Some inequalities related to $2\times 2$ block sector partial transpose matrices

Abstract

In this article, two inequalities related to $2\times 2$ block sector partial transpose matrices are proved, and we also present a unitarily invariant norm inequality for the Hua matrix which is sharper than an existing result.

1 Introduction

We denote by $\mathbb{M}_{n}$ the set of $n\times n$ complex matrices. $\mathbb{M}_{n}(\mathbb{M}_{k})$ is the set of $n\times n$ block matrices with each block in $\mathbb{M}_{k}$. The $n\times n$ identity matrix is denoted by $I_{n}$. We use $\|\cdot \|$ for an arbitrary unitarily invariant norm. A positive semidefinite matrix A will be expressed as $A\geq 0$. Likewise, we write $A>0$ to refer that A is a positive definite matrix. The singular values of A, denoted by $s_{1}(A), s_{2}(A),\ldots, s_{n}(A)$, are the eigenvalues of the positive semidefinite matrix $|A|=(A^{*}A)^{1/2}$, arranged in decreasing order and repeated according to multiplicity as $s_{1}(A)\geq s_{2}(A)\geq \cdots \geq s_{n}(A)$. When A is Hermitian, we enumerate eigenvalues of A in nonincreasing order $\sigma _{1}(A)\geq \sigma _{2}(A)\geq \cdots \geq \sigma _{n}(A)$. Recall that $C\in \mathbb{M}_{m\times n}$ is (strictly) contractive if ($I_{n}>C^{*}C$) ${ I_{n} \geq C^{*}C}$. The geometric mean of two positive definite matrices $A, B\in \mathbb{M}_{n}$, denoted by $A\sharp B$, is the positive definite solution of the Riccati equation $XB^{-1}X=A$ and has the explicit expression $A\sharp B=A^{\frac{1}{2}}(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^{ \frac{1}{2}}A^{\frac{1}{2}}$. More details on the matrix geometric mean can be found in [2, Chap. 4].

The numerical range of $A\in \mathbb{M}_{n}$ is defined by

$$\begin{aligned} W(A)=\bigl\{ x^{*}Ax|x\in \mathbb{C}^{n}, x^{*}x=1 \bigr\} . \end{aligned}$$

For basic properties of numerical range, see [5]. Also, we define a sector on the complex plane

$$\begin{aligned} S_{\alpha }=\bigl\{ z\in \mathbb{C }|\mathfrak{R}z\geq 0, \vert \mathfrak{I}z \vert \leq (\mathfrak{R}z)\tan (\alpha )\bigr\} ,\quad \alpha \in\biggl[0, \frac{\pi }{2}\biggr). \end{aligned}$$

Actually, the class of matrices T with $W(T)\subseteq S_{\alpha }$ and the class of T with positive definite real part (i.e. accretive matrices) are both called sector matrices. Sector matrices have been the subject of a number of recent papers [3, 8, 14].

A matrix $H=(H_{ij})_{i,j=1}^{n}\in \mathbb{M}_{n}(\mathbb{M}_{k})$ is said to be positive partial transpose (i.e. PPT) if H is positive semidefinite and its partial transpose $H^{\tau }=(H_{ji})_{j,i=1}^{n}$ is also positive semidefinite. Inspired by PPT, Kuai [6] defined a new conception called sectorial partial transpose (i.e. SPT). That is, if $W(A)\subseteq S_{\alpha }$ for $A=(A_{ij})_{i,j=1}^{n}\in \mathbb{M}_{n}(\mathbb{M}_{k})$, then $W(A^{\tau })\subseteq S_{\alpha }$. Thus, it is natural to extend the results for PPT matrices to SPT matrices.

Hiroshima [4, Theorem 1] proved the following result.

Theorem 1.1

Let$H=\left(\begin{array}{cc}A& X\\ X^{\ast }& B\end{array}\right)\in {\mathbb{M}}_2({\mathbb{M}}_n)$be PPT. Then

$$\begin{aligned} \Vert H \Vert \leq \Vert A+B \Vert . \end{aligned}$$

(1)

As the application of Theorem 1.1, Lin and Hiroshima [10, Theorem 3.3] presented a relation between the norm of diagonal blocks of the Hua matrix, e.g., [12] and the norm of its off diagonal blocks.

