数理解析研究所講究録 1427 巻 2005 年 65-75

65
EIGENVALUES, SINGULAR VALUES, AND LITTLEWOOD-RICHARDSON COEFFICIENTS
Chi-Kwong Li
Department of Mathematics, College of William and Mary,
Williamsburg, Virginia 23187-8795, USA

E-mail: ckli@math.wm.edu

We briefly describe some recent results on inequalities relating the eigenvalues of the sum of
Hermitian or real matrices, and how to use these them inequalities relating the eigenvalues andsingular values of a matrix and its submatrices. These results are joint work with Poon, Fomin,
and Fulton [4, 14, 15]. Some open problems and remarks are also mentioned.

Sum of Hermitian (Real Symmetric) Matrices

1Let$\mathrm{H}_{n}$be the set of$n\mathrm{x}$$n$Hermitian matrices. Denote the vector of eigenvalues of$X\in \mathrm{H}_{n}$byA(X) $=(\lambda_{1}(X), \ldots, \lambda_{n}(X))$

with$\lambda_{1}(X)$$\geq\cdots\geq\lambda_{n}(X)$

Problem Let ,those of and $E$ .
$A$$E\in \mathrm{H}_{n}$. There has been a great deal of interest in studying the following.

and $\tilde{A}=A+E$ . Determine inequalities relating the eigenvalues of$\tilde{A}$

and

$A$

One can regard as a small perturbation of the matrix . So, we are interested in the relations
between the eigenvalues of the original matrix and the perturbed matrix .
One may see [6] for an excellent survey on this problem and its solution. To motivate our
discussion, we collect several well known results by Weyl, Liskii, Mirsky, Wielandt, Thompson and
Freede; see $[1, 17]$ .
$A$

$E$

$\tilde{A}$

$A$

.

.

For

$\mathrm{i}=1$

Suppose,$\ldots$, ,$n$$\lambda_{n}(E)\leq\lambda_{i}(\tilde{A})-\lambda_{i}(A)\leq\lambda_{1}(E)$$1\leq \mathrm{i}_{1}<\cdots<\mathrm{i}_{m}\leq n$.

and $1\leq j_{1}<\cdots<j_{m}\leq n$ . Then$\sum_{s=1}^{m}\lambda_{n-s+1}(E)\leq\sum_{s=1}^{m}(\lambda_{j_{s}}(\tilde{A})-\lambda_{j_{s}}(A))\leq\sum_{s=1}^{m}\lambda_{s}(E)$Consequently, for any unitarily invariant norm.

$||\cdot$$||$

we have$-||E||\leq||\overline{A}||-||A||\leq||E||$

Suppose$1\leq \mathrm{i}_{1}<\cdots<\mathrm{i}_{m}\leq n$.

.

and $1\leq j_{1}<\cdots<j_{m}\leq n$ . If $i_{m}+j_{m}-m\leq n$ , then$\sum_{s=1}^{m}\lambda_{i_{S}+j_{S}-s}(\tilde{A})\leq\sum_{s=1}^{m}\lambda_{i_{s}}(A)+\sum_{s=1}^{m}\lambda_{j_{s}}(E)$.

One may apPly the result to $-\overline{A}=-A-E$ to get a dual set of inequalities

6G
All the above and many more other early results suggest that there are inequalities of the form$\sum_{j\in J_{0}}\lambda_{j}(\tilde{A})\leq\sum_{j\in J_{1}}\lambda_{\mathrm{J}}(A)+\sum_{j\in J_{2}}\lambda_{j}(E)$, . It turns out that a complete set of inequalitiesof 1,
for some suitable subsets Jo, ,can be described in this way; see [9] and also [6].
$J_{2}$$J_{1}$,$\mathrm{X}(\mathrm{C})=(\mathrm{c}\mathrm{i}, \ldots, c_{n})$$\ldots$$n\}$satisfying $C=A+B$ with $\lambda(A)=(a_{1}, \ldots : a_{n})$ , $\lambda(B)=$
if and only if we have the trace equality

