }^2\}{s}_j(f\left(A{A}^{\ast }\right)\oplus f\left(B{B}^{\ast }\right)).\end{array}$$

 □

Corollary 1

Let $A,B,X,Y\in \mathbb{M}_{n}(\mathbb{C})$ such that $\max \{ \Vert X \Vert , \Vert Y \Vert \}\leq 1$. Then, for $r\geq 1$,

$$ s_{j} \bigl( XAB^{\ast }Y^{\ast } \bigr) ^{r} \leq \max \bigl\{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \bigr\} s_{j} \bigl( \bigl(AA^{\ast } \bigr)^{r}\oplus \bigl(BB^{\ast }\bigr)^{r} \bigr) $$

(16)

and

$$ s_{j} \bigl( e^{ \vert XAB^{\ast }Y^{\ast } \vert }-I \bigr) \leq \max \bigl\{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \bigr\} s_{j} \bigl[ \bigl( e^{AA^{\ast }}-I \bigr) \oplus \bigl( e^{BB^{\ast }}-I \bigr) \bigr] $$

(17)

for $j=1,2,\ldots,n$.

Proof

Letting $f(t)=t^{r}$, $r\geq 1$, and $f(t)=e^{t}-1$ in Theorem 1 gives inequalities (16) and (17), respectively. □

By using Theorem 1, we here by present the following theorem.

Theorem 2

Let $A,B,C,X,Y\in \mathbb{M}_{n}(\mathbb{C})$ such that $P=\left[\begin{array}{cc}A& B\\ B^{\ast }& C\end{array}\right]\ge 0$, $\max \{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \} \leq 1$, and let f be a nonnegative increasing convex function on $[ 0,\infty )$ satisfying $f(0)=0$. Then

$$ s_{j} \bigl( f\bigl( \bigl\vert XBY^{\ast } \bigr\vert \bigr) \bigr) \leq \max \bigl\{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \bigr\} s_{j} \bigl( f(A)\oplus f(C) \bigr) $$

(18)

for $j=1,2,\ldots,n$.

Proof

Let $P=\left[\begin{array}{cc}A& B\\ B^{\ast }& C\end{array}\right]\ge 0$. Then there exists a matrix $Q= [ ( H ) ^{\ast } (L)^{\ast } ] \in M_{n,2n}(\mathbb{C} )$ such that $P=Q^{\ast }Q$ for some $H,L\in M_{n}(\mathbb{C} )$. Then $A=HH^{\ast }$, $C=LL^{\ast }$ and $B=HL^{\ast }$. Applying inequality (4) gives

$$\begin{aligned} s_{j} \bigl( f\bigl( \bigl\vert XBY^{\ast } \bigr\vert \bigr) \bigr) =&s_{j} \bigl( f\bigl( \bigl\vert XHL^{\ast }Y^{\ast } \bigr\vert \bigr) \bigr) \\ \leq &\max \bigl\{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \bigr\} s_{j} \bigl( f\bigl(HH^{\ast }\bigr) \oplus f\bigl(LL^{ \ast }\bigr) \bigr) , \quad \text{(by Theorem 1)} \\ =&\max \bigl\{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \bigr\} s_{j} \bigl( f ( A ) \oplus f ( C ) \bigr) . \end{aligned}$$

Inequality (18) has thus been proved. □

Remark 1

Letting $X=Y=I$ and $f(t)=t$ in inequality (18) gives inequality (4). In that sense, inequality (18) is certainly a generalization of inequality (4).

Corollary 2

Let $A,B,C,X,Y\in \mathbb{M}_{n}(\mathbb{C})$ such that $P=\left[\begin{array}{cc}A& B\\ B^{\ast }& C\end{array}\right]\ge 0$ and $\max \{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \} \leq 1$. Then, for $r\geq 1$,

$$ s_{j} \bigl( XBY^{\ast }\bigr)^{r} ) \leq \max \bigl\{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \bigr\} s_{j} \bigl( A^{r}\oplus C^{r} \bigr) $$

(19)

and

$$ s_{j}\bigl(e^{ \vert XBY^{\ast } \vert }-I\bigr)\leq \max \bigl\{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \bigr\} s_{j} \bigl( \bigl( e^{A}-I\bigr)\oplus \bigl(e^{C}-I\bigr) \bigr) $$

(20)

for $j=1,2,\ldots,n$.

Proof

Letting $f(t)=t^{r}$, $r\geq 1$, and $f(t)=e^{t}-1$ in Theorem 2 gives inequalities (19) and (20), respectively. □

Using the proper incites of Theorem 2 gives the following inequality, which is a generalization of inequality (1).