Theorem 1.2

If the Hua matrix is given by

$$\begin{aligned} H: = \begin{pmatrix} (I-A^{*}A)^{-1} & (I-B^{*}A)^{-1} \\ (I-A^{*}B)^{-1} & (I-B^{*}B)^{-1} \end{pmatrix}, \end{aligned}$$

where$A,B \in \mathbb{M}_{m\times n}$are strictly contractive, then

$$\begin{aligned} 2 \bigl\Vert \bigl(I-A^{*}B\bigr)^{-1} \bigr\Vert \leq \bigl\Vert \bigl(I-A^{*}A\bigr)^{-1}+ \bigl(I-B^{*}B\bigr)^{-1} \bigr\Vert \end{aligned}$$

(2)

for any unitarily invariant norm.

Actually, it was only recently observed that H is PPT; see [1].

Lin [7] obtained a singular value inequality for PPT matrices related to a linear map.

Theorem 1.3

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

$$ M= \begin{pmatrix} A & X \\ X^{*} & B \end{pmatrix} $$

is PPT, then for the linear map$\varPhi: C\rightarrow C+\operatorname{Tr}(C)I$,

$$\begin{aligned} s_{j}\bigl(\varPhi (X)\bigr) \leq s_{j}\bigl(\varPhi (A \sharp B)\bigr),\quad j=1,\ldots, n. \end{aligned}$$

In this paper, we extend Theorem 1.1 and Theorem 1.3 to SPT matrices and show a stronger inequality than (2).

2 Main results

We start with some lemmas. The first three lemmas are quite standard in matrix analysis.

Lemma 2.1

([13, p. 63])

If$H\in \mathbb{M}_{n}$, then

$$\begin{aligned} \sigma _{j}(\operatorname {Re}H) \leq s_{j} (H),\quad j=1, \ldots,n. \end{aligned}$$

(3)

Lemma 2.2

([11, Theorem 1])

Let$H=\left(\begin{array}{cc}A& X\\ X^{\ast }& B\end{array}\right)\in {\mathbb{M}}_{m+n}$be positive semidefinite with$A\in \mathbb{M}_{m}, B\in \mathbb{M}_{n}$. Then

$$\begin{aligned} 2s_{j}(X) \leq s_{j}(H),\quad j=1,\ldots, \min \{m, n\}. \end{aligned}$$

(4)

Lemma 2.3

([2, p.106])

Let$A, B\in \mathbb{M}_{n}$be positive definite matrices. Then, for all$X\in M_{n}$,

$$\begin{aligned} X^{*}(A\sharp B)X \leq \bigl(X^{*}AX\bigr) \sharp \bigl(X^{*}BX\bigr). \end{aligned}$$

(5)

The next lemma is due to Zhang [14, Lemma 3.1].

Lemma 2.4

Let$A\in \mathbb{M}_{n}$have$W(A)\subseteq S_{\alpha }$for some$\alpha \in [0, \frac{\pi }{2})$. Then

$$\begin{aligned} \Vert A \Vert \leq \sec (\alpha ) \Vert \operatorname {Re}A \Vert \end{aligned}$$

(6)

for any unitarily invariant norm.

The following result about geometric mean has been proved by Lin and Sun [9].

Lemma 2.5

Let$A, B\in \mathbb{M}_{n}$be matrices with positive semidefinite real part. Then

$$\begin{aligned} (\operatorname {Re}A)\sharp (\operatorname {Re}B) \leq \operatorname {Re}(A\sharp B). \end{aligned}$$

(7)

Now we are ready to present our results. The first theorem is an extension of Theorem 1.1.

Theorem 2.6

Let$H_{11}, H_{12}, H_{21}, H_{22}\in \mathbb{M}_{n}$. If$H=\left(\begin{array}{cc}{H}_{11}& H_{12}\\ H_{21}& H_{22}\end{array}\right)$is SPT, then

$$\begin{aligned} \Vert H \Vert \leq \sec (\alpha ) \Vert H_{11}+H_{22} \Vert \end{aligned}$$

for any unitarily invariant norm.

Proof

Since H is a sector partial transpose matrix, then we know that

$$\operatorname {Re}H= \begin{pmatrix} \frac{ H_{11}+H_{11}^{*}}{2} & \frac{ H_{12}+H_{21}^{*}}{2} \\ \frac{ H_{21}+H_{12}^{*}}{2} &\frac{ H_{22}+H_{22}^{*}}{2} \end{pmatrix} $$

is PPT.

