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Matricial and vectorial norms Kahlon, Gurdeep Singh 1972

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MATRICIAL AND VECTORIAL NORMS by Gurdeep Singh Kahlon A Thesis Submitted i n P a r t i a l F u l f i l m e n t of The Requirements f o r the Degree of Master of Arts i n the Department of Mathematics We accept t h i s t h e s i s as conforming t o the req u i r e d standard The U n i v e r s i t y of B r i t i s h Columbia, J u l y , 1972 In presenting t h i s t h e s i s i n p a r t i a l f u l f i l m e n t of the requirements f o r an advanced degree at t h i s U n i v e r s i t y of B r i t i s h Columbia, I agree that the L i b r a r y s h a l l make i t f r e e l y a v a i l a b l e f o r reference and study. I f u r t h e r agree that permission f o r extensive copying of t h i s t h e s i s f o r s c h o l a r l y purposes may be granted by the Head 6f my Department or by h i s r e p r e s e n t a t i v e s . I t i s understood that copying or p u b l i c a t i o n of t h i s t h e s i s f o r f i n a n c i a l gain s h a l l not be allowed without my w r i t t e n permission. Department of JYiaJtM^ju^jiM^ The U n i v e r s i t y of B r i t i s h Columbia Vancouver 8, Canada ( i i ) S upervisor: Dr. R. Westwick Abstract M a t r i c i a l norms, minimal m a t r i c i a l norms, v e c t o r i a l norms and v e c t o r i a l norms subordinate to m a t r i c i a l norms, which are r e s p e c t i v e l y g e n e r a l i z a t i o n s of matrix norms , minimal matrix norms , v ector norms and v e c t o r i a l norms subordinate to matrix norms , are defined and t h e i r various a p p l i c a t i o n s and p r o p e r t i e s are discussed. ( i i i ) ACKNOWLEDGEMENTS "I .slntjeTely'^tJiaffk Dr. "R. "Westwick f o r h i s comments and suggestions, and Dr. R. Ree f o r reading t h i s t h e s i s . I a l s o thank the Mathematics Department, U n i v e r s i t y of B r i t i s h Columbia, f o r g i v i n g me f i n a n c i a l a s s i s t a n c e . (iv) Dedicated to my parents (v) I n t r o d u c t i o n Chapter 1 Chapter 2 Chapter 3 B i b l i o g r a p h y TABLE OF CONTENTS M a t r i c i a l norms Zeroes of polynomials V e c t o r i a l norms ( v i ) I n t r o d u c t i o n The concept o f m a t r i c i a l norm was mentioned by Wielandt as a suggestion and l a t e r s t u d i e d by Bauer and Robert, and r e c e n t l y by Deutsch [1,2]. C e r t a i n types of mappings which s a t i s f y m a t r i c i a l norm axioms were also s t u d i e d by Ostrowski [12] and Robert [14]. The advantage of i n t r o d u c i n g m a t r i c i a l norm i s tha t i t gives us a b e t t e r upperbound f o r the s p e c t r a l r a d i u s of the n ~ 1 n companion matrix o f f(z ) = a 0 + a i z + ... a n.^z +z , a^ e C. In f a c t , w i t h the a p p l i c a t i o n of m a t r i c i a l norm, upperbounds have been found f o r the s p e c t r a l radius which are b e t t e r than the bounds due to Cauchy [11], W i l f [ 1 7], Kojima [11] and Fuj i v a r a [11]. V e c t o r i a l norm was introduced by K a u t o r o v i t c h [10] and l a t e r s t u d i e d by Robert [13,14], S t o e r , and r e c e n t l y by Deutsch [3,4]. Other norms such as m a t r i c i a l norm subordinate to v e c t o r i a l norms and matrix norm subordinate to v e c t o r norm are disc u s s e d i n Chapter 3. Various p r o p e r t i e s such as r e g u l a r i t y and d u a l i t y of d i f f e r e n t norms are al s o discussed i n Chapter 3. In t h i s t h e s i s many of the p r o p o s i t i o n s and r e s u l t s l e f t unproved i n the recent papers where they appear are proved here and examples are worked out i n d e t a i l . Many examples are given ( v i i ) and to make t h i s paper more complete, examples, d e f i n i t i o n s , and r e s u l t s have been added wherever necessary. In f a c t , every e f f o r t has been made t o make things complete and e x p l i c i t . Chapter I M a t r i c i a l norms In t h i s chapter the d e f i n i t i o n of a m a t r i c i a l norm introduced by Deutsch [1] i s given. Various p r o p e r t i e s of minimal m a t r i c i a l norms and g e n e r a l i z e d m a t r i c i a l norms are d i s c u s s e d , and a proof f o r the ex i s t a n c e of a minimal m a t r i c i a l norm i s i n c l u d e d . Theorem 2 g e n e r a l i z e s a r e s u l t of Householder [8].. Since zeroes of f (z) = a 0 + a 1z+ .. . a n _ 1 z I l " 1 + z n e C are eigen values of i t s companion matrix so d i f f e r e n t bounds f o r the s p e c t r a l radius are obtained. Enough examples are given to see t h a t these bounds f o r the s p e c t r a l radius are b e t t e r than the bounds due t o Kojima [11], Cauchy [11], Fujwara [11] and W i l f [17]. Let M„ denote the set of a l l nxn matrices over the complex f i e l d C. We know tha t M R i s a non-commutative complex algebra w i t h i d e n t i t y . A l s o , l e t R + be the set of a l l non-negative r e a l numbers and M + be the set of a l l nxn matrices w i t h e n t r i e s from n R +. Then i t i s easy t o see th a t M R + i s a p a r t i a l l y ordered set elementwise i . e . , For A = (a 4.)eM„ +, B = (b..) eM +, d e f i n e * ij n ' I j n (a i ; J-) 4 ( b i j ) fff a i j 4 b i ; j • A matrix norm i s a mapping from M n R + such t h a t f o r a l l A ,B e M and ae C, we have i ) A f 0 =>||A|| f 0 i i ) l l a A M = 1.1 • ||A|| i i i ) ||A+B|| < ||A|| • ||B|| i v ) ||AB|| < ||A|| • l|B|| /•••••A-'imappi*T^ '~-wMx'h-'-^ a^t4'S#i>e'S;"'i-') , i i ) , i d i ) -but not i v ) i s called a generalized matrix norm. Our object in this chapter i s . t o g e n e r a l i z e matrix norm to the m a t r i c i a l norm and see the a p p l i c a t i o n s of the l a t t e r . Define the matrix norms ty . ty . row' c o l . ** d * e u c . o n M n b y n row M - max. . M a ^ l n c o l . ( A ) = » » i i j l a i j . ' W . < * > . • • j j J U i J | a ) - r V A = C a l j " e V That these are a c t u a l l y matrix norms i s easy t o v e r i f y . We c a l l these norms, r e s p e c t i v e l y , row norm, column norm, and e u c l i d i a n norm. D e f i n i t i o n ; - A m a t r i c i a l norm i s a mapping y : M n -*• which s a t i s f i e s the f o l l o w i n g axioms. - (M-l) A 1 0 =>u(A) ? 