Some determinantal inequalities for Hadamard and Fan products of matrices

Abstract

In this note, we generalize some determinantal inequalities which are due to Lynn (Proc. Camb. Philos. 60:425-431, 1964), Chen (Linear Algebra Appl. 368:99-106, 2003) and Ando (Linear Multilinear Algebra 8:291-316, 1980).

1 Introduction

Let $C^{m\times n}$ $(R^{m\times n})$ be the set of all complex (real) matrices and let $\mathbb{M}_{n}^{+}$ be the positive definite Hermitian matrices. Let $Z^{n\times n}=\{A=(a_{ij})\in R^{n\times n}:a_{ij} \leq0, i\neq j, i, j \in\{1,2,\ldots,n\}\}$. For any $A=(a_{ij}) \in C^{n\times n}$, its associated matrix is defined by $A^{\prime}=( \alpha_{ij})$, where $\alpha_{ii}=\vert a_{ii}\vert$, $\alpha_{ij}=-\vert a_{ij}\vert$ ($i\neq j$). For $A=(a_{ij})$, $B=(b_{ij})$ $\in C^{m\times n}$, the Hadamard product of A and B is $A \circ B =(a_{ij}b_{ij})\in C ^{m\times n}$ while their Fan product $A*B=(c_{ij})$ is defined by $c_{ii}=a_{ii}b_{ii}$ and $c_{ij}=-a_{ij}b_{ij}$ for $i\neq j$.

If $A=(a_{ij}) \in C^{n\times n}$, then the $k \times k$ leading principal submatrix of A is denoted by $A_{k}$ ($k\in\{1,2,\ldots,n\}$). $A_{\alpha}$ denotes the principal submatrix of A, with indices in $\alpha\subseteq\{1,2,\ldots,n\}$. $A\in R^{n\times n}$ is called an M-matrix if $A\in Z^{n\times n}$ and $\det A_{k}>0$ ($\forall k\in\{1,2,\ldots,n\}$), and we denote it by $A\in M_{n}$. A matrix $A\in C^{n\times n}$ is called an H-matrix if $A^{\prime}$ is an M-matrix, and we denote it by $A\in H_{n}$.

Lynn [1], Theorem 3.1, proved the following determinantal inequality for H-matrices: if $A, B\in H_{n}$, then

$$\det(A\circ B)^{\prime}+\det A^{\prime}\det B^{\prime} \geq \prod_{i=1}^{n}\vert b _{ii} \vert \det A^{\prime}+\prod_{i=1}^{n} \vert a_{ii}\vert \det B^{\prime}, $$

i.e.

$$\begin{aligned} \det(A\circ B)^{\prime} \geq& \det A^{\prime}\det B^{\prime} \biggl( \frac{\prod_{i=1} ^{n}\vert a_{ii}\vert }{\det A^{\prime}}+\frac{\prod_{i=1}^{n}\vert b_{ii}\vert }{\det B^{\prime}}-1 \biggr). \end{aligned}$$

(1.1)

Chen [2], Theorem 2.7, obtained a determinantal inequality for positive definite matrices: if $A=(a_{ij})$, $B=(b_{ij})\in \mathbb{M}_{n}^{+}$, then

$$\begin{aligned} \det(A\circ B) \geq&\det A\det B \prod _{k=2}^{n} \biggl( \frac{a_{kk} \det A_{k-1}}{\det A_{k}}+ \frac{b_{kk}\det B_{k-1}}{\det B_{k}}-1 \biggr). \end{aligned}$$

(1.2)

Lin [3] recently proved that a similar result to the block positive definite matrices holds for the block Hadamard product.

Ando [4], Theorem 5.3, has given the following result: if $A=(a_{ij})$, $B=(b_{ij})$ are M-matrices, then

$$\begin{aligned}& \det(A*B)+\det A\cdot\det B \\& \quad \geq \Biggl( \prod_{i=1}^{n}a_{ii} \Biggr) \cdot\det B+\det A \cdot \Biggl( \prod_{i=1}^{n}b_{ii} \Biggr), \end{aligned}$$

i.e.

$$\begin{aligned} \det(A*B) \geq& \det A \det B \biggl( \frac{\prod_{i=1}^{n}a_{ii}}{ \det A} + \frac{\prod_{i=1}^{n}b_{ii}}{\det B}-1 \biggr). \end{aligned}$$

(1.3)

In this paper, we will present some determinantal inequalities for matrices which are generalizations of (1.1), (1.2), and (1.3).