Theorem 3

Let $A,B,X,Y\in \mathbb{M}_{n}(\mathbb{C})$ such that A is self-adjoint, $B\geq 0$, $\pm A\leq B$, $\max \{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \} \leq 1$, and let f be a nonnegative increasing convex function on $[ 0,\infty )$ satisfying $f(0)=0$. Then

$$ s_{j} \bigl( f\bigl( \bigl\vert XAY^{\ast } \bigr\vert \bigr) \bigr) \leq \max \bigl\{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \bigr\} s_{j}\bigl(f(B)\oplus f(B)\bigr) $$

(21)

for $j=1,2,\ldots,n$.

Proof

Let $P=\left[\begin{array}{cc}B& A\\ A& B\end{array}\right]$. Since $\left[\begin{array}{cc}B& A\\ A& B\end{array}\right]$ is unitarily equivalent to $\left[\begin{array}{cc}B+A& 0\\ 0& B-A\end{array}\right]$ and since $\pm A\leq B$, it follows that P is a positive matrix. Applying inequality (18) to the operator matrix P gives inequality (21). □

Remark 2

Letting $f(t)=t$ and $X=Y=I$ in inequality (21) gives inequality (1). In that sense, inequality (21) is a generalization of inequality (1).

Corollary 3

Let $A,B,X,Y\in \mathbb{M}_{n}(\mathbb{C})$ such that A is self-adjoint, $B\geq 0$, $\pm A\leq B$, and $\max \{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \} \leq 1$. Then

$$ s_{j} \bigl( XAY^{\ast }\bigr)^{r} ) \leq \max \bigl\{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \bigr\} s_{j}\bigl(B^{r} \oplus B^{r}\bigr) $$

(22)

and

$$ s_{j}\bigl(e^{ \vert XAY^{\ast } \vert }-I\bigr)\leq \max \bigl\{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \bigr\} s_{j} \bigl( \bigl( e^{B}-I\bigr)\oplus \bigl(e^{B}-I\bigr) \bigr) $$

(23)

for $j=1,2,\ldots,n$.

Proof

Letting $f(t)=t^{r}$, $r\geq 1$, and $f(t)=e^{t}-1$ in Theorem 3 gives inequalities (22) and (23), respectively. □

The following lemma, which was proved in [12], is necessary to prove the next result.

Lemma 4

Let $A\in \mathbb{M}_{n}(\mathbb{C})$. Then

$$\left[\begin{array}{cc}|A|& \pm A^{\ast }\\ \pm A& |A^{\ast }|\end{array}\right]\ge 0.$$

(24)

Theorem 4

Let $A,B,X,Y\in \mathbb{M}_{n}(\mathbb{C})$ such that $\max \{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \} \leq 1$, and let f be a nonnegative increasing convex function on $[ 0,\infty )$ satisfying $f(0)=0$ and

$$ P=A \vert C \vert A^{\ast }+B \bigl\vert C^{\ast } \bigr\vert B^{\ast }. $$

Then

$$ s_{j} \bigl( f \bigl( \bigl\vert X\bigl(AC^{\ast }B^{\ast }+BCA^{\ast } \bigr)Y^{ \ast } \bigr\vert \bigr) \bigr) \leq \max \bigl\{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \bigr\} s_{j} \bigl( f(P)\oplus f(P) \bigr) $$

(25)

for $j=1,2,\ldots,n$.

Proof

Let $Z=\left[\begin{array}{cc}A& B\\ 0& 0\end{array}\right]$ and $Y=\left[\begin{array}{cc}|C|& \pm C^{\ast }\\ \pm C& |C^{\ast }|\end{array}\right]$. Then $Y\geq 0$ and

$$ZY{Z}^{\ast }=\left[\begin{array}{cc}A|C|A^{\ast }\pm BC{A}^{\ast }\pm A{C}^{\ast }{B}^{\ast }+B|C^{\ast }|B^{\ast }& 0\\ 0& 0\end{array}\right]\ge 0,$$

this implies that

$$ A \vert C \vert A^{\ast }+B \bigl\vert C^{\ast } \bigr\vert B^{\ast }\geq \pm \bigl(AC^{\ast }B^{\ast }+BCA^{\ast } \bigr). $$

(26)

Applying the conclusion of inequality (21) to the operator matrix $ZYZ^{\ast }$ gives

$$ s_{j} \bigl( f \bigl( \bigl\vert X\bigl(AC^{\ast }B^{\ast }+BCA^{\ast } \bigr)Y^{ \ast } \bigr\vert \bigr) \bigr) \leq \max \bigl\{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \bigr\} s_{j} \bigl( f(P)\oplus f(P) \bigr) $$

which is precisely (25). □

Remark 3

Substituting $f(t)=t$, $X,Y,C=I$, and $R=(AA^{\ast }+BB^{\ast })$ in Theorem 4 gives the following inequality, which is a generalization of inequality (2):

$$ s_{j}\bigl(f\bigl( \bigl\vert AB^{\ast }+BA^{\ast } \bigr\vert \bigr)\bigr)\leq s_{j}\bigl(f(R) \oplus f(R)\bigr) $$

(27)

for $j=1,2,\ldots,n$.

