Inequalities for partial determinants of accretive block matrices

Abstract

Let $A=[A_{i,j}]^{m}_{i,j=1}\in \mathbf{M}_{m}(\mathbf{M}_{n})$ be an accretive block matrix. We write det₁ and det₂ for the first and second partial determinants, respectively. In this paper, we show that

$$\begin{aligned} \bigl\Vert \operatorname{det}_{1} (\operatorname{Re}A) \bigr\Vert \leq \biggl\Vert \biggl(\frac{\operatorname{tr}( \vert A \vert )}{m} \biggr)^{m}I_{n} \biggr\Vert \end{aligned}$$

and

$$\begin{aligned} \bigl\Vert \operatorname{det}_{2} (\operatorname{Re}A) \bigr\Vert \leq \biggl\Vert \biggl( \frac {\operatorname{tr}( \vert A \vert )}{n} \biggr)^{n}I_{m} \biggr\Vert \end{aligned}$$

hold for any unitarily invariant norm $\|\cdot \|$. The two inequalities generalize some known results related to partial determinants of positive-semidefinite block matrices.

1 Introduction

The set of $n\times n$ complex matrices is denoted by $\mathbf{M}_{n}$. $I_{n}$ is $n\times n$ identity matrix. Let $\mathbf{M}_{m}(\mathbf{M}_{n})$ be the set of all $m\times m$ block matrices with each block in $\mathbf{M}_{n}$. If $A\in \mathbf{M}_{n}$ is positive-semidefinite (definite), then we write $A \geq 0$ ($A>0$). For two Hermitian matrices A, B of the same size, $A\geq B$ ($A> B$) means that $A-B\geq 0$ ($A-B>0$). For $A\in \mathbf{M}_{n}$, 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 $\vert A \vert =(A^{*}A)^{1/2}$, arranged in nonincreasing order and repeated according to multiplicity as $s_{1}(A)\geq s_{2}(A)\geq \cdots \geq s_{n}(A)$. If A is Hermitian, we enumerate eigenvalues of A in nonincreasing order $\lambda _{1}(A)\geq \lambda _{2}(A)\geq \cdots \geq \lambda _{n}(A)$. We denote by $A^{T}$ and $A^{*}$ the transpose and conjugate transpose of A, respectively. Recall that a norm $\Vert \cdot \Vert$ is unitarily invariant if $\Vert UAV \Vert = \Vert A \Vert$ for any unitary matrices $U, V\in \mathbf{M}_{n}$ and any $A\in \mathbf{M}_{n}$. The Ky Fan k-norms, a special class of unitarily invariant norms, are defined as $\Vert \cdot \Vert _{(k)}=\sum_{j=1}^{k} s_{j}(A)$, $1\leq k\leq n$. The Schatten p-norms ($p\geq 1$) are defined as

$$\begin{aligned} \Vert A \Vert _{p} = \bigl(\operatorname{tr}\bigl( \vert A \vert ^{p} \bigr) \bigr)^{\frac{1}{p}}= \Biggl[\sum _{j=1}^{n}s_{j}^{p}(A) \Biggr]^{\frac{1}{p}}. \end{aligned}$$

The Schatten p-norms ($p\geq 1$) are also typical examples of unitarily invariant norms. We say that $A\in \mathbf{M}_{m}(\mathbf{M}_{n})$ is an accretive block matrix if its real part $\operatorname{Re}A:=\frac{A+A^{*}}{2}$ is positive-semidefinite.

