Abstract
Let \(A_{i}, B_{i}, X_{i}\) and \(Y_{i}\) be \(n \times n\) complex matrices such that \(X_{i}\) and \(Y_{i}\) are positive semidefinite for \(i = 1,2, \ldots ,n\). We show that
for all \(p\in [0,1]\), where
We also give other results that represent a generalizations of the celebrated arithmetic–geometric mean inequality.
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1 Introduction
On the space of \(n\times {n}\) complex matrices, denoted by \(M_{n}(\mathbb {C})\), \({\left| \left| \left| . \right| \right| \right| }\) stands for the unitarily invariant norm satisfying the property that \({\left| \left| \left| A \right| \right| \right| }={\left| \left| \left| UAV \right| \right| \right| }\) for all \(A,U,V \in {M_{n}(\mathbb {C})},\) where U and V are unitary. For \(A\in {M_{n}(\mathbb {C})}\), let \(s_{i}(A)\), for \(i = 1,2, \ldots ,n\), be the eigenvalues of the matrix \({\left| A \right| }=(A^{*}A)^{\frac{1}{2}}\), arranged so that \({s_i}(A) \ge {s_{i + 1}}(A)\), for \(i = 1,2, \ldots ,n - 1\). It is well known that for \(A\in {M_{n}(\mathbb {C})}\), \(s_{i}(A^{*})=s_{i}(A)\) and \(s_{i}({\left| A \right| })=s_{i}(A)\) for \(i=1,2,\ldots ,n\). For \(A\in {M_{n}(\mathbb {C})}\), the celebrated norm \({\left| \left| A \right| \right| }_{(k)}=\sum _{j=1}^{k} s_{j}(A)\) where \(k=1,2,\ldots ,n\) is called the Ky Fan norm of A, according to the Fan dominance theorem (see [4, p. 93]), \({\left| \left| \left| A \right| \right| \right| } \le {\left| \left| \left| B \right| \right| \right| }\) if and only if \({\left| \left| A \right| \right| }_{(k)}\le {\left| \left| B \right| \right| }_{(k)}\) for \(k=1,2,\ldots ,n\), where \(A,B\in {M_{n}(\mathbb {C})}\).
Throughout this paper, we confine our results to matrices regarded as operators acting on finite dimensional Hilbert space. Inequalities involving unitarily invariant norms have been of special interest; see e.g., [11].
The authors, in [7], obtained a matrix version of the arithmetic–geometric mean inequality; this version asserts that if \(A, B\in {M_{n}(\mathbb {C})}\), then
for \(i=1,2,\ldots ,n\). Using the Fan dominance theorem and inequality (1), we get that
In [5, 9], a generalization of inequality (2) has been obtained, which asserts that if \(A, B, X\in {M_{n}(\mathbb {C})}\), then
The Cauchy–Schwarz inequality for matrices says that if \(A, B\in {M_{n}(\mathbb {C})}\), then
A generalization of inequality (4) asserts that if \(A, B, X\in {M_{n}(\mathbb {C})}\), then
We refer to [6, 8] and the references therein for more general forms of the Cauchy–Schwarz inequality.
Audenaert proved in [3] an inequality that interpolates between inequality (2) and inequality (4); this inequality says that if \(A, B\in {M_{n}(\mathbb {C})}\), then
for all \(p\in [0,1]\).
In [10], the authors obtained an interpolation inequality between inequality (3) and inequality (5). This inequality, which is a generalization of inequality (6), asserts that if \(A, B, X\in M_{n}(\mathbb {C})\), then
for all \(p\in [0,1]\).