2 Main results and some remarks

We give some lemmas before we present the main theorems of this paper.

Lemma 1

[4], Corollary 4.1.2

Let $A=(a_{ij})\in R^{n\times n}$ be an M-matrix. If $\alpha_{i}\subseteq\{1,2,\ldots,n\}$ ($i=1,2,3, \ldots,N$) satisfies $\alpha_{i}\cap\alpha_{j}=\phi$ for $i\neq j$ and $\bigcup_{j=1}\alpha_{j}=\{1,2,\ldots,n\}$, then

$$\begin{aligned} \det A \leq& \prod_{i=1}^{N} \det(A_{\alpha_{i}}). \end{aligned}$$

In particular,

$$\begin{aligned} \det A \leq& \prod_{i=1}^{n}a_{ii}. \end{aligned}$$

(2.1)

Lemma 2

[1], Theorem 3.1

If A, B are H-matrices and $C=A\circ B$, then C is H-matrix.

Lemma 3

[5], Theorem 5.2.1

If A, B are positive definite matrices and $C=A\circ B$, then C is positive definite matrix.

Lemma 4

[6]

If A, B are M-matrices and $C=A* B$, then C is M-matrix.

Now we present the main results.

First of all, we give a determinantal inequality for the Hadamard product of finite number of H-matrices as follows:

Theorem 5

If $A_{1}=(a_{1}^{kl}), A_{2}=(a_{2}^{kl}),\ldots, A_{m}=(a_{m}^{kl})$ ($k,l=1,\ldots,n$) are H-matrices, then

$$\begin{aligned} \det(A_{1}\circ\cdots\circ A_{m})^{\prime} \ge&\det(A_{1}\cdots A_{m}) \\ &{}\times \biggl( \frac{\prod_{i=1}^{n}\vert a_{1}^{ii}\vert }{\det A_{1}^{\prime}}+ \cdots+\frac{\prod_{i=1}^{n}\vert a_{m}^{ii}\vert }{\det A_{m}^{\prime }}-(m-1) \biggr). \end{aligned}$$

(2.2)

Proof

By Lemma 2, it is straightforward to observe that the Hadamard product $A_{1}\circ\cdots\circ A_{m}$ is an H-matrix. Use induction on k. When $k=2$, the result is (1.1). Suppose that (2.2) holds when $k=m-1$

$$\begin{aligned} \det(A_{1}\circ\cdots\circ A_{m-1})^{\prime} \ge&\det\bigl(A_{1}^{\prime} \cdots A _{m-1}^{\prime} \bigr) \\ &{} \times \biggl( \frac{\prod_{i=1}^{n}\vert a_{1}^{ii}\vert }{ \det A_{1}^{\prime}}+\cdots+\frac{\prod_{i=1}^{n}\vert a_{m-1}^{ii}\vert }{\det A _{m-1}^{\prime}}-(m-2) \biggr). \end{aligned}$$

When $k=m$, we need to show

$$\begin{aligned} \det(A_{1}\circ\cdots\circ A_{m})^{\prime} \ge&\det\bigl(A_{1}^{\prime} \cdots A _{m}^{\prime}\bigr) \\ &{} \times \biggl( \frac{\prod_{i=1}^{n}\vert a_{1}^{ii}\vert }{\det A _{1}^{\prime}}+\cdots+\frac{\prod_{i=1}^{n}\vert a_{m}^{ii}\vert }{\det A_{m}^{\prime}}-(m-1) \biggr). \end{aligned}$$

By (1.1), we have

$$\begin{aligned} \det\bigl((A_{1}\circ\cdots\circ A_{m-1}) \circ A_{m}\bigr)^{\prime} \ge&\det\bigl( \bigl(A _{1}^{\prime}\circ\cdots\circ A_{m-1}^{\prime} \bigr)A_{m}^{\prime}\bigr) \\ &{}\times \biggl( \frac{ \prod_{i=1}^{n}( \vert a_{1}^{ii}\vert \cdots \vert a_{m-1}^{ii}\vert )}{\det (A_{1}^{\prime} \circ\cdots\circ A_{m-1}^{\prime})}+\frac{\prod_{i=1}^{n} \vert a_{m}^{ii}\vert }{ \det A_{m}^{\prime}} -1 \biggr). \end{aligned}$$

