数理解析研究所講究録
第 674 巻 1988 年 127-133

On the Norm of Block Products of Matrices(Yoshihiro Nakamura)

中村美浩

北大応電研

1. Introduction and Preliminaries
coinplex matrices, and set $M_{n}=M_{n,n}$ . For each
Let $M_{m,n}$ be the space of all$A\in M_{m,n}$ the vector of singular values of(i.e. eigenvalues of $|A|=(A^{*}A)^{1/2}\in M_{n}$ )arranged in decreasing order is denoted by$m\cross n$$A$$\sigma(A)=(\sigma_{1}(A), \sigma_{2}(A),$For $1\leq p<\infty$ , we denote the p-norm of$A$by$\cdots,$$||A||_{p}$$\sigma_{n}(A))$.

, i.e.

$||A||_{p}=[ tr(|A|^{p})]^{1/p}=[\sum_{i=1}^{n}\sigma_{i}(A)^{p}]^{1/p}$,.
and the spectral norm (or operator norm) by
It is well-known that for $A,$ $B\in M_{n}$ the following H\"older-type norm inequality holds:$||A||_{\infty}=\sigma_{1}(A)$whenever$||AB||_{r}\leq||A||_{p}||B||_{q}$$\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$(1).

This can be implied from the inequalities$\sum_{i=1}^{k}\sigma_{i}(AB)\leq\sum_{i=1}^{k}\sigma_{i}(A)\sigma_{i}(B)$$k=1,2,$$\cdots,$ $n$.

(2)

for $k=1,2,$$\cdots,$ $n$

.

(3)

for

Furthermore, stronger inequalities hold:$\prod_{i=1}^{k}\sigma_{i}(AB)\leq\prod_{i=1}^{k}\sigma_{i}(A)\sigma_{i}(B)$For $A=[a_{ij}],$ $B=[b_{ij}]\in M_{n}$ , their Schur product (or Hadamard product)defined by the entrywise multiplicationA

$A$

$oB$

is

$oB=[a_{ij}b_{ij}]_{i}^{n_{j=1}},\cdot$

Recently it has shown that the following similar inequalities hold ([3], [5]):$\sum_{i=1}^{k}\sigma_{i}$(A $oB$ )$\leq\sum_{i=1}^{k}\sigma_{i}(A)\sigma_{i}(B)$1for$k=1,2,$$\cdots,$ $n$.

(4)

These imply the H\"older-type norm inequality
whenever$||AoB||_{r}\leq||A||_{p}||B||_{q}$$\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$.

(5)(See [1], [2] and [6] for related results.)

In the present article, we are interested in the problem to find a product (of twomatrices) which unifies the ordinary matrix product and the Schur product and satisfies
the H\"older-type norm inequalities. There are two quite natural candidates called box$B=[B_{ij}]_{i}^{N_{j=1}}$products: let $A,$ $B\in M_{n}$ be partitioned intoblocks;$B_{ij}\in M_{p}(n=Np)$ . We define block products $A\circ B$ andbywith$N^{2}$$A=[A_{ij}]_{i_{J}}^{N_{j=1}},$$A\blacksquare B$$A_{ij},$and$A\circ B=[A_{ij}B_{ij}]_{i}^{N_{j=1}}$$A \blacksquare B=[\sum_{k=1}^{N}A_{ik}oB_{kj}]_{i}^{N_{j=1}}$.

lf we consider the trivial partition $N=n,$ $p=1$ , then $A\circ B=AoB$ and $A\blacksquare B=AB$ ,$\circ B=AB$ and $A\blacksquare B=A$ $oB$ . We investigate thesewhile if $N=1,p=n$ , thenproducts in the next section.
For later use, we explain a notion and elementary facts of majorization. Let $\xi=$andbe vectors in. We denote the decreasing
rearrangements of the components of by.
is said to besubmajorized by (in symbols) if$A$$\eta=(\eta_{1}, \eta_{2}, \cdots, \eta_{n})$$(\xi_{1}, \xi_{2}, \cdots, \xi_{n})$$\mathbb{R}^{n}$$\xi_{[1]}\geq\xi_{[2]}\geq$

$\xi$

–

$\geq\xi_{[n]}$$\xi$$\xi\prec_{w}\eta$$\eta$

for$\sum_{i=1}^{k}\xi_{[i]}\leq\sum_{i=1}^{k}\eta_{[i]}$$k=1,2,$$\cdots,$ $n$.