Very recently, it has been obtained in [2] that if \(A,B,S,T\in M_{n}(\mathbb {C})\), then
for \(i = 1,2, \ldots ,n\). The norm version of this inequality asserts that
It has been concluded in the same paper that if \(A,B,S,T\in M_{n}(\mathbb {C})\) such that S and T are positive semidefinite, then
for \(i = 1,2, \ldots ,n\). The norm version of inequality (10) says that
Moreover, the author in this paper obtained that if \({A_i},{B_i},{S_i},{T_i}\in M_{n}(\mathbb {C})\) such that \({S_i}\) and \({T_i}\) are positive semidefinite for \(i = 1,2, \ldots ,n\), then
for \(i = 1,2, \ldots ,n\), where \(W = \left[ {\begin{array}{*{20}{c}} {{A_1}{S_1}^{\frac{1}{2}}}&{}{{A_2}{S_2}^{\frac{1}{2}}}&{} \cdots &{}{{A_n}{S_n}^{\frac{1}{2}}}\\ {{B_1}{T_1}^{\frac{1}{2}}}&{}{{B_2}{T_2}^{\frac{1}{2}}}&{} \cdots &{}{{B_n}{T_n}^{\frac{1}{2}}} \end{array}} \right]\). Inequality (12) has the following norm version
In [1], the authors obtained that if \(A_{i}, B_{i}, S_{i}\in {M_{n}(\mathbb {C})}\) for \(i=1, 2,\ldots , n\) such that each \(S_{i}\) is positive semidefinite and if \(p\in [0,1]\), then
In the second section, we give a generalizations of inequalities (9), (11), and a generalization of inequality (13) in the same manner in inequality (14).
2 Main results
Before we start our work, we need the following lemmas; the first can be found in [4], while the second is a direct consequence of Weyl’s monotonicity theorem (see [4, p. 63]) and Fan dominance theorem.
Lemma 1
If \(X \in {M_{n}(\mathbb {C})}\) is Hermitian, then
Lemma 2
Let \(A,B \in {M_{n}(\mathbb {C})}\) be positive semidefinite such that \(A \le B\). Then,
We begin this section with our first result which can be considered as a generalization of inequality (9).
Lemma 3
Let \(A, B, X, Y \in {M_{n}(\mathbb {C})}\). Then,
for all \(p\in [0,1]\).
Proof
Replace A and B by \(X^{*}A^{*}\) and \(Y^{*}B^{*}\) respectively in inequality (6) to get the result.
Our second result is a generalization of inequality (11).
Lemma 4
Let \(A, B, X, Y \in {M_{n}(\mathbb {C})}\) be such that X and Y are positive semidefinite. Then,
for all \(p\in [0,1]\).
Proof
Replace X and Y by \(X^\frac{1}{2}\) and \(Y^\frac{1}{2},\) respectively, in inequality (15) to get the result.
Note that substituting \(p=\frac{1}{2}\) and \(X=Y=I\) in inequalities (15) and (16) leads to the arithmetic–geometric mean inequality (2). Moreover, substituting \(p=0\) or 1 and \(X=Y=I\) in inequalities (15) and (16) leads to the Cauchy–Schwarz inequality (4).
Corollary 1
Let \(A, B, X, Y \in {M_{n}(\mathbb {C})}\) be positive semidefinite. Then,
for all \(p\in [0,1]\).
Proof
Substitute \(A=A^\frac{1}{2}\) and \(B=B^\frac{1}{2}\) in inequality (16) to get the result.
Now, we will introduce the main result of our paper which can be considered as a generalization of inequality (13).