Theorem 1.1 There exist $A,B$ ,$(b_{1}, \ldots, \mathrm{b}\mathrm{n})$$\{$$C\in \mathrm{H}_{n}$$\sum_{s=1}^{n}c_{s}=\sum_{s=1}^{n}(a_{s}+b_{s})$and for any (Jo,$J_{1}$, J2)$\in LR_{m}^{n}$,with $m<n$$\sum_{j\in J_{0}}c_{j}\leq\sum_{j\in J_{1}}a_{j}+\sum_{j\in J_{2}}b_{j}$.

. A good reference
In the theorem, we use the concepts of Littlewood-Richardson sequences
for this concept is [5]. Here we describe the formal definition and give a simple example.
$LR_{m}^{n}$,$\ldots<j_{m}\leq n$ . Define
Let $[n]=$ 1,$($$\ldots$$n)$

and $J=(j_{1}, \ldots,j_{m})$ be an increasing subsequences of $[n]$ , i.e., $1\leq j_{1}<$$\mu(J)=(j_{m}-m, \ldots ,j_{1}-1)$.

)
are increasing subsequences of $[n]$ . Then (Jo,according to the Littlewood-Richardson rules:
andgenerated from

Suppose $J0$ ,$J_{1}$,$J_{1},$ $J_{2}$$J_{2}$$\mu(J_{1})$$\in LR_{m}^{n}$if$\mu(J_{0})$can be$/\mathrm{i}(\mathrm{J}_{2})$as Young diagrams.
, and,Display)
to generate
Add $t_{1}+\cdots+t_{m}$ entries from 1, .. ., $m\}$ to the rows of the Young diagram ofso that:the Young diagram of$\mathrm{f}\mathrm{x}\{\mathrm{J}\mathrm{o}$$=(\mathrm{c}\mathrm{i}, \ldots, \mathrm{r}\mathrm{m})$$\mu(J_{1})=(\mathrm{c}\mathrm{i}, \ldots, s_{m})$$\mathrm{p}(\mathrm{J}2)=(t_{1}, \ldots, t_{m})$$\mu(J_{1})$

$\{$

.
.
.
.

$\mu(J_{0})$The entries$\mathrm{i}$occurs exactly$t_{i}$

so many times.

The entries in each row is weakly increasing from left to right.
The entries in each column is strictly increasing from top to bottom.
For any$p$with$1 \leq p\leq\sum_{j=1}^{m}t_{f7}$define $p(i)$ to be the number of in the first$\mathrm{i}$counting from right to left and top to bottom, we have
In such a case, the Littlewood-Richardson coefficientand$\mu(J_{2})$is positive$p(\mathrm{i})\geq p(\mathrm{i}+1)$$c_{\mu(J_{1})\mu(J_{2})}^{\mu J_{0})}$$p$

assigned values.

of the three partitions$\mu(J_{0}),\mu(J_{1})$,87
Example 1.2 Suppose $\mu(J_{0})=(5,$4,3, 1),$\mu(J_{1})=(3,2,1,0)$ ,$\mu(J_{2})=(3,2,$2,0).

Here are three

examples of good constructions:
$*$

$*$

$*$

$*$

$*$

1
2 3
$*$

1

1
2

$*$

$*$

$*$

$*$

$*$

1

1
2 2
3
$*$

1

3

$*$

$*$

$*$

$*$

3

1
2 2
3 3

$*$

$*$

1

$*$

I1

1

Here are three examples of bad constructions:
$*$

$*$

$*$

$*$

$*$

2

$*$

2 3

1
1

1

$*$

$*$

$*$

1

$*$

$*$

1

$*$

2 3
2 1

$*$

$*$

$*$

3

3

$*$

1 2
3 3
$*$

2

One can use the LR rules to explain the Weyl inequalities, and the standard inequalities ofThompson.
[Weyl’s inequalities] $((j_{0}), (j_{1})$ , (j2) \in LR% if and only if$)$[Thompson’s standard inequalities] If $J_{1}=$ (ii . . . ,$i_{m}$$j\mathrm{O}$$=j_{1}+j_{2}-1$.