Remark 4

Letting $f(t)=t$ in inequality (27), we give inequality (2). In that sense, inequality (27) is certainly a generalization of inequality (2).

The following result is a direct consequence of Theorem 2.

Corollary 4

Let $A,B,X,Y\in \mathbb{M}_{n}(\mathbb{C})$ where $\max \{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \} \leq 1$, and let f be a nonnegative increasing convex function on $[ 0,\infty )$ satisfying $f(0)=0$ and $K=M\oplus N$, where

$$ M= \vert A \vert + \vert B \vert \quad \textit{and}\quad N= \bigl\vert A^{\ast } \bigr\vert + \bigl\vert B^{\ast } \bigr\vert . $$

Then

$$ s_{j} \bigl( f \bigl( \bigl\vert X(A+B)Y^{\ast } \bigr\vert \bigr) \bigr) \leq \max \bigl\{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \bigr\} s_{j}\bigl(f(M)\oplus f(N)\bigr) $$

(28)

for $j=1,2,\ldots,n$.

Proof

It was shown in [15] that the matrix

$$K=\left[\begin{array}{cc}|A|+|B|& A+B\\ A^{\ast }+B^{\ast }& |A^{\ast }|+|B^{\ast }|\end{array}\right]$$

is positive semidefinite. Now, inequality (28) is a direct consequence of Theorem 2. □

Remark 5

Substituting $X=Y=I$ in inequality (28) leads to inequality (5). In that sense, inequality (28) is certainly a generalization of inequality (5).

The following inequality is a generalization of inequality (6).

Theorem 5

Let $A,B,C,X,Y\in \mathbb{M}_{n}(\mathbb{C})$ such that $\left[\begin{array}{cc}A& B\\ B^{\ast }& C\end{array}\right]\ge 0$, $\max \{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \} \leq 1$, and let f be a nonnegative increasing convex function on $[ 0,\infty )$ satisfying $f(0)=0$. Then

$$2s_j\left(f\left(\right|XB{Y}^{\ast }\left|\right)\right)\le max\{{\parallel X\parallel }^2,{\parallel Y\parallel }^2\}{s}_j\left(f\left(\left[\begin{array}{cc}A& B\\ B^{\ast }& C\end{array}\right]\right)\right)$$

(29)

for $j=1,2,\ldots,n$.

Proof

Let $P=\left[\begin{array}{cc}A& B\\ B^{\ast }& C\end{array}\right]\ge 0$. Then there exist $H,L\in \mathbb{M}_{n}(\mathbb{C})$ such that $P=Q^{\ast }Q$, where $Q= [ ( H ) ^{\ast } (L)^{\ast } ]$. This means that $A=HH^{\ast }$, $C=LL^{\ast }$, and $B=HL^{\ast }$.

$$\begin{aligned} 2s_{j}\bigl(f \bigl( \bigl\vert XBY^{\ast } \bigr\vert \bigr) \bigr) =&2s_{j}\bigl(f \bigl( \bigl\vert XHL^{\ast }Y^{\ast } \bigr\vert \bigr) \bigr) \\ =&2f\bigl(s_{j} \bigl( XHL^{\ast }Y^{\ast } \bigr) \bigr) \\ \leq &f \bigl( s_{j}(W) \bigr) \end{aligned}$$

for $j=1,2,\ldots,n$, where $W=H^{\ast }X^{\ast }XH+L^{\ast }Y^{\ast }YL$. Now, letting $Z=\left[\begin{array}{cc}{X}^{\ast }X& 0\\ 0& Y^{\ast }Y\end{array}\right]$ gives