In the following, two partial traces [6, p. 12] of $A=[A_{i, j}]_{i, j=1}^{m}\in \mathbf{M}_{m}(\mathbf{M}_{n})$ are defined by

$$ \operatorname{tr}_{1} A= \sum_{i=1}^{m} A_{i, i}\in \mathbb{M}_{n}\quad \text{and} \quad \operatorname{tr}_{2} A= [\operatorname{tr}A_{i, j} ]_{i,j=1}^{m} \in \mathbb{M}_{m}. $$

Assume that $A=[A_{i, j}]_{i, j=1}^{m}\in \mathbf{M}_{m}(\mathbf{M}_{n})$, where $A_{i,j}=[a_{l,k}^{i,j}]_{l,k=1}^{n}$. We introduce two partial determinants $\operatorname{det}_{1} A\in \mathbf{M}_{n}$ and $\operatorname{det}_{2} A\in \mathbf{M}_{m}$ analogous to the two partial traces as follows [2]:

$$ \operatorname{det}_{1} A=[\det G_{l,k}]_{l,k=1}^{n}, $$

where $G_{l,k}=[a_{l,k}^{i,j}]_{i,j=1}^{m}$, and

$$ \operatorname{det}_{2} A=[\det A_{i,j}]_{i,j=1}^{m}. $$

For $A=[[a_{l,k}^{i,j}]_{l,k=1}^{n}]_{i,j=1}^{m}\in \mathbf{M}_{m}( \mathbf{M}_{n})$, we will denote by $\tilde{A}\in \mathbf{M}_{n}(\mathbf{M}_{m})$ and $A^{\tau}\in \mathbf{M}_{m}(\mathbf{M}_{n})$ the matrices

$$ \widetilde{A}=[G_{l,k}]_{l,k=1}^{n}= \bigl[ \bigl[a_{l,k}^{i,j} \bigr]_{i,j=1}^{m} \bigr]_{l,k=1}^{n} \quad \text{and} \quad A^{\tau} =[A_{j,i}]^{m}_{i,j=1}= \bigl[ \bigl[a_{l,k}^{j,i} \bigr]^{n}_{l,k=1} \bigr]^{m}_{i,j=1}. $$

Note that $\tilde{\tilde{A}}=A$ and $\operatorname{det}_{1} A=\operatorname{det}_{2} \tilde{A}$ and therefore also $\operatorname{det}_{2} A=\operatorname{det}_{1} \tilde{A}$.

Recently, Xu et al. [8] presented the following unitarily invariant norm inequalities for two partial determinants of positive-semidefinite block matrices.

Theorem 1.1

Let $A=[A_{i,j}]^{m}_{i,j=1}\in \mathbf{M}_{m}(\mathbf{M}_{n})$ be positive-semidefinite. Then, the inequalities

$$\begin{aligned} \Vert \operatorname{det}_{1} A \Vert \leq \biggl\Vert \biggl(\frac{\operatorname{tr}A}{m} \biggr)^{m}I_{n} \biggr\Vert \end{aligned}$$

and

$$\begin{aligned} \Vert \operatorname{det}_{2} A \Vert \leq \biggl\Vert \biggl(\frac {\operatorname{tr}A}{n} \biggr)^{n}I_{m} \biggr\Vert \end{aligned}$$

hold for any unitarily invariant norm $\Vert \cdot \Vert$.

This theorem is inspired by a determinantal inequality for partial traces given by Lin [5, Theorem 1.2]. Actually, the two unitarily invariant norm inequalities for partial determinants of $A^{\tau}$ in Theorem 1.1 also hold; see [8, Theorem 2.12].

The main goal of this paper is to extend the above two inequalities to accretive block matrices that is a larger class of matrices than the class of positive-semidefinite block matrices. At the same time, some related results are obtained.

2 Partial determinant inequalities

We begin this section with some lemmas that are useful to present our main results. The following two results will be used in Theorem 2.6.

Lemma 2.1

[2, Theorem 7 and Remark 9] For $A=[A_{i, j}]_{i, j=1}^{m}\in \mathbf{M}_{m}(\mathbf{M}_{n})$,

• 1. A and à are unitarily similar;

• 2. if A is positive-semidefinite, so is Ã.

The next lemma is standard in matrix analysis.

Lemma 2.2

[4, p. 511] Let $A, B\in \mathbf{M}_{n}$ be positive-semidefinite. Then,

$$\begin{aligned} \det (A)+\det (B) \leq \det (A+B) . \end{aligned}$$

For the convenience of proofs, we also need to list some recent results as lemmas.