Theorem 1
Let \(A_{i}, B_{i}, X_{i}, Y_{i} \in {M_{n}(\mathbb {C})}\) be such that \(X_{i}\) and \(Y_{i}\) are positive semidefinite for \(i = 1,2, \ldots ,n\). Then,
for all \(p\in [0,1]\), where
Proof
Consider the \(n^{2}\times n^{2}\) matrices
\(A = \left[ {\begin{array}{*{20}{c}} {{A_1}}&{}{{A_2}}&{} \cdots &{}{{A_n}}\\ 0&{}0&{} \cdots &{}0\\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0&{}0&{} \cdots &{}0 \end{array}} \right]\), \(B = \left[ {\begin{array}{*{20}{c}} {{B_1}}&{}{{B_2}}&{} \cdots &{}{{B_n}}\\ 0&{}0&{} \cdots &{}0\\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0&{}0&{} \cdots &{}0 \end{array}} \right]\), \(X = \left[ {\begin{array}{*{20}{c}} {{X_1}}&{}0&{} \cdots &{}0\\ 0&{}{{X_2}}&{} \cdots &{}0\\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0&{}0&{} \cdots &{}{{X_n}} \end{array}} \right]\), and \(Y = \left[ {\begin{array}{*{20}{c}} {{Y_1}}&{}0&{} \cdots &{}0\\ 0&{}{{Y_2}}&{} \cdots &{}0\\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ 0&{}0&{} \cdots &{}{{Y_n}} \end{array}} \right]\), and let \({M_{ij}} = p{X_i}^{\frac{1}{2}}{A_i}^*{A_j}{X_j}^{\frac{1}{2}} + (1 - p){Y_i}^{\frac{1}{2}}{B_i}^*{B_j}{Y_j}^{\frac{1}{2}}\). Then
and
where \(L=\left[ {\begin{array}{*{20}{c}} {\sqrt{p} {A_1}{X_1}^{\frac{1}{2}}}&{}{\sqrt{p} {A_2}{X_2}^{\frac{1}{2}}}&{} \cdots &{}{\sqrt{p} {A_n}{X_n}^{\frac{1}{2}}}\\ {\sqrt{1 - p} {B_1}{Y_1}^{\frac{1}{2}}}&{}{\sqrt{1 - p} {B_2}{Y_2}^{\frac{1}{2}}}&{} \cdots &{}{\sqrt{1 - p} {B_n}{Y_n}^{\frac{1}{2}}} \end{array}} \right]\).
Thus,
Similarly,
Our desired result follows from applying inequality (16) to A, B, X and Y.
Corollary 2
Let \(A, B, X, Y \in {M_{n}(\mathbb {C})}\) such that X and Y are positive semidefinite. Then,
for all \(p\in [0,1]\).
Proof
Let \(n=2, A_{1}=B_{2}=A, A_{2}=B_{1}=B, X_{1}=X_{2}=X\) and \(Y_{1}=Y_{2}=Y\) in inequality (18) to get the result.
Remark 1
Letting \(Y=X\) in inequality (19) leads to the following inequality:
Replacing X by I in inequality (20) leads to the following inequality:
Corollary 3
Let \(A, B, X, Y \in {M_{n}(\mathbb {C})}\) be positive semidefinite matrices. Then,
for all \(p \in [0,1]\), where \(S={A^{\frac{1}{2}}}{X^{\frac{1}{2}}}{Y^{\frac{1}{2}}}{A^{\frac{1}{2}}}\), \(T={B^{\frac{1}{2}}}{X^{\frac{1}{2}}}{Y^{\frac{1}{2}}}{B^{\frac{1}{2}}}\), \(H(p)=p{X^{\frac{1}{2}}}A{X^{\frac{1}{2}}} + \left( {1 - p} \right) {Y^{\frac{1}{2}}}A{Y^{\frac{1}{2}}}\), \(L(p)=p{X^{\frac{1}{2}}}B{X^{\frac{1}{2}}} + \left( {1 - p} \right) {Y^{\frac{1}{2}}}B{Y^{\frac{1}{2}}}\), and \(K(p)=p{X^{\frac{1}{2}}}{A^{\frac{1}{2}}}{B^{\frac{1}{2}}}{X^{\frac{1}{2}}} + \left( {1 - p} \right) {Y^{\frac{1}{2}}}{A^{\frac{1}{2}}}{B^{\frac{1}{2}}}{Y^{\frac{1}{2}}}\). In particular, if \(Y=X\), then
where \(M = {X^{\frac{1}{2}}}A{X^{\frac{1}{2}}}\), \(N = {X^{\frac{1}{2}}}B{X^{\frac{1}{2}}}\), and \(Q = {X^{\frac{1}{2}}}{A^{\frac{1}{2}}}{B^{\frac{1}{2}}}{X^{\frac{1}{2}}}\). Moreover, if \(X=I\), then
Proof
Let \(n=2\), \(A_{1}=B_{1}=A^{\frac{1}{2}}\), \(A_{2}=B_{2}=B^{\frac{1}{2}}\) , \(X_{1}=X_{2}=X\), and \(Y_{1}=Y_{2}=Y\) in inequality (18), to get that
But,
Similarly,
Combining inequality (25) with the equations (26) and (27) leads to
Using Lemmas 1 and 2, we can get that
Similarly, we can get that
Combining inequality (28) with inequalities (29) and (30) leads to the desired result.
