An arithmetic–geometric mean inequality for products of three matrices
An arithmetic–geometric mean inequality for products of three matrices
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三个矩阵乘积的算术几何平均不等式
DOI：
10.1016/j.laa.2015.09.013
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发表时间：
2014
影响因子：
1.1
通讯作者：
Rachel A. Ward
中科院分区：
文献类型：
--
作者：
Arie Israel;F. Krahmer;Rachel A. Ward
考虑以下非交换算术几何平均不等式：给定正半定矩阵a1，…，a1，对于每个整数m≤n，以下成立：1 n m∑j 1， j 2，…，j m= 1 n⦀A j 1 A j 2…A j m⦀≥(n - m)！n !∑j 1, j 2，…，j m= 1均不同n⦀A j 1 A j 2…A j m⦀，其中⦀⋅⦀表示一个酉不变范数，包括算子范数和Schatten p-范数作为特殊情况。虽然这个不等式的完全普遍性仍然是一个猜想，但我们证明了该不等式对最多三个矩阵的乘积成立，m≤3。m= 1,2的证明很简单；为了推导m= 3的证明，我们借助于经典的Araki-Lieb-Thirring不等式的一个变体来证明矩阵乘积的置换。
Consider the following noncommutative arithmetic–geometric mean inequality: Given positive-semidefinite matrices A 1,…, A n, the following holds for each integer m≤ n: 1 n m∑ j 1, j 2,…, j m= 1 n⦀ A j 1 A j 2… A j m⦀≥(n− m)! n!∑ j 1, j 2,…, j m= 1 all distinct n⦀ A j 1 A j 2… A j m⦀, where⦀⋅⦀ denotes a unitarily invariant norm, including the operator norm and Schatten p-norms as special cases. While this inequality in full generality remains a conjecture, we prove that the inequality holds for products of up to three matrices, m≤ 3. The proofs for m= 1, 2 are straightforward; to derive the proof for m= 3, we appeal to a variant of the classic Araki–Lieb–Thirring inequality for permutations of matrix products.