0 . (M-2) y(ctA) = | a | y (A) (M-3) y(A+B) < y ( A ) y (B) ¥ A , B e M R (M-4) y (AB) < y (A) y (B) a e C. k i s c a l l e d the order of the m a t r i c i a l norm y. Remark: I f k=l Then y i s matrix norm on M n. Remark: From (M*3) and (M* 4) one gets by i n d u c t i o n y(Ai+A 2+ . - . A N ) i P ( A I ) + P ( A 2 ) + ... u ( A n ) y( A i A 2 . . . . A n ) <y(A x) y(A 2) ... y(A n) In p a r t i c u l a r y(nA) 4 ny(A) and y (A n) ^ ( A ) 1 1 V A eM n > A mapping y :iM -»-M1 + which s a t i s f i e s ( M * l ) , (M*2), and n *t (M"3) i s c a l l e d a generalized matricial norm on Mn . Certain mappings which s a t i s f y (M*2) and (M*3) are c a l l e d pseudo-norms on M„. Example 1. Define y: M n -*• M£ such that n(A) = (\&ij\y, A = ( a i j ) e M n * T h e n v i s a h* i j " * . v 13 m a t r i c i a l norm of order n on M. (M«l) and (M-2) are obvious. To see (M #3) and (M'4) l e t A = ( a i ; j ) , B ( b i ; j ) , i , j , = 1,2, .... n. Then A+B = ( a i j + b i j ) -> v (A+B) = C|a i j+b i j |D < C l a ^ l + j b ^ l ) = + ( I b ^ l ) = v(A) +y(B) and AB = (C..) where C.. = E a . b, .. Therefore (£ n=l P (AB) = ( I C ^ I ) < E | a i k | | b k j | ) =y(A)y(B). Example 2. Given a matrix norm * on M n and a kxk matrix A such t h a t A 2 >. A > 6, de f i n e y :Mn-»- M.J' such that y(A) = *(A )A ¥ A c M n <Then y i s a m a t r i c i a l norm on M R of order k. Again (M«l) and (M«2) are obvious. To check (M .3) and (M«4) we have y ( A + B ) = • (A+B) A 4 (* (A) + *(B)>=^(A) A+ ^ ( B ) A = y ( A ) + y ( B ) and y (AB) = * ( A B ) A < ( ^ ( A ) ^ ( B ) ) A 4 * ( A ) A * ( B ) A s i n c e A 2 >_ ,A . Therefore y (AB) 4 .y (A) y ( B ) ¥ A , B e M n . P r o p o s i t i o n 1. Let y: ~Mn -*• My* be a l a t f i c i a l norm."'Thyen y * : + t t t M n "*• M v such that y (A) = ( y ( A ) ) , A e M i s a l s o a m a t r i c i a l norm on M n < (A* denotes transpose of matrix A ) . Proof. To check •. (M-l) and (M-2) we note that A ? 0 =>At f 0 =>y (A*) i 0 s i n c e y i s a m a t r i c i a l norm. Therefore y t ( A ) f 0 , i and y t ( a A ) = ( y ( ( a A ) t ) ) t = ( y C a A * ) ) * = ( l a J p C A * ) ) * = | a | ( y ( A t ) ) t = | o | y t ( A ) . For (M-3) we have y t ( A + B ) = ( y ( ( A + B ) * ) ) 1 = ( y ( A t + B t ) ) t < ( P C A * ) + y ( B t ) j t = ( p ( ' F ' ) ) t + ( v C B * ) ) * - " ^ ( A ) + y t ( B ) . To see (M«4) , y t ( A B ) = ( y ( A B ) t ) t = ( y ( B t A t ) t 4 ( y ( A 1 ) y ( B 1 ) ) t • ( v ( A t ) ) t ( y ( B t ) ) t = y t ( A ) y t ( B ) and t h i s completes the p r o o f . P r o p o s i t i o n 2. Let y be a m a t r i c i a l norm on M n and l e t G be an i n v e r t i b l e n x n ma t r i x . Then the f u n c t i o n U r j : M n M k + such th a t y ~ ( A ) = y ( G A G _ 1 ) V A e NT i s a m a t r i c i a l norm, (j n Remark: y G i s c a l l e d G-transform of y . Proof: - Let A f 0 . Then G A G " 1 i 0 and t h e r e f o r e y ( G A G ~ x ) f 0 s i n c e y i s a m a t r i c i a l norm. Thus y G (A) = y(GAG _ 1 ) ^ 0 . A l s o y G ( a A ) = y ( G a A G _ 1 ) = ^ ^ ( G A G " 1 ) = | a | y G ( A ) . F u r t h e r , f o r A.B e M we have n y G ( A + B ) = y ( G ( A + B ) G _ 1 ) = y ( G A G ^ + G B G " 1 ) < y ( G A G " 1 ) + y ( G B G ~ 1 ) = y G ( A ) + y G ( B ) . and y G ( A B ) = y ( G A B G ~ 1 ) = y ( G A G " 1 G B G " 1 ) < y ( G A G _ 1 ) y ( G B G _ 1 ) = y G ( A ) y G ( B ) which proves the p r o p o s i t i o n . Remark: P r o p o s i t i o n 2 shows that s i m i l a r matrices have the same m a t r i c i a l norm. The next p r o p o s i t i o n shows that given a direct-sum decomposition of C n = Xj© ... @Xk of C n (vector space of n-tuples over C) and a matrix norm on M R, we can generate a m a t r i c i a l norm of order k on M . n P r o p o s i t i o n 5: Let Ei,E2, . . . E ^ be the p r o j e c t i o n s a s s o c i a t e d w i t h the direct-sum decomposition C n = Xi© ©X^ and l e t ty be a matrix norm on M_ . Then the mapping u :M •+ M v + such t h a t v(A) = (*(E i A E.)) i ( j = 1 > 2 ... k A . ^ i s a m a t r i c i a l norm of order k on M n. Remark: The f u n c t i o n y generatedv i n t h i s way i s c a l l e d the m a t r i c i a l norm induced by the direct-sum decomposition Cn=Xi@ ... ©X^ and the matrix norm ty. Proof: To prove (M«l) , suppose y ( A ) » 0. Then • ( E J A E J ) • 0 f o r a l l i , j , = 1,2, ... k and th e r e f o r e E ^ A E ^ = 0 . Now s i n c e A = .E. E . A E . we have A = 0 and so y s a t i s f i e s (M«1). I t i s »J ^ J easy t o see (M»2) . To check (M«3) we have f o r a l l A , B e M n y ( A + B ) = (ty(E±(A+B)E..)) = C * ( E I A E . + E . B E . ) ) I J = 1 > 2 > ... k < ( • ( E . A E - J ) ' + * ( E . B E . D ( * ( E I A E . ) + O K E . B E . ) ) ^ „ 1 > 2 _ K . H A ) + H B ) / Since E i , E 2 , . . . E , are a s s o c i a t e d p r o j e c t i o n s , K k E . 2 = E . , E . E . = E . E . = 0 and Z E.=l. Let p..(AB) denote the absolute value of ( i , j ) element of matrix AB. Then k k W^CAB) = p..(AZ E.B) < Z p. (AE B) 1 3 ^ i = l 1 a=l 1 3 a CD Now consider the element p. . (AE B)=ij-(E. AE BE.) = ty(E. AE^BE.) i j a l a j i a j < <,(E iAE a)^(E aBE j) = M i a ( A ) u a j ( B ) Hence from (1) we have P i :j(AB) < S p i. f l(A)p a i (B) and.so p(AB) < p(A)p(B) a =1 l a a J and t h i s completes the proof. P r o p o s i t i o n 4 : Let p :Mn .-*• be a g e n e r a l i z e d m a t r i c i a l norm and l e t (A^} be a sequence of complex n x n m a t r i c e s . Then Lim A =0 i f f Lim v f A ) = 0 . nn-t00 m-»-+00 Proof: Let A = Za. - E . . where { E . . } i s a standard bases f o r — — - m i ; j 13 13 13 Since p ^ i s a pseudo-norm on M n, t h e r e f o r e v^.( ^°ijEij) ^ . I" jo. . I p. • ( E . .) . Now i f A -> 0 then p. . (A ) -*• 0 and so i f ; i j i ] m • . i j m p(A m) + 0 . ; :;7-p(E iAE^) =p Conversely suppose u(A m) .-»• 0. I f E^= 0 y ( a i j E i j ) = j a j j jp.CEjj) and p^AE..) = 0 1 J - « o 1 0 , then v ( E i ) p ( A ) p ( E j ) . Therefore y (E i) y (A) y (E.) = | • | y (E^ •) . Hence y(A m) •> 0 and A m -»• 0 f o l l o w s . Let X 1,X 2» ••• x n be the eigenvalues of the matrix A .. •. Define the s p e c t r a l radius of matrix A as the maximum of the absolute value of the eigenvalues of A, i . e . , r(A) = s p e c t r a l radius of A = max.|A.|. i 1 Theorem 1 Let y: Mn+ be a m a t r i c i a l norm. Then r(A) 4 r ( y ( A ) ) f o r a l l A e M n. "Proof: 'Note that " i f "r CA) > 1 then A n" does - not tend t o the zfero matrix as n increases w h i l e i f r(A) < 1 then A n + 0 as n + ». Hence i f r (y(A)) < 1 then y ( A ) n ^  0 and so y ( A n ) + 0. Therefore by p r o p o s i t i o n 4, A n 0 =>r(A n) < 1. Thus r ( v ( A ) ) < 1 ^ r ( A ) < 1. On the other hand i f r ( y ( A m ) ) < 1 then a l s o we have r(A) < 1. v Let e > 0. Write B=A/ r ( y ( A ) ) + .e. Then y(B) ».-p{ A(r( p ( A ) ) +"e)'i}.'.- (rCy(A)) + e ) " 1 p ( A ) . Therefore r ( y ( B ) ) = ( r ( y ( A ) ) + e ) _ 1 r ( y ( A ) ) = r ( y ( A ) ) r ( y ( A ) ) + e < 1 s i n c e e > 0 . Hence by the above argument we have r(B) < 1 . S 0 r ( A ) / r ( y ( A ) ) + e < 1 =>r(A) < r ( y ( A ) ) +e. Since e i s a r b i t r a r y so r(A) 4 r ( y ( A ) ) . The next p r o p o s i t i o n gives us a b e t t e r bound f o r the s p e c t r a l radius than given by Theorem "1. We have seen that f o r a l l A e M n, r(A) < r ( y ( A ) ) . Therefore r ( A ) p = r ( A p ) < r ( y ( A p ) ) and we have r(A) <_ [ r ( y ( A p ) ) ] p". On the other h a n d ^ y C A ^ < y ( A ) p =>r(y(A p)) < r ( y ( A ) p ) = [ r ( y ( A ) ) ] p =>[r(y(A p))] p < r ( y ( A ) ) . Hence r(A) < [ r ( y ( A p ) ) ] p" 4 r ( p ( A ) ) . Thus we have proved ?  