If in addition
holds, then is said to be majorized by. Inequalities (2) and (4) can be expressed by submajorization$\sum_{i=1}^{n}\xi_{i}=\sum_{i=1}^{n}\eta_{i}$$\xi$$\eta$(in symbols$\xi\prec\eta)$$\sigma(AB)\prec_{w}\sigma(A)\cdot\sigma(B)$and$\sigma(AoB)\prec_{w}\sigma(A)$where we denotes the coordinatewise product of vectors
Submajorization for the sum of matrices is also known:$\sigma(A)$$\sigma(A+B)\prec_{w}\sigma(A)+\sigma(B)$and.

$\sigma(B)$$\sigma(B)$,by$\sigma(A)\cdot\sigma(B)$

.

(6)

.

It is a basic fact that submajorization is preserved by the increasing convex functions:if, then $f(\xi)\prec_{w}f(\eta)$ for all increasing convex function , where $f(\xi)$ denotes
the vector $(f(\xi_{1}), f(\xi_{2}),$. In particular, if
and$\xi\prec_{w}\eta$$f$$\cdots,$ん$f(\xi_{n}))$$\xi,$$\eta\in \mathbb{R}_{+}^{n}$ん$\prod_{i=1}\xi_{[i]}\leq\prod_{i=1}\eta_{[i]}$for2$k=1,2,$$\cdots$$n$,

then$\xi\prec_{w}\eta$. See [4] for further details.

2. Results

First we consider the box product $A\circ B$ .
Lemma 1. For any $A,$ $B\in M_{n}$$[^{\mathcal{E}(B_{o^{*}}B)}AB$

where$\mathcal{E}$$(A\circ B_{*})\mathcal{E}(AA)^{*}]\geq 0$,(7): $M_{n}arrow M_{n}d$en otes the pinching, i.e.
$\mathcal{E}(X)=[\delta_{ij}X_{ij}]_{i}^{N_{j=1}}$

Proof. Take any vectors$\xi=[\xi_{j}]_{j=1}^{N},$for$X=[X_{ij}]_{i}^{N_{j=1}}\in M_{n}$$\eta=[\eta_{j}]_{j}^{N_{=1}}\in \mathbb{C}^{n}$with$\xi_{j},$.

$\eta_{j}\in \mathbb{C}^{p}$. Then$|<(A \circ B)\xi|\eta>|^{2}=|\sum_{i,j=1}^{N}<A_{ij}B_{ij}\xi_{j}|\eta_{i}>|^{2}$$=| \sum_{i)j=1}^{N}<B_{ij}\xi_{j}|A_{ij}^{*}\eta_{i}>|^{2}$$\leq$$\{\sum_{i,j=1}^{N}||B_{ij}\xi_{j}||\cdot||A_{ij}^{*}\eta_{i}||\}^{2}$$\leq\{\sum_{i,j=1}^{N}||B_{ij}\xi_{j}||^{2}\}\cdot$$\{\sum_{i,j=1}^{N}||A_{ij}^{*}\eta_{i}||^{2}\}$$= \{\sum_{j=1}^{N}<(\sum_{i=1}^{N}B_{ij}^{*}B_{ij})\xi_{j}|\xi_{j}>\}\cdot\{\sum_{i=1}^{N}<(\sum_{j=1}^{N}A_{ij}A_{ij}^{*})\eta_{i}|\eta_{i}>\}$$=<\mathcal{E}(B^{*}B)\xi|\xi>\cdot<\mathcal{E}(AA^{*})\eta|\eta>$

which shows that (7) holds.

,I

Using this lemma we have the following.
Theorem 2. For any $A,$ $B\in M_{n}$$\sum_{j=1}^{k}\sigma_{j}$$(A \circ B)^{2}\leq\sum_{j=1}^{k}\sigma_{j}(A)^{2}\sigma_{j}(B)^{2}$3

for$k=1,2,$$\cdots,$ $n$.

(8)

Proof. By (7) there is$C\in M_{n}$such thatand$||C||_{\infty}\leq 1$$A\circ B=\mathcal{E}(AA^{*})^{1/2}\cdot C\cdot \mathcal{E}(B^{*}B)^{1/2}$.