By the inductive assumption, the above inequality is

$$\begin{aligned} \det(A_{1}\circ\cdots\circ A_{m})^{\prime} \ge&\det\bigl(A_{1}^{\prime} \cdots A _{m-1}^{\prime}\bigr)\det A_{m}^{\prime} \\ &{}\times \biggl( \frac{\prod_{i=1}^{n}\vert a _{1}^{ii}\vert }{\det A_{1}^{\prime}}+\cdots+\frac{\prod_{i=1}^{n}\vert a_{m-1}^{ii}\vert }{ \det A_{m-1}^{\prime}}-(m-2) \biggr) \\ &{}\times \biggl( \frac{\prod_{i=1}^{n}( \vert a _{1}^{ii}\vert \cdots \vert a_{m-1}^{ii}\vert )}{\det(A_{1}^{\prime}\circ\cdots \circ A_{m-1}^{\prime})}+\frac{\prod_{i=1}^{n} \vert a_{m}^{ii}\vert }{\det A_{m}^{\prime}} -1 \biggr). \end{aligned}$$

(2.3)

Let

$$\begin{aligned}& a = \biggl( \frac{\prod_{i=1}^{n}\vert a_{1}^{ii}\vert }{\det A_{1}^{\prime }}+\cdots +\frac{\prod_{i=1}^{n}\vert a_{m-1}^{ii}\vert }{\det A_{m-1}^{\prime }}-(m-2) \biggr), \\& b = \biggl( \frac{\prod_{i=1}^{n}( \vert a_{1}^{ii}\vert \cdots \vert a_{m-1}^{ii}\vert )}{ \det(A_{1}^{\prime}\circ\cdots\circ A_{m-1}^{\prime})}+\frac {\prod_{i=1}^{n} \vert a_{m}^{ii}\vert }{\det A_{m}^{\prime}} -1 \biggr). \end{aligned}$$

By (2.1), we have

$$\begin{aligned}& \frac{\prod_{i=1}^{n}\vert a_{j}^{ii}\vert }{\det A_{j}^{\prime}} \geq 1,\quad j=1, \ldots, m, \\& \frac{\prod_{i=1}^{n}( \vert a_{1}^{ii}\vert \cdots \vert a_{m-1}^{ii}\vert )}{\det(A _{1}^{\prime}\circ\cdots\circ A_{m-1}^{\prime})} \geq 1, \end{aligned}$$

and so

$$a,b\geq1. $$

Thus by $ab\ge a+b-1$ for $a, b\ge1$, the above inequality (2.3) is

$$\begin{aligned} \det(A_{1}\circ\cdots\circ A_{m})^{\prime} \ge&\det\bigl(A_{1}^{\prime} \cdots A _{m-1}^{\prime}\bigr)\det A_{m}^{\prime} \times a\times b \\ \ge&\det \bigl(A_{1}^{\prime}\cdots A_{m}^{\prime} \bigr)\times(a+b-1) \\ \geq&\det\bigl(A_{1}^{\prime}\cdots A_{m}^{\prime} \bigr) \\ &{}\times \biggl( \frac{\prod_{i=1}^{n}\vert a_{1}^{ii}\vert }{\det A_{1}^{\prime}}+ \cdots+\frac{\prod_{i=1}^{n}\vert a_{m}^{ii}\vert }{\det A_{m}^{\prime }}-(m-1) \biggr). \end{aligned}$$

This completes the proof. □

Remark 6

The above inequality in Theorem 5 is a generalization of the inequality (1.1).