) and $J_{2}=(j_{1}, \ldots, j_{m})$ satisfy $\mathrm{i}_{m}+j_{m}-m\leq n$ ,then $J_{0}=$ $(\mathrm{i}_{1}+j_{1}-1, \ldots, \mathrm{i}_{m}+j_{m}-m)$ is admissible.
Note that one can do a good construction by adding $j_{r}-r$ to theget$\mu(J_{2})$$(m-r+1)\mathrm{s}\mathrm{t}$row of$\mu(J_{1})$to.

In general, it is not easy to solve the following.

Problem How to generate all the$(J_{0}, J_{1}, J_{2})$sequences, and do it efficiently?

By the result in [12], one can focus on
fromi.e., there is a unique construction of$(J_{0}, J_{1}, J_{2})$$\mu(J_{0})$$\mu(J_{1})$sequences with LR coefficient equal to one,.
and$\mu(J_{2})$However it is hard to determine when the LR coefficient is positive or equals one. In particular,
it is difficult to write a computer program to generate all LR sequences. in some situations, one
may prefer to generate a class of sequences systematically even though the class may contain many
redundant sequences. Taking this approach, one can use the Horn’s consistent sequences $(R, S, T)$ ,which is defined recursively as follows,
Let $R=$.
.

$(r_{1}, \ldots, r_{m})$, $S=(s_{1}, \ldots, s_{m}),T=(t_{1}, \ldots, t_{m})\in[n]$ .

For $m\geq 1$ , $\sum_{l=1}^{m}(r\ell-\ell)=\sum_{\ell=1}^{m}(s\ell+t\ell-2l)$ .

If $m>1$ , then for any consistent triple $(U, V, W)$ :$U=$ $(u_{1}$ ,...,$u_{m’})$, $V=(v_{1}$ , . . . ,$v_{m’})$, $W=(w_{1}$ , ... ,$w_{m’})$

88with $m’\in\{1, \ldots , m-1\}$ , we have$\sum_{\ell=1}^{m^{J}}(r_{u_{l}}-\ell)\geq\sum_{\ell=1}^{m’}(s_{vp}+t_{w_{t}}-2\ell)$One can extend Theorem 1.1 to the sum ofquaternions; see [6].
Theorem 1.3 There are$\lambda(\sum A_{j})=(a_{1}^{(0)}, \ldots, a_{n}^{(0)})$$A_{1}$,$\ldots$,$A_{r}\in \mathrm{H}_{n}$

$r$

with

.

Hermitian matrices over real, complex, or real$\lambda(A_{s})=(a_{1}^{(s\rangle}, \ldots, a_{n}^{(s)})$for $s=1$ ,$\ldots$, , and$r$

if and $on/y$ if$\sum_{j}a_{j}^{(0)}=\sum_{j}a_{j}^{\langle 1)}+\cdots+\sum_{j}a_{j}^{(f)}$and for any $(J_{0}, J_{1}, ., ., J_{r})\in LR_{m}^{n}(r)$ with $m<n$$a_{j}^{(r\}}$$\sum_{j\in J_{0}}a_{J}^{(0)}\leq\sum_{j\in J_{1}}a_{j}^{(1)}+\cdots+\sum_{j\in J},$.

It is interesting that the same set of inequalities govern the eigenvalues of the sum of Hermitianmatrices over real, complex, or real quaternions. To elaborate this comment, note that for every,,
there arewith the same eigenvalues as such that

$A=[a_{ij}]\in \mathrm{H}_{n}$$A_{1}$$\ldots$$A$$A_{n}\in \mathrm{H}_{n}$diag $(a_{11}, \ldots , a_{nn})=\frac{1}{n}(A_{1}+\cdots+A_{n})$ .
In fact, if $w=e^{2\pi i/n}$ and $D=$ diag $(1, w, \ldots, w^{n-1})$ , thendiag $(a_{11}, \ldots, ann)$$= \frac{1}{n}(\sum_{j=1}^{n}D^{j}A(D^{j})^{*})$.

