2 Some lemmas

Let $X^{*}$ denote the conjugate transpose of matrix or vector X. For a square matrix A, we write $A\ge 0$ if A is positive semidefinite and $A>0$ if A is positive definite. For Hermitian matrices $A, B\in \Bbb C^{n\times n}$ , we write $A\ge B$ if $A-B\ge 0$ .

Throughout the paper, let n be a positive integer. Let $\mathcal {I}=\{i_{1}, \dots , i_{k}\}$ , where $1\le i_{1} < \cdots < i_{k}\le n$ is any subsequence of $1, \dots , n$ , $k=1, \dots , n$ . If $A\in \mathbb {C}^{n\times n}$ is positive semidefinite, then $\lambda _{i_{1}}(A)\geq \cdots \geq \lambda _{i_{k}}(A)$ are k eigenvalues of A indexed by $\mathcal {I}=\{i_{1}, \dots , i_{k}\}$ . For $t=1, \dots , k$ , setting $i_{t}=t$ gives the first k largest eigenvalues of A; putting $i_{t}=n-k+t$ gives the last k smallest eigenvalues of A.

Lemma 2.1. Hoffman [Reference Amir-Moéz1, Corollary 2.5]

Let $A\in \mathbb {C}^{n\times n}$ be positive semidefinite. Then,

$$ \begin{align*} \prod\limits_{t=1}^{k}\lambda_{i_{t}}(A) = \max\limits_{\substack{\mathbb{S}_{1}\subset\cdots\subset \mathbb{S}_{k}\subset \mathbb{C}^{n} \\ \dim \mathbb{S}_{t}=i_{t}}} \min\limits_{\substack{{x}_{t}\in \mathbb{S}_{t}, ({x}_{r},{x}_{s})=\delta_{rs}\\ U_{k}=\left(x_{1},\dots,x_{k}\right)}}\det({U_{k}^{*}AU_{k}}), \end{align*} $$

where $\delta _{rs}$ is the Kronecker delta, i.e., $\delta _{rs}=1$ if $r=s$ , or 0 otherwise.

Lemma 2.2. Lidskiǐ [Reference Lidskiǐ11]

Let $A, B \in \mathbb {C}^{n\times n}$ be positive semidefinite. Then,

(2.1) $$ \begin{align} \prod\limits_{t=1}^{k}\lambda_{i_{t}}(A)\lambda_{n-t+1}(B) \le \prod\limits_{t=1}^{k}\lambda_{i_{t}}(AB)\le\prod\limits_{t=1}^{k}\lambda_{i_{t}}(A)\lambda_{t}(B). \end{align} $$

More general inequalities of (2.1) for singular values are due to Gelfand and Naimark (see, e.g., [Reference Marshall, Olkin and Arnold13, p. 340]). The inequalities on the left-hand side of (2.1) can be found explicitly in [Reference Wang and Zhang17].

Lemma 2.3. Let $a_{1}, \dots , a_{n}$ and $b_{1}, \dots , b_{n}$ be non-negative real numbers. Then,

(2.2) $$ \begin{align} \prod\limits_{t=1}^{n} (a_{t}+b_{t})\geq \prod\limits_{t=1}^{n} a_{t}+\prod\limits_{t=1}^{n} b_{t} +(2^{n}-2) \left [ \prod\limits_{t=1}^{n} (a_{t}b_{t})\right ]^{\frac12}. \end{align} $$

In particular, for $x, y\geq 0$ ,

(2.3) $$ \begin{align} \big(x^{\frac{1}{n}}+y^{\frac{1}{n}}\big)^{n} \ge x+y+\left(2^{n}-2\right)(xy)^{\frac{1}{2}}. \end{align} $$

We remark that Lemma 2.3 implies immediately a result of Hartfiel [Reference Hartfiel8]:

(2.4) $$ \begin{align} \det(A + B)\ge \det A + \det B+(2^{n}-2)[\det A\det B]^{\frac12} \end{align} $$

because of the fact that any two positive semidefinite matrices of the same size are simultaneously ${}^{*}$ -congruent to diagonal matrices (see, e.g., [Reference Zhang23, p. 209]).

3 Main results

We begin with a result that is important to the proof of our main theorem.

