Properties of symmetric matrices
8
J. Math. Anal. Appl. 305 (2005) 219–226 www.elsevier.com/locate/jmaa Properties of symmetric matrices Jaesung Lee ∗,1 , Kyung Soo Rim Department of Mathematics, Sogang University, Seoul 121-742, South Korea Received 12 September 2002 Available online 7 January 2005 Submitted by J. Diestel Abstract We find the extreme points and the smooth points of the unit ball of the Banach space ℘ n of all real symmetric (n × n)-matrices. 2004 Elsevier Inc. All rights reserved. Keywords: Extreme point; Smooth point 1. Introduction Let B X and ∂B X denote the closed unit ball and the unit sphere of a real Banach space X, respectively. A point x ∈ B X is said to be an extreme point of B X if x lies on no segment both of whose end points lie in B X and are different from x . We denote the set of all extreme points of B X by E(B X ). A point x ∈ B X is said to be a smooth point of B X if there exists a unique Λ in the dual space X ∗ of X such that Λ= Λx = 1. Geomet- rically, a point x ∈ B X is a smooth point if there exists a unique supporting tangent plane at x . We denote the set of all smooth points of B X by S(B X ). Recall that a continuous m-homogeneous polynomial on a real Banach space X is a function p : X → R for which there exists a necessarily unique continuous, symmetric m-linear form F : X m → R with the property that p(x) = F(x,...,x) for every x ∈ X. We denote by ℘( m X) the Banach * Corresponding author. E-mail addresses: [email protected] (J. Lee), [email protected] (K.S. Rim). 1 The author was partially supported by grant R14-2002-044-01001-0 (2003) from ABRL of KOSEF. 0022-247X/$ – see front matter 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2004.11.011
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Transcript of Properties of symmetric matrices
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J. Math. Anal. Appl. 305 (2005) 219–226
www.elsevier.com/locate/jma
Properties of symmetric matrices
Jaesung Lee∗,1, Kyung Soo Rim
Department of Mathematics, Sogang University, Seoul 121-742, South Korea
Received 12 September 2002
Available online 7 January 2005
Submitted by J. Diestel
Abstract
We find the extreme points and the smooth points of the unit ball of the Banach space℘n of allreal symmetric (n × n)-matrices. 2004 Elsevier Inc. All rights reserved.
Keywords: Extreme point; Smooth point
1. Introduction
Let BX and ∂BX denote the closed unit ball and the unit sphere of a real BaspaceX, respectively. A pointx ∈ BX is said to be an extreme point ofBX if x lies onno segment both of whose end points lie inBX and are different fromx. We denote the seof all extreme points ofBX by E(BX). A point x ∈ BX is said to be a smooth point ofBX
if there exists a uniqueΛ in the dual spaceX∗ of X such that‖Λ‖ = Λx = 1. Geomet-rically, a pointx ∈ BX is a smooth point if there exists a unique supporting tangent pat x. We denote the set of all smooth points ofBX by S(BX). Recall that a continuoum-homogeneous polynomial on a real Banach spaceX is a functionp : X → R for whichthere exists a necessarily unique continuous, symmetricm-linear formF : Xm → R withthe property thatp(x) = F(x, . . . , x) for everyx ∈ X. We denote by℘(mX) the Banach
* Corresponding author.E-mail addresses: [email protected] (J. Lee), [email protected] (K.S. Rim).
1 The author was partially supported by grant R14-2002-044-01001-0 (2003) from ABRL of KOSEF.
0022-247X/$ – see front matter 2004 Elsevier Inc. All rights reserved.doi:10.1016/j.jmaa.2004.11.011
220 J. Lee, K.S. Rim / J. Math. Anal. Appl. 305 (2005) 219–226
y
y
of
Johne 2-reewithe.2
space of all continuousm-homogeneous polynomials onX, where the norm is given b‖p‖ = sup{|p(x)|: ‖x‖ � 1} [2]. Then, for a positive integern,
℘(2
Rn) =
{p:
p(x) = xtP x, whereP is a real symmetric(n × n)-matrix andx ∈ R
n
}.