So by (6) we have

$$\begin{aligned} \Vert H \Vert &\leq \sec (\alpha ) \Vert \operatorname {Re}H \Vert \\ &\leq \sec (\alpha ) \Vert \operatorname {Re}H_{11}+ \operatorname {Re}H_{22} \Vert \quad\bigl(\text{by (1)}\bigr) \\ &\leq \sec (\alpha ) \Vert H_{11}+H_{22} \Vert . \end{aligned}$$

 □

Remark 2.7

When $H_{12}=H_{21}^{*}$ and $\alpha =0$, then H is PPT in Theorem 2.6. Thus, our result is Hiroshima’s inequality (1).

Next will give a stronger inequality than Theorem 1.2.

Theorem 2.8

Let the Hua matrix be given by

$$\begin{aligned} H:= \begin{pmatrix} (I-A^{*}A)^{-1} & (I-B^{*}A)^{-1} \\ (I-A^{*}B)^{-1} & (I-B^{*}B)^{-1} \end{pmatrix}, \end{aligned}$$

where$A,B \in \mathbb{M}_{m\times n}$are strictly contractive. Then

$$\begin{aligned} \bigl\Vert \bigl(I-A^{*}B\bigr)^{-1} \bigr\Vert \leq \bigl\Vert \bigl(I-A^{*}A\bigr)^{-1}\sharp \bigl(I-B^{*}B\bigr)^{-1} \bigr\Vert \end{aligned}$$

for any unitarily invariant norm.

Proof

Since H is PPT, then

$$\begin{aligned} \begin{pmatrix} (I-A^{*}A)^{-1} & (I-B^{*}A)^{-1} \\ (I-A^{*}B)^{-1} & (I-B^{*}B)^{-1} \end{pmatrix},\qquad \begin{pmatrix} (I-A^{*}A)^{-1} & (I-A^{*}B)^{-1} \\ (I-B^{*}A)^{-1} & (I-B^{*}B)^{-1} \end{pmatrix} \end{aligned}$$

are both positive semidefinite matrices.

Hence,

$$\begin{aligned} \bigl(I-B^{*}B\bigr)^{-1} \geq \bigl(I-A^{*}B\bigr)^{-1}\bigl(I-A^{*}A\bigr) \bigl(I-B^{*}A\bigr)^{-1} \end{aligned}$$

and

$$\begin{aligned} \bigl(I-B^{*}B\bigr)^{-1} \geq \bigl(I-B^{*}A\bigr)^{-1} \bigl(I-A^{*}A\bigr) \bigl(I-A^{*}B\bigr)^{-1}. \end{aligned}$$

(8)

Clearly, by unitary similarity transformation,

$$ \begin{pmatrix} (I-B^{*}B)^{-1} & (I-A^{*}B)^{-1} \\ (I-B^{*}A)^{-1} & (I-A^{*}A)^{-1} \end{pmatrix} $$

is also positive semidefinite.

Therefore,

$$\begin{aligned} \bigl(I-A^{*}A\bigr)^{-1} \geq \bigl(I-B^{*}A\bigr)^{-1} \bigl(I-B^{*}B\bigr) \bigl(I-A^{*}B\bigr)^{-1}. \end{aligned}$$

(9)

Thus,

$$\begin{aligned} &\bigl(I-B^{*}B\bigr)^{-1}\sharp \bigl(I-A^{*}A \bigr)^{-1} \\ &\qquad{}- \bigl(I-B^{*}A\bigr)^{-1}\bigl(\bigl(I-A^{*}A \bigr)^{-1}\sharp \bigl(I-B^{*}B\bigr)^{-1} \bigr)^{-1}\bigl(I-A^{*}B\bigr)^{-1} \\ &\quad\geq \bigl(I-B^{*}B\bigr)^{-1}\sharp \bigl(I-A^{*}A \bigr)^{-1} \\ &\qquad{}-\bigl(\bigl(I-B^{*}A\bigr)^{-1}\bigl(I-A^{*}A \bigr) \bigl(I-A^{*}B\bigr)^{-1}\bigr)\sharp \bigl( \bigl(I-B^{*}A\bigr)^{-1}\bigl(I-B^{*}B\bigr) \bigl(I-A^{*}B\bigr)^{-1}\bigr) \\ & \qquad\bigl(\text{by (5) and monotonicity}\bigr) \\ &\quad\geq \bigl(I-B^{*}B\bigr)^{-1}\sharp \bigl(I-A^{*}A \bigr)^{-1}-\bigl(I-B^{*}B\bigr)^{-1}\sharp \bigl(I-A^{*}A\bigr)^{-1} \quad \bigl(\text{by (8) and (9)}\bigr) \\ &\quad=0. \end{aligned}$$