Lemma 2.3

[7] Let $A\in \mathbf{M}_{m}(\mathbf{M}_{n})$ be positive-semidefinite. Then,

$$\begin{aligned} \operatorname{det}_{2} A \geq 0. \end{aligned}$$

Lemma 2.4

[2, Theorem 6] Let $A\in \mathbf{M}_{m}(\mathbf{M}_{n})$ be positive-semidefinite. Then,

• 1. $\operatorname{det}_{1} A \geq 0$,

• 2. $\det (\operatorname{tr}_{2} A)\geq \operatorname{tr}(\operatorname{det}_{1} A)$.

Lemma 2.5

[3, Proposition 2.1] Let $A\in \mathbf{M}_{m}(\mathbf{M}_{n})$ be positive-semidefinite. Then,

$$ \det (\operatorname{tr}_{1} A) \leq \biggl(\frac{\operatorname{tr}A}{n} \biggr)^{n}\quad \textit{and} \quad \det (\operatorname{tr}_{2} A) \leq \biggl(\frac{\operatorname{tr}A}{m} \biggr)^{m}. $$

As an analog of Theorem 1.1, we prove the following inequalities for unitarily invariant norms.

Theorem 2.6

Let $A=[A_{i,j}]^{m}_{i,j=1}\in \mathbf{M}_{m}(\mathbf{M}_{n})$ be a sector block matrix. Then, the inequalities

$$\begin{aligned} \bigl\Vert \operatorname{det}_{1} (\operatorname{Re}A) \bigr\Vert \leq \biggl\Vert \biggl(\frac{\operatorname{tr}( \vert A \vert )}{m} \biggr)^{m}I_{n} \biggr\Vert \end{aligned}$$

and

$$\begin{aligned} \bigl\Vert \operatorname{det}_{2} (\operatorname{Re}A) \bigr\Vert \leq \biggl\Vert \biggl( \frac {\operatorname{tr}( \vert A \vert )}{n} \biggr)^{n}I_{m} \biggr\Vert \end{aligned}$$

hold for any unitarily invariant norm $\Vert \cdot \Vert$.

Proof

To prove the desired results, by Ky Fan’s dominance theorem [1, p. 93], we just need to show that for all $k=1,\ldots,n$,

$$ \bigl\Vert \operatorname{det}_{1} (\operatorname{Re}A) \bigr\Vert _{(k)}\leq \biggl\Vert \biggl( \frac{\operatorname{tr}( \vert A \vert )}{m} \biggr)^{m}I_{n} \biggr\Vert _{(k)} $$

and for all $k=1,\ldots,m$,

$$ \bigl\Vert \operatorname{det}_{2} (\operatorname{Re}A) \bigr\Vert _{(k)}\leq \biggl\Vert \biggl( \frac {\operatorname{tr}( \vert A \vert )}{n} \biggr)^{n}I_{m} \biggr\Vert _{(k)}. $$

Compute

$$\begin{aligned} \bigl\Vert \operatorname{det}_{1} (\operatorname{Re}A) \bigr\Vert _{(k)} =& \sum_{j=1}^{k} s_{j} \bigl( \operatorname{det}_{1} (\operatorname{Re}A) \bigr) \\ =& \sum_{j=1}^{k} \lambda _{j} \bigl(\operatorname{det}_{1} (\operatorname{Re}A) \bigr) \quad (\text{by Lemma 2.4}) \\ \leq & \operatorname{tr}\bigl(\operatorname{det}_{1} (\operatorname{Re}A) \bigr) \\ \leq & \det \bigl(\operatorname{tr}_{2} (\operatorname{Re}A) \bigr)\quad (\text{by Lemma 2.4}) \\ \leq & \biggl(\frac {\operatorname{tr}(\operatorname{Re}A)}{m} \biggr)^{m} \quad (\text{by Lemma 2.5}) \\ =& \frac{1}{k}\sum_{j=1}^{k} s_{j} \biggl( \biggl( \frac {\operatorname{tr}(\operatorname{Re}A)}{m} \biggr)^{m}I_{n} \biggr) \\ =& \frac{1}{k} \biggl\Vert \biggl(\frac {\operatorname{tr}(\operatorname{Re}A)}{m} \biggr)^{m}I_{n} \biggr\Vert _{(k)} \\ \leq & \biggl\Vert \biggl(\frac {\operatorname{tr}( \vert A \vert )}{m} \biggr)^{m}I_{n} \biggr\Vert _{(k)}, \quad k=1,\ldots, n, \end{aligned}$$