Now, by Theorem 1.3, the same result holds for real symmetric matrices. However, even for $n=3$ ,,it is hard to construct
Let us consider the following.
$B_{1}$$B_{2},B_{3}!$Example 1.4 Let,$A=(\begin{array}{lll}4 2 12 3 11 1 1\end{array})$$D=(\begin{array}{lll}1 0 00 e^{i2\pi}/3 00 0 e^{i4\pi}/3\end{array})$.

Then $B_{1}=D^{*}AD$ , $B_{2}=(D^{2})^{*}AD^{2},B_{3}=A\in \mathrm{H}_{3}$ satisfy(a)(b)$\lambda(A)=\lambda(B_{1})=\lambda(B2)=\lambda(B_{3})$, and$(\begin{array}{lll}4 0 00 3 00 0 1\end{array})=\frac{1}{3}(B_{1}+B_{2}+ \mathrm{S}_{3})$.

Even for this specific example, it is not easy to constructhold.

$B_{1}$,$B_{2}$,$B_{3}\in \mathrm{S}_{3}$such that (a) and (b)69Principal Submatrices of a Hermitian Matrix

2

Using the result on the sum of Hermitian matrices, we can obtain inequalities relating the eigenvalues of a Hermitian matrix and those of the principal submatrices. Here is the specific problem.
and those of its principal subProblem Study the relations between the eigenvalues of$A\in \mathrm{H}_{n}$matrices,

Again, let us begin by describing some well known results; see $[1, 3]$ .

.
.

,with the vector of diagonal entries ( ,There is, i.e., $\sum_{j=1}d_{j}=\sum_{j=1j}^{n}\lambda(A)$ and for $k=1$ , , $n-1$ .
by$d_{1}$$A\in \mathrm{H}_{n}$$\lambda(A)$$d_{n}\rangle$$\ldots$

if and only if it is majorized$\ldots$$m$ principal submatrix
with anand $\lambda(B)=(b_{1}, \ldots, b_{m})$ if and only if

There is$m\cross$$A\in \mathrm{H}_{n}$$a_{j}\geq b_{j}\geq a_{n-m+j}$$B\in \mathrm{H}_{m}$for $j=1$ ,$\ldots$such that $\lambda(A)=(a_{1}$ , . . . ,$a_{n})$, $m$ .

We have the following result; see [14].
Theorem 2.1 There is $A=(_{*}^{A_{1}}$$A_{2}*)\in \mathrm{H}_{n}$with $\lambda(A)=(a_{1}, \ldots, a_{n})$ ,$A_{1}\in \mathrm{H}_{k}$and$A_{2}\in \mathrm{H}_{n-k}$such that$\mathrm{A}(\mathrm{A})=(a_{1}^{(1)}, \ldots, a_{k}^{(1)})$and$\lambda(A_{2})=(a_{1}^{(2)}, \ldots, a_{n-k}^{(2)})$

if and only if$\sum_{j}a_{j}=\sum_{j}a_{j}^{(1)}+\sum_{j}a_{j}^{(2)}$and for any (

$J0$

,

$J_{1}$, J2)$\in LR_{m}^{n}$with $m<n$$\sum_{\mathrm{i}\in J_{0}}(a_{j}-a_{n})\leq\sum_{j\in J_{1}}(a_{j}^{(1)}-a_{n})+\sum_{j\in J_{2}}(a_{j}^{(2)}-a_{n})$here$a_{j}^{\langle s)}=a_{n}$,whenever $j>n_{s}$ .

More generally, we have the following.