Lemma 3.1. Let $D=\text{diag}\big (\lambda _{1},\lambda _{2},\dots ,\lambda _{n}\big )$ , where $\lambda _{1}\ge \lambda _{2}\ge \cdots \ge \lambda _{n}\ge 0$ , and let $U_{k}=\left (u_{1},\dots ,u_{k}\right )=\left (u_{ij}\right )$ be such an $n\times k$ matrix that $U^{*}_{k}U_{k}=I_{k}$ . If $m\geq 1$ and $u_{i1}=0$ for all $i> m$ , then $\det (U^{*}_{k}DU_{k})\ge \lambda _{m}$ for $k=1$ , and

(3.1) $$ \begin{align} \det(U^{*}_{k}DU_{k})\ge\lambda_{m}\det\left [\left(u_{2},\dots,u_{k}\right)^{*}D_{m}\left(u_{2},\dots,u_{k}\right)\right ] \end{align} $$

for $k>1$ , where $D_{m}=\text{diag}\big (\underbrace {\lambda _{m},\dots ,\lambda _{m}}_{m}, \lambda _{m+1},\dots ,\lambda _{n}\big ).$

Proof. When $k=1$ , $U_{k}$ is a unit vector with ith component 0 for all $i>m$ . Then,

$$ \begin{align*}U_{k}^{*}DU_{k}= \sum_{i=1}^{n}\lambda_{i} |u_{i1}|^{2} =\sum_{i=1}^{m}\lambda_{i} |u_{i1}|^{2} \geq \lambda_{m}.\end{align*} $$

Let $k\geq 2$ and $V=\left (u_{2},\dots ,u_{k}\right )$ . Because $D\ge D_{m}$ and $u_{i1}=0$ for $i>m$ , we have

$$ \begin{align*} \det(U_{k}^{*}DU_{k})& \ge \det(U_{k}^{*}D_{m}U_{k}) =\det\begin{pmatrix} \lambda_{m}u^{*}_{1}u_{1} & \lambda_{m}u^{*}_{1}V \\ \lambda_{m}V^{*}u_{1} & V^{*}D_{m}V \end{pmatrix}\\ & =\det\begin{pmatrix} \lambda_{m} & 0 \\ 0 & V^{*}D_{m}V \end{pmatrix} =\lambda_{m}\det\left(V^{*}D_{m}V\right)\\ & = \lambda_{m}\det\left [\left(u_{2},\dots,u_{k}\right)^{*}D_{m}\left(u_{2},\dots,u_{k}\right)\right ].\end{align*} $$

▪

We are ready to present our main result.

Theorem 3.1. Let A and B be $n\times n$ positive semidefinite matrices. Then,

(3.2) $$ \begin{align} \prod\limits_{t=1}^{k}\lambda_{i_{t}}^{\frac1k}(A + B) \ge\prod\limits_{t=1}^{k}\lambda_{i_{t}}^{\frac1k}(A) + \prod\limits_{t=1}^{k}\lambda_{n-t+1}^{\frac1k}(B) \end{align} $$

and

(3.3) $$ \begin{align} {\small \prod\limits_{t=1}^{k}\lambda_{i_{t}}(A + B) \ge\prod\limits_{t=1}^{k}\lambda_{i_{t}}(A) + \prod\limits_{t=1}^{k}\lambda_{n-t+1}(B) +(2^{k}-2)\prod\limits_{t=1}^{k}\left[\lambda_{i_{t}}(A)\lambda_{n-t+1}(B)\right]^{\frac{1}{2}}\!.\;}\notag\\ \end{align} $$

Proof. If $n=1$ , the inequalities become trivial equalities. If $k=1$ , then inequalities (3.2) and (3.3) are the same, and it is the known result (see, e.g., [Reference Zhang23, p. 274])

$$ \begin{align*}\lambda_{i_{1}}(A + B)\geq \lambda_{i_{1}}(A)+\lambda_{n}(B).\end{align*} $$

Let $n\ge 2$ , $k\geq 2$ . By spectral decomposition, there exists a unitary matrix $U=(u_{1}, u_{2}, \dots , u_{n})\in \mathbb {C}^{n\times n} $ , where $u_{1}, u_{2}, \dots , u_{n}$ are orthonormal eigenvectors associated with $\lambda _{1}(A), \lambda _{2}(A), \dots , \lambda _{n}(A)$ , respectively, such that

$$ \begin{align*} A=U\,\mathrm{{diag}}(\lambda_{1}(A), \lambda_{2}(A), \dots,\lambda_{n}(A))U^{*}. \end{align*} $$