Moreover, by Spectral theorem, forp ∈ ℘(2R
n) we can write
p(x) = xtUt
λ1 · · · 0
.... . .
...
0 · · · λn
Ux,
whereU ∈ O(n), then-dimensional orthogonal group. Thus if‖x‖ � 1, then we have∣∣∣∣∣∣xtUt
λ1 · · · 0
.... . .
...
0 · · · λn
Ux
∣∣∣∣∣∣ � max1�i�n
|λi |.
Also for a 2-homogeneous polynomialp with p(x) = xt (aij )n×nx onRn, it is a necessar
condition that|aij | � 1 in order forp to have norm 1.Hence we get
‖p‖ = the maximum absolute value of eigenvalues ofP .
Now we denote the space of all real symmetric (n× n)-matrices by℘n. From the abovementioned remark we see that℘n is a Banach space under the norm‖P ‖ = the maximumabsolute value of eigenvalues ofP .
The purpose of this paper is to classify the extreme points and the smooth pointsB℘n
by finding out relevant properties of symmetric matrices. Indeed, we will prove:
Theorem 1.1. P ∈ E(B℘n) if and only if the absolute values of eigenvalues of P are iden-tically 1, i.e., the set of all extreme points of B℘n is the following:
Ut
λ1 · · · 0
.... . .
...
0 · · · λn
U : U ∈ O(n), |λi | = 1, and 1� i � n
.
Theorem 1.2. P ∈ S(B℘n) if and only if P has exactly one eigenvalue whose absolutevalue is 1.
To mention some of the previous results related with this, R. Grzaslewicz and K.characterized the extreme points of the unit ball of all bilinear operators on thdimensional real Hilbert space�2
2 [5]. Z. Hu and B. Lin obtain a result on extremal structuof the unit ball ofLp(µ,X)∗ [6]. R.M. Aron and M. Klimek characterized the extrempoints of the unit ball in the space of all quadratic polynomials of one real variablesupremum norm over the interval[−1,1] [1]. Y.S. Choi and S. Kim characterized thextreme points and smooth points ofB℘(2�2
2)[4]. Therefore, our Theorems 1.1 and 1
generalize and extend the result of [4] to then-dimensional real Hilbert space�n2 = R
n.
J. Lee, K.S. Rim / J. Math. Anal. Appl. 305 (2005) 219–226 221
mma
es
tions-f
. We
Also, Y.S. Choi, H. Ki, and S. Kim [3] classified the extreme points ofB℘(2�21)
for the real
space�21.
In Section 2, we prove Theorem 1.1, and in Section 3, we introduce a useful leand then we prove Theorem 1.2. We openly use capital lettersP , Q, . . . as the symmetricmatrices which are uniquely determined by 2-homogeneous polynomialsp,q, . . . , respec-tively.
2. Extreme points of B℘n
To understand the geometry ofB℘n , first of all, we have to investigate invariancand fixed points, etc., under the suitable actions. The unit sphere∂B℘n is isomorphicto Sn(n+1)/2−1 ⊂ R
n(n+1)/2. Thus, the orthogonal groups can be considered as acon the ball. From Spectral theorem, we easily see thatP is invariant under the composition of O(n), i.e., ‖P ‖ = ‖PU‖ for U ∈ O(n). Using this, we provide a proof oTheorem 1.1.
Proof of Theorem 1.1. We may put
P = Ut
λ1 · · · 0
.... . .
...
0 · · · λn
U,
whereU ∈ O(n). SupposeP has an eigenvalue whose absolute value is less than 1assume|λ1| < 1. Then chooseµ andν such that
−1< µ < λ1 < ν < 1.
Now we taker = (λ1 − µ)/(ν − µ). Then 0< r < 1 and
P = rP1 + (1− r)P2,
where
P1 = Ut
ν 0 · · · 00 λ2 · · · 0...