Second, we achieve a determinantal inequality for the Hadamard product of positive definite matrices as follows:

Theorem 7

If $A_{i}$ ($i=1,\ldots,m$) ($m\ge2$) are $n\times n$ positive definite matrices, the Hadamard product of $A_{i}=(a_{i}^{lt})$ and $A_{j}=(a_{j}^{lt})$ ($i\neq j$) is denoted by $A_{i}\circ A_{j}$, and $A_{i}^{(k)}$ is the $k\times k$ ($k=1,2,\ldots,n$) leading principal submatrix of $A_{i}$, then

$$\begin{aligned} \det(A_{1}\circ\cdots\circ A_{m}) \ge&\det(A_{1}\cdots A_{m}) \\ & {} \times\prod_{\mu=2}^{n} \biggl( \frac{ a_{1}^{\mu\mu}\det A^{(\mu-1)} _{1}}{\det A^{(\mu)}_{1}} +\cdots+\frac{ a_{m}^{\mu\mu}\det A^{( \mu-1)}_{m}}{\det A_{m}^{(\mu)}}-(m-1) \biggr). \end{aligned}$$

(2.4)

Proof

By Lemma 3, it is straightforward to see that the Hadamard product $A_{1}\circ\cdots\circ A_{m}$ is a positive definite matrix. Use induction on m. When $k=2$, the result is (1.2). Suppose that (2.4) holds when $k=m-1$. We have

$$\begin{aligned} \det(A_{1}\circ\cdots\circ A_{m-1}) \ge&\det(A_{1} \cdots A_{m-1}) \\ &{}\times\prod_{\mu=2}^{n} \biggl( \frac{ a_{1}^{\mu\mu}\det A^{(\mu-1)} _{1}}{\det A^{(\mu)}_{1}} +\cdots+\frac{ a_{m-1}^{\mu\mu}\det A ^{(\mu-1)}_{m-1}}{\det A_{m-1}^{(\mu)}}-(m-2) \biggr). \end{aligned}$$

When $k=m$, we need to show

$$\begin{aligned} \det(A_{1}\circ\cdots\circ A_{m}) \ge&\det(A_{1}\cdots A_{m}) \\ &{} \times\prod_{\mu=2}^{n} \biggl( \frac{ a_{1}^{\mu\mu}\det A^{(\mu-1)} _{1}}{\det A^{(\mu)}_{1}} +\cdots+\frac{ a_{m}^{\mu\mu}\det A^{( \mu-1)}_{m}}{\det A_{m}^{(\mu)}}-(m-1) \biggr). \end{aligned}$$

By (1.2), we have

$$\begin{aligned}& \det\bigl((A_{1}\circ\cdots\circ A_{m-1}) \circ A_{m}\bigr) \\& \quad \ge\det\bigl((A _{1}\circ\cdots\circ A_{m-1})A_{m} \bigr) \\& \qquad {}\times\prod_{\mu=2}^{n} \biggl( \frac{(a_{1}^{\mu\mu} \cdots a_{m-1}^{\mu\mu})\det(A_{1} \circ\cdots\circ A_{m-1})^{(\mu-1)}}{\det(A_{1}\circ\cdots \circ A_{m-1})^{(\mu)}}+\frac{a_{m}^{\mu\mu}\det A^{(\mu-1)}_{m}}{ \det A^{(\mu)}_{m}} -1 \biggr). \end{aligned}$$

By the inductive assumption, the above inequality is such that

$$\begin{aligned}& \det(A_{1}\circ\cdots\circ A_{m-1})\det A_{m} \\& \qquad {}\times\prod_{ \mu=2}^{n} \biggl( \frac{(a_{1}^{\mu\mu} \cdots a_{m-1}^{\mu\mu}) \det(A_{1}\circ\cdots\circ A_{m-1})^{(\mu-1)}}{\det(A_{1}\circ \cdots\circ A_{m-1})^{(\mu)}}+\frac{a_{m}^{\mu\mu}\det A^{(\mu-1)} _{m}}{\det A^{(\mu)}_{m}} -1 \biggr) \\& \quad \ge\det(A_{1}\cdots A_{m-1}) \det A_{m} \\& \qquad {} \times\prod_{\mu=2}^{n} \biggl( \frac{ a_{1}^{\mu \mu}\det A^{(\mu-1)}_{1}}{\det A^{(\mu)}_{1}} +\frac{ a_{2}^{ \mu\mu}\det A^{(\mu-1)}_{2}}{\det A^{(\mu)}_{2}}+\cdots+\frac{ a _{m}^{\mu\mu}\det A^{(\mu-1)}_{m-1}}{\det A_{m-1}^{(\mu)}}-(m-2) \biggr) \\& \qquad {}\times\prod_{\mu=2}^{n} \biggl( \frac{(a_{1}^{\mu\mu} \cdots a _{m-1}^{\mu\mu})\det(A_{1}\circ\cdots\circ A_{m-1})^{(\mu-1)}}{ \det(A_{1}\circ\cdots\circ A_{m-1})^{(\mu)}}+\frac{a_{m}^{\mu \mu}\det A^{(\mu-1)}_{m}}{\det A^{(\mu)}_{m}} -1 \biggr). \end{aligned}$$