$$\begin{array}{rcl}f(s_j(W))& =& f\left(s_j\left(\left[\begin{array}{cc}{H}^{\ast }& L^{\ast }\\ 0& 0\end{array}\right]Z\left[\begin{array}{cc}H& 0\\ L& 0\end{array}\right]\right)\right)\\ & =& f\left({\lambda }_j\left(\left[\begin{array}{cc}{H}^{\ast }& L^{\ast }\\ 0& 0\end{array}\right]Z\left[\begin{array}{cc}H& 0\\ L& 0\end{array}\right]\right)\right)\\ & =& f\left({\lambda }_j\left(Z^{1/2}\left[\begin{array}{cc}H& 0\\ L& 0\end{array}\right]\left[\begin{array}{cc}{H}^{\ast }& L^{\ast }\\ 0& 0\end{array}\right]Z^{1/2}\right)\right)\\ & =& f\left({\lambda }_j\left(Z^{1/2}\left[\begin{array}{cc}A& B\\ B^{\ast }& C\end{array}\right]Z^{1/2}\right)\right)\\ & =& {\lambda }_j\left(f\left(Z^{1/2}\left[\begin{array}{cc}A& B\\ B^{\ast }& C\end{array}\right]Z^{1/2}\right)\right)\\ & \le & {\lambda }_j\left(Z^{1/2}f\left(\left[\begin{array}{cc}A& B\\ B^{\ast }& C\end{array}\right]\right)Z^{1/2}\right)\\ & =& s_j\left(Z^{1/2}f\left(\left[\begin{array}{cc}A& B\\ B^{\ast }& C\end{array}\right]\right)Z^{1/2}\right)\\ & \le & \parallel Z\parallel s_j\left(f\left(\left[\begin{array}{cc}A& B\\ B^{\ast }& C\end{array}\right]\right)\right)\\ & =& max\{{\parallel X\parallel }^2,{\parallel Y\parallel }^2\}{s}_j\left(f\left(\left[\begin{array}{cc}A& B\\ B^{\ast }& C\end{array}\right]\right)\right).\end{array}$$

Inequality (29) has thus been substantiated. □

Remark 6

Letting $f(t)=t$ and $X=Y=I$ in Theorem 5 gives inequality (6). In that sense inequality (29) is a generalization of inequality (6).

At this stage of our discussion, we provide a considerable generalization of inequality (3).

Theorem 6

Let $A,B,X,Y\in \mathbb{M}_{n}(\mathbb{C})$ such that A is self-adjoint, $B\geq 0$, $\pm A\leq B$, $\max \{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \} \leq 1$, and let f be a nonnegative increasing convex function on $[ 0,\infty )$ satisfying $f(0)=0$. Then

$$ 2s_{j}\bigl(f \bigl( \bigl\vert XAY^{\ast } \bigr\vert \bigr) \bigr)\leq \max \bigl\{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \bigr\} s_{j}\bigl(f \bigl( (B+A)\oplus f(B-A) \bigr) \bigr) $$

(30)

for $j=1,2,\ldots,n$.

Proof

Since $\pm A\leq B$, it follows that

$$P=\left[\begin{array}{cc}B+A& 0\\ 0& B-A\end{array}\right]\ge 0.$$

If $U=\frac{1}{\sqrt{2}}\left[\begin{array}{cc}I& -I\\ I& I\end{array}\right]$, then

$$Q=\left[\begin{array}{cc}B& A\\ A& B\end{array}\right]=UP{U}^{\ast },$$

which is equivalent to stating that $Q\geq 0$. Now applying Theorem 5 to the operator matrix Q leads to

$$\begin{array}{rcl}2s_j\left(f\left(\right|XA{Y}^{\ast }\left|\right)\right)& \le & max\{{\parallel X\parallel }^2,{\parallel Y\parallel }^2\}{s}_j\left(f\left(\left[\begin{array}{cc}B& A\\ A& B\end{array}\right]\right)\right)\\ & =& max\{{\parallel X\parallel }^2,{\parallel Y\parallel }^2\}{s}_j\left(f\left(\left[\begin{array}{cc}B+A& 0\\ 0& B-A\end{array}\right]\right)\right)\\ & =& max\{{\parallel X\parallel }^2,{\parallel Y\parallel }^2\}{s}_j\left(f((B+A)\oplus f(B-A))\right).\end{array}$$

Inequality (30) has thus been substantiated. □

Corollary 5

Let $A,B,X,Y\in \mathbb{M}_{n}(\mathbb{C})$ such that A is self-adjoint, $B\geq 0$, $\pm A\leq B$, and $\max \{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \} \leq 1$. Then

$$ 2s_{j}\bigl(XAY^{\ast }\bigr)^{r}\leq \max \bigl\{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \bigr\} s_{j}\bigl((B+A)^{r}\oplus (B-A)^{r}\bigr) $$

(31)

and

$$ 2s_{j}\bigl(e^{ \vert XAY^{\ast } \vert }-I\bigr)\leq \max \bigl\{ \Vert X \Vert ^{2}, \Vert Y \Vert ^{2} \bigr\} s_{j} \bigl( \bigl(e^{B+A}-I\bigr)\oplus \bigl( e^{B-A}-I \bigr) \bigr) $$

(32)

for $j=1,2,\ldots,n$.