P r o p o s i t i o n 5. I f u:Mn •> M v + i s a m a t r i c i a l norm, then r(A) o 1 < [ r ( p ( A " ) ) ] p <. r ( y ( A ) ) sf A eM and p any p o s i t i v e i n t e g e r . — n Example 3 Let B = r 2 0 -1 -1 0 2 -2 0 2 The eigenvalues of matrix B are x = 0,0,3 and th e r e f o r e r(B) = 3 Now consider the m a t r i c i a l norm y :M3 •*• M 2 d e f i n e d by ii(A) = I a i 1 1 | a 1 2 | + | a 1 3 | m a x ( | a 2 1 | , | a 3 1 | ) m a x ( | a 2 2 I + I a 2 3 1 , 1 a 3 2 | + | a 3 3 1 ) ¥ A • (a. .) e M 3 » •9-Therefore y(B) = and r (y(B)) » 4.73. Again B 2 = 4 •1 •2 •4 1 2 •8 2 4 , y ( B 2 ) -4 2 12 6 , r ( y ( B 2 ) ) -. 10 B3 = 12 -12 -24 -3 3 6 -6 6 12 y ( B 3 ) 3Y = 12 36 6 18 r(y(B3)) =30 l . i _ Therefore [ r ( y ( B 2 ) ) ] 2 = 3.16 and [ r ( y ( B 3 ) ) ] 3 = 3.11. i_ S i m i l a r l y [r(y (B1*) ) ] 4 =3.08. This agrees w i t h the conclusions of P r o p o s i t i o n 5. We w i l l see i n the next P r o p o s i t i o n t h a t as n + +», [ r ( y ( A n ) ) ] n r ( A ) . I t i s also i n t e r e s t i n g to = 4 v e r i f y that * r o w ( B ) = 4 , [ * r 0 W C B 2 ) ] 2 ^ r o w ^ l 3 = 3 ' 6 3 and [at ^ ( B * * ) ] 1 * = 3.46. Thus we see that i n the above cases 1 row r ( ^ ( B ) ) i s the weakest upperbound f o r r ( B ) . The f o l l o w i n g Lemma i s needed f o r P r o p o s i t i o n 6 and Theorem 2. I t s proof can be found i n [5, P.113]. Lemma: S p e c t r a l radius of a matrix cannot exceed the value of any of i t s norms. P r o p o s i t i o n 6; Let y:M •»• M v + be a m a t r i c i a l norm. Then f o r a l l A eM n, r(A) = ^ i m [ r ( p ( A m ) ) ] m . Proof:. Let 4»:M^-* R + be the euclidean matrix norm on M^. Define p0 :Mn -*• R + such that Vo(A) = *(y(A)) , V" A e M n. Then i t i s r o u t i n e to verify that u<> i s a matrix norm on M n» Now by the above Lemma r ( p ( A m ) ) <_ p 0 ( A m ) from which we have [ r ( p ( A m ) ) ] m '< [u 0 ( A m ) ] m < [ p o(A) m]m . Therefore by P r o p o s i t i o n 5, r(A) < [ r ( p ( A m ) ) ] m < [ p 0 ( A m ) ] m . Now i n general i f || || i s matrix norm on M n, then [15,P.312] r(A) = nlim ||An||n- . Therefore r(A) 4 [ r ( p ( A ))]m 4 r(A) as m •*•*• «=>r(A) = m l i m [r(y(A m))]m- . The next Theorem g e n e r a l i z e s a well-known theorem of matrix norms [8,P.46] . Theorem 2. Let j, denote the set of a l l m a t r i c i a l norms on M of orders k where 1 < k < n. Then n = = r(A) = i n f ( r ( p ( A ) ) } V- A e M . / e>"n,k Proof: By Theorem 1 r(A) 4 r ( p ( A ) ) ¥ P E Yft v . Now C, the f i e l d o f complex numbers^ i s a l g e b r a i c a l l y c l o s e d . Therefore a l l the eigenvalues of matrix A l i e i n C and so there e x i s t s an i n v e r t i b l e matrix B such that BAB"1 i s i n the Jordan Form. Let e>0 and T=BAB _ 1. Let X x ,X2 , ...Xn be the ei g e n -values o f A (not n e c e s s a r i l y d i s t i n c t ) . Consider the diagonal matrix -11-D = 1 0 0 0 0 e 0 0 0 0 0 0 .. .e det D-*-e 1 ;e 2 .e 3 ..... .e 1 1" 1 and .0 .0 n-DTD l = D-* = 1 0 0 .... 0 0 ^ 0 0 0 0 I * 0 0 0 • • • • • 0 0 0 . i cn- 1 Xl Y I 0 0 0 0 A2 Y2 0 • • • • 0 ' n - i Where y^-J^' , i - l , 2 , . . n - l Y T > 0. Let e^ be the vecto r s i n C n such t h a t e^-1 i n the i - t h p l a c e . Let X i , X 2 , be .the subspaces of C n spanned by the vectors ei»e 2» ••• ek- 1 a n (* \ spanned by e^, e j ^ , e n . Then ob v i o u s l y C11 = Xj © X 2 i m a t r i c i a l norm induced by t h i s direct-sum decomposition and the ...© X k. Let p Q : M n be the -12-column norm on M n (Prop.3). Therefore u (DTD-' ) = (* (e. DTD'1 e .)) - . 7 v 0 d o o T l o o 0 I X 3 I • • • • • • 0 • • • • • • 0 • • • * • • 0 0 0 0 0 0 0 . l x k - 1 ' Y k - j 0 0 0 . . 0 o where a = max. { | | , j | + Y ^ . J } i=k+l, ...n By t a k i n g the column norm of the above matrix and a p p l y i n g the above Lemma we have r ( y 0 D T D - 1 ) ) < max. U i + i •'TTj I I « +T k. 11. i=l,2,...k-2 < max | Y 1 | + max | x i + 1 | i = l , 2 , . . n - l < 2- + max. IX. I =2- + r(A) (1) e. 1=1..n A E Now de f i n e the mapping v*«Mn •*• My* such that V(Q) r P 0(DBQB" 1D" 1) V-Q e 1^. Since u i s a DB - transform of y Q [P ropos i t ion 2] -'u i s a m a t r i c i a l norm on M . Set Q=A. Therefore n -13-y(A) = p o(DBAB' 1D" 1) = y Q(DTD' 1) => r ( y ( A ) ) = r ( y 0 ( D T D " i ) ) < I + r(A) by ( 1 ) . 2. Therefore r(A) 4 r ( y ( A ) ) 4 e + r(A) => r(A) = i n f r ( y ( A ) ) . ^ ^ e ^ n ,k Let v'Hn •* m£ be a ^generalized . m a t r i c i a l norm.. Define K i j ( y ) = {A e M n I v ± . (A) = 0} W i j ( w ) = { KaB ^ ) K a » ^ * C i , j ) } W(y) = I W (y) i j = l ^ Claim K ^ ^ ( y ) , W_ (y) and W(y) are subspaces of M n. Let A and B e (y) . Then y.^ (A) = 0 = v^. (B) , and s i n c e y ^ i s a pseudo-norm on M n, we have V i j (A+B) < p _ (A) + y i ; j (B) = 0 and y^j(aA) = |a| y^j(A) = 0. Therefore K^ .. (y) i s of M n. Now s i n c e sum and i n t e r s e c t i o n of subspaces i s a subspace, t h e r e f o r e W^(y). and W(y) are subspaces of M n > . These subspaces were introduced by Robert [14] i n h i s i n v e s t i g a t i o n of v e c t o r i a l norms. P r o p o s i t i o n 7: Let y:Mn be a m a t r i c i a l norm such that (y) ? 0,j = 1,2 ... k. Then i ) A e W(y) and A = Z A . . , A . , e W . . ( y ) => y . . ( A ) - j = 1 ? r 33 33 33 i i ) There e x i s t s a n x n matrix A such that y(A) = 1^. i i i ) v ( I n ) > I k . Proof: i ) -v.. i s -pseudo-norm -on Mn. Therefore P j j ( A ) = Mjj (Aj ! + A 22 + A k k ) = \i..(A.p s i n c e i n general •M.i(b..) = 0 => y..(a..+b..) < y..(a..) + y..(b..) = y..(a..) 33 33 3 3 33 3 3 = 33 3 3 33 33 33 ^ 33 On the other hand y..(a..) = y..fa..+b..-b..) < y..(a..+b..) 33 33 33 33 33 33 = 13 31 33 + y . . (b . .) = y..(a..+b..) => y . . (a..+b . .) = y . . ( a . . ) • " • 33K 33 33 33 33 33K 33 33 33 33 i i ) Since W..(y) i s subspace of M n and y.. i s ' n o r m on , we have v.. 33 W 33(y) 13 i s a norm. Let A... e W..(y) such t h a t 3 3 3 3 A . , f 0. Let A = Z 3 = 1 ..(A..-) V33 3 3 Then y ^ (A) = 1 => y(A) = I k -i i i ) y ( I n ) = y ( I n ) I k = y ( I n ) - y ( A ) > y ( I n A ) - y(A) = I k => Remark: In P r o p o s i t i o n 7, the assumption that VL..(y) f 0 cannot be dropped. For example l e t ty be some matriz norm on M . ' Then the mappings y,v: M '•-*• M v d e f i n e d by y ( A ) = • (A) •2*(A) v ( A ) = *(A) V A e , n 7 -15-are m a t r i c i a l norms of order 2. But there e x i s t s no ma t r i x A e M n such that U(A) = I 2 and we do not have v ( I n ) >, I 2 • We introduce the f o l l o w i n g m i n i m a l i t y n o t i o n on ma t r i x norms. A matrix norm ty i s s a i d to be minimal f o r a complex n x»n matrix A i f the s p e c t r a l r adius of A i s equal to the m a t r i x norm of A i . e . , r(A) = <|>(A). I t i s obvious that there e x i s t s matrices f o r which no matrix norm i s minimal. For example, n i l p o t e n t matrices(because Ar*0 => ^,(A)^0, and i f A i s n i l p o t e n t then r ( A m ) = 0 and so ty(k) f r(A)).. Our object i s to extend the n o t i o n of minimal matrix norm to that of minimal m a t r i c i a l norm. We w i l l see that f o r any n x n complex matrix A there e x i s t s a minimal m a t r i c i a l norm f o r A. D e f i n i t i o n : - A m a t r i c i a l norm v ;M n -*• M^* i s s a i d to be minimal f o r the matrix A c M n, i f r(A) = r ( y ( A ) ) . Example 4: Let A = (12 0 0 2 4 0 0 0 1 1 2 