(2.5)

Let

$$\begin{aligned}& a_{\mu}=\frac{ a_{1}^{\mu\mu}\det A^{(\mu-1)}_{1}}{\det A^{( \mu)}_{1}} +\frac{ a_{2}^{\mu\mu}\det A^{(\mu-1)}_{2}}{\det A^{( \mu)}_{2}}+\cdots+ \frac{ a_{n}^{\mu\mu}\det A^{(\mu-1)}_{m-1}}{ \det A_{m-1}^{(\mu)}}-(m-2), \\& b_{\mu}=\frac{(a_{1}^{\mu\mu} \cdots a_{m-1}^{\mu\mu})\det(A _{1}\circ\cdots\circ A_{m-1})^{(\mu-1)}}{\det(A_{1}\circ\cdots \circ A_{m-1})^{(\mu)}}+\frac{a_{m}^{\mu\mu}\det A^{(\mu-1)}_{m}}{ \det A^{(\mu)}_{m}} -1. \end{aligned}$$

By Fischer’s inequality [5], p.506, we have

$$\begin{aligned}& \frac{a_{i}^{\mu\mu}\det A^{(\mu-1)}_{i}}{\det A^{(\mu)}_{i}} \geq1,\quad i=1,\ldots,m, \\& \frac{(a_{1}^{\mu\mu} \cdots a_{m-1}^{\mu\mu})\det(A_{1}\circ \cdots\circ A_{m-1})^{(\mu-1)}}{\det(A_{1}\circ\cdots\circ A_{m-1})^{( \mu)}}-1 \geq 0, \end{aligned}$$

and so

$$a_{\mu}, b_{\mu}\geq1. $$

Thus by $a_{\mu}b_{\mu}\ge a_{\mu}+b_{\mu}-1$ for $a_{\mu}, b _{\mu}\ge1$, the above inequality (2.5) is

$$\begin{aligned}& \det(A_{1}\circ\cdots\circ A_{m}) \\& \quad \geq\det(A_{1}\cdots A_{m-1}) \det A_{m}\times\prod_{\mu=2}^{n} a_{\mu}b_{\mu} \\& \quad \geq\det(A _{1}\cdots A_{m})\times\prod _{\mu=2}^{n}( a_{\mu}+b_{\mu}-1) \\& \quad \geq\det(A_{1}\cdots A_{m}) \prod _{\mu=2}^{n} \biggl( \frac{ a_{1} ^{\mu\mu}\det A^{(\mu-1)}_{1}}{\det A^{(\mu)}_{1}} +\cdots+ \frac{a _{m}^{\mu\mu}\det A^{(\mu-1)}_{m}}{\det A^{(\mu)}_{m}} -(m-1) \biggr). \end{aligned}$$

This completes the proof. □

Remark 8

The inequality in Theorem 7 is a generalization of the inequality (1.2).

Finally, a result on Fan product of M-matrices is obtained in the following theorem.

Theorem 9

If $A_{1}=(a_{1}^{kl}), A_{2}=(a_{2}^{kl}),\ldots, A_{m}=(a_{m}^{kl})$ ($k,l=1,\ldots,n$) are M-matrices, then

$$\begin{aligned} \det(A_{1}* \cdots * A_{m}) \ge &\det(A_{1}\cdots A_{m}) \\ &{}\times \biggl( \frac{\prod_{i=1}^{n}a_{1}^{ii}}{\det A_{1}}+\cdots+\frac{ \prod_{i=1}^{n}a_{m}^{ii}}{\det A_{m}}-(m-1) \biggr). \end{aligned}$$

(2.6)

