An extension of Harnack type determinantal inequality

TL;DR

This paper revisits a classical Harnack type determinantal inequality for contractive matrices and extends it to include multiple positive semidefinite matrices, broadening its applicability.

Contribution

It provides a new extension of the Harnack type determinantal inequality involving multiple positive semidefinite matrices, enhancing previous results.

Findings

Extended the inequality to multiple matrices

Provided new bounds for positive semidefinite matrices

Clarified the conditions under which the inequality holds

Abstract

We revisit and comment on the Harnack type determinantal inequality for contractive matrices obtained by Tung in the nineteen sixtieth and give an extension of the inequality involving multiple positive semidefinite matrices.

Equations

$$\prod _{k=1}^n\frac{1-r_k}{1+r_k}\le \frac{\det (I-Z^{\ast }Z)}{|\det (I-UZ){|}^2}\le \prod _{k=1}^n\frac{1+r_k}{1-r_k}.$$

$$\prod _{k=1}^n(1-r_k)\le |\det (I-A)|\le \prod _{k=1}^n(1+r_k)$$

$$\prod _{k=1}^n\frac{|1-r_k|}{1+r_k}\le \frac{|\det (I-Z^{\ast }Z)|}{|\det (I-UZ){|}^2}.$$

$$|\det (I-A)|\le \prod _{k=1}^n(1+r_k).$$

$$\prod _{k=1}^n\frac{|1-r_k|}{1+r_k}\le \frac{|\det (I-Z^2)|}{|\det (I-UZ){|}^2}$$

$$\frac{\det (I-Z^2)}{|\det (I-UZ){|}^2}\le \prod _{k=1}^n\frac{1+r_k}{1-r_k}$$

$$\mbox{If\quad$x\prec_{\rm log}y,$ \quad then \quad$\tilde{x}\prec_{\rm\rm wlog}\tilde{y}$ \;\; {but}\quad$\tilde{x}\not\prec_{\rm log}\tilde{y}$}.$$

$$\prod _{k=1}^n(1+x_k)<\prod _{k=1}^n(1+y_k).$$

$$\prod _{k=1}^n(1-x_k)>\prod _{k=1}^n(1-y_k).$$

$$|\det (I-UZ)|=\prod _{k=1}^n(1+r_k).$$

$$|\lambda (UZ)|{\prec }_{\log }\sigma (UZ)=\sigma (Z)=\lambda (Z),$$

$$0<|\det (I-UZ)|=\prod _{k=1}^n|1-{\lambda }_k(UZ)|\le \prod _{k=1}^n(1+|{\lambda }_k(UZ)|)\le \prod _{k=1}^n(1+r_k).$$

$$\prod _{k=1}^n|1-{\lambda }_k(UZ)|\ge \prod _{k=1}^n(1-|{\lambda }_k(UZ)|)\ge \prod _{k=1}^n(1-{\sigma }_k(Z))=\prod _{k=1}^n(1-r_k).$$

$$\prod _{k=1}^n\prod _{i=1}^m{\left(\frac{1-r_{ik}}{1+r_{ik}}\right)}^{w_i}\le \frac{\det (I-({\sum }_{i=1}^m{w}_i{Z}_i{)}^2)}{|\det (I-U{\sum }_{i=1}^m{w}_i{Z}_i){|}^2}\le \prod _{k=1}^n\prod _{i=1}^m{\left(\frac{1+r_{ik}}{1-r_{ik}}\right)}^{w_i}.$$

$$\frac{1+{\sum }_{i=1}^n{\alpha }_i{x}_i}{1-{\sum }_{i=1}^n{\alpha }_i{x}_i}\le \prod _{i=1}^n{\left(\frac{1+x_i}{1-x_i}\right)}^{{\alpha }_i},$$

$$\lambda (W)\prec \sum _{i=1}^m{w}_i\lambda (Z_i),\text{mbox}i.e.,\sum _{k=1}^{\ell }{s}_k\le \sum _{k=1}^{\ell }\sum _{i=1}^m{w}_i{r}_{ik}^{\downarrow },\ell =1,2,\dots ,n.$$

$$\sum _{k=1}^n\ln \frac{1+s_k}{1-s_k}\le \sum _{k=1}^n\ln \frac{1+{\sum }_{i=1}^m{w}_i{r}_{ik}^{\downarrow }}{1-{\sum }_{i=1}^m{w}_i{r}_{ik}^{\downarrow }},$$