0 0 2 4 The eigenvalues of A are given by the equation x1* - 10x 3 + 25x 2= 0 => x = 0,0,5,5 =>r(A) = 5. Define the m a t r i c i a l norm p :M4 -> M 2 + such t h a t f o r a l l B = ( b ^ ) e y(B) = ' ( I b i i |* +ibi 2 F + l i t ! I 2 + | b 2 | 2 ) i - d h a ( 2 + N 2 + M 2 + l b * I2 >* ( l b * I 2 +|ba>l 2 + lb,! | 2+ lb*])*- d b 3 3 l 2 + |b*P+ |h»3 I2* K I 2 ) 2 Therefore y (A) f5 1 0 5 => r ( y ( A ) ) = 5 = r(A) which shows t h a t y i s minimal f o r A. P r o p o s i t i o n 8: Let M ;M n be a m a t r i c i a l norm. Then y i s minimal f o r A =>y i s minimal f o r A m, m=l,2,3 Proof: - Since y i s minimal f o r A, t h e r e f o r e r(A) = r ( y ( A ) ) . A l s o by Theorem 1 ; ¥ A e M n, r(A) < r ( y ( A ) ) =>r(A m) < r ( y ( A m ) ) < r ( y ( A ) m ) = [ r ( y ( A ) J ] m = r ( A ) m = r ( A m ) . Therefore r ( A m ) = f'(yTA m)), and hence the " P r o p o s i t i o n . P r o p o s i t i o n 9: Let y:Mn M^* be m a t r i c i a l norm. Suppose v ( A m ) = y ( A ) m ¥ m = 1,2,3 . Then y i s minimal f o r A. Proof : Since y ( A ) m - y(A m) we have r ( y ( A m ) ) = r ( y ( A ) m ) . Therefore r ( y ( A ) ) = [ r ( y ( A m ) ) ] m f o r m=l,2,3 . Now by P r o p o s i t i o n 6 we have r(A) = [ r ( y ( A m ) ) ] m = r ( y ( A ) ) , and t h e r e f o r e y i s minimal f o r A. Remark: The c o n d i t i o n i n P r o p o s i t i o n 9 i s necessary but not s u f f i c i e n t . For example, i t i s shown i n example 5 tha t y i s minimal f o r A. However, y(A) f5 01 1 5 II (A) : f 25 0 10 25 -17-5 10 0 0 ' 25 0 10 20 0 0 => y(A2) = m 25 2 5 5 10 0 2 10 20 ,y (A) 2 • We show below that i f y ( A m ) i s then the c o n d i t i o n i s s u f f i c i e n t . D e f i n i t i o n : Square matrix A i s s a i d t o be r e d u c i b l e , i f there e x i s t s a permutation matrix P such that P*AP = l i l v2 1 2 2 where diagonal matrices are square matrices and P denotes the transpose of P . Matrices which are not r e d u c i b l e are c a l l e d i r r e d u c i b l e . P r o p o s i t i o n 10: Let y :MR -> M k +bea minimal m a t r i c i a l norm f o r a matrix A. Suppose y(A m) i s i r r e d u c i b l e where m i s some p o s i t i v e i n t e g e r . Then u ( A m ) = y ( A ) m . Proof: Suppose v ( A m ) i y ( A ) m . Then u ( A m ) < y ( A ) m . Therefore [16, P.30], r C y ( A m ) ) < r ( y ( A ) J U ) = [r (y (A).) ] m = r ( A ) m s i n c e y m _ i s minimal. Also r ( A m ) < r ( y ( A m ) ) . Thus r ( y ( A )) < r ( A m ) < r ( y ( A m ) ) which i s not p o s s i b l e of course. Hence we must have y(A m ) = y ( A ) m . Remark: In the above P r o p o s i t i o n the c o n d i t i o n y(A m) i s i r r e d u c i b l e cannot be dropped. In the above remark, y ( A 2 ) i s r e d u c i b l e , y i s minimal f o r A and we do not have y( A 2 ) f y ( A ) 2 . P r o p o s i t i o n 11: Let y be a minimal m a t r i c i a l norm. Suppose G i s an i n v e r t i b l e n x n complex m a t r i x . Then the G-transform of y i s minimal f o r G"1AG. m Proof: Since s i m i l a r matrices have the same c h a r a c t e r i s t i c p o l y n o m i a l s , t h e r e f o r e r(A) - rCG'VAG). Define Pg"M n •*• M^* such t h a t P g ( A ) = u(GAG _ 1) f o r a l l A e Mn> By P r o p o s i t i o n 2 y^ i s a m a t r i c i a l norm. Now the G-transform .of .y ..for G"1 AG i s .p G(G _ 1AG) = y(GG'^GG" 1) = y(A) and t h e r e f o r e r ( y g ( G ~ 1 AG)) = r ( y ( A ) ) = r(A) = r(G" 1AG). Hence the r e s u l t . The next p r o p o s i t i o n guarantees the ex i s t a n c e of a m a t r i c i a l norm of order n which i s minimal f o r A. P r o p o s i t i o n 12: Given A e M n, there e x i s t s a m a t r i c i a l norm y :M •*• M * of order n which i s minimal f o r A. n n Proof: Given A e M_ we can f i n d an i n v e r t i b l e m a t rix G such n t h a t GAG"1 i s i n Jordan form. Define the m a t r i c i a l norm y: M n M n + s u c h t h a t y ( B ) = ( l b i j l ) £°r a l l B e M n. Now i f B i s a t r i a n g u l a r m a t r i x , then, s i n c e the eigen-values of t r i a n g u l a r matrix l i e on the main d i a g o n a l , y i s minimal f o r B. In p a r t i c u l a r y i s minimal f o r GAG"1. So by the above P r o p o s i t i o n , y i s minimal f o r A. D e f i n i t i o n : Let k be the l e a s t p o s i t i v e i n t e g e r such t h a t there e x i s t s a m a t r i c i a l norm y :M„ -> M,+ which i s minimal f o r n k A. Then A i s s a i d to be of c l a s s k. The exis t a n c e of such a p o s i t i v e i n t e g e r i s guaranteed by P r o p o s i t i o n 12. -19-Remark: The matrix of example 5 i s of Class 2. Remark: Matrices of Class 1 are a l s o c a l l e d of Class M [8,P.46] Example 5: Di a g o n a l i z a b l e matrices i . e . , matrices s i m i l a r t o diagonal matrices are of Class 1. Let A = A i l A i 2 A A21 A2 2 ... A Vk 2k A-i, • • • • A-kk be any p a r t i t i o n of the matrix A e M n« Then i t i s easy to check d i r e c t l y t h a t the mapping <j> :M + M? such t h a t $ (A) = (<j> (A. .')') n -. 1 J i j = l , 2 , . . . k f o r a l l A e M , where $ i s matrix norm i s a m a t r i c i a l norm on M n. The next theorem i s a well-known theorem, and i t s proof can be found i n most books on l i n e a r a lgebra. Theorem 3: Let V be a f i n i t e dimensional vector space over C . Let T be the l i n e a r t r a n s f o r m a t i o n on V. Suppose th a t f ( z ) = n - ^  n a© + a x z + .... a n_^z " + z i s the minimal polynomial of T over C. Then there e x i s t s bases of V i n which the mat r i x of T i s -20-F = 0 0 1 0 0 1 0 0 0 0 0 0 a 0 a 2 0 0 n-and t h a t the roots of f ( z ) are p r e c i s e l y the eigenvalues of the m a t r i x F. Note: For convenience, we have taken f ( z ) to be monic. The m a t r i x F i s c a l l e d the companion matrix of f ( z ) . Let e ( f ) be the l a r g e s t of the absolute values of the zeroes of f ( z ) . Consider the matrix D = "0 0 0 0 0 0 0 0 a o n- 2 0 0 0 0 0 0 0 1 o > Det D = aoaj an_2 -21-D~1 = i _ ao 0 0 l _ <*1 0 0 0 0 0 0 0 0 0 0 0 0 n- 2 0 1 Therefore D"1FD = 0 0 ....... - a 0 / O 0 24. 0 " a i / 0 i 0 0 P a r t i t i o n t h i s matrix as f o l l o w s D"1FD = 0 0 0 0 0 " a o / a 0 o_ l 0 0 0 0 0 a l a 2 0 . 0 o -a2/„ 0 0 • 0 °n-3 * * • * tt 0 " a n - 2 / „ °n- 2 0 0 0 0 n- 2 " a n - i Now s i n c e F and D~lFD are s i m i l a r m a t r i c e s , t h e r e f o r e they have -22-the same s p e c t r a l r a d i u s . Thus, e ( f ) = r(F) = r(D _ 1FD) For the p a r t i t i o n given by (1) , define the m a t r i c i a l norm (2) <t»: M n M 2 such that <j.(D_1FD) = n- 2 n - i where 3 = Y _ max{ «1 ' <*2 max{ -L-—L, J — -an~-3 } n- 2 a n-2 a 0 <*1 } (3) (4) (5) n-2 • • • « Now by Theorem 1 r(A) < r ( y ( A ) ) => r(D _ 1FD) < r ( * ( D _ 1 F D ) ) , and t h e r e f o r e e ( f ) < rC^'CD'^FD)) Let a be any matrix norm on M 2. Since the s p e c t r a l radius cannot exceed any of the matrix norms, t h e r e f o r e by ( 1 6 ) , e ( f ) < o(*(D"iFD)) .. The upperbound given by (7) cannot be b e t t e r than ( 6 ) . Let us evaluate (6) and ( 7 ) . Eigenvalues of the matrix given by (3) are the zeroes of x 2 - (3+1 a |)x + 8 l a I -a Y =0, namely 1 n - j 1 . 