关于矩阵Frobenius范数的一个猜想_英文_
A note on a conjecture on the Frobenius no rm of matrices
1 1 2ZOU L i2m in, J IAN G You2yi, HU X ing2ka i
( 1. Co llege of M a them a tic s and Comp u te r Sc ience, Chongq ing Th ree Go rge s U n ive rsity, Chongq ing 404000, Ch ina;
)2. Co llege of M a them a tic s and Physic s, Chongq ing U n ive rsity, Chongq ing 400044, Ch ina
A b stra c t: A con jec tu re on the no rm of m a trices is d iscussed. The con jec tu re on the no rm of m a trices p rop osed by
S loane and H a rw it a re p roved fo r som e sp ec ia l m a trices.
Key word s: no rm ; bound; p os itive def in ite m a trices; sp ec tra l rad ius
关于矩阵 F r oben i us范数的一个猜想
1 1 2邹黎敏 , 姜友谊 , 胡兴凯
( )1. 重庆三峡学院
与计算机科学学院 , 重庆 404000; 2. 重庆大学数理学院 , 重庆 400044
摘要 :讨论矩阵范数的一个猜想 。对于一些特殊矩阵 , 由 S loane 和 H a rw it提出的关于矩阵范数的这个猜想被证
明 。
关键词 :范数 ; 界 ; 正定矩阵 ; 谱半径
中图分类号 : O 15112 文献标志码 : A
1 P re lim ina ries
n ×n n ×n L e t D and C be the co llec tions of a ll n ×n m a trices w ith en tries in the in te rva l [ 0, 1 ] and n ×n com p lex
> ρ() λm a trices, resp ec tive ly. The sp ec tra l rad ius of a squa re m a trix A is the nonnega tive rea l num be rA m ax { || :λ is an e igenva lue of A }. L e t ‖?‖deno te the F roben ius no rm. In [ 1 ] , P rofesso r Zhan desc ribes the h is to ry F
and cu rren t s ta te of som e op en p rob lem s in m a trix theo ry. In 1976 S loane and H a rw it m ade the fo llow ing con jec2
[ 1 22 ] tu re .
C on jec ture If A is a nons ingu la r m a trix of o rde r n a ll of w hose en tries a re in the in te rva l[ 0, 1 ] , then
2 n - 1 ‖A ‖ ? .F n + 1
Equa lity ho lds if and on ly if A is an S 2m a trix.
( ) ( ) A n S 2m a trix of o rde r n is a 0, 1 2m a trix fo rm ed by tak ing a H andam a rd m a trix of o rde r n + 1 in w h ich the en tries in the f irs t row and co lum n a re 1, chang ing 1 ’s to 0 ’s and - 1 ’s to 1 ’s, and de le ting the f irs t row and co lum n.
P rofesso r Zhan p o in ts ou t tha t he does no t know of any resu lts on th is con jec tu re. Th is p rob lem a rises f rom
[ 1 ] w e igh ing des igns in op tics and s ta tis tics . In th is p ap e r, w e p rove th is con jec tu re fo r som e sp ec ia l m a trices.
Rece ived da te: 2009 204 207
第 4期49 ZOU L i2m in, e t a l: A no te on a con jec tu re on the F roben ius no rm of m a trices
2 L em m as and resu lts
[ 1, 3 ]n ×n L emm a 211 L e t A ?Cbe a H e rm itian m a trix, r be an in tege r w ith 1?r?n, and A deno te any r2 r
( ( )) by 2r p rinc ip a l sub 2m a trix of A ob ta ined by de le ting n - r row s and the co rresp ond ing co lum ns f rom A . Fo r
each in tege r k such tha t 1 ?k?r w e have
λ() () () λλA ?A ?A . k k r k + n - r 2 ×2 Theorem 211 L e t A ?D and supp ose tha t A is nons ingu la r, then
- 1 ‖A ‖ ? 2. F
2 ×2 Proof S ince A ?D , w rite A as
a b A = . c d
S ince
d - b 1 - 1 A = , a d - bc - c a
w e have
1 2 2a d + bc - 1 22 + b+ c ‖A ‖ = + d? 2 .a F | a d - bc |a d - bc B e low the p rob lem can be c lass if ied in to tw o cases.