Proof

Letting $f(t)=t^{r}$, $r\geq 1$, and $f(t)=e^{t}-1$ in Theorem 6 gives inequalities (31) and (32), respectively. □

3 Norm inequalities

We begin this section by the following lemmas that are essential in our study. The first lemma is folk lemma, the second lemma was introduced in [22], and the third lemma was obtained in [3].

Lemma 5

Let $A,B\in \mathbb{M}_{n}$, then

$$\left|\left|\left|\left[\begin{array}{cc}A& B\\ B& A\end{array}\right]\right|\right|\right|=\left|\left|\left|\left[\begin{array}{cc}A+B& 0\\ 0& A-B\end{array}\right]\right|\right|\right|.$$

(33)

Lemma 6

Let $A,B\in \mathbb{M}_{n}$, where AB is Hermitian. Then

$$ |\!|\!|AB|\!|\!|\leq \big|\!\big|\!\big|\operatorname{Re}(BA)\big|\!\big|\!\big|. $$

(34)

Let $A,B,X\in B(H)$ such that X is positive. Then

$$ 2\big|\!\big|\!\big|AXB^{\ast }\big|\!\big|\!\big|\leq \big|\!\big|\!\big|X^{1/2} \vert A \vert ^{2}X^{1/2}+X^{1/2} \vert B \vert ^{2}X^{1/2}\big|\!\big|\!\big|. $$

(35)

The following theorem can be obtained from norm inequality (13).

Theorem 7

Let $A_{i},B_{i},X_{i}\in \mathbb{M}_{n}$, $i=1,2,\ldots,n$, such that every $X_{i}$ is positive,

$$C=\left[\begin{array}{ccc}{A}_1^{\ast }{A}_1+B_1^{\ast }{B}_1& \cdots & A_1^{\ast }{A}_n+B_1^{\ast }{B}_n\\ A_2^{\ast }{A}_1+B_2^{\ast }{B}_1& \cdots & A_2^{\ast }{A}_n+B_2^{\ast }{B}_n\\ ⋮& \ddots & ⋮\\ A_n^{\ast }{A}_1+B_n^{\ast }{B}_1& \cdots & A_n^{\ast }{A}_n+B_n^{\ast }{B}_n\end{array}\right],$$

and

$$X=\left[\begin{array}{cccc}{X}_1& 0& \dots & 0\\ 0& X_2& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & X_n\end{array}\right],$$

then

$$ 2\Bigg|\!\Bigg|\!\Bigg|\Biggl( \sum_{i=1}^{n}A_{i}X_{i}B_{i}^{ \ast } \Biggr) \Bigg|\!\Bigg|\!\Bigg|\leq \big|\!\big|\!\big|C^{1/2}XC^{1/2}\big|\!\big|\!\big|. $$

(36)

Proof

On $\oplus \mathbb{M}_{n}$, define

$$A=\left[\begin{array}{cccc}{A}_1& A_2& \dots & A_n\\ 0& 0& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & 0\end{array}\right],B=\left[\begin{array}{cccc}{B}_1& B_2& \dots & B_n\\ 0& 0& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & 0\end{array}\right].$$

Then

$$AX{B}^{\ast }=\left[\begin{array}{cccc}{\sum }_{i=1}^n{A}_i{X}_i{B}_i^{\ast }& 0& \dots & 0\\ 0& 0& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & 0\end{array}\right],$$

and

$$\begin{array}{rcl}{X}^{1/2}|A{|}^2{X}^{1/2}+X^{1/2}|B{|}^2{X}^{1/2}& =& X^{1/2}(|A{|}^2+|B{|}^2)X^{1/2}\\ & =& X^{1/2}C{X}^{1/2}\\ & =& \left[\begin{array}{cccc}{V}_{11}& V_{12}& \dots & V_{1n}\\ V_{21}& V_{22}& \cdots & V_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ V_{n1}& V_{n2}& \cdots & V_{nn}\end{array}\right],\end{array}$$

where

$$ V_{i,j}=X_{i}^{1/2}A_{i}^{\ast }A_{j}X_{j}^{1/2}+X_{i}^{1/2}B_{i}^{ \ast }B_{j}X_{j}^{1/2}. $$

Applying inequality (35) gives

$$\begin{aligned} 2\Bigg|\!\Bigg|\!\Bigg|\Biggl( \sum_{i=1}^{n}A_{i}X_{i}B_{i}^{ \ast } \Biggr) \Bigg|\!\Bigg|\!\Bigg| \leq &\big|\!\big|\!\big|X^{1/2}CX^{1/2}\big|\!\big|\!\big|\\ =&\big|\!\big|\!\big|X^{1/2}C^{1/2}C^{1/2}X^{1/2} \big|\!\big|\!\big|\\ =&\big|\!\big|\!\big|C^{1/2}XC^{1/2}\big|\!\big|\!\big|, \\ &{}\big(\text{by applying inequality (8)}\big). \end{aligned}$$

Thus, inequality (36) has been substantiated. □

Remark 7

Inequality (36) is a general norm inequality that involves inequality (13). To see this, substitute $A_{i}=X_{i}=B_{i}=0$ for $i=2,\ldots,n$ in inequality (36), which leads to inequality (13).