For each $i_{t}$ in the sequence $i_{1} < i_{2} < \cdots < i_{k}$ , let $\mathbb {S}^{(0)}_{t}=\text{Span}(u_{1},u_{2}, \ldots ,u_{i_{t}}), t=1, 2, \dots , k.$ Then, $\mathbb {S}^{(0)}_{1}\subset \mathbb {S}^{(0)}_{2}\subset \cdots \subset \mathbb {S}^{(0)}_{k}$ and $\dim \big (\mathbb {S}^{(0)}_{t}\big )={i_{t}}, t=1, 2, \dots , k.$

Let $\{x_{1}, \dots , x_{k}\}$ be any set of orthonormal vectors, where $x_{t}\in \mathbb {S}^{(0)}_{t}, t=1, 2, \dots , k.$ Let $x_{t}=a_{1t}u_{1}+\cdots +a_{i_{t} t} u_{i_{t}}+0u_{i_{t}+1}+\dots + 0 u_{i_{k}}=(u_{1}, \dots , u_{i_{k}})\alpha _{t}$ , where $\alpha _{t} = (a_{1t}, \dots , a_{i_{t} t}, 0, \dots , 0)^{\tiny T}$ (here ${}^{\tiny T}$ is for transpose), $t=1, 2, \dots , k$ . Then,

$$ \begin{align*}U_{k}=(x_{1}, \dots, x_{k})= (u_{1}, \dots, u_{i_{k}})(\alpha_{1}, \dots, \alpha_{k}).\end{align*} $$

Let $V_{k}=(\alpha _{1}, \dots , \alpha _{k})\in \Bbb C^{i_{k}\times k}$ . Because $\{x_{1}, \dots , x_{k}\}$ and $\{u_{1}, \dots , u_{i_{k}}\}$ are orthonormal sets, we see that $V_{k}^{*}V_{k}=I_{k}$ . The components of the first column of $V_{k}$ are zeros except the first $i_{1}$ components (i.e., the last $i_{k}-i_{1}$ components of $\alpha _{1}$ are all equal to zero).

In Lemma 3.1, setting $n=i_{k}$ , $m=i_{1}$ , $U_{k}=V_{k}$ , and applying (3.1) to the $i_{k}\times i_{k}$ matrix $D_{i_{1}}=\mathop {\text {diag}} (\underbrace {\lambda _{i_{1}}(A),\dots , \lambda _{i_{1}}(A)}_{i_{1}}\,, \lambda _{{i_{1}}+1}(A), \dots , \lambda _{i_{k}}(A)),$ we obtain

$$ \begin{align*} \det (U^{*}_{k}AU_{k}) & = \det \big [ (\alpha_{1}, \dots, \alpha_{k})^{*}(u_{1}, \dots, u_{i_{k}})^{*}A (u_{1}, \dots, u_{i_{k}})(\alpha_{1}, \dots, \alpha_{k})\big ]\\ & = \det \big [(\alpha_{1}, \dots, \alpha_{k})^{*}\mathop{\text{diag}} (\lambda_{1}(A), \dots, \lambda_{i_{k}}(A))(\alpha_{1}, \dots, \alpha_{k})\big]\\ &= \det \big [ V_{k}^{*} \mathop{\text{diag}} (\lambda_{1}(A), \dots, \lambda_{i_{k}}(A)) V_{k}\big ]\\ & \geq \lambda_{i_{1}}(A) \det \big [(\alpha_{2}, \dots, \alpha_{k})^{*}D_{i_{1}}(\alpha_{2}, \dots, \alpha_{k}) \big ]. \end{align*} $$

Repeatedly using Lemma 3.1, we get

(3.4) $$ \begin{align} \det(U_{k}^{*}AU_{k})\ge \prod\limits_{t=1}^{k}\lambda_{i_{t}}(A). \end{align} $$