.... . .
...
0 0 · · · λn
U and P2 = Ut
µ 0 · · · 00 λ2 · · · 0...
.... . .
...
0 0 · · · λn
U
have the norm 1. ThusP is not an extreme point.Suppose|λi | = 1 for all i. For some 0< r < 1, put
P = rQ + (1− r)R.
ThenQ andR have the norm 1 from‖P ‖ = 1. Further, note thatrQ + (1 − r)R has the
valueλi atU−1ei whereei = (0, . . . ,ith1 , . . . ,0). Since|λi | = 1 for all i, Q andR have the
same valueλi ’s atU−1ei ’s, respectively. Hence we get
UQUt =
λ1 a12 · · · a1n
a12 λ2 · · · a2n...
.... . .
...
and URUt =
λ1 b12 · · · b1n
b12 λ2 · · · b2n...
.... . .
...
,
a1n a2n · · · λn b1n b2n · · · λn
222 J. Lee, K.S. Rim / J. Math. Anal. Appl. 305 (2005) 219–226
ove
are less
tively.
en
we
whereraij + (1 − r)bij = 0 for i �= j . We claim thataij = bij = 0 for i �= j . First weconsider the case ofλi = 1 for all i. By Spectral theorem, there existU1 andU2 ∈ O(n)
such that
1 · · · 0
.... . .
...
0 · · · 1
= rUt
1U
µ1 · · · 0
.... . .
...
0 · · · µn
UtU1
+ (1− r)Ut2U
ν1 · · · 0
.... . .
...
0 · · · νn
UtU2.
We multiply UtU1 andUt1U on the left-hand side and the right-hand side of the ab
equation, respectively. Then
1 · · · 0
.... . .
...
0 · · · 1
= r
µ1 · · · 0
.... . .
...
0 · · · µn
+ (1− r)UtU1Ut2U
ν1 · · · 0
.... . .
...
0 · · · νn
UtU2U
t1U.
Let us recall that the absolute values of elements of matrices in the above equationthan or equal to 1. And, necessarily we haveµi = 1 for all i. We multiplyUtU2U
t1U and
UtU1Ut2U on the left-hand side and the right-hand side of the above equation, respec
Then1 · · · 0
.... . .
...
0 · · · 1
= r
1 · · · 0
.... . .
...
0 · · · 1
+ (1− r)
ν1 · · · 0
.... . .
...
0 · · · νn
.
Since |νi | � 1, we getνi = 1 for all i. Thus the claim holds. By the same way, whλi = −1 for all i, the claim also holds.
Next, we assume thatP has 1 and−1 as its eigenvalues. Without loss of generalitymay assume that for a positive integerk, λi = 1 for 1� i � k andλi = −1 for k + 1 �i � n. By Spectral theorem, and the same method as before by consideringO(k) andO(n − k), we have
1 · · · 0 0 · · · 0...
. . ....
.... . .
...
0 · · · 1 0 · · · 00 · · · 0 −1 · · · 0...
. . ....
.... . .
...
0 · · · 0 0 · · · −1
J. Lee, K.S. Rim / J. Math. Anal. Appl. 305 (2005) 219–226 223
e
.if
= r
1 · · · 0 a1k+1 · · · a1n
.... . .
......
. . ....
0 · · · 1 akk+1 · · · akn
a1k+1 · · · akk+1 −1 · · · 0...
. . ....
.... . .
...
a1n · · · akn 0 · · · −1
+ (1− r)
1 · · · 0 b1k+1 · · · b1n
.... . .
......
. . ....
0 · · · 1 bkk+1 · · · bkn
b1k+1 · · · bkk+1 −1 · · · 0...
. . ....
.... . .