Theorem 2.2 Suppose $n_{1}+\cdots+n_{r}=n$ . There exists$(a_{1}, \ldots, a_{n})$, and$A_{jj}\in \mathrm{H}_{n_{j}}$with$\lambda(A_{jj})=(a_{1}^{(j)}, \ldots, a_{n_{\mathrm{j}}}^{(j)})$$A=(A_{ij})_{1\leq i,j\leq r}\in \mathrm{H}_{n}$for $j=1$ ,$\sum_{j}a_{J}=\sum_{s}\sum_{j}a_{j}^{(s)}$$\ldots$,$r$such that

if and only if$\lambda(A)$

$=$

70
and for any (Jo,$J_{1}$,$\ldots$,$J_{r}$) $\in LR_{m}^{n}(r)$ with $m<n$,$\sum_{J\in J_{0}}(a_{j}-a_{n})\leq\sum_{s=1j}^{r}\sum_{\in J_{s}}(a_{j}^{(s)}-a_{n})$here$a_{j}^{(s)}=a_{n}$whenever $j>n_{s}$ .

Similar to the results on the sum of matrices, one would like to reduce the list of inequalities.

or the lastsuch that either, only consider $J_{s}=$For each $s=1$ ,, $n_{s}+p$ . Also, we did the case when $nj\leq 2$ . To describe
terms have the form: $n_{s}+1$ , $n_{s}+2$ ,the result, we need some more notation.
$\ldots$$j_{m}^{(s)}\leq n_{s}$$(j_{1}^{(s)}, \ldots, j_{m}^{(s)})$

$r$

$p$

$\dot,$$\ldots$

Suppose$A_{ii}\in \mathrm{H}_{2}$$m+1\leq \mathrm{i}\leq n-m$

has eigenvalues$a_{1}^{(i)}\geq a_{2}^{(i)}$for $1\leq \mathrm{i}\leq m$ , andLet $(1, \cdots, im)$ be a permutation of 1,$($

For any subset $R\subseteq\{1, \cdots , m\}$ with $|R|=r$ , let$\cdots$$A_{ii}=[a_{1}^{(i)}]\in \mathrm{H}_{1}$, $m)$ such that$b_{1}^{R}\geq\cdots\geq b_{n-m-2r}^{R}$$a_{2}^{(i_{1})}\geq\cdots\geq a_{2}^{(x_{m})}$be the eigenvalues ofTheorem 2.3 There exists $A=(A_{ij})\in \mathrm{H}_{n}$ with eigenvalues $c_{1}\geq\cdots\geq c_{n}$ , such thathas eigenvalues$a_{1}^{(i)}\geq a_{2}^{(i)}$for$1\leq \mathrm{i}\leq m$, and$A_{ii}=[a_{1}^{(i\rangle}]\in \mathrm{H}_{1}$

for

for

$m+1\leq \mathrm{i}\leq n$-$m$$\oplus_{i\not\in R}A_{\dot{\mathrm{t}}}$.
.

$A_{ii}\in \mathrm{H}_{2}$if and onlyif$\sum_{i=1}^{n}c_{i}=\sum_{i=1}^{n-m}a_{1}^{(i)}+\sum_{i=1}^{m}a_{2}^{(i)}$and for any $(s, t)\in\{0, \cdots, m\}$ $\mathrm{x}\{0, \ldots, n-2s\}$ with $0<s+t<n$ and any
with $\ell=\min\{m, s+t\}$ , we have$s$element subset$S\subseteq\{\mathrm{i}_{1}, \cdots, \mathrm{i}\ell\}$$\sum_{l=1}^{t}c_{\iota}+\sum_{i=t+2}^{s+t+1}c_{i}\geq\sum_{j\in S}a_{2}^{(j)}+\sum_{i=1}^{t}b_{i}^{S}$

3.

OfF-diagonal blocks

In this section, we study the following.