By (3) this implies$\prod_{j=1}^{k}\sigma_{j}(A\coprod B)^{2}\leq\prod_{j=1}^{k}\sigma_{j}(\mathcal{E}(AA^{*}))\sigma_{j}(\mathcal{E}(B^{*}B))$for$k=1,2,$$\cdots,$ $n$,

for$k=1,2,$$\cdots,$ $n$.

and consequently$\sum_{j=1}^{k}\sigma_{j}$Let$M_{n}$$(A \circ B)^{2}\leq\sum_{j=1}^{k}\sigma_{j}(\mathcal{E}(AA^{*}))\sigma_{j}(\mathcal{E}(B^{*}B))$be a primitive $N$ th root of 1, and define the unitary matrix. Since the pinching can be written in the form$\omega$$U=[\delta_{ij}\omega^{j}I_{p}]_{i}^{N_{j=1}}\in$$\mathcal{E}$$\mathcal{E}(X)=\frac{1}{N}\sum_{k=1}^{N}U^{*k}XU^{k}$for$X\in M_{n}$,(9)we get by (6)$\sum_{j=1}^{\text{ん}}\sigma_{j}(\mathcal{E}(AA^{*}))\leq\sum_{j=1}^{k}\sigma_{j}(AA^{*})=\sum_{j=1}^{k}\sigma_{j}(A)^{2}$and$\sum_{j=1}^{k}\sigma_{j}(\mathcal{E}(B^{*}B))\leq\sum_{j=1}^{k}\sigma_{j}(B^{*}B)=\sum_{j=1}^{k}\sigma_{j}(B)^{2}$Hence, by elementary calculation, we have (8).

.

ElAs the consequence of the last theorem we have the norm inequalities.
Corollary 3. Whenever$p,$ $q,$$r\geq 2$

sa tisfy $1/r=1/p+1/q$ ,$||A\circ B||_{r}\leq||A||_{p}||B||_{q}$

In$p$

(10)

.

articul $ar$$||A\circ B||_{\infty}\leq||A||_{\infty}||B||_{\infty}$Note that Lemma 1 and norm inequality (11) remain valid in theIn fact, we can obtain$\Vert[A_{ij}B_{ij}]_{i}^{N_{j=1}}\Vert\leq\Vert[A_{ij}]_{i}^{N_{j=1}}\Vert\cdot\Vert[B_{ij}]_{i}^{N_{j=1}}\Vert)$4

(11)

.

$C^{*}$-algebra setting.

(12)where $A=[A_{ij}]_{i}^{N_{j=1}},$ $B=[B_{ij}]_{i}^{N_{j=1}}\in M_{n}(A)$ with a. Let
Next we consider the box product, and define the unitary matrix $V\in M_{n}$ by$C^{*}$-algebra .
be the cannonical basis of$A$$A\blacksquare B$$\{e_{i}\}_{i=1}^{n}$for $j=1,2,$$\cdots$$\mathbb{C}^{n}$$Ve_{N(k-1)+j}=e_{p(j-1)+k}$, $N,$ $k=1,2,$$\cdots,p$.

For $A,$ $B\in M_{n}$ , let $C=V^{*}AV,$ $D=V^{*}BV$ . Then we have
$A\blacksquare B=V(C\circ D)V^{*}$,(13)

where the block product in the right hand side is the one with respect to the partition
blocks; $C=[C_{kt}]_{k,t=1}^{p},$ $D=[D_{kt}]_{k,1=1}^{p}$ with
intoん $t\in M_{N}$ .
The next theorem follows from (13) and Theorem 2.
$\circ$$p^{2}$$C_{kl},$$D$

Theorem 4. For any $A,$ $B\in M_{n}$for$\sum_{j=1}^{\text{ん}}\sigma_{j}(A\blacksquare B)^{2}\leq\sum_{j=1}^{\text{ん}}\sigma_{j}(A)^{2}\sigma_{j}(B)^{2}$$k=1,2,$$\cdots,$ $n$.

(14)

The following is a consequence of this theorem.
Corollary 5. Whenever$p,$ $q,$ $r\geq 2$

sa tisfy $1/r=1/p+1/q$ ,$||A\blacksquare B||_{r}\leq||A||_{p}||B||_{q}$

(15)

.

In particular$||A\blacksquare B||_{\infty}\leq||A||_{\infty}||B||_{\infty}$.

(16)

Finally we remark that there is another approach to the norm inequalities of thebox products. The idea is the following: letbe a bilinear map fromto. If there are linear maps
fromto $M_{n,m}$ andfrom
to $M_{m,n}$ (for somem) satisfying$M_{n}\cross M_{n}$$\Phi(\cdot, \cdot)$$M_{n}$$\Phi_{\ell}$$M_{n}$$\Phi_{r}$$\Phi(A, B)=\Phi_{t}(A)\Phi_{r}(B)$$||\Phi_{t}(A)||_{\infty}\leq||A||_{\infty}$

and,$||\Phi_{r}(B)||_{\infty}\leq||B||_{\infty}$

for any $A,$ $B\in M_{n}$ , then$||\Phi(A, B)||_{\infty}\leq||A||_{\infty}||B||_{\infty}$5$M_{n}$.