A minimax eigenvalue result of Fan (see [Reference Fan4, Reference Fan5] or [Reference Marshall, Olkin and Arnold13, A.3.a, p. 787]) ensures

$$ \begin{align*} \det(U_{k}^{*}BU_{k})\ge \min_{U^{*}U=I_{k}}\det (U^{*}BU)=\prod\limits_{t=1}^{k}\lambda_{n-t+1}(B).\end{align*} $$

An application of the Minkowski inequality (1.3) together with (3.4) reveals

$$ \begin{align*} \det\big [ {U_{k}^{*}(A + B)U_{k}}\big ] & = \det({U_{k}^{*}AU_{k} + U_{k}^{*}BU_{k}})\\ & \ge \left[\big (\det({U_{k}^{*}AU_{k} }) \big )^{\frac{1}{k}}+\big(\det({U_{k}^{*}BU_{k}})\big)^{\frac{1}{k}}\right]^{k}\notag\\ & \ge \left [\prod\limits_{t=1}^{k}\lambda^{\frac{1}{k}}_{i_{t}}(A)+\prod\limits_{t=1}^{k}\lambda^{\frac{1}{k}}_{n-t+1}(B)\right]^{k}. \end{align*} $$

Thus, by Lemma 2.1, we derive (3.2) and (3.3) as follows:

$$ \begin{align*} \prod\limits_{t=1}^{k}\lambda_{i_{t}}(A + B) & = \max\limits_{\substack{\mathbb{S}_{1}\subset\cdots\subset \mathbb{S}_{k}\subset \mathbb{C}^{n}\\ \dim \mathbb{S}_{t}=i_{t}}} \min\limits_{\substack{{y}_{t}\in \mathbb{S}_{t}, \,({y}_{r},{y}_{s})=\delta_{rs}\\ W_{k}=\left(y_{1},\dots,y_{k}\right)}} \det \big [{W_{k}^{*}(A + B)W_{k}}\big ]\\ & \ge \min\limits_{\substack{{x}_{t}\in \mathbb{S}^{(0)}_{t}\, ({x}_{r},{x}_{s})=\delta_{rs}\\ U_{k}=\left(x_{1},\dots,x_{k}\right)}} \det\big [{U_{k}^{*}(A + B)U_{k}}\big ] \\ & \ge \left [\prod\limits_{t=1}^{k}\lambda^{\frac{1}{k}}_{i_{t}}(A) +\prod\limits_{t=1}^{k}\lambda^{\frac{1}{k}}_{n-t+1}(B)\right]^{k}\\ & \geq \prod\limits_{t=1}^{k}\lambda_{i_{t}}(A) + \prod\limits_{t=1}^{k}\lambda_{n-t+1}(B)\\ &\quad +(2^{k}-2)\prod\limits_{t=1}^{k}\left[\lambda_{i_{t}}(A) \lambda_{n-t+1}(B)\right]^{\frac{1}{2}}. \end{align*} $$

The last inequality is by (2.3) with $x= \prod \limits _{t=1}^{k}\lambda _{i_{t}}(A)$ and $y= \prod \limits _{t=1}^{k}\lambda _{n-t+1}(B)$ .▪

In (3.2), letting $i_{t}=n-k+t$ , $t=1, 2, \ldots , k$ , we arrive at the inequalities (1.2) of Oppenheim for the product of k smallest eigenvalues. Setting $i_{t}=t$ , $t=1, 2, \ldots , k$ , we obtain analogous inequalities of (1.2) for the product of k largest eigenvalues.

Corollary 3.1. Let A and B be $n\times n$ positive semidefinite matrices. Then,

(3.5) $$ \begin{align} \prod\limits_{t=1}^{k}\lambda_{t}^{\frac1k}(A + B) \ge\prod\limits_{t=1}^{k}\lambda_{t}^{\frac1k}(A) + \prod\limits_{t=1}^{k}\lambda_{n-t+1}^{\frac1k}(B). \end{align} $$

What follows is a lower bound for the product of any two eigenvalues of the sum in terms of the eigenvalues of individual matrices.