$$\begin{array}{rlr}\prod _{k=1}^n\frac{1+s_k}{1-s_k}& \le & \prod _{k=1}^n\frac{1+{\sum }_{i=1}^m{w}_i{r}_{ik}^{\downarrow }}{1-{\sum }_{i=1}^m{w}_i{r}_{ik}^{\downarrow }}\\ & \le & \prod _{k=1}^n\prod _{i=1}^m{\left(\frac{1+r_{ik}^{\downarrow }}{1-r_{ik}^{\downarrow }}\right)}^{w_i}\\ & =& \prod _{k=1}^n\prod _{i=1}^m{\left(\frac{1+r_{ik}}{1-r_{ik}}\right)}^{w_i},\end{array}$$

$$\frac{\det (I-({\sum }_{i=1}^m{w}_i{Z}_i{)}^2)}{|\det (I-U{\sum }_{i=1}^m{w}_i{Z}_i){|}^2}\le \prod _{k=1}^n\left(\frac{1+s_k}{1-s_k}\right).$$

$$\prod _{k=1}^n\frac{1-s_k}{1+s_k}\ge \prod _{k=1}^n\prod _{i=1}^m{\left(\frac{1-r_{ik}}{1+r_{ik}}\right)}^{w_i}.$$

$$\frac{\det (I-({\sum }_{i=1}^m{w}_i{Z}_i{)}^2)}{|\det (I-U{\sum }_{i=1}^m{w}_i{Z}_i){|}^2}\ge \prod _{k=1}^n\left(\frac{1-s_k}{1+s_k}\right).$$

$$\frac{\det (I-{\sum }_{i=1}^m{w}_i{Z}_i^{\ast }{Z}_i)}{|\det (I-U{\sum }_{i=1}^m{w}_i|Z_i|){|}^2}\le \prod _{k=1}^n\prod _{i=1}^m{\left(\frac{1+r_{ik}}{1-r_{ik}}\right)}^{w_i},$$

$$\prod _{k=1}^n\prod _{i=1}^m{\left(\frac{1-r_{ik}}{1+r_{ik}}\right)}^{w_i}\le \frac{\det (I-{\sum }_{i=1}^m{w}_i{Z}_i^{\ast }{Z}_i)}{|\det (I-U{\sum }_{i=1}^m{w}_i|Z_i|){|}^2}.$$

$$Z_{1}=\left(\begin{array}[]{cc}0.34&-0.15\\ -0.15&0.07\end{array}\right),\;\;Z_{2}=\left(\begin{array}[]{cc}0.02&-0.01\\ -0.01&0.01\end{array}\right),\;\;U=\left(\begin{array}[]{cc}-0.60&0.80\\ 0.80&0.60\end{array}\right).$$

$$\prod _{k=1}^n\prod _{i=1}^m{\left(\frac{1-r_{ik}}{1+r_{ik}}\right)}^{w_i}\le \frac{\det (I-{\sum }_{i=1}^m{w}_i{Z}_i^{\ast }{Z}_i)}{|\det (I-{\sum }_{i=1}^m{w}_i{Z}_i){|}^2}\le \prod _{k=1}^n\prod _{i=1}^m{\left(\frac{1+r_{ik}}{1-r_{ik}}\right)}^{w_i}.$$

Full text

In memory of Marvin Marcus.

In 1964, Tung [11] established the following Harnack type determinantal inequality.

[TABLE]

Soon after the appearance of the Tung’s paper, Marcus [8] gave another proof of (1) and pointed out that (1) is equivalent to

[TABLE]

for any $n\times n$ matrix $A$ with the same singular values as the contractive matrix $Z$.

Marcus’s proof of (2) makes use of majorization theory and singular value (eigenvalue) inequalities of Weyl. This approach is still very fruitful today in deriving determinantal inequalities; see, for example, [6, 7]. At about the same time of Marcus’s proof, Hua [2] gave a proof of (2) using the determinantal inequality he had previously obtained in [3].

• (i).

In the well-known book [9] of Marshall, Olkin and Arnold, Tung’s theorem is cited as result E.3 on page 319 in which the condition that $A$ be contractive is missing; that is to say, Theorem 1 need not be true in general if $Z$ is not contractive. Take, for example, $Z=2iI$ with odd $n$ (the matrix size) and appropriate $U$. We see neither the left nor the right inequality in (1) holds. However, the following inequality holds true for any $n\times n$ matrix $Z$ and any $n\times n$ unitary matrix $U$ (a fraction with zero denominator is viewed as $\infty$)

[TABLE]

• (ii).

Inequalities (1) and (2) are not equivalent for general matrices. The right-hand side inequality in (2) is true for all $n\times n$ matrices $A$; that is,

[TABLE]

Using the polar decomposition (see, e.g., [13, p. 83]), we restate and slightly generalize Theorem 1 as follows with discussions on the equality cases.