which means that

$$ \bigl\Vert \operatorname{det}_{1} (\operatorname{Re}A) \bigr\Vert \leq \biggl\Vert \biggl(\frac {\operatorname{tr}( \vert A \vert )}{m} \biggr)^{m}I_{n} \biggr\Vert . $$

By Lemma 2.1, we have $\operatorname{tr}(\operatorname{Re}A)=\operatorname{tr}(\widetilde{\operatorname{Re}A})=\operatorname{tr}(\operatorname{Re}\widetilde{A})$. Therefore, by $\widetilde{\operatorname{Re}A}=\operatorname{Re}\widetilde{A}$,

$$ \bigl\Vert \operatorname{det}_{2} (\operatorname{Re}A) \bigr\Vert = \bigl\Vert \operatorname{det}_{1} (\widetilde{\operatorname{Re}A}) \bigr\Vert = \bigl\Vert \operatorname{det}_{1} (\operatorname{Re}\widetilde{A}) \bigr\Vert \leq \biggl\Vert \biggl( \frac{\operatorname{tr}( \vert \widetilde{A} \vert )}{n} \biggr)^{n}I_{m} \biggr\Vert = \biggl\Vert \biggl(\frac{\operatorname{tr}( \vert A \vert )}{n} \biggr)^{n}I_{m} \biggr\Vert . $$

 □

Remark 1

When $A=[A_{i,j}]^{m}_{i,j=1}\in \mathbf{M}_{m}(\mathbf{M}_{n})$ is positive-semidefinite in Theorem 2.6, our result is Theorem 1.1. Thus, our result is a generalization of Theorem 1.1.

Next, we will prove two determinantal inequalities for accretive block matrices involving partial determinants.

Theorem 2.7

Let $A=[A_{i,j}]^{m}_{i,j=1}\in \mathbf{M}_{m}(\mathbf{M}_{n})$ be an accretive block matrix. Then,

$$\begin{aligned} \det \bigl(\operatorname{det}_{1} ( \operatorname{Re}A) \bigr) \leq \frac{(\operatorname{tr}( \vert A \vert ))^{mn}}{m^{mn}n^{n}} \end{aligned}$$

and

$$\begin{aligned} \det \bigl(\operatorname{det}_{2} (\operatorname{Re}A) \bigr) \leq \frac{(\operatorname{tr}( \vert A \vert ))^{mn}}{n^{mn}m^{m}}. \end{aligned}$$

Proof

Let $\lambda _{j}$, $j=1,\ldots, m$, be the eigenvalues of $\operatorname{det}_{2} (\operatorname{Re}A)$. Then, by the AM-GM inequality and Lemma 2.5, we have the following result:

$$\begin{aligned} \det \bigl(\operatorname{det}_{2} (\operatorname{Re}A) \bigr) =&\lambda _{1}\cdots \lambda _{m} \\ \leq & \biggl( \frac{\lambda _{1}+\lambda _{2}+\cdots + \lambda _{m}}{m} \biggr)^{m} \\ =& \biggl(\frac{\operatorname{tr}(\operatorname{det}_{2} (\operatorname{Re}A))}{m} \biggr)^{m} \\ =& \biggl(\frac{\sum_{i=1}^{m}\det (\operatorname{Re}A)_{ii}}{m} \biggr)^{m} \\ \leq & \biggl(\frac{\det (\sum_{i=1}^{m} (\operatorname{Re}A)_{ii})}{m} \biggr)^{m} \quad (\text{by Lemma 2.2}) \\ =& \biggl(\frac{\det (\operatorname{tr}_{1} (\operatorname{Re}A))}{m} \biggr)^{m} \\ \leq & \biggl(\frac{ (\frac{\operatorname{tr}(\operatorname{Re}A)}{n} )^{n}}{m} \biggr)^{m} \quad (\text{by Lemma 2.5}) \\ \leq & \frac{(\operatorname{tr}( \vert A \vert ))^{nm}}{n^{nm}m^{m}}, \end{aligned}$$

which means that

$$\begin{aligned} \det \bigl(\operatorname{det}_{2} (\operatorname{Re}A) \bigr) \leq \frac{(\operatorname{tr}( \vert A \vert ))^{nm}}{n^{nm}m^{m}}. \end{aligned}$$

On the other hand, by $\operatorname{det}_{1} (\operatorname{Re}A)=\operatorname{det}_{2} (\widetilde{\operatorname{Re}A})$ and Lemma 2.1, we have

$$\begin{aligned} \det \bigl(\operatorname{det}_{1} (\operatorname{Re}A) \bigr) = \det (\operatorname{det}_{2} \widetilde{\operatorname{Re}A} )=\det \bigl( \operatorname{det}_{2} (\operatorname{Re}\widetilde{A}) \bigr)\leq \frac{(\operatorname{tr}( \vert \tilde{A} \vert ))^{mn}}{m^{mn}n^{n}}= \frac{(\operatorname{tr}( \vert A \vert ))^{mn}}{m^{mn}n^{n}}. \end{aligned}$$

 □

We would like to know whether or not the inequalities above hold in the case of replacing A with $A^{\tau}$. Now, we will present a result on sector block matrices that is the same as it was under the positive-semidefinite condition.

Theorem 2.8

If $A=[A_{i,j}]^{m}_{i,j=1}\in \mathbf{M}_{m}(\mathbf{M}_{n})$ is a sector block matrix, then

$$ \operatorname{det}_{1} \bigl(A^{\tau} \bigr)= \operatorname{det}_{1} A\quad \textit{and}\quad \operatorname{det}_{2} \bigl(A^{\tau} \bigr)=(\operatorname{det}_{2} A)^{T}= \operatorname{det}_{2} \bigl(A^{T} \bigr). $$

Proof

Since $A^{\tau}=[A_{j,i}]^{m}_{i,j=1}$ and $\widetilde{A}=[G_{l,k}]_{l,k=1}^{n}$, we have $\widetilde{A^{\tau}}=[G_{l,k}^{\;\;\;T}]^{n}_{l,k=1}$.

Hence,

$$\begin{aligned} \operatorname{det}_{1} \bigl(A^{\tau} \bigr) =& \operatorname{det}_{2} \bigl( \widetilde{A^{\tau}} \bigr) \\ =& \bigl[\det G_{l,k}^{\;\;\; T} \bigr]_{l,k=1}^{n} \\ =& \bigl[\det \bigl[a_{l,k}^{j,i} \bigr]_{i,j=1}^{m} \bigr]_{l,k=1}^{n} \\ =& \bigl[\det \bigl[a_{l,k}^{i,j} \bigr]_{i,j=1}^{m} \bigr]_{l,k=1}^{n} \\ =&[\det G_{l,k}]_{l,k=1}^{n} \\ =& \operatorname{det}_{1} A. \end{aligned}$$