Proof

By Lemma 4, it is straightforward to see that the Hadamard product $A_{1}\ast\cdots\ast A_{m}$ is an M-matrix. Use induction on k. When $k=2$, the result is (1.3). Let $k=m-1$, (2.6) holds:

$$\begin{aligned} \det(A_{1}\circ\cdots \circ A_{m-1}) \ge&\det(A_{1} \cdots A_{m-1}) \\ &{} \times \biggl( \frac{\prod_{i=1}^{n}a_{1}^{ii}}{\det A_{1}}+\cdots +\frac{ \prod_{i=1}^{n}a_{m-1}^{ii}}{\det A_{m-1}}-(m-2) \biggr). \end{aligned}$$

When $k=m$, we need to show

$$\begin{aligned} \det(A_{1}* \cdots * A_{m}) \ge& \det(A_{1}\cdots A_{m}) \\ &{} \times \biggl( \frac{\prod_{i=1}^{n}a_{1}^{ii}}{\det A_{1}}+\cdots+\frac{ \prod_{i=1}^{n}a_{m}^{ii}}{\det A_{m}}-(m-1) \biggr). \end{aligned}$$

By (1.3), we have

$$\begin{aligned} \det\bigl((A_{1}* \cdots * A_{m-1}) \ast A_{m}\bigr) \ge&\det\bigl((A_{1}* \cdots * A_{m-1})A_{m}\bigr) \\ &{}\times \biggl( \frac{\prod_{i=1}^{n}( a_{1} ^{ii} \cdots a_{m-1}^{ii})}{\det(A_{1}* \cdots * A_{m-1})}+\frac{ \prod_{i=1}^{n} a_{m}^{ii}}{\det A_{m}} -1 \biggr). \end{aligned}$$

By the inductive assumption, the above inequality is

$$\begin{aligned} \det(A_{1}* \cdots * A_{m}) \ge&\det(A_{1}\cdots A_{m-1})\det A _{m} \\ &{} \times \biggl( \frac{\prod_{i=1}^{n}a_{1}^{ii}}{\det A_{1}}+ \cdots+\frac{\prod_{i=1}^{n}a_{m-1}^{ii}}{\det A_{m-1}}-(m-2) \biggr) \\ &{}\times \biggl( \frac{\prod_{i=1}^{n}( a_{1}^{ii} \cdots a_{m-1}^{ii})}{ \det(A_{1}* \cdots * A_{m-1})}+\frac{\prod_{i=1}^{n} a_{m}^{ii}}{ \det A_{m}} -1 \biggr). \end{aligned}$$

(2.7)

Let

$$\begin{aligned}& a = \biggl( \frac{\prod_{i=1}^{n}a_{1}^{ii}}{\det A_{1}}+\cdots+\frac{ \prod_{i=1}^{n}a_{m-1}^{ii}}{\det A_{m-1}}-(m-2) \biggr), \\& b = \biggl( \frac{\prod_{i=1}^{n}( a_{1}^{ii} \cdots a_{m-1}^{ii})}{ \det(A_{1}\circ\cdots\circ A_{m-1})}+\frac{\prod_{i=1}^{n} a_{m} ^{ii}}{\det A_{m}} -1 \biggr). \end{aligned}$$

By (2.1), we have

$$\begin{aligned}& \frac{\prod_{i=1}^{n}a_{j}^{ii}}{\det A_{j}}\geq1,\quad j=1,\ldots, m, \\& \frac{\prod_{i=1}^{n}( a_{1}^{ii} \cdots a_{m-1}^{ii})}{\det(A_{1} \circ\cdots\circ A_{m-1})} \geq 1, \end{aligned}$$

and so

$$a,b\geq1. $$

So by $ab\ge a+b-1$ for $a, b\ge1$, the above inequality (2.7) is

$$\begin{aligned} \det(A_{1}*\cdots * A_{m}) \ge& \det(A_{1}\cdots A_{m}) \\ &{}\times \biggl( \frac{\prod_{i=1}^{n}a_{1}^{ii}}{\det A_{1}}+\cdots+\frac{ \prod_{i=1}^{n}a_{m}^{ii}}{\det A_{m}}-(m-1) \biggr). \end{aligned}$$

This completes the proof. □

Remark 10

The inequality in Theorem 9 is a generalization of the inequality (1.3).