[TABLE]

with equality if and only if $Z$ has an eigenvalue $1$ or $UZ$ has eigenvalues $-r_{1},-r_{2},$ $\dots,-r_{n}$. If both $Z$ and $I-Z$ are nonsingular, the strict inequality holds for $U\neq-I$.

Moreover, if $0\leq r_{k}<1$, $k=1,2,\ldots,n$, then

[TABLE]

with equality if and only if $UZ$ has eigenvalues $r_{1},r_{2},\dots,r_{n}$. If $Z$ is nonsingular, then the strict inequality in (4) holds if $U\neq I$.

If we use ${\rm Spec}(X)$ to denote the spectrum of the matrix $X$, then equality holds in (3) if and only if $1\in{\rm Spec}(Z)$ or ${\rm Spec}(UZ)={\rm Spec}(-Z)$; and equality holds in (4) if and only if ${\rm Spec}(UZ)={\rm Spec}(Z)$.

To prove this theorem, we need a lemma which is of interest in its own right. We proceed with adoption of standard notation in majorization theory (see, e.g., [13]).

[TABLE]

Consequently,

[TABLE]

Under the lemma’s condition, if all $x_{i}$ and $y_{i}$ are further in $[0,1)$, a similar but reversal log-majorization inequality can be derived for $(1-x_{k})$ and $(1-y_{k})$. In particular,

[TABLE]

In fact, this can be proved by applying [13, Theorem 10.14, p. 343] to $\ln x\prec\ln y$ with $g(t)=-\ln(1-e^{t})$, which is strictly increasing and convex when $t\in(-\infty,0)$.

Proof of Theorem 3. We only need to show the equality cases. For (3), if $Z$ has a singular (eigen-) value 1, then both sides vanish. If $UZ$ has eigenvalues $-r_{1},-r_{2},\dots,-r_{n}$, then $\det(I-UZ)=\prod_{k=1}^{n}(1+r_{k})$. Equality is readily seen. Conversely, suppose equality occurs in (3). We further assume that no $r_{k}$ ($k=1,2,\dots,n$) equals $1$. Since $|\det(I-Z^{2})|=\prod_{k=1}^{n}|1-r_{k}|(1+r_{k})$, we have

[TABLE]

Moreover, by Weyl majorization inequality (see, e.g., [13, Corollary 10.2, p. 353]),

[TABLE]

where $\lambda(X)$ and $\sigma(X)$ denote the vectors of the eigenvalues and singular values of matrix $X$, respectively. With $\lambda_{k}(X)$ denoting the eigenvalues of the $n\times n$ matrix $X$, $k=1,2,\dots,n$, by the lemma, we have

[TABLE]

Thus, (7) yields $|1-\lambda_{k}(UZ)|=1+|\lambda_{k}(UZ)|$ for all $k$, which implies $\lambda_{k}(UZ)\leq 0$ for $k=1,2,\dots,n$, i.e., all eigenvalues of $-UZ$ are nonnegative. If $|\lambda(UZ)|=\lambda(-UZ)$ is not a permutation of $\lambda(Z)$, then, by strict inequality (5), we have $\prod_{k=1}^{n}(1+|\lambda_{k}(UZ)|)<\prod_{k=1}^{n}(1+\lambda_{k}(Z))=\prod_{k=1}^{n}(1+r_{k})$, a contradiction to (7). It follows that $UZ$ has the eigenvalues $-r_{1},-r_{2},\dots,-r_{n}$.

For the equality in (4), it occurs if and only if $\prod_{k=1}^{n}(1-r_{k})=|\det(I-UZ)|$. Note that $|\lambda(UZ)|\prec_{\rm log}\sigma(Z)$ and

[TABLE]

The first equality in (8) occurs if and only if all $\lambda_{k}(UZ)$ are in $[0,1)$; the second equality occurs if and only if $\lambda(UZ)$ is a permutation of $\sigma(Z)$, i.e., ${\rm Spec}(UZ)={\rm Spec}(Z)$.

Now assume that $Z$ is nonsingular and suppose that equality holds in (4). Then $UZ$ has eigenvalues $r_{1},r_{2},\dots,r_{n}$. Moreover, the singular values of $UZ$ are $r_{1},r_{2},\dots,r_{n}$. Let $P=UZ$. Then the eigenvalues of $P$ are just the singular values of $P$. So $P$ is positive definite. It follows that $U=PZ^{-1}$ has only positive eigenvalues. Since $U$ is unitary, $U$ has to be the identity matrix. The case for (3) is similar. $\Box$

In what follows, we prove an extension of Theorem 1 that involves multiple matrices.