Corollary 3.2. Let A and B be $n\times n$ positive semidefinite matrices. Then,

$$ \begin{align*} \lambda_{i}(A+B)\lambda_{j}(A+B) & \ge \lambda_{i}(A)\lambda_{j}(A) +\lambda_{n-1}(B)\lambda_{n}(B)\\ &\quad+2\big [\lambda_{i}(A)\lambda_{j}(A) \lambda_{n-1}(B)\lambda_{n}(B)\big ]^{\frac{1}{2}}. \end{align*} $$

Remark: In view of Fiedler’s (1.1) and inequalities (3.3), it is tempting to have

(3.6) $$ \begin{align} \prod\limits_{t=1}^{k}\lambda_{i_{t}}(A + B) \ge\prod\limits_{t=1}^{k} \big [ \lambda_{i_{t}}(A) + \lambda_{n-t+1}(B)\big ]. \end{align} $$

However, this need not be true. Take $A=B=\left ( \begin {smallmatrix} 1 & 0 \\ 0 & 0 \end {smallmatrix}\right )$ , $k=2, i_{1}=1, i_{2}=2$ . Then, the left-hand side of (3.6) is 0, whereas the right-hand side is 1. (Note: it is always true that $\lambda _{i}(A + B)\geq \lambda _{i}(A) + \lambda _{n}(B)$ . See, e.g., [Reference Zhang23, p. 274]).

The following inequality in (3.7) is Fiedler’s first inequality in (1.1). Inequality in (3.8) is stronger than Fiedler’s second inequality. Inequality in (3.8) is proved by Lemma 2.3. The inequalities in (3.9) is immediate from the inequalities in (3.3) in the theorem.

Corollary 3.3. Let $A, B\in \mathbb {C}^{n\times n}$ be positive semidefinite matrices. Then,

(3.7) $$ \begin{align} \prod\limits_{t=n-k+1}^{n}\lambda_{t}(A + B) &\ge \prod\limits_{t=n-k+1}^{n} \big [\lambda_{t}(A)+\lambda_{t}(B)\big ] \end{align} $$

(3.8) $$ \begin{align} & \ge\prod\limits_{t=n-k+1}^{n}\lambda_{t}(A) + \prod\limits_{t=n-k+1}^{n}\lambda_{t}(B) \\ &+\left(2^{k}-2\right)\prod\limits_{t=n-k+1}^{n}\left [\lambda_{t}(A)\lambda_{t}(B)\right ]^{\frac{1}{2}}\nonumber \end{align} $$

and

(3.9) $$ \begin{align} \prod\limits_{t=1}^{k}\lambda_{t}(A + B) & \ge\prod\limits_{t=1}^{k}\lambda_{t}(A) + \prod\limits_{t=1}^{k}\lambda_{n-t+1}(B) \\ &\quad +\left(2^{k}-2\right)\prod\limits_{t=1}^{k} \left [\lambda_{t}(A)\lambda_{n-t+1}(B)\right ] {{}}^{\frac{1}{2}}. \notag \end{align} $$

Remark: In view of (3.7), it is appealing in the display (3.9) to have

$$ \begin{align*}\prod\limits_{t=1}^{k}\lambda_{t}(A + B)\;\; \ge \;\; \prod\limits_{t=1}^{k} \big [\lambda_{t}(A)+\lambda_{t}(B)\big ]\;\; \mbox{or}\;\; \prod\limits_{t=1}^{k} \big [\lambda_{t}(A)+\lambda_{n-t+1}(B)\big ] .\end{align*} $$

But neither one is true. Let $A=\left ( \begin {smallmatrix} 1 & 0 \\ 0 & 0 \end {smallmatrix}\right )$ , $B=\left ( \begin {smallmatrix} 0& 0 \\ 0 & 1 \end {smallmatrix}\right )$ , $k=1$ . Then, $\lambda _{1}(A+B)=1<2= \lambda _{1}(A)+\lambda _{1}(B).$ See the counterexample below (3.6) for the second case.

4 Hua–Marcus inequalities for contractions

A matrix $A\in \mathbb {C}^{m\times n}$ is said to be (strictly) contractive if $I_{n}-A^{*}A\geq 0$ ( $>$ 0); equivalently, the largest singular value (i.e., the spectral norm) of A is less than or equal to (resp. $<$ ) $1$ . We use $\mathcal {C}_{m\times n}$ to denote the set of $m\times n$ contractive matrices and $\mathcal {S}\mathcal {C}_{m\times n}$ to denote the set of $m\times n$ strictly contractive matrices.