On the other hand,

$$\begin{aligned} \operatorname{det}_{2} \bigl(A^{T} \bigr) =& \bigl[\det A^{T}_{\;\;\; i,j} \bigr]^{m}_{i,j=1} \\ =& \bigl[\det A_{j,i}^{\;\;\; T} \bigr]^{m}_{i,j=1} \\ =& \bigl[\det \bigl[a_{k,l}^{j,i} \bigr]^{n}_{l,k=1} \bigr]_{i,j=1}^{m} \\ =& \bigl[\det \bigl[a_{l,k}^{j,i} \bigr]^{n}_{l,k=1} \bigr]_{i,j=1}^{m} \\ =&\operatorname{det}_{2} \bigl(A^{\tau} \bigr) \\ =&[\det A_{j,i}]_{i,j=1}^{m} \\ =& \bigl([\det A_{i,j}]_{i,j=1}^{m} \bigr)^{T} \\ =&(\operatorname{det}_{2} A)^{T}. \end{aligned}$$

 □

The following result for sector block matrices involving partial determinants of $A^{\tau}$ can be regarded as a complement of Theorem 2.7.

Theorem 2.9

Let $A=[A_{i,j}]^{m}_{i,j=1}\in \mathbf{M}_{m}(\mathbf{M}_{n})$ be a sector matrix. Then,

$$\begin{aligned} \det \bigl(\operatorname{det}_{1} (\operatorname{Re}A)^{\tau} \bigr) \leq & \frac{(\operatorname{tr}( \vert A \vert ))^{mn}}{m^{mn}n^{n}} \end{aligned}$$

and

$$\begin{aligned} \det \bigl(\operatorname{det}_{2} (\operatorname{Re}A)^{\tau} \bigr) \leq & \frac {(\operatorname{tr}( \vert A \vert ))^{mn}}{n^{nm}m^{m}}. \end{aligned}$$

Proof

The proof is similar to that of Theorem 2.7. □

Remark 2

In fact, the analogous inequalities below for partial traces are also valid using a similar idea to that of Lemma 2.5:

$$ \det \bigl(\operatorname{tr}_{1} \bigl(\operatorname{Re}A^{\tau} \bigr) \bigr) \leq \biggl(\frac{\operatorname{tr}(\operatorname{Re}A)}{n} \biggr)^{n}\leq \biggl(\frac{\operatorname{tr}( \vert A \vert )}{n} \biggr)^{n} $$

and

$$ \det \bigl(\operatorname{tr}_{2} \bigl(\operatorname{Re}A^{\tau} \bigr) \bigr) \leq \biggl(\frac{\operatorname{tr}(\operatorname{Re}A)}{m} \biggr)^{m}\leq \biggl(\frac{\operatorname{tr}( \vert A \vert )}{m} \biggr)^{m}. $$

Next, we give inequalities for partial determinants of $A^{\tau}$ involving unitarily invariant norms.

Theorem 2.10

Let $A=[A_{i,j}]^{m}_{i,j=1}\in \mathbf{M}_{m}(\mathbf{M}_{n})$ be a sector matrix. Then, the inequalities

$$ \bigl\Vert \operatorname{det}_{1} \bigl(\operatorname{Re}A^{\tau} \bigr) \bigr\Vert \leq \biggl\Vert \biggl( \frac{\operatorname{tr} \vert A \vert }{m} \biggr)^{m}I_{n} \biggr\Vert $$

and

$$ \bigl\Vert \operatorname{det}_{2} \bigl(\operatorname{Re}A^{\tau} \bigr) \bigr\Vert \leq \biggl\Vert \biggl( \frac{\operatorname{tr} \vert A \vert }{n} \biggr)^{n}I_{m} \biggr\Vert $$

hold for any unitarily invariant norm $\Vert \cdot \Vert$.

Proof

Note that $\operatorname{det}_{1} ((\operatorname{Re}A)^{\tau})=\operatorname{det}_{1} (\operatorname{Re}A)$, $\operatorname{det}_{2}( \operatorname{Re}A^{\tau})=\operatorname{det}_{2}( (\operatorname{Re}A)^{T})$ by Theorem 2.8 and $\operatorname{tr}A=\operatorname{tr}(A^{T})$, hence the proof is similar to that of Theorem 2.6. □