Theorem 8

Let $A_{i},X_{i},B_{i}\in \mathbb{M}_{n}$, $i=1,2,\ldots,n$. Then

$$ 2\Bigg|\!\Bigg|\!\Bigg|\sum_{i=1}^{n}A_{i}X_{i}B_{i}^{ \ast } \Bigg|\!\Bigg|\!\Bigg|\leq \big|\!\big|\!\big|A^{\ast }AX+XB^{\ast }B\big|\!\big|\!\big|, $$

(37)

where

$$A=\left[\begin{array}{cccc}{A}_1& A_2& \dots & A_n\\ 0& 0& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & 0\end{array}\right],B=\left[\begin{array}{cccc}{B}_1& B_2& \dots & B_n\\ 0& 0& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & 0\end{array}\right]$$

and

$$X=\left[\begin{array}{cccc}{X}_1& 0& \cdots & 0\\ 0& X_2& \cdots & ⋮\\ ⋮& ⋮& \ddots & 0\\ 0& \cdots & 0& X_n\end{array}\right].$$

Proof

Replacing the operator matrices A, B, and X in inequality (9) gives inequality (37). □

The next special case of Theorem 8 was proved by Bhatia and Davis in [13].

Corollary 6

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

$$ 2\big|\!\big|\!\big|AXB^{\ast }\big|\!\big|\!\big|\leq \big|\!\big|\!\big|A^{\ast }AX+XB^{ \ast }B \big|\!\big|\!\big|. $$

(38)

Proof

Inequality (38) follows from inequality (37) by substituting $A_{i}=B_{i}=X_{i}=0$ for $i=2,3,\ldots,n$. □

Another conclusion of Theorem 8 is a generalization of arithmetic–geometric mean inequality for unitarily invariant norms.

Corollary 7

Let $A_{i},B_{i}\in \mathbb{M}_{n}$, $i=1,2,\ldots,n$, such that

$$A=\left[\begin{array}{cccc}{A}_1& A_2& \dots & A_n\\ 0& 0& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & 0\end{array}\right],B=\left[\begin{array}{cccc}{B}_1& B_2& \dots & B_n\\ 0& 0& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & 0\end{array}\right].$$

Then

$$ 2\Bigg|\!\Bigg|\!\Bigg|\Biggl( \sum_{i=1}^{n}A_{i}B_{i}^{ \ast } \Biggr) \Bigg|\!\Bigg|\!\Bigg|\leq \big|\!\big|\!\big|A^{\ast }A+B^{\ast }B\big|\!\big|\!\big|. $$

(39)

Proof

Substituting $X=I$ in inequality (37) gives inequality (39). □

Remark 8

Letting $A_{i}=B_{i}=0$ for $i=2,3,\ldots,n$ in inequality (39) gives

$$ 2\big|\!\big|\!\big|A_{1}B_{1}^{\ast }\big|\!\big|\!\big|\leq \big|\!\big|\!\big|A_{1}^{ \ast }A_{1}+B_{1}^{\ast }B_{1} \big|\!\big|\!\big|, $$

which is the arithmetic–geometric mean inequality for unitarily invariant norms.

The following inequality is a generalization and more general than inequality (24).

Corollary 8

Let $A_{i}\in \mathbb{M}_{n}$, $i=1,2,\ldots,n$, where

$$A=\left[\begin{array}{cccc}{A}_1& A_2& \dots & A_n\\ 0& 0& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & 0\end{array}\right].$$

Then

$$ \Bigg|\!\Bigg|\!\Bigg|\Biggl( \sum_{i=1}^{n}A_{i}A_{i}^{ \ast } \Biggr) \Bigg|\!\Bigg|\!\Bigg|\leq \big|\!\big|\!\big|A^{\ast }A\big|\!\big|\!\big|. $$

(40)

Proof

Substituting $B_{i}=A_{i}$ in inequality (39) gives inequality (40). □

Remark 9

By making use of inequality (10), we note that when we specify inequality (40) to the usual spectral norm, it is sharper than inequality (12).

The following inequality is a generalization and more general than inequality (11).