In this section, we apply our main theorem to derive some inequalities of Hua–Marcus type for contractive matrices. We begin by citing Hua’s results in [Reference Hua10].

Theorem 4.1. Hua [Reference Hua10, Theorems 1 and 2]

Let $A,B\in \mathcal {S}\mathcal {C}_{n\times n}$ . Then,

(4.1) $$ \begin{align} \det(I-A^{*}A) \det(I-B^{*}B) + \left|\det(A-B)\right|{{}}^{2}\le \left|\det(I-A^{*}B)\right|{{}}^{2}. \end{align} $$

Consequently,

(4.2) $$ \begin{align} \det(I-A^{*}A) \det(I-B^{*}B)\le \left|\det(I-A^{*}B)\right|{{}}^{2}. \end{align} $$

Equality in (4.1) holds if and only if $A=B$ .

The following [Reference Zhang23, Theorem 7.18] is a reversal of the Hua inequality (4.1) for general $n\times n$ matrices A and B that need not be contractive:

(4.3) $$ \begin{align} \left|\det(I-A^{*}B)\right|{{}}^{2}\le\det(I+A^{*}A)\det(I+B^{*}B)-\left|\det(A+B)\right|{{}}^{2}. \end{align} $$

Marcus [Reference Marcus12] extended the Hua inequality (4.2) to the inequalities of eigenvalues.

Theorem 4.2. Marcus [Reference Marcus12, Theorem]

Let $A,B\in \mathcal {C}_{n\times n}$ . Then,

(4.4) $$ \begin{align} \prod\limits_{t=1}^{k}|\lambda_{n-t+1}(I-A^{*}B)|^{2}\ge \prod\limits_{t=1}^{k}\big[ 1-\lambda_{t}(A^{*}A)\big]\big[1-\lambda_{t}(B^{*}B)\big]. \end{align} $$

Because $|\det (I-A^{*}B)|= \prod \limits _{t=1}^{n}|\lambda _{n-t+1}(I-A^{*}B)|=\prod \limits _{t=1}^{n}\sigma _{n-t+1}(I-A^{*}B)$ , by Weyl’s log-majorization inequality ([Reference Weyl18] or [Reference Marshall, Olkin and Arnold13, p. 317]), the eigenvalues in absolute values are log-majorized by the singular values, and we get the stronger inequality than (4.4):

(4.5) $$ \begin{align} \prod\limits_{t=1}^{k}\sigma^{2}_{n-t+1}(I-A^{*}B)\ge \prod\limits_{t=1}^{k}\big[ 1-\lambda_{t}(A^{*}A)\big]\big[1-\lambda_{t}(B^{*}B)\big]. \end{align} $$

Related inequalities of Hua–Marcus type are seen in [Reference Ando3, Reference Oppenheim14–Reference Thompson and Freede16, Reference Xu, Xu and Zhang21, Reference Xu, Xu and Zhang22].

The following result is stronger and more general than (4.5).

Theorem 4.3. Let $A, B\in \mathcal {S}\mathcal {C}_{n\times n}$ . Then,

(4.6) $$ \begin{align} \prod\limits_{t=1}^{k}\sigma^{2}_{i_{t}}\left(I-A^{*}B\right) & \ge \prod\limits_{t=1}^{k} \big [1-\lambda_{t}(A^{*} A)\big ] \big [1-\lambda_{n-i_{t}+1}(B^{*}B) \big ] \end{align} $$

(4.7) $$ \begin{align} &\qquad\qquad\qquad\qquad\qquad +\prod\limits_{t=1}^{k} \frac{\big [1-\lambda_{t}(A^{*}A)\big ]\,\sigma^{2}_{n-t+1}(A-B)}{1-\lambda_{n-t+1}(A^{*}A)}\notag\\ &\qquad\qquad\qquad\qquad\qquad + (2^{k}-2)\prod\limits_{t=1}^{k} \big [ 1-\lambda_{t}(A^{*}A)\big ] \big [ \lambda_{i_{t}}(F)\lambda_{n-t+1}(H)\big ]^{\frac{1}{2}}\!,\; \end{align} $$

where $F=I-B^{*}B$ and $ H=(A-B)^{*}(I-AA^{*})^{-1}(A-B).$

Proof. For $A, B\in \mathcal {S}\mathcal {C}_{n\times n}$ , let

$$ \begin{align*}F=I-B^{*}B\;\;\; \mbox{and}\;\;\; H=(A-B)^{*}(I-AA^{*})^{-1}(A-B).\end{align*} $$