Problem Determine when a matrix $X\in M_{k,n-k}$ can be the off-diagonal block of a matrix
with prescribed eigenvalues.
Observation There is $C=(_{X^{*}}^{*}$$X*)\in \mathrm{H}_{n}$with$\mathrm{X}(\mathrm{C})=(1, \ldots,Ca)$$C\in \mathrm{H}_{n}$if and only if there are (forany) unitary matrices $U\in M_{k}$ and $V\in M_{n-k}$ the matrix$\tilde{C}=$($UXV*)\in \mathrm{H}_{n}$w.th$\lambda(\tilde{C})=(c_{1}$,...,$c_{n})$.

Denote by $s(X)=(s_{1}(X), \ldots, s_{k}(X))$ the vector of singular values of $X\in M_{k,n-k}$ with entries
arranged in descending order71
We have the following result; [15].
Theorem 3.1 Let $c_{1}\geq\cdots\geq c_{n}$ and $s_{1}\geq\cdots\geq s_{k}\geq 0$ be given, where $k\leq n/2$ . The following
are equivalent.
(a) There is $C=(_{X^{*}}^{*}$$(s_{1}, \ldots, s_{k})$such that $X\in M_{k,n-k}$ , $\lambda(C)=(c_{1}, \ldots,c_{n})$ and $s(X)=$.

(b) There exist$C_{1}$,$(c_{n-k+1}, \ldots, c_{n})$(c) For each ($X*)\in \mathrm{H}_{n}$$J_{0}$,$J_{1}$$C_{2}$,$S\in \mathrm{H}_{k}$such that, and A(S) $=(s_{1}$ , . . .,, J2)$\in LR_{m}^{k}$$C_{1}$$s_{k})$–$C_{2}\geq 2S$, where.
4$\lambda(C_{2})=$.

$\sum_{j\in J_{0}}s_{j}\leq\sum_{\mathrm{j}\in J_{1}}c_{j}-\sum_{j\in J_{2}}c_{n-\mathrm{J}}+1$

There are some interesting consequences of this theorem. Let.

,with $m\leq k$
2matrices for the existence of$\lambda(C_{1})=(c_{1}, \ldots,c_{k})$$C=(_{X^{*}}^{*}$$X*)\in \mathrm{H}_{n}$.

$S_{k}(c)$be the set of$k\mathrm{x}$$(n-k)$

with $\lambda(C)=c=(c_{1}, \ldots, c_{n})$ .

Suppose $X\in S_{k}(c)$ . Then for any contractions $R\in M_{k}$ and $T\in M_{n-k}$ , we have $RXT\in S_{k}(c)$ .
In particular, if $X\in S_{k}(c)$ then $tX\in S_{k}(c)$ for any $t\in[0,1]$ . So,
the zero matrix as the star-center.
,...’
The set $S_{k}(c)$ is convex if and only ifsions with the same common difference.
$(c_{1}$$c_{k})$$S_{k}(c)$is star-shaped withand $(c_{n-k+1}, \ldots, c_{n})$ are arithmetic progres-Complex Symmetric Matrices

In this section, we consider the following.

Problem Study the singular values of the off-diagonal blocks of complex symmetric and generalmatrices.
It turns out that there are not much difference between complex symmetric or general matrices
with real symmetric matrices! We have the following result; see [4].

Theorem 4.1 Let
are equivalent.

$\gamma_{1}\geq\cdots\geq\gamma_{n}$and $s_{1}\geq\cdots\geq s_{k}\geq 0$ be given, where $k\leq n/2$ . The following(a) There is a symmetric matrix $C=(_{X}^{*}t$and $s(X)$ $=(s_{1}, \ldots, s_{k})$ .

$X*)\in M_{n}$wilh $X\in M_{k,n-k}$ ,$s(C)=(\gamma_{1}, \ldots, \gamma_{n})$72(b) There is $C=$$(\begin{array}{ll}* YZ *\end{array})$

with$\in M_{n}$combined list of singular values: si,$s_{1}$

,

$s2$

,

$s_{2}$,$\ldots$,$s_{k}$,$s_{k}$(c) There exists a real symmetric matr$rixC=(_{X^{t}}^{*}$$(s_{1}, \ldots, s_{k})$, ansuch that$s(C)=(\gamma_{1}, \ldots, \gamma_{n})$$C_{1}$

such that $X\in M_{k,n-k}$ , $s(X)=$

$X*)$

$d$

,

$C_{2}$,$J_{1},$$J_{2}$).

such that $C_{1}+C_{2}\geq 2S$ with A(S) $=(s_{1}, \ldots , sk)$ ,$S\in \mathrm{H}_{k}$$\lambda(C_{1})=(\gamma_{1}, \gamma_{3}\ldots, c_{2k-1})$(e) For each (Jo,$a$.