...
b1n · · · bkn 0 · · · −1
= rQ + (1− r)R,
say. It suffices to prove thataij = bij = 0 for i �= j . Supposeaij > 0. We consider a spherelement
x = (0, . . . ,1− 1/m︸ ︷︷ ︸
ith
, . . . ,
√2/m − 1/m2︸ ︷︷ ︸
j th
, . . . ,0)
for a positive integerm. Then
xt Qx = 1+ 2√
2/m − 1/m2(aij (1− 1/m) −
√2/m − 1/m2
).
Since
limm→1
(aij (1− 1/m) −
√2/m − 1/m2
) = −1,
limm→∞
(aij (1− 1/m) −
√2/m − 1/m2
) = aij > 0,
we can select a positive integerm > 1 so that|xt Qx| > 1. Thusaij � 0 for all i �= j .Similarly bij � 0 for all i �= j . This fact and the identity,raij + (1 − r)bij = 0 yieldaij = bij = 0. Henceaij = bij = 0 for all i �= j , and the proof is complete.�
The unit sphere∂B℘n has two fixed pointsI and−I under actions ofO(n), whereI isthe identity matrix. In fact,E(℘n) is a strong extreme set since it is of finite dimension
A Banach spaceX is called rotund if∂BX contains no nontrivial line segments, i.e.,‖(x + y)/2‖ < 1 wheneverx, y ∈ X, x �= y and‖x‖ = ‖y‖ = 1.
Corollary 2.1. The Banach space ℘n is not rotund.
224 J. Lee, K.S. Rim / J. Math. Anal. Appl. 305 (2005) 219–226
eo-
t,prove
y
trix ishose
Proof. It suffices to show that∂B℘n has a line segment. For example, letU ∈ O(n). Thefollowing line, for 0< r < 1,
rI + (1− r)Ut
−1 0 · · · 00 1 · · · 0...
.... . .
...
0 0 · · · 1
U = Ut
2r − 1 0 · · · 00 1 · · · 0...
.... . .
...
0 0 · · · 1
U
is contained in∂B℘n . �
3. Smooth points of B℘n
Let us recall that ifπ is an isometry onX, thenx ∈ S(BX) is equivalent toπ(x) ∈S(BX). p ∈ ℘(mX) is called norm attaining if there exists a pointx ∈ X, ‖x‖ = 1 suchthat‖p‖ = |p(x)|. Such a pointx is called a norming point ofp.
Lemma 3.1. Suppose p ∈ B℘(mX) is norm attaining, and suppose there are at most k Λi ’sin the dual space ℘(mX)∗ of ℘(mX) satisfying
(a) Λi ’s are linearly independent,(b) Λip(x) = ‖Λi‖ = 1 for all i.
Then there are at most k norming points which are linearly independent.
Proof. Suppose there arek + 1 linearly independent norming pointsxi ’s of p. Forx ∈ X,let Ex ∈ ℘(mX)∗ be an evaluation functional atx. Putαi = p(xi)/|p(xi)|, then we have‖αiExi
‖ = 1 andαiExip(x) = 1. Sincexi ’s are linearly independent, Hahn–Banach th
rem yields that there isΛ ∈ X∗ such thatΛx1 = 1 andΛxi = 0 for 2� i � k + 1. Thenα1Ex1Λ
m = α1 �= 0 andαiExiΛm = 0 for 2� i � k + 1. So,α1Ex1 is not contained in the
set of linear combinations ofαiExi’s (2 � i � k +1). In similar way, we know thatαiExi
’sare linearly independent andαiExi
p(x) = 1. Thusp has at leastk + 1 linear functionalsatisfying (a) and (b), which is a contradiction.�
From Lemma 3.1, we know that ifk = 1, i.e.,p is norm attaining and is a smooth pointhenp has exactly one linearly independent norming point. Now we are ready toTheorem 1.2.
Proof of Theorem 1.2. Let p ∈ ℘(2R
n) be defined byp(x) = xtP x for P ∈ ℘n.For U ∈ O(n), we defineπU on ℘(2
Rn) by πUp(x) = p(Ux). Then one can easil
check thatπU is an isometry. Thus Theorem 1.2 is equivalent to
A 2-homogeneous polynomial in the ball, which is generated by a diagonal maa smooth point if and only if the diagonalized matrix has exactly one element w
absolute value is 1.
