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. 1.

    A and à are unitarily similar;

  2. 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. 1.

    \(\operatorname{det}_{1} A \geq 0\),

  2. 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. □