,(16)

$A$.
$\iota$When we consider the bilinear map $\Phi(A, B)=A\square B$ , we can find nice maps
fordefineand$\Phi_{t}$and$\Phi_{r}$:$B=[B_{ij}]_{i}^{N_{j=1}}$$A=[A_{ij}]_{i}^{N_{j=1}}$$\Phi_{t}(A)=[\tilde{A}_{1},\tilde{A}_{2}, \cdots,\tilde{A}_{n}]$,$\Phi_{r}(B)=\{\begin{array}{l}\hat{B}_{2}\hat{B}^{1}|\hat{B}_{n}\end{array}\}$

,

where
$\tilde{A}_{k}=[\delta_{ij}A_{ik}]_{i}^{N_{j=1}}$

Then we can check that$\Phi_{1}$,for$\hat{B}_{k}=[\delta_{\text{ん}j}B_{ij}]_{ij=1}^{N_{)}}\in M_{n}$and$\Phi_{r}$$k=1,2,$$\cdots,$ $n$.

satisfy (16). This nice idea was discovered by P. Nylen.

3. Counterexample
For the box products, desired inequalities are the following:$\sum_{j=1}^{k}\sigma_{j}(A\square B)\leq\sum_{j=1}^{k}\sigma_{j}(A)\sigma_{j}(B)$for$k=1,2,$$\cdots,$ $n$.

(17)

Though inequalities (8) hold, (17) or even the weaker inequalities$\sum_{j=1}^{k}\sigma_{j}$$(A \circ B)\leq\{\sum_{j=1}^{k}\sigma_{j}(A)\}\cdot\sigma_{1}(B)$for

do not hold. A counterexample is the following: taking the$A=\{\begin{array}{ll}E_{1l} E_{12}E_{21} E_{22}\end{array}\}$,$k=1,2,$$4\cross 4$$B=\{\begin{array}{ll}E_{11} E_{21}E_{12} E_{22}\end{array}\}$$\cdots,$

(18)

$n$

matrices,

whereismatrix whose $(i, j)$ -entry is equal to 1 and all other entries are , wecan easily compute the block product$E_{ij}$$2\cross 2$$0$$A\circ B=\{\begin{array}{ll}E_{11} E_{11}E_{22} E_{22}\end{array}\}$Hence we have.

$\sigma(A)=\{2,0,0,0\}$ ,$\sigma(B)=\{1,1,1,1\}$ ,$\sigma(A\circ B)=\{\sqrt{2}, \sqrt{2},0,0\}$6,

which do not satisfy (18). In view of (13) the inequalities$\sum_{j=1}^{k}\sigma_{j}(A\blacksquare B)\leq\sum_{j=1}^{k}\sigma_{j}(A)\sigma_{j}(B)$for$k=1,2,$$\cdots,$ $n$(19)

or even the weaker inequalities$\sum_{j=1}^{k}\sigma_{j}(A\blacksquare B)\leq\{\sum_{j=1}^{\text{ん}}\sigma_{j}(A)\}\cdot\sigma_{1}(B)$for$k=1,2,$$\cdots,$ $n$(20)

do not hold.
Finally the box products do not meet our request. Our purpose does not have been
attained. But we do not have another candidate.

References[1] T. Ando, R. A. Horn and C. R. Johnson, The singular values of a Hadamard product:a basic inequality, Linear and Multilinear Algebra 21 (1987), 345-365.

[2] R. Bapat and V. Sunder, On majorization and Schur products, Linear Algebra Appl.
72 (1985), 107-117.
[3] R. A. Horn and C. R. Johnson, Hadamard and conventional submultiplicativity forunitarily invariant norms on matrices, Linear and Multilinear Algebra 20 (1987),91-106.
[4] A. W. Marshall and I. Olkin, Inequalities: Theory of Majorization and Its Applications, Academic, 1979.
[5] K. Okubo, H\"older-type norm inequalities for Schur products of matrices, LinearAlgebra Appl. 91 (1987), 13-28.
[6] M. Walter, On the norm of a Schur product, Linear Algebra Appl. 79 (1986), 209-213.