[TABLE]

Equality on the left-hand side occurs if and only if all $Z_{i}$ are equal to $Z$, say, and $Z$ has an eigenvalue 1 or Spec($UZ$)=Spec($-Z$) (in which $U=-I$ if $Z$ is nonsingular); Equality on the right-hand side occurs if and only if all $Z_{i}$ are equal to $Z$, say, and Spec($UZ$)=Spec($Z$) (in which $U=I$ if $Z$ is nonsingular).

Clearly, if $m=1$, then (9) reduces to (1).

Our additional ingredients in proving (9) are Fan’s majorization relation that $\lambda(H+S)\prec\lambda(H)+\lambda(S)$ for $n\times n$ Hermitian matrices $H$ and $S$ (see, e.g., [13, p. 356]) and Lewent’s inequality [4, 5] which asserts, for $x_{i}\in[0,1)$, $i=1,2,\dots,n$,

[TABLE]

where $\sum_{i=1}^{n}\alpha_{i}=1$ and all $\alpha_{i}>0$. Equality holds if and only if $x_{1}=x_{2}=\cdots=x_{n}.$

Let $r_{ik}^{\downarrow}$ be the $k$th largest eigenvalue of $Z_{i}$, $i=1,2,\ldots,m$, and $s_{k}$ be the $k$th largest eigenvalue of $W:=\sum_{i=1}^{m}w_{i}Z_{i}$, $k=1,2,\ldots,n$. Fan’s majorization relation implies

[TABLE]

(Note that the components of $\lambda(\cdot)$ are in nonincreasing order.) Now the convexity and the monotonicity of the function $f(t)=\ln\frac{1+t}{1-t}$, $0\leq t<1$, imply (see, e.g., [13, p. 343])

[TABLE]

where equality holds if and only if $s_{k}=\sum_{i=1}^{m}w_{i}r_{ik}^{\downarrow}$ for all $k$; that is, $\lambda(W)=\sum_{i=1}^{m}w_{i}\lambda(Z_{i})$. It follows that all $Z_{i}$ are simultaneously unitarily diagonalizable with their eigenvalues on the main diagonals in the same order (nonincreasing, say).

Applying the exponential function to both sides and using Lewent’s inequality yield

[TABLE]

in which equality occurs in the second inequality if and only if $r_{1k}=r_{2k}=\cdots=r_{mk}$ for $k=1,2,\dots,n$. Thus both equalities in (10) hold if and only if $Z_{1}=Z_{2}=\cdots=Z_{m}$.

By (4), we have

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Combining (10) and (11) gives the second inequality of (9).

Note that the inequalities in (10) reverse by taking reciprocals, which implies

[TABLE]

Then by (3), we have

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Combining (12) and (13) yields the first inequality of (9).

If either equality holds in (9), then all $Z_{i}$ are equal to $Z$, say. The conclusions are immediate from Theorem 3. ∎

The absolute value of a complex matrix $Z$, denoted by $|Z|$, is the positive semidefinite square root of $Z^{*}Z$. The following result extends the second inequality in (1).

[TABLE]

where $w_{i}>0$, $i=1,2,\ldots,m$, such that $\sum_{i=1}^{m}w_{i}=1$. Equality occurs if and only if all $Z_{i}$ have the same absolute value, say $Z$, and Spec($UZ$)=Spec($Z$) (in which $U=I$ if $Z$ is nonsingular).

In view of (9), it is tempting to have the lower bound inequality

[TABLE]

However, this is not true. Set $m=n=2$, $w_{1}=w_{2}=1/2$ and take

[TABLE]

One may check that the left hand side is $0.6281$, while the right hand side is $0.6250$.

The unitary matrix in Theorem 1 seems superfluous, as one could replace $Z$ with $U^{*}Z$ and leave the singular values unchanged. So perhaps a more natural extension of Theorem 1 to more matrices is giving an upper bound and lower bound, in terms of the singular values of individual matrices, for the quantity ${\displaystyle\frac{\det(I-\sum_{i=1}^{m}w_{i}Z_{i}^{*}Z_{i})}{|\det(I-\sum_{i=1}^{m}w_{i}Z_{i})|^{2}}}$, where $Z_{i}$, $i=1,2,\ldots,m$, are general contractive matrices. We would guess

[TABLE]

The first inequality in (14) is untrue in general as it is disproved by substituting $Z_{1}$ and $Z_{2}$ in (14) with $U|Z_{1}|$ and $U|Z_{2}|$, respectively, in the previous example. However, simulation seems to support the second inequality which is unconfirmed yet.

Acknowledgments

The work was partially supported by National Natural Science Foundation of China (NNSFC) No. 11601314 and No. 11571220.