1 n - 1 1 n - 2 ' • x = e + l a n . J + / T g ~ n ' a n . i h - - 4CB|a n. i| -° n. 2YT 2 (6) (7) V 7 11* + K - J / C B - | a n . i l ) 2 • 4« n_ 2Y ] Taking p o s i t i v e value to get the maximum, we have by (6) 1 : • e ( f ) < T [3 + | a n . i | + / ( I a n _ x | - 3 ) 2 + 4 « N _ 2 Y ] . -23-To f i n d ( 7 ) , we take r e s p e c t i v e l y row norm, column norm and eucl i d e a n norm of (3 ), we have e ( f ) <. max. {B +y, o_ • | a | } — n 2 1 1 1 e ( f ) <. max. {3 + a , y + |a_ |> — " 2 1 1 1 e ( f ) ^ ( B 2 + y 2 • I ^ . J 2 * a 2 n _ 2 ) | P u t t i n g a l l the pieces together, we obt a i n Theorem 4: Let f ( z ) = 'a 0+a 1z + .... * n z 1 1" 1 + z n , a i e C and . l e t <xo ,ai , • ••<*„ be p o s i t i v e r e a l numbers. Then n- 2 e ( f ) < ' ]r [B + I a n _ i | • / C | a n . i | - B )< + 4» n. 2Y] . . . . (8) ^ ?(.f) .< . Jnax{.B +.Y ,, ctn _ 2 +.. .j,an . J ..) .... (9) e ( f ) < max{g+ a , Y+ I a n |} (10) ""2 , l " 1 e ( f ) < (B 2 •+ Y 2 + |a n_ | 2 + a 2 )*" . . . . ( 1 1 ) i i j n 2 where B and y are given by (4) and (5) . Since ( 9 ) , (10) and (11) are obtained by applying d i f f e r e n t norms, t h e r e f o r e these i n e q u a l i t i e s cannot give b e t t e r bounds than t h a t given by ( 8 ) . S u b s t i t u t e f o r B and y i n ( 9 ) , we have 1 a 0 | , | ai l+q 0 , I a n _ 2 +• .n-a-e ( f ) < max{ which i s W i l f ' s i n e q u a l i t y [ 1 7 ] . Another proof T i w i l l be given i n the next chapter u s i n g Frobenius theory. » -24-Taking <x0 = ax = .... a = 1 i . e . , when the diagonal n _ 2 matrix D i s the i d e n t i t y m a t r i x , .we ob t a i n from Theorem 4, Cor.1 Let f ( z ) = a 0 +a!Z + .... 2 1 + 2 , a^ e C. Set ct^ » 1 V i = 0,1 ... n-2, Then we have e ( f ) < L [1 + |a_ |+ /ITS |-D 2 + 4M ] .... (12) - 2 1 1 l n ' l e ( f ) < max{ | a 0 1 ,1+ |ax|, l+|a | } (13) • — n" l e ( f ) <. max{2,|a 0| +|a |, |a | + |a | } (14) ~ n " l n*2 n " l . . . e ( f ) < (2+M2+|a l 2 ) ^ .... (15) n" l where M = max{|a 0I , | a i | , ... | a I ) • . n~2 (12) i s Rehman's i n e q u a l i t y [9] and (13) i s Cauchy's i n e q u a l i t y [11]. Another proof of these i n e q u a l i t i e s can be found i n the next chapter. The f o l l o w i n g example shows that (14) can give a b e t t e r upper bound than Cauchy's (13). Example 6: Consider f ( z ) = - 2 - 2z - 0.5z 2 + z 3 Cauchy's (13) gives e ( f ) 4 3. (14) gives e ( f ) < 2.5. I t i s noteworthy to see that a l l i n e q u a l i t i e s from (8) to (15) con t a i n c o e f f i c i e n t s of z 1 1" 1 and i f we set a n_^ =0 the i n e q u a l i t i e s are s i m p l i f i e d . In p a r t i c u l a r , i f I a n |= 1 i n Rehman's i n e q u a l i t y , we have e ( f ) 4 1+/M. Assume that a l l the c o e f f i c i e n t s i n f ( z ) are non-zero and r e a l . Set = a^ +^ i= 0,1, ... n-2, we have from theorem 4. -25-Cor.2: Let f ( z ) = ao+ajz + ... a z n " l + z n a A + 0 i . Set otj_ = ^i+i f° r i = o , l , ... n-2. we ob t a i n e ( f ) < £ [ B' + | a n _ J + / ( | a n . i | - B ' ) 2 • < V I a ^ J ] . . . . . . (16) e ( f ) < max{3» + Y ' , | a n _ | + a n } . . . . . (17) e ( f ) < max{3 » + a , Y * + ! a n _ J } . . . . . (18) e ( f ) < (B'2 + Y ' 2 • 2 | a n . J2) 2 ~ (19) where 3 ' = . max{ [^} J *21 , . . . J a n - 2 * } I a 2 | I a 3 1 l a n - J Y ' - max{la°l , l a i l , ....| an- 2 | > f a i t TaTT l a n - i I Observe t h a t 3 ' < Y ' but i f 3 ' = Y ' then (16) and (18) c o i n c i d e provided a n l > 0, f o r then i n each case e ( f ) 4 3 ' + A N • S u b s t i t u t e f o r 3 ' and Y 1 i n ( 1 7 ) , and set = | a ^ + 1 | i= 0 , 1 , ... n-2. We obtain the f o l l o w i n g Kojima's i n e q u a l i t y [11] e ( f ) < max{ 2 K | , | a o 1 , 2 1a> 1 , .. . 2 1a n - 2 1 ).' For another proof of t h i s i n e q u a l i t y see next chapter. I n e q u a l i t i e s (16) t o (19) can s t i l l be modified i f a — a r e r e a l and some of the c o - e f f i c i e n t s are zero. In that case, set -26-«i ° a i + 1 i f a i + 1 1 0 and ai ~ * i f a i + i = 0 • The f o l l o w i n g example shows that (18) can give a b e t t e r upper bound than Kojima's i n e q u a l i t y . Example 7: f ( z ) = -1 + 2z + 3 z 2 - 4 z 3 + z1* (18) gives e ( f ) < 4 + max{ i-,2-,^-} = 4.75 2 3 «• Kojima's i n e q u a l i t y gives e ( f ) ± 8 . n - i - 1 Let t? 1 be any p o s i t i v e number and set a^=t . „ - . > i = 0 , l , ... n-2 i n Theorem 4, we o b t a i n Cor. 3 Let f ( z ) = ao+a^ + ... a n _ ^ z n _ 1 + z , a^ e C. Let n - i - 1 t be a p o s i t i v e * n u ^ Set a'£-t i s O , l , ... n-2. We obta i n e ( f ) < 2-[t+|a n. i|+ / ( | a n . i | - t ) ^ 4 6 t ] .... (20) e ( f ) < m a x { i | f J - , t + i ^ j - , ... t + i a i ^ , l t + | a n . 1 | } t t t • (21) e ( f ) < m a x { 2 t > l a n , . . . | a n _ 1 h | a n - ? | } , z t ... ' \ :~[ . .... (22) e ( f ) < ( 2 t 2 + 6 2 + | a n _ 1 | 2 ) 2 ' .... (23) where 6 = m a x ( 1 ^ f , ^  , 1 *"-2 1 >. t 1 t 1 t (21) i s W i l f ' s i n e q u a l i t y [17], We w i l l give another proof i n Chapter 2. The next example shows th a t (22) can give a -27-b e t t e r bound than W i l f ' s (21). Example 8: Let f ( z ) = -24-38z-13z 2-f2z 3 + z't (22) gives e ( f ) < max{6,2+|4r,2 + 3^-,2+i|} =6.34 (21) gives e ( f ) < max{| 4 r' 7' 2 2' 7- 3 3' 5} = 7 ' 3 3 In f a c t i t can be e a s i l y v e r i f i e d that W i l f ' s bound (21) i s the weakest of the above bounds. Let l a | ^ 0 . I f t i s so chosen that t = |a | ' n - j ' • 1 n - i • then we have from Cor. 3 a o | e ( f ) • ± max{ a_ ,n-x n-i •ha. n - i n-e ( f ) 4 |  an-1 I + m a * { n - i 5 ' a. e ( f ) < (3 where 6' = ; max{ n-! ao 2+ _ A l L _ )2 n-n-.... (24) » ••• 2| an- 1|} ( 2 5^ .... (26) • .... (27) la a | n " 2 ' |a I " " 3 ' n - i | | n - i n-2 Let N= max{|a 0J n , l a i | n " \ • • • l a n . 2 l > Then N n _ i ^. J a^ ' | V i = 0,1 ...n-2 w i t h e q u a l i t y f o r at l e a s t one i as N i s maximum. I f t i s so chosen that t=N, then from (28) Cor. 3 we have -28-e ( f ) < L { N + | a n . i | + •( ( |-N)'+4N' } .... (29) e(£) < max { M L . N + M , ... N + j a ^ J , N n - i N n - 2 maxN ( M _ , ••• 1 + l a ^ - i l > N ma* 1,1+1, ... 1+1 > as 1 ai | < 1 N n - i = 2N (30) e(£) < N+max. { N, l a n . J > .... (31) e ( f ) < (3N 2 + | a n _ J 2 ) ^ ....(32) where N i s given by (28) and 5 = N, From (31) we have e ( f ) < = m a x . | a n + max. {I a n ,|a n_ |} i=2..n i=2 . .n 1 1 maxr"~ i l ^ - i l 1 } • i K - i l 1 * i=2.,n i = l . .n . i _ < 2 max{|a n. i| } i = l ,2. .n which i s Fujwara's i n e q u a l i t y [11] . Another proof can be found i n the next chapter. The f o l l o w i n g example shows that (29) can give a b e t t e r bound than Fujwara's i n e q u a l i t y . But before we give an example, we note that i f the maximum i n (28) i s I a n | then a l l the four i n e q u a l i t i e s c o i n c i d e and i n each case we have e ( f ) 4 2N = 2|a n. i|. -29-Example 9: f ( z ) = -2-2z-0 . 