Corollary 9

Let $A_{i}\in \mathbb{M}_{n}$, $i=1,2,\ldots,n$. Then

$$ \Bigg|\!\Bigg|\!\Bigg|\sum_{i=1}^{n}A_{i} \Bigg|\!\Bigg|\!\Bigg|\leq \big|\!\big|\!\big|\bigl( A_{i}^{1/2}A_{j}^{1/2} \bigr) _{1\leq i,j\leq n} \big|\!\big|\!\big|. $$

(41)

Proof

Replacing A by $A^{1/2}$ in inequality (40) gives inequality (41). □

Remark 10

By using inequality (27), we note that when we specify inequality (41) to the spectral norm, it is sharper than inequality (21).

Because of the large number of papers that discuss unitarily invariant norms for $2\times 2$ operator matrices, we specialize inequality (37) for $n=2$. This special case contains several remarkable inequalities.

Corollary 10

Let $A_{i},X_{i},B_{i}\in \mathbb{M}_{n}$, $i=1,2$. Then

$$ 2\big|\!\big|\!\big|A_{1}X_{1}B_{1}^{\ast }+A_{2}X_{2}B_{2}^{ \ast } \big|\!\big|\!\big|\leq |\!|\!|Z|\!|\!|, $$

(42)

$where$

$$Z=\left[\begin{array}{cc}{A}_1^{\ast }{A}_1{X}_1+X_1{B}_1^{\ast }{B}_1& A_1^{\ast }{A}_2{X}_2+X_1{B}_1^{\ast }{B}_2\\ A_2^{\ast }{A}_1{X}_1+X_2{B}_2^{\ast }{B}_1& A_2^{\ast }{A}_2{X}_2+X_2{B}_2^{\ast }{B}_2\end{array}\right].$$

Proof

Substituting $A_{i}=X_{i}=B_{i}=0$ for $i=3,4,\ldots,n$ in inequality (37) gives inequality (42). □

By making use of inequality (42), we provide the following inequality.

Corollary 11

Let $A_{i},B_{i}\in \mathbb{M}_{n}$, $i=1,2$. Then

$$ 2\big|\!\big|\!\big|A_{1}B_{1}^{\ast }+A_{2}B_{2}^{ \ast } \big|\!\big|\!\big|\leq |\!|\!|L\oplus M|\!|\!|, $$

(43)

where

$$ L=A_{1}^{\ast }A_{1}+B_{1}^{\ast }B_{1}+ \bigl\vert A_{2}^{\ast }A_{1}+B_{2}^{ \ast }B_{1} \bigr\vert $$

and

$$ M=A_{2}^{\ast }A_{2}+B_{2}^{\ast }B_{2}+ \bigl\vert A_{1}^{\ast }A_{2}+B_{1}^{ \ast }B_{2} \bigr\vert . $$

Proof

Throughout the proof of this theorem, let

$$\begin{array}{c}S=\left[\begin{array}{cc}{A}_1^{\ast }{A}_1+B_1^{\ast }{B}_1& 0\\ 0& A_2^{\ast }{A}_2+B_2^{\ast }{B}_2\end{array}\right],\hfill \\ T=\left[\begin{array}{cc}0& A_1^{\ast }{A}_2+B_1^{\ast }{B}_2\\ A_2^{\ast }{A}_1+B_2^{\ast }{B}_1& 0\end{array}\right].\hfill \end{array}$$

Substituting $X_{1}=X_{2}=I$ in inequality (42) gives

$$\begin{array}{rcl}2\left|\right||A_1{B}_1^{\ast }+A_2{B}_2^{\ast }|\left|\right|& \le & \left|\left|\left|\left[\begin{array}{cc}{A}_1^{\ast }{A}_1+B_1^{\ast }{B}_1& A_1^{\ast }{A}_2+B_1^{\ast }{B}_2\\ A_2^{\ast }{A}_1+B_2^{\ast }{B}_1& A_2^{\ast }{A}_2+B_2^{\ast }{B}_2\end{array}\right]\right|\right|\right|\\ & =& \left|\right||(S+T)|\left|\right|\\ & =& \left|\right|\left|(|S+T|)\right|\left|\right|\\ & \le & \left|\right|\left|(|S|+|T|)\right|\left|\right|\\ & =& \left|\left||S+\left[\begin{array}{cc}|A_2^{\ast }{A}_1+B_2^{\ast }{B}_1|& 0\\ 0& |A_1^{\ast }{A}_2+B_1^{\ast }{B}_2|\end{array}\right]|\right|\right|\\ & =& \left|\left|\left|\left[\begin{array}{cc}L& 0\\ 0& M\end{array}\right]\right|\right|\right|.\end{array}$$

Inequality (43) has thus been substantiated. □

In turn, inequality (43) gives us the following finding.