Corollary 4
Let \(A, B, X_{1}, Y_{1}, X_{2}, Y_{2} \in {M_{n}(\mathbb {C})}\) such that \(X_{1}, Y_{1}, X_{2}\) and \(Y_{2}\) are positive semidefinite matrices. Then,
for all \(p \in [0,1]\), where \(C = A{X_{1}}^{\frac{1}{2}}{Y_{1}}^{\frac{1}{2}}{A^{*}}\), \(D = B{X_{2}}^{\frac{1}{2}}{Y_{2}}^{\frac{1}{2}}{B^{*}}\), \(L(p) = p{X_1}^{\frac{1}{2}}{\left| A \right| ^2} {X_1}^{\frac{1}{2}} + (1 - p){Y_1}^{\frac{1}{2}}{\left| A \right| ^2}{Y_1}^{\frac{1}{2}}\), \(H(p) = p{X_2}^{\frac{1}{2}}{\left| B \right| ^2}{X_2}^{\frac{1}{2}} + (1 - p){Y_2}^{\frac{1}{2}}{\left| B \right| ^2}{Y_2}^{\frac{1}{2}}\) and \(K(p) = p{X_1}^{\frac{1}{2}}{A^*}B{X_2}^{\frac{1}{2}} - (1 - p){Y_1}^{\frac{1}{2}}{A^*}B{Y_2}^{\frac{1}{2}}\).
Proof
Letting \(n=2\), \(A_{1}=B_{1}=A\), and \(A_{2}=-B_{2}=B\) in inequality (18) leads to
But,
Similarly, we can get that
Combining inequalities (33) and (34) with inequality (32) leads to our result.
References
Al-Khlyleh, M., Kittaneh, F.: Interpolating inequalities related to a recent result of Audenaert. Linear Multilinear Algebra 65(5), 922–929 (2017)
Audeh, W.: Generalizations for singular value and arithmitic–geometric mean inequalities of operators. J. Math. Anal. Appl. 489(2), 124184 (2020)
Audenaert, K.M.R.: Interpolating between the arithmetic–geometric mean and Cauchy–Schwarz matrix norm inequalities. Oper. Matrices 9, 475–479 (2015)
Bhatia, R.: Matrix Analysis, GTM169. Springer, New York (1997)
Bhatia, R., Davis, C.: More matrix forms of the arithmetic–geometric mean inequality. SIAM J. Matrix Anal. Appl. 14, 132–136 (1993)
Bhatia, R., Davis, C.: A Cauchy–Schwarz inequality for operators with applications. Linear Algebra Appl. 223(224), 119–129 (1995)
Bhatia, R., Kittaneh, F.: On the singular values of a product of operators. SIAM J. Matrix Anal. Appl. 11, 272–277 (1990)
Horn, R.A., Mathias, R.: Cauchy–Schwarz inequalities associated with positive semidefinite matrices. Linear Algebra Appl. 142, 63–82 (1990)
Kittaneh, F.: A note on the arithmetic–geometric mean inequality for matrices. Linear Algebra Appl. 171, 1–8 (1992)
Limin, Z., Jiang, Y.: A note on interpolation between the arithmetic-geometric mean and Cauchy–Schwarz matrix norm inequalities. J. Math. Inequal. 10(4), 1119–1122 (2016)
Matharu, J.S., Moslehian, M.S.: Grüss inequality for some types of positive linear maps. J. Oper. Theory 73(1), 265–278 (2015)
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Communicated by Qingxiang Xu.
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Al-khlyleh, M., Alrimawi, F. Generalizations of the unitarily invariant norm version of the arithmetic–geometric mean inequality. Adv. Oper. Theory 6, 2 (2021). https://doi.org/10.1007/s43036-020-00106-1
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DOI: https://doi.org/10.1007/s43036-020-00106-1