Then (see [Reference Hua10, Theorem 1] or [Reference Zhang23, pp. 230–231]),

(4.8) $$ \begin{align} F+H= (I-B^{*}A)(I-A^{*}A)^{-1}(I-A^{*}B). \end{align} $$

Applying the left inequality of (2.1) with $i_{t}=n-t+1$ to H reveals

(4.9) $$ \begin{align} \prod\limits_{t=1}^{k}\lambda_{n-t+1}(H) & = \prod\limits_{t=1}^{k} \lambda_{n-t+1}\big [ (I-AA^{*})^{-1} (A-B)(A-B)^{*}\big ] \notag \\ & \ge \prod\limits_{t=1}^{k}\lambda_{n-t+1}\big [ (I-AA^{*})^{-1} \big ] \sigma^{2}_{n-t+1}\big(A-B\big)\notag \\ & = \prod\limits_{t=1}^{k}[1-\lambda_{n-t+1}(A^{*}A)]^{-1}\sigma^{2}_{n-t+1}\big(A-B\big). \end{align} $$

Applying the right inequality of (2.1) to $F+H$ in the product form in (4.8) yields

(4.10) $$ \begin{align} \prod\limits_{t=1}^{k}\lambda_{i_{t}}\left(F+H\right)\leq \prod\limits_{t=1}^{k}\left[1-\lambda_{t}(A^{*}A)\right]^{-1}\sigma^{2}_{i_{t}}\left(I-A^{*}B\right). \end{align} $$

Thus,

(4.11) $$ \begin{align} \prod\limits_{t=1}^{k}\sigma^{2}_{i_{t}}\left(I-A^{*}B\right) \ge \prod\limits_{t=1}^{k}\left[1-\lambda_{t}(A^{*}A)\right]\lambda_{i_{t}}\left(F+H\right). \end{align} $$

By inequalities (3.3) and (4.9), we obtain

(4.12) $$ \begin{align} &\prod\limits_{t=1}^{k}\lambda_{i_{t}}\left(F+H\right)\nonumber \\ & \geq \prod\limits_{t=1}^{k}\lambda_{i_{t}}(F) + \prod\limits_{t=1}^{k}\lambda_{n-t+1}(H) +(2^{k}-2)\prod\limits_{t=1}^{k}\left[\lambda_{i_{t}}(F)\lambda_{n-t+1}(H)\right]^{\frac{1}{2}} \nonumber \\ & \ge \prod\limits_{t=1}^{k}\left[1-\lambda_{n-i_{t}+1}(B^{*}B)\right] + \prod\limits_{t=1}^{k}\left[1-\lambda_{n-t+1}(A^{*}A)\right]^{-1}\sigma^{2}_{n-t+1}(A-B)\nonumber \\ &\quad + (2^{k}-2)\prod\limits_{t=1}^{k}\left[\lambda_{i_{t}}(F)\lambda_{n-t+1}(H)\right]^{\frac{1}{2}}. \end{align} $$

Combining (4.11) and (4.12), we obtain the desired inequalities. ▪

Setting $k=n$ in Theorem 4.3 gives a stronger version of the Hua’s inequality.

Corollary 4.1. Let $A, B\in \mathcal {S}\mathcal {C}_{n\times n}$ . Then,

$$ \begin{align*} &\left|\det\left(I-A^{*}B\right)\right|{{}}^{2} \ge\det\left(I-A^{*}A\right)\det\left(I-B^{*}B\right) +\left|\det\left(A-B\right)\right|{{}}^{2}\notag\\ &\qquad \qquad\qquad \qquad+(2^{n}-2)\big[\det\left(I-A^{*}A\right)\det\left(I-B^{*}B\right)\big]^{\frac{1}{2}}\left|\det\left(A-B\right)\right|. \end{align*} $$

Below is a reversal inequality of the previous theorem.