Corollary 12

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

$$ 2\big|\!\big|\!\big|AB^{\ast }+BA^{\ast }\big|\!\big|\!\big|\leq |\!|\!|X\oplus X|\!|\!|, $$

(44)

where

$$ X=A^{\ast }A+B^{\ast }B+ \bigl\vert B^{\ast }A+A^{\ast }B \bigr\vert . $$

Proof

Substituting $A_{1}=B_{2}=A$ and $A_{2}=B_{1}=B$ in inequality (43) gives inequality (44). □

Another attractive special case of inequality (43) is the following result, which was shown in [21].

Corollary 13

Let $A,B\in \mathbb{M}_{n}$ be positive. Then

$$ |\!|\!|A+B|\!|\!|\leq \big|\!\big|\!\big|\bigl( A+ \bigl\vert B^{1/2}A^{1/2} \bigr\vert \bigr) \oplus \bigl( B+ \bigl\vert A^{1/2}B^{1/2} \bigr\vert \bigr) \big|\!\big|\!\big|. $$

(45)

Proof

Substituting $A_{1}=B_{1}=A^{1/2}$ and $A_{2}=B_{2}=B^{1/2}$ in inequality (43) gives inequality (45). □

If we look at inequality (42) from another side, we obtain the following inequality, which was proven in [20].

Corollary 14

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

$$ 2\big|\!\big|\!\big|AB^{\ast }+BA^{\ast }\big|\!\big|\!\big|\leq \big|\!\big|\!\big|(A+B)^{\ast }(A+B) \oplus (A-B)^{\ast }(A-B)\big|\!\big|\!\big|. $$

(46)

Proof

Throughout this proof, let

$$ P=A^{\ast }A+B^{\ast }B+A^{\ast }B+B^{\ast }A $$

and

$$ Q=A^{\ast }A+B^{\ast }B-A^{\ast }B-B^{\ast }A. $$

Substituting $A_{1}=B_{2}=A$ and $A_{2}=B_{1}=B$ in inequality (42) gives

$$\begin{array}{rcl}2\left|\right||A{B}^{\ast }+B{A}^{\ast }|\left|\right|& \le & \left|\left|\left|\left[\begin{array}{cc}{A}^{\ast }A+B^{\ast }B& A^{\ast }B+B^{\ast }A\\ B^{\ast }A+A^{\ast }B& A^{\ast }A+B^{\ast }B\end{array}\right]\right|\right|\right|\\ & =& \left|\left|\left|\left[\begin{array}{cc}P& 0\\ 0& Q\end{array}\right]\right|\right|\right|\\ & & \left(\text{by applying equation (33)}\right)\\ & =& \left|\left|\left|\left[\begin{array}{cc}{(A+B)}^{\ast }(A+B)& 0\\ 0& {(A-B)}^{\ast }(A-B)\end{array}\right]\right|\right|\right|,\end{array}$$

which is precisely inequality (46). □

The next inequality is a special case of inequality (42), which was shown in [18].

Corollary 15

Let $X_{1}, X_{2}\in \mathbb{M}_{n}$. Then

$$ |\!|\!|X_{1}+X_{2}|\!|\!|\leq 2|\!|\!|X_{1} \oplus X_{2}|\!|\!|. $$

(47)

Proof

Substituting $A_{1}=A_{2}=B_{1}=B_{2}=I$ in inequality (42) gives

$$\begin{array}{rcl}2|||X_1+X_2|||& \le & \left|\left|\left|\left[\begin{array}{cc}2X_1& X_1+X_2\\ X_1+X_2& 2X_2\end{array}\right]\right|\right|\right|\\ & \le & 2\left|\left|\left|\left[\begin{array}{cc}{X}_1& 0\\ 0& X_2\end{array}\right]\right|\right|\right|+\left|\left|\left|\left[\begin{array}{cc}0& X_1+X_2\\ X_1+X_2& 0\end{array}\right]\right|\right|\right|\\ & =& 2|||X_1\oplus X_2|||+|||X_1+X_2|||.\end{array}$$

This is equivalent to saying that

$$ |\!|\!|X_{1}+X_{2}|\!|\!|\leq 2|\!|\!|X_{1} \oplus X_{2}|\!|\!|. $$

 □

Another special case of inequality (42) has been established by using a completely different technique in [25].

Corollary 16

Let $A, B \in \mathbb{M}_{n}$ be positive. Then

$$ |\!|\!|A-B|\!|\!|\leq |\!|\!|A\oplus B |\!|\!|. $$

(48)

Proof

Letting $A_{1}=B_{1}=A^{1/2}$, $A_{2}=-B_{2}=B^{1/2}$ in inequality (42) gives

$$\begin{array}{rcl}2|||A-B|||& \le & \left|\left|\left|\left[\begin{array}{cc}2A& 0\\ 0& 2B\end{array}\right]\right|\right|\right|\\ & =& 2|||A\oplus B|||,\end{array}$$

which is inequality (48). □