Ca se 1 If a d > bc, then w e have
a d + bc a d a d - 1 ? 2‖A ‖ ? 2 ? 2? 2. F a d - bc a d - bc a d - bc Ca se 2 If a d < bc, then w e have
a d + bc bc bc - 1 ‖A ‖ ? 2 ? 2? 2.? 2 F bc - a d bc - a d bc - a d Thus the resu lt is p roved.
2 ×2 C oro lla ry 211 L e t A ?D and supp ose tha t A is nons ingu la r, then
42 ×2- 1 = ‖A ‖ ? . F 3 2 + 1
n ×n 2 ρ() then Theorem 212 L e t A ?D be a p os itive def in ite m a trix and supp ose tha tA ? n,4
- 1 2 n ‖A ‖ ? .F n + 1
n ×n Proof S ince A ?D is a p os itive def in ite m a trix, w e w rite A as
αB A = , T αa
( ) ( ) α ( ) w he re B is n - 1 by n - 1 p rinc ip a l lead ing sub 2m a trix of A ,is a co lum n vec to r of d im ens ion n - 1 a ll
of w hose en tries a re in the in te rva l [ 0, 1 ] and a ?[ 0, 1 ]. O bv ious ly, B is a p os itive def in ite m a trix of d im en2
s ion n - 1. B e low w e p rove Theo rem 212 by m a them a tica l induc tion.
Fo r k = 2, the resu lt ho lds by Theo rem 211
N ow , w e assum e tha t the resu lt is va lid fo r p os itive def in ite m a trix a ll of w hose en tries a re in the in te rva l[ 0,
λμλμ 1 ] of o rde r k fo r a ll k?n - 1. L e t? ?be the o rde red e igenva lues of A , le t? ?be the o rde redn 1 n - 1 1 [ 1, 3 ] e igenva lues of B. B y the induc tion hyp o thes is and S chu r Theo rem , w e have 2 n - 1 ( )2 n - 1 1- 1 2 = ? ? ‖B ‖. F 2i = 1n - 1 + 1 μ i
()50 第 44卷山 东 大 学 学 报 理 学 版
B ecause of in te rlac ing resu lt, w e a lso have
μλλμλ ?????, n n - 1 n - 1 1 1 hence
k k + 1 11 ? ?? fo r a ll k = 2,, n - 1. 22i = 2i = 1λ μ i i
Thus 2 2 2 n n - 1 n - 2 2 n - 2 1 2 n - 2 8 2 1 1 1 1 - 1 2 = ? ?? + ?+ ?+ ?+ . ‖A ‖ F 2 22222 i = 1i = 1n n n λμ λ λ ρ() nA i i 1 1 M eanw h ile 2 2 2 n - 2 ( ) 8 2 n 4 4 n + 3 + - = > 0. 22 2n + 1 n )( n n + 1 n
The refo re, w e have
2 n - 1 ‖A ‖ ? .F n + 1 Thus the resu lt is p roved.
3 N um e rica l exam p le
In o rde r to illus tra te the resu lts a num e rica l exam p le is g iven be low. Exam p le 311 L e t
1 1 1 1 n 2 3
1 1 1 1
2 3 4 n + 1 A = .
ω
1 1 1 1
+ 1 n + 2 2 n - 1 n n
O bv ious ly, A is a H ilbe rt m a trix. If one chooses n = 10, then one can ob ta in
2 ρ() A ?21929 0 ? ×10 = 31535 5.4 H ence
2 ×10- 1 ‖A ‖ ? . F 10 + 1 B u t, if one chooses n = 3, then one can ob ta in
ρ() A = 11408 3. hence
2 11408 3 ?11060 7 = ×3.4 S o, Theo rem 212 can no t be used. H ow eve r, one has
2 ×3- 1 ‖A ‖ = 3721205 6 ? = 11500 0. F 3 + 1 In th is p ap e r, w e on ly p rove th is con jec tu re ho lds fo r som e sp ec ia l m a trices by s tanda rd m a trix ana lys is.
Referen ce s:
[ 1 ] ZHAN X ingzh i. M a trix theo ry[M ]. B e ijing: H ighe r Educa tion P re ss, 2008. [ 2 ] SLOAN E N J A , HARW IT M. M a sk s fo r H adam a rd tran sfo rm op tic s and we igh ing de sign s[ J ]. App l Op tic s, 1976 , 15: 1072114.
[ 3 ] HORN R A , JOHN SON C R. M a trix ana lysis[M ]. Cam b ridge: Cam b ridge U n ive rsity P re ss, 1985.
()编辑 :陈丽萍