Theorem 4.4. Let $A,B\in \mathbb {C}^{n\times n}$ . Then,

(4.13) $$ \begin{align} \prod\limits_{t=1}^{k}\sigma^{2}_{i_{t}}\big(I-A^{*}B\big) & \leq \prod\limits_{t=1}^{k}\big[1+\lambda_{i_{t}}(A^{*}A )\big]\big[1+\lambda_{t}(B^{*}B)\big]-\prod\limits_{t=1}^{k}\sigma^{2}_{n-t+1}\big(A+B\big) \notag \\ &\quad - (2^{k}-2)\prod\limits_{t=1}^{k}\sigma_{i_{t}}\big(I-A^{*}B\big)\sigma_{n-t+1}\big(A+B\big). \end{align} $$

Proof. For $A, B\in \mathbb {C}^{n\times n}$ , the following matrix identity holds (see, e.g., [Reference Zhang23, p. 228]):

$$ \begin{align*}I+A^{*}A=P+Q,\end{align*} $$

where

$$ \begin{align*}P=(A+B)^{*}(I+BB^{*})^{-1}(A+B).\end{align*} $$

$$ \begin{align*}Q=(I-A^{*}B)(I+B^{*}B)^{-1}(I-A^{*}B)^{*}. \end{align*} $$

By Lemma 2.2, we have

(4.14) $$ \begin{align} \prod\limits_{t=1}^{k}\lambda_{n-t+1}(P)&\ge \prod\limits_{t=1}^{k}\big[1+\lambda_{t}(BB^{*})\big]^{-1}\sigma^{2}_{n-t+1}\big(A+B\big), \notag\\ \prod\limits_{t=1}^{k}\lambda_{i_{t}}(Q) & \ge \prod\limits_{t=1}^{k}\big[1+\lambda_{t}(B^{*}B)\big]^{-1}\sigma^{2}_{i_{t}}\big(I-A^{*}B\big). \end{align} $$

Using (3.3) in Theorem 3.1, we derive

$$ \begin{align*} &\prod\limits_{t=1}^{k}\big[1+\lambda_{i_{t}}(A^{*}A )\big]=\prod\limits_{t=1}^{k}\lambda_{i_{t}}(P+Q)) \\ & \ge \prod\limits_{t=1}^{k}\lambda_{n-t+1}(P) +\prod\limits_{t=1}^{k}\lambda_{i_{t}}(Q)+(2^{k}-2) \prod\limits_{t=1}^{k}[\lambda_{n-t+1}(P)\lambda_{i_{t}}(Q)]^{\frac{1}{2}}\\ & \ge \prod\limits_{t=1}^{k}\big[1+\lambda_{t}(B^{*}B)\big]^{-1} \bigg [\prod\limits_{t=1}^{k}\sigma^{2}_{i_{t}}\big(I-A^{*}B\big)+\prod\limits_{t=1}^{k}\sigma^{2}_{n-t+1}\big(A+B\big) \\ &\quad +(2^{k}-2)\prod\limits_{t=1}^{k}\sigma_{i_{t}}\big(I-A^{*}B\big)\sigma_{n-t+1}\big(A+B\big)\bigg ]. \end{align*} $$

Multiplying by $\prod \limits _{t=1}^{k}\big [1+\lambda _{t}(B^{*}B)\big ]$ , we obtain the desired inequalities (4.13). ▪

As (4.3) is a reversal of Hua’s (4.1), the following result, as a special case of Theorem 4.4 by setting $k=n$ , may be viewed as a counterpart of Corollary 4.1.

Corollary 4.2. Let $A,B\in \mathbb {C}^{n\times n}$ . Then,

$$ \begin{align*} \big|\det\left(I-A^{*}B\right)\big|^{2} & \le \det\big(I+A^{*}A\big)\det\left(I+B^{*}B\right)-\big|\det\left(A+B\right)\big|^{2} \\ &\quad -(2^{n}-2)\big|\det\left(I-A^{*}B\right)\big|\big|\det\left(A+B\right)\big|. \end{align*} $$

Consequently,

$$ \begin{align*} \big|\det\left(I-A^{*}B\right)\big|^{2}&\le\det\left(I+A^{*}A\right)\det\left(I+B^{*}B\right) -\big|\det\left(A+B\right)\big|^{2}. \end{align*} $$