J. Lee, K.S. Rim / J. Math. Anal. Appl. 305 (2005) 219–226 225
.
e
y
We will prove the sufficiency by constructing the unique tangent planeΛ at p. Supposethere exists exactly one eigenvalue ofp whose absolute value is 1. Then we may write
p(x) = xt
1 0 · · · 00 λ2 · · · 0...
.... . .
...
0 0 · · · λn
x,
where|λi | < 1 for all i. Let Λ ∈ ℘(2R
n)∗ such that‖Λ‖ = Λp(x) = 1. PutΛxixj = Λij
for i � j . Note that ifΛij ’s are uniquely determined, then alsoΛ is determined uniquelyWe can chooseδn > 0 such that|λn ± δn| < 1. Then
pn(x) ≡ xt
1 0 · · · 00 λ2 · · · 0...
.... . .
...
0 0 · · · λn ± δn
x
lies in the sphere. However, since
∣∣Λpn(x)∣∣ =
∣∣∣∣∣1+n−1∑i=2
λiΛii + (λn + δn)Λnn
∣∣∣∣∣ = |1± δnΛnn| � 1,
we obtainΛnn = 0. In a similar way,Λii = 0 for 2� i � n − 1. If we show thatΛij = 0for i �= j , then the proof is complete. Now consider the sphere element,
pj (x) ≡ xt
1− 1/m2 0 · · ·j th
±1/m · · · 00 0 · · · · · · · · · 0...
.... . .
...j th
±1/m...
. . . 0...
.... . .
...
0 0 · · · · · · · · · 0
x.
Then|Λpj (x)| = |1− 1/m2 ± 2/mΛ1j | � 1. Thus|Λ1j | � 1/2m and we getΛ1j = 0 asm → ∞. For the others, consider the sphere element,
pij (x) ≡ xt
1 0 · · · · · · 00 0 · · · · · · 0...
.... . .
(i,j)thδ
......
...(j,i)th
δ. . .
...
0 0 · · · · · · 0
x,
where 0< δ ≈ 0. Then|Λpij (x)| = |1± 2δΛij | � 1. ThusΛij ’s have to vanish. Hencepis a smooth point with the unique tangent planeΛ = Ex1x1 whereEx1x1 is an evaluationfunctional atx1x1.
Conversely supposep generated by a diagonal matrixP has two eigenvalues whosabsolute values are 1. We may assume that the eigenvectors of such values aree1 ande2,respectively. Thene1 ande2 are norming points ofp and linearly independent. Thus b
Lemma 3.1,p is not a smooth point. �
226 J. Lee, K.S. Rim / J. Math. Anal. Appl. 305 (2005) 219–226
d 1.2
)
Remark 3.2. In [4], authors showed that the unit sphere of℘(2�22) consists of extreme
points and smooth points of its unit ball [4, Corollary 3.3], but our Theorems 1.1 anshow that the same is true for℘(2�n
2) exactly whenn = 2.
References
[1] R.M. Aron, M. Klimek, Supremum norms for polynomials, Arch. Math. 76 (2001) 73–80.[2] S.B. Chae, Holomorphy and Calculus in Normed Spaces, Dekker, New York, 1985.[3] Y.S. Choi, H. Ki, S. Kim, Extreme polynomials and multilinear forms on�1, J. Math. Anal. Appl. 228 (1998
467–482.[4] Y.S. Choi, S. Kim, The unit ball of℘(2�2
2), Arch. Math. 71 (1998) 472–480.
[5] R. Grzaslewicz, K. John, Extreme elements of the unit ball of bilinear operators on�22, Arch. Math. 50 (1988)
264–269.p ∗
[6] Z. Hu, B. Lin, Extremal structure of the unit ball ofL (µ,X) , J. Math. Anal. Appl. 200 (1996) 567–590.