$\lambda(C)=(\gamma_{1}, -\gamma_{2}, \gamma_{3}\ldots, (-1)^{n}\gamma_{n})$(d) There exist, $Z\in M_{k,n-k}$ have$Y$$\in LR_{m}^{k}$,and$\lambda(C_{2})=(\gamma_{2}, \gamma_{4}, \ldots, \gamma_{2k})$.

with $m\leq k$2$j \sum_{\in J_{0}}s_{j}\leq\sum_{j\in J_{1}}\gamma_{2j-1}+\sum_{j\in J_{2}}\gamma_{2j}$.

Again, there are some interesting consequences of this result.

.

If $X\in M_{k,n-k}$ is the $(1, 2)$ block of$|c_{1}|\geq\cdots\geq|c_{n}|$.

, then $X$ is the $(1, 2)$ block of$|c_{1}|,$

If

$A$

, $B$ ,

$C\in \mathrm{H}_{n}$$\gamma_{1}\geq\cdots\geq\gamma_{2n}$$C\in \mathrm{H}_{n}$$-|c_{2}|$$-|c_{4}|$,$\ldots$$\tilde{A}_{2}$$\tilde{A}_{3}$$\mathrm{c}_{n}$, then$C=\tilde{A}+\tilde{B}$such thatandhas eigenvalues

has eigenvalues
has eigenvalues$\gamma_{1},\gamma_{r+1},\gamma_{2r+1}$$\gamma_{2},\gamma_{r+2},\gamma_{2r+2}$$\gamma_{3},\gamma_{r+3}$,$\gamma 2r+3$,,,$\ldots$$\ldots$$\ldots$, such that.

$\lambda(\tilde{B})=(\gamma_{2},$$\gamma_{4}$,$\ldots$,$\gamma_{2n^{\}}},$.

Same result work for $A_{0}=A_{1}+\cdots+A_{r}$ , we can rearrange the eigenvalues:$\tilde{A}_{1}$,with eigenvalues$\tilde{C}\in \mathrm{H}_{n}$$|c_{3}|,$$c_{1;}\ldots$satisfies $C=A+B$ and the combined list of eigenvalues of$\lambda(\overline{A})=(\gamma_{1}, \gamma_{3}, \ldots, \gamma_{2n-1})$.

,

with eigenvalues values$A$and $B$ is73
Future research

5

There are many interesting problems deserve further study. We mention a few of them in thefollowing.

.

.

Determine numerical algorithms to construct the matrices with desired properties.
Study the relations between the singular values of complementary blocks; see $[4, 16]$ .
Study the relations between the singular values of $C$ and those of , or those of$S$$\bullet$$C=(\begin{array}{ll}R 0S T\end{array})$

$R$

and

$T$

, for

;

see [13].
$\bullet$Study the implications of the results in the real world!

Acknowledgement
Li is an honorary professor of the Heilongjiang University, and also an honorary professor of
the Un iversity of Hong Kong. His research was partially supported by a USA NSF grant and aHK
RCG grant. He would like to thank Professor K. Okubo for hosting his visit to Sapporo and Kyoto
in 2004, the support of the Hokkaido University of Education in Sapporo, the Hokkaido University,and the Kyoto University, and also the Grant-Aid for Science Research (number 15540148) from
the Ministry of Education in Japan

74
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. Thompson and. Freede, Eigenvalues of paxtitioned Hermitian matrices, Bulletin
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