5 z 2 + z 3 N = /2~ (29) gives e ( f ) < 2.36 Fujwara's i n e q u a l i t y gives e ( f ) 4 3.46 Even (32) gives a b e t t e r bound than Fujwara. -30-Chapter I I  Zeroes of Polynomials The main object i n t h i s chapter i s to deal w i t h the l o c a t i o n of the zeroes of the polynomial f(z)=a 0+a 1z+ ... a n _ i z n _ 1 + z n , 'a- e C and i n p a r t i c u l a r w i t h the t r i n o m i a l equation l - z + xz n=0 ,X^0. Proofs of d i f f e r e n t i n e q u a l i t i e s promised e a r l i e r are a l s o given. The f o l l o w i n g theorem i s a c l a s s i c a l r e s u l t due to Cauchy [11]. Theorem 1: A l l the zeroes of the polynomial f(z)=a 0+a 1z+ ...a z n _ 1 + z n a. e C l i e i n the c i r c l e I z l < 1 + M where "TV--j 1 ' M = max | a- | , 0 <. i <_ n-1. i I - -Proof: We have | z | n = | f (z) - (ao + aj z+ .. .a n_ ^  z 1 1" -1) | < | f ( z ) | + |a 0+aiz+ ... a n . i z n ' 1 | < | f ( z ) | + | a 0 | :+ l a i l j z j + . . - l a n . J l z 1 1 " 1 ! i f ( z ) '."> |z|» . ? 11 a ± | | z | 1 = | z n | (1- z 1 j^ J_ ; ) i=0 i=0 ,•„ I n - i n > l z | n C l " i l i M l z l " 1 ) s i n c e M=max | a i | >|z|n (1- i M|z|~i) = | z | n (l - | fp i ) i f |z| >1 H f ( o i > | ? J T i C | Z | - 1 ' M ) -Now i f |z| ^  1+M then | f ( z ) | > 0. Therefore no root of f ( z ) l i e s i n the c i r c l e |z| ^  1+M, and so a l l roots l i e i n |z| < 1+M. -31-The next theorem i s due .to'Rehman [9]'.' I t gives a b e t t e r ( s m a l l e r ) c i r c l e than given i n Theorem 1. Theorem 2: A l l zeroes of the polynomial f ( z ) = a.Q ,alz- + ... ^ .»•>«» l i . ' i n the c i r c l e | z | < \ t 1 + 1 • » - , I * ' U V , ! ' 1 1 " 4 M ' where M = max{|a .J 0 <_ i '<_'n-D. • Proof: Suppose I z l > i - [1+1 a 1+ / ( l a , , I-1)2+4M ] . • 1 1 z 1 n - ^ 1 1 n - ^  1 Then ( 2 | z | - l - | a n _ i | ) 2 > |-1)2* 4M, and (|z|-l) ( | z | - | a n - i P " M > 0'"» l 2 | - | a n - i | - M > °-| z | - l M u l t i p l y by | z| n _ 1we have | z | n - | a n _ H z j 1 1 " 1 - M| z ) " " 1 > 0. | z j - l . . . d ) Consider M\z\n~l = M l z j " " 1 = M | z | n " 2 ( l + 1 + 1 + ...» ) | z | - i | z | ( i - i ) T*T 1*V > M | z | n ' 2 ( l + 1 + ... 1 ) > M(l+|z|+ ...|z| n" 2) > I ao I + l a x H z| + ...|a | | z | n _ 2 ^ | a 0 + aj z + . . . a n - 2 z 1 1 " I • A l s o | z | n - | a n _ 1 | | z | n " 1 < | z n + a n ^ 1 z n " 1 | . Hence from (1) we have J z n + a n _ 1 z n _ 1 | - |ao+aiz+ ... a n _ z n " 2 | > 0 => | f ( z ) | >0 Thus no root -of f ( z ) l i e s i n the c i r c l e |z| > I a n- 1I + / ( | a n _ i | - 1 ) 2 + 4M ] . Therefore a l l zeroes l i e i n |z| < \ [ l + | a n . i | + / ( | a n . i j - l ) 2 + 4M ] . I f we replac e z b y - i - i n f (z) , we have z n f ( i } > l + a n . i z + ... a i z n _ 1 + z n r | a 0 | = 1. Let N = max| a | . By applying Theorem 2 to z n f ( \ ) , we get a l l l < i < n - l zeroes of l+a n_^z + ... a i z n ~ l + z n to l i e i n the c i r c l e 1 < 1 [1+|ai|+ / ( 1 - | a i | ) 2 + 4N ] . 2 Therefore | zj >. ~—— we have proved [1+| a i |+ / ( l - | a | ) 2 . + 4N] Cor. 1 Let N = max{|a-| }. Then the polynomial g(z') = • . l<i < n - l 1 l + a n ^ z + ... a i z n - 1 + z n has no root i n the c i r c l e |z| < 2 • [l+| a i|+ / ( 1 - l a x l ) 2 + 4N] I f we apply theorem 2 to the polynomial ( z - a n ^ ) f ( z ) , we have Cor.2 A l l the zeroes of f ( z ) = a 0 + a i z + ... a z n _ 1 + z n l i e . * n -1 i n the c i r c l e |z| < j [1+ /1+4N' ] where N' = m a x | a i _ i " a n . J a i I 0<l<n-l a =0. The f o l l o w i n g theorem has been proved i n d i f f e r e n t ways -33-by Zedek and Ru b i n s t i e n s e p a r a t e l y . The f o l l o w i n g proof i s sim p l e r than the both. Theorem 3 : Let f ( z ) = a 0 + a i z + ... a m _ ^ z m ' 1 + z m , g(z) = bo+biZ + ... b n _ ^ z n - 1 + z n . Suppose t h e i r zeroes l i e i n the c i r c l e s | z - c i | <_ X\ and ,|,z-r.c2| 4 , r 2 respectively,. Let m > n ^ 1 and ef be the unique p o s i t i v e root of the equation h(z) = z m - IXI (z+1c 2-Ci|+r! + r 2 ) n . Then the m zeroes o f f ( z ) + Xg(z) l i e i n the c i r c l e |z-c1| £ r i + e i . Proof: Let 5 i , S 2 , ••• £ m and n1 ,n 2 > ••• n n be the zeroes of f ( z ) and g(z) r e s p e c t i v e l y . So |£^ - c1^| 4 r i and |n^-c 2| £ r 2 . Suppose I z-cx I = e + r j . Then |z- i | = | z - c 1 + c 1 - c 2 - c 2 - n i | < | z - c i I + I c 2 - C ! | + | n ± - c 2 I j = c+ri+lca-cx|+|n i-c 2| < r!+r 2+e+|c 2-c x|. n Thus |xg(z)|= |x| n |z-n.| < | X | (ri+r 2+e+ | c 2 - C i | ) n .... (1) - i = l 1 m A l s o | f ( z ) | = . n | z - 5 . | = n C l z - C i + c x - c . I > n ( | z - c i | - | C 4 - c i | ) . • i - 1 . 1 i 1 ~ i 1 < ^ nCri + e - r J = e m .... (2) i Therefore i f oey then s i n c e e i (but note ) i s the root of h(z) we have e m > |X| ( e + | c 2 - c i | + r i + r 2 ) n . Thus from (1) and (2) we have |xg(z)| < em < | f ( z ) | . Thus when | z - c x | = e+rj and e > e1 then Xg(z)+f(z) cannot vanish f o r any value of z i n | z - c j j > ri + t i . Therefore a l l the zeroes must l i e i n the c i r c l e j z - c j | 4 r i + e j . -34-Theorem 4 : If a > a„ > . . . a i > a 0 . Then f(z) = a 0 + a,z + •. n = n-j = i = u . <• w > . . . a„ z n _ 1+a z n has a l l the zeroes in the ci r c l e Izl < n -1 n I I = (a n-a 0 + |a0|) an Proof: - Consider-the polynomial (l-z)f(z) = - a n z n + 1 + z n ( a n - a n _ i ) + ...(a 1-a 0)z +a0 = g ( z ) - a n z n * i where g(z) = U n-a n_^)z n +\ + ... (a 1-a 0)z+a 0. Now on the unit c i r c l e |z|=l, we have |g(z)| 4 |a n-a n_ |+ ... I a i _ a o I + l ao I and therefore |g(z)| 4 a ^ a ^ + a ^ " a n - 2 + * ** aj-a 0+J a 0 I =an-a0 +1 a 0 I . Again on the unit cir c l e |z|=l, we have |z ng(l_)| 4 a -a 0 + |a 0|. Since z ng(l) is polynomial i t is z z analytic in |z| 4 1. Thus |z n g ( l ) | < a n - a 0 + |a 0| for |z| < 1 => | g(z) | 4(a n-a 0 + | a 0 | ) | z| n . z — n " n — for |z| > 1. Hence | ( l - z ) f ( z ) | = |g(z) - a n z n + i | > | a n z n + i | - | g ( z ) | > |a n l | z | n + 1 - | z | n C a n - a 0 + |a0|) - 1 a j | z [ n ([ z j-a n-a 0 + | a 0 | )>0 i f |z| > a n-a 0 + |a 0|. Thus for |z| > a n-a 0 + |a 0| f(zj does not | an| l a n l ' vanish. Hence a l l the zeroes of f(z) l i e in the c i r c l e |z| < a n ~ a 0 + J a 0 J Remark: Suppose a 0 ^ 0. Then |z| 4 a -a 0+a 0 Therefore the above theorem reduces to the f o l l o w i n g Kakeya and Enestrom theorem. Theorem 4' : I f a n v a n_^ >, ... . a x ^ a 0 ^ Oj then a l l the zeroes cf f ( z ) = a 0 + a 1z + . . «a n z n _ 1 + a n z n l i e i n the u n i t c i r c l e . We now consider a s p e c i a l case, the t r i n o m i a l equation l-z+Az n=0,x f* 0. We consider f i r s t the case n=2, i . e . , l-z+Xz 2=0. S e t t i n g z = l+£, we have X£2+(2x-l) 5 +x = 0. Let Si and £ 2 be i t s r o o t s . Then Ci C2 = 1 = > *>i= 1_ => Sjand s |- are the roots and t h e i r product i s u n i t y . Hence these are in v e r s e p o i n t s w i t h respect t o the u n i t c i r c l e . Being i n v e r s e p o i n t s both cannot l i e i n | 5 | 4 1 i . e . , | z - l | 4 I . Thus the equation must have a zero i n | z - l | > 1. Now consider the case n >^  3. Suppose, i f p o s s i b l e , a l l the zeroes of