In a similar way, we can prove

$$\begin{aligned} &\bigl(I-B^{*}B\bigr)^{-1}\sharp \bigl(I-A^{*}A \bigr)^{-1} \\ &\quad{}- \bigl(I-A^{*}B\bigr)^{-1}\bigl(\bigl(I-A^{*}A \bigr)^{-1}\sharp \bigl(I-B^{*}B\bigr)^{-1} \bigr)^{-1}\bigl(I-B^{*}A\bigr)^{-1} \geq 0. \end{aligned}$$

So

$$ K:= \begin{pmatrix} (I-A^{*}A)^{-1}\sharp (I-B^{*}B)^{-1} & (I-A^{*}B)^{-1} \\ (I-B^{*}A)^{-1} & (I-B^{*}B)^{-1}\sharp (I-A^{*}A)^{-1} \end{pmatrix} $$

is PPT.

Therefore,

$$\begin{aligned} 2 \bigl\Vert \bigl(I-A^{*}B\bigr)^{-1} \bigr\Vert \leq{}& \Vert K \Vert \quad\bigl(\text{by (4)} \bigr) \\ \leq{}& \bigl\Vert \bigl(\bigl(I-A^{*}A\bigr)^{-1}\sharp \bigl(I-B^{*}B\bigr)^{-1}\bigr)+\bigl(\bigl(I-B^{*}B \bigr)^{-1} \sharp \bigl(I-A^{*}A\bigr)^{-1} \bigr) \bigr\Vert \\ & \bigl(\text{by (1)}\bigr) \\ ={}&2 \bigl\Vert \bigl(I-B^{*}B\bigr)^{-1}\sharp \bigl(I-A^{*}A\bigr)^{-1} \bigr\Vert . \end{aligned}$$

 □

Remark 2.9

Obviously, our result is sharper than (2).

Finally, we present an extension of Theorem 1.3.

Theorem 2.10

Let$A, B, X, Y\in \mathbb{M}_{n}$. If$M=\left(\begin{array}{cc}A& X\\ Y^{\ast }& B\end{array}\right)$is SPT, then

$$\begin{aligned} s_{j} \biggl(\varPhi \biggl(\frac{X+Y}{2} \biggr) \biggr) \leq s_{j} \bigl(\varPhi (A\sharp B ) \bigr), \end{aligned}$$

(10)

where$\varPhi: C\rightarrow C+\operatorname{Tr}(C)I$.

Proof

Since M is SPT, then

$$\begin{aligned} \operatorname {Re}M = \begin{pmatrix} \operatorname {Re}A & (X+Y)/2 \\ (X+Y)^{*}/2 &\operatorname {Re}B \end{pmatrix} \end{aligned}$$

and

$$\begin{aligned} \operatorname {Re}\bigl(M^{\tau }\bigr) = \begin{pmatrix} \operatorname {Re}A &(X+Y)^{*}/2 \\ (X+Y)/2 &\operatorname {Re}B \end{pmatrix}=(\operatorname {Re}M)^{\tau } \end{aligned}$$

are both positive semidefinite matrices. Thus, ReM is PPT.

By Theorem 1.3, we have

$$\begin{aligned} s_{j} \biggl(\varPhi \biggl(\frac{X+Y}{2} \biggr) \biggr) \leq s_{j} \bigl(\varPhi \bigl((\operatorname {Re}A)\sharp (\operatorname {Re}B) \bigr) \bigr). \end{aligned}$$

Compute

$$\begin{aligned} s_{j} \biggl(\varPhi \biggl(\frac{X+Y}{2} \biggr) \biggr)&\leq s_{j} \bigl(\varPhi (\operatorname {Re}A\sharp \operatorname {Re}B ) \bigr) \\ &\leq s_{j} \bigl(\varPhi \bigl(\operatorname {Re}(A\sharp B ) \bigr) \bigr) \quad\bigl(\text{by (7)}\bigr) \\ &=s_{j} \bigl(\operatorname {Re}\bigl(\varPhi (A\sharp B)\bigr) \bigr) \\ &\leq s_{j}\bigl(\varPhi (A\sharp B)\bigr) \quad \bigl(\text{by (3)}\bigr). \end{aligned}$$

 □

Remark 2.11

If M is PPT, then (10) becomes Lin’s result in Theorem 1.3.