l-z+Xz n = 0 l i e i n | z - l | 4 1. Then by Gaus-Lucas theorem a l l the zeroes of the d e r i v e d equation X n z n _ 1 - 1 = must a l s o l i e i n | z - l | 4 1. But t h i s i s impossible f o r h 3. (by the above argument). Hence the t r i n o m i a l equation must have a root i n | z - l | > 1. S t i l l i t remains to see that i t has a root i n I z-11 4 1. Let C = z-1. Therefore we are to show that x ( t;+l) n -£ =0 has a root i n 1. Change? tc^- . So we are t o show th a t A ( z + l ) n - z n ~ 1 = 0 has a root i n | 5 | > 1 or e q u i v a l e n t l y -36-X z n - ( z - l ) n - 1 = 0 has a root i n | z - l | ^  1. To show t h i s suppose X z n - ( z - l ) n " 1 = 0 has a l l the toots i n | z - l | < 1. Then again by Gauss-Lucas theorem the successive d e r i v e d equations have a l l roots i n | z - l | < 1. In p a r t i c u l a r l-z+yz 2=0 has a l l the roots i n | z - l | < 1 what ever the c o e f f i c i e n t of z 2may be. But we have seen above f o r n=2 tha t the qu a d r a t i c equation has a root i n | z - l | ^ 1 . Hence we have proved Theorem 5: The t r i n o m i a l equation l-z+xz n=0 has a root i n both the regions | z - l | >_ 1 and | z - l ' | - 4 1. In f a c t , i t has been proved t h a t every general t r i n o m i a l equation a 0+a 1z + a n z n = 0, a x a n ^ 0 , n ^ 2 has at l e a s t one zero i n the c i r c l e |.z,| 2 and ,that ,e,ve.ry quadrinomial equation a 0 + a i z + a z m+a„z n = 0, a i a „ a „ ^ 0 2 < m < n has at l e a s t one U A m n ' 1 m n = 17 zero i n the c i r c l e |z| 4 - j For d e t a i l e d study of t h i s type see Marden [11]. Let us see the a p p l i c a t i o n of Perron-Fzobenius theorem to f i n d the upperbounds f o r the zeroes of the polynomial f ( z ) = a 0 + a i Z + ... A z n _ 1 + z n . We need the f o l l o w i n g Lemma A n -1 and Lemma B. T h e i r proofscan be found i n Marden [11] and Gantmacher [7] respec t ive ly . Lemma 1 A l l the zeroes of the polynomial f ( z ) = aQ+a.1z+ ... a n i z n - 1 + z n , a 0 f 0 l i e i n the c i r c l e |z| 4 r where r i s the r e a l p o s i t i v e root of the equation g(z) = - z n + |a n_ |z n _ 1 + ... | a i | z + |a 0|. n i .e ., r = j a n-r n - i + | an._2 | rn"2+ . -1 ajlT+laol . i s Lemma 2 Let A and B be square matrices of order n where A i r r e d u c i b l e and B + < A. Then | y | 4 r where y i s any eigenvalue of B and r i s the l a r g e s t eigenvalue of the matrix A. The companion matrix of f ( z ) = a 0+aiz.. .n- • n- + zn i s B = B n-i 1 0 I a n-• a • ..« ~ a Q n" 2 0 0 6 i o o o 0 1 o n- 2 1 0 0 ao By Lemma 1 the eigenvalues of B(zeroes o f f (z ) ) l i e i n the c i r c l e |z| 4 r where r i s the zero of g ( z ) . Now ao/ 0 => f ( z ) i s i r r e d u c i b l e => B + i s i r r e d u c i b l e . Also B < B +. Therefore by Lemma 2 , the eigenvalues of matrix B are dominated by the l a r g e s t eigenvalue of B + i . e . , the r e a l p o s i t i v e root o f the equation h(z) = -|ao|-|aj|z ... n-i Now apply Perron-Fzobenius theorem to the matrix B we have Theorem 6: The polynomial h(z) has a zero r which is real, .. s i m p l e , and exceeds i n modulus a l l other eigenvalues of h(z) I f p r e c i s e l y p zeroes of h(z) have modulus r , then each of these s a t i s f i e s the polynomial z p - r p = 0 and then the set of a l l zeroes of h(z) is carried into itse l f by a r o t a t i o n of the complex plane through an angle 2_n_ . P The number r i s given by r = mm • > 0 l<i<n (max ( [ B X ] i , X, ¥ 0>;> 1 -• - X, 1 where X i s an n-tuple of non-negative entries-Now replace B + by i t s transpose, namely n-2 ail a 0 I we have Cor. The number r ! 1 n-x I a. 1 0 0 1 0 0 .. 0 .. 0 . 0 1 . 0 0 min (max f Xi=0 l<i<n 1 a X 1 n - i I i s the r e a l p o s i t i v e dominated root of h ( z ) . I f z i s a zero of f ( z ) then s i n c e a l l the zeroes of f ( z ) l i e i n |z| < r , we have -39-1< i<n z| <max { | a n - i | 1 + i + i } , = 0 (3) X i A p p l i c a t i o n s (a) Take X^ = 1, i = l , 2 , ... n i n ( 3 ) , we have | Z| < maxa + | a n . i | , l + | a n _ 2 | , ...1 + | ai | U a 0 | > • . which i s W i l f ' s i n e q u a l i t y .proved on Page 23. z , < max *1 , i= = 1,2, .. . n a 0 I,2 ai 2 , • • • a n - 2 a>l a 2 a n - , 2 ^ which i s Kojima's [11] i n e q u a l i t y proved on Page 25. (c) Take X^^ = t * , t > 0, i = 1,2 , ... n | 2 | < m a x { t + | a n _ J , t + | a n . J , ... t + | ^ J , | a 0 | } ' t t n - 2 j j F T which i s W i l f ' s i n e q u a l i t y (21) , proved on Page 26. (4) i s a s p e c i a l case of (6) From (6) we have |z| 4 t + max{ a_ (4) (5) ( 6 ) - T — i - 1= 1,2, . . . n K Choose t such that two terms are equal. Therefore |z| <t + maxVla^J 1"} £ 2 . B a x . { | ^ . j * > i = 1 > 2 , . . .„', which i s Fujiwara i n e q u a l i t y [11] proved on Page 28. -40-Chapter I I I  V e c t o r i a l Norms The present chapter i s devoted to the study of v e c t o r i a l norms, dual of v e c t o r i a l norms, r e g u l a r v e c t o r i a l norms, v e c t o r i a l pseudo - norms , -matrix -norms -subordinate to - v e c t o r norms and m a t r i c i a l norms subordinate to v e c t o r i a l norms. Recently these norms have been s t u d i e d by Deutsch [3,4]. The main object i s to di s c u s s the d i f f e r e n t p r o p e r t i e s such as r e g u l a r i t y and d u a l i t y . Theorem 3 e s t a b l i s h e s the r e l a t i o n between two m a t r i c i a l norms, one generated from a v e c t o r i a l norm and the second generated from a matrix norm. Let C n denote the vector space of n-tuples over the k complex numbers and R + be k-tuples of non-negative r e a l numbers p a r t i a l l y ordered component-wise. n k D e f i n i t i o n : A v e c t o r i a l norm of order k i s a mapping v:C •+ R + such t h a t (V. l ) v(ax) = |a|v(x) ¥ o e C, x e C n. (V-2) v(x+y) < v(x) + v(y) ¥x ,y e C n. (V.3) x f 0 => v(x) f 0. Remark: I f k=l then v i s a ve c t o r norm on C n, so the n o t i o n of a v e c t o r i a l norm i s a g e n e r a l i z a t i o n of v e c t o r norm. Remark: A g e n e r a l i z e d m a t r i c i a l norm of order k on M i s ' a 2 v e c t o r i a l norm of order k 2on C . TI k A mapping v:C -*• R^ which s a t i s f i e s (V«l) and (V*2) i s c a l l e d a v e c t o r i a l pseudo-norm of order k on C . Let V i ( x ) , ... v k (x) be the components of v(x) . P r o p o s i t i o n 1: v i s v e c t o r i a l pseudo-norm i f f the mapping x •*• v^(x) (x e C n) i s a pseudo-norm on C n f o r each i = l , 2 , . . . k Proof: Let v f ( x ) be a pseudo-norm f o r each i . Define the n k mapping "v: C -*R+ 'such that v(x) = ( v i t x ) , v 2 ( x ) , ... v k ( x ) ) ¥ x e C n. Then v(x+y) < ( v x (x) + vi(y) , . .. v k ( x ) + v k ( y ) ) = ( v i ( x ) , ... v k ( x ) ).-•+. ( v i ( y ) . v k ( y ) ) = v ( x ) + v ( y ) . v(ax) = ( v i ( a x ) , ... v k ( a x ) ) = ( | o | v i ( x ) , '•... | o | v (x)) | o | v(x Conversely, i f v i s a v e c t o r i a l pseudo-norm, then v^(x+y) 4 v^(x) + v^(y) and v^(ax) = |a|v^(x)., so v i i s a v e c t o r i a l pseudo-norm a l s o . Let v: C n -> be v e c t o r i a l norm on C n of order k. Define KjCv) » {x e C n | V j ( x ) = 0 } K(v). = f)KAv) W (v) » 0 K.(v) i 1 ' \ i * j 1 k W(v) = l W.(v) i = l 1 P r o p o s i t i o n 2: i ) K . . ( v ) , K ( v ) ( v ) , W ( v ) are subspaces of C n. i i ) v i s v e c t o r i a l norm i f f K(v) = {0} . Proof: i ) since sum and i n t e r s e c t i o n of subspaces i s a subspace, i t i s enough t o show that (v) i s a subspace of C n. Let x ,y e K.. (v) . Then si n c e Vj i s a pseudo-norm, we have v.(x+y) 4 v.(x) and v.(y) = 0 and v.(ax) ~|a|v.(x) = 0 => - 4 2 -ax + By E Kj (v) => (v) i s subspace of C n. i i ) Let v be v e c t o r i a l norm. Then v(x) = 0 => v^(x) = 0 ¥i = 1 , 2 , . . . k => K i(v) = 0 => K(v) = 0 . On the other hand i f v(x) f 0 then v^(x) f 0 f o r at l e a s t one i . Therefore K(v) r 0 . D e f i n i t i o n : A vectorial norm v is said to be a regular v e c t o r i a l norm i f C n = Wx(v)@ . ..@Wk(y) and a v e c t o r i a l pseudo norm v i s s a i d to be a r e g u l a r v e c t o r i a l pseudo-norm i f C n = W!(v) + ... +W k(v). The f o l l o w i n g example shows that not every r e g u l a r v e c t o r i a l pseudo-norm i s a r e g u l a r v e c t o r i a l norm. Example .1: Consider the mapping v:C 3 -*• R 2 such t h a t v(a,B,y) = ( | a | , | Y | ) ¥ a , B , Y e .C. Then v i s a v e c t o r i a l pseudo-norm. v ( ( a , B , Y ) + (x,y,z)) = V(a+X ,B+y , Y + Z) = (|a+x|,|.Y + z|) < ( | a | , | Y | ) + ( I X | , | Z | ) = v ( a , B , Y ) + v ( x , y , z ) and v(xx,xy,xz) = (|xx|,|xz|) = |x| v(x,y,z) Now i q C v ) = { (a ,B ,Y) I v I (a ,B ,Y) = 0 }.= { (0 ,3 , Y ) | B^ yeC }=W 2(v) 7 and K 2(v)= { ( O , 3 , Y ) | v 2 ( a , B ,Y ) = 0 } = {(a ,3 ,0)|a,BeC} =W :(v). Now C 3 = (v) + W 2(v) and so v i s a r e g u l a r v e c t o r i a l pseudo-norm. However v i s not a r e g u l a r v e c t o r i a l norm s i n c e C V Wj(v) © W 2(v) . D e f i n i t i o n : Two v e c t o r i a l pseudo-norms u,v are s a i d t o be eq u i v a l e n t i f W.(v) = W.(u) * j = 1 , 2 , ... k. -43-P r o p o s i t i o n 3 : Let v be v e c t o r i a l pseudo-norm on C and G an n k n x n complex ma t r i x . Then the mapping v ^ : C •* R + d e f i n e d by v G ( x ) = v(Gx) ¥ x e C n i s a v e c t o r i a l pseudo-norm on C n. Remark: v G i s c a l l e d the G-transform of v. Note the s i m i l a r i t y of v G w i t h y G . Proof: V G ( a x ) » v(Gax) = |a|v(Gx) = | a|v G C x ) ¥ x e C n , a e C , v G(x+y) = v(Gx+Gy) < v(Gx)+v(Gy) = V GM *'-VQW ¥ X>Y e C n . n k P r o p o s i t i o n 4: Let v : C + R + be a v e c t o r i a l norm and G an i n v e r t i b l e n x n complex ma t r i x . Then 1) v G i s v e c t o r i a l norm. 2) ^ ( v G ) = G " l K j Cv) * j - l , 2 , . . . k . 3) W 5(v G) = G'lW^Cv) V j = 1,2, . . . k . 4) W(v G) = G _ 1W(v). 5) v i s a r e g u l a r v e c t o r i a l norm => v G i s a r e g u l a r v e c t o r i a l norm. Proof: (1) x f 0 => G x f 0 => v(Gx) i 0 => v G ( x ) f 0. Therefore by P r o p o s i t i o n 3 ,v G i s a v e c t o r i a l norm. (2) K.(v r) = {x e C n | v . (x) = 0} ={x e C n|v.(Gx) = 0 } J b 3G 3 — {x e Cn|Gx e (v) } = {x e C n | X E G " lK. (v) } =G _ 1K..(v). £3) W.(v r) =0 K-(v r) =0 G'iK.(v) = G'lO K £ ( v ) = G _ 1W.(v). k k k (4) W(v r) = E W.(v r) = E G _ 1W.(v) = G - 1E W.(v) = G _ 1W(v). G i = l 1 G i - 1 ' 1 i = l 1 -44-k (5) v i s a r e g u l a r v e c t o r i a l norm ==> W(v) = z @W. (v) ==> i = l 1 . k k G _ 1W(v)- - Z © G_1W.(v)=> Z ©W.(v r) = W(v r) (using i = l 1 i = l 3 b \ (3) and (4) ).=> v^ i s a r e g u l a r v e c t o r i a l norm. P r o p o s i t i o n 5: Let p and q be v e c t o r i a l pseudo-norms on C n. Assume p < q ( i . e . , p(x) 4 q(x) ¥x e C n) . Then ( i ) p i s a v e c t o r i a l norm => q i s a v e c t o r i a l norm. ( i i ) q i s a r e g u l a r v e c t o r i a l pseudo-norm => p i s a r e g u l a r v e c t o r i a l pseudo-norm. Proof: ( i ) Let x e K..(q) => q..(x) = 0 => p^(x) = 0 s i n c e P <. q» t h e r e f o r e x e K.(p) => K.(q) C K.(p) =>OK.(q) £ 3 ~ 3 j 3 p^^ *K*r(ipOt^^ K'(*q<)','C^ K"'(*pO . 'Therefore-by - P r o p o s i t i o n 2 ( i i ) we 5 3 have K(q) = 0 => q i s a v e c t o r i a l norm. ( i i ) x e W.(q) = 0 K.(q) => q.(x) = 0 ¥ i ¥ j => p- (x) = 0 s i n c e p <. q => x e A K. (p) => x E W. (p) V ~ i r j 1 3 => W. (q) C W. (p) ¥j = 1,2, ... k. Since q i s r e g u l a r we have J J p i s r e g u l a r . n k P r o p o s i t i o n 6: Let v:C -»- R + be a r e g u l a r v e c t o r i a l norm. Then K. (v) = Z © W-(v) . ¥i = 1,2, ... k. ^ 3 Proof: x E Kj.(v). => v^ (x) = 0 ^  ¥ x E C n. Since v i s a r e g u l a r v e c t o r i a l norm, t h e r e f o r e C n = W(v) = Wj(v) © ... © w k ( v ) . Then x = x 1+x 2+ x k , x^ ^ E W. (v) => v^ (x) = -45-v.(xi+x2+ . . . x v) = v.(x.) [since i f v(y) = 0 then v(x+y) 3 K J J 4 v(x) , on the other hand v(x) = v(x+y-y) 4 v(x+y)«-> v(x+y = v(x)] . Hence 0 = Vj (x) = v^(x^) «=> Xj = 0. Now x = Xi+x2+ . . . x. +^x. + x. + . . . x. • xi+ . . . x. + x . + , + . . . x v e E W.(v), J I 3 3 + 1 K 3 1 3*1 K j ^ j 1 'S ince ' v i s r e g u i a r t h e r e f o r e x E ZT*©W."'CV) . Thus K.(v) "C. i f j 1 y ~ E © W. (v) . On the other hand we always have E © W. ( v ) C K . ( v ) i^3 1 i « 1 3 >K.(v) = E © W. ( v ) . Remark: The converse of the above P r o p o s i t i o n i s f a l s e . Consider the v e c t o r i a l norm v:C** -*• R 3 such t h a t v ( a 1 . , a 2 . , a 3 . , a M ) = (J a . J + | a k J. . ,J a 2 J + | at,J | a 3 | + | \ ) I t i s r o u t i n e t o see that v so defi n e d i s a v e c t o r i a l norm. Now Kx(v) = {x e C n|v 1(x) = 0 } ->ai=a4= 0 => K x(v) = { ( 0 , a 2 , a 3,0) | a 2 , a 3 e C } . S i m i l a r l y K 2 (v) = { ( a i ,0 ,a 3 ,0) | a i ,a 3 e C } K 3(v) = { -(ox ,a 2 ,0,0) | a i ,a 2 e c}. Now • W r(v) = K 2 ( v ) 0 K 3(v) = (ai ,0,0,0) W 2(v) = K . i(v ) n K 3(v) = (0,a 2,0,0) W 3(v) = K 1 ( v ) 0 K 2(v) = (0,0,a 3,0). Obviously Kx (v) = W2 (v) @W 3(v),K 2(v) = W: (v) ©W 3(v),K 3(v) = W 2(v) + W r(v). But v i s not r e g u l a r s i n c e C1* r" W.j (v) © W 2(v) © W 3(v). -46-Theorem 1: Let p be a v e c t o r i a l norm on C n. The the f o l l o w i n g are e q u i v a l e n t :-(1) p i s a r e g u l a r v e c t o r i a l norm. (2) There e x i s t s a norm v on C n and a direct-sum decomposition C n = Xi© ... ©X k w i t h a s s o c i a t e d p r o j e c t i o n s Ex,E 2, . . . E k such t h a t p(x) = ( v C E j x ) , v ( E 2 x ) , ... v ( E k x ) ) ¥ x e C n. k (3) I f p(x) = u + a,u,a e R ,- then there e x i s t s v e c t o r s y and z e C n such that x = y+z, p(y) = u and p(z) = a. (4) C n = K.(p) © W.(p) Vj = 1,2, ... k. Proof: (1) => ( 2 ) . Suppose p i s a r e g u l a r v e c t o r i a l norm. Let X i = W i(p). Therefore C n = W x(p) © ... © w k ( p ) . Let Ei , ... E k be the p r o j e c t i o n s a s s o c i a t e d w i t h t h i s d i r e c t -sum decomposition. Then p-(x) = p . ( E i x + E 2x + ... E,x) = p^(E^ x ) . Now consider the mapping v on C n defined by x -*- v(x) = max p. (x). i Then i t i s easy to see that t h i s mapping defines a norm on C n. Now v(E.x) = max{pr, (E.x)} = max ' {p, (E, E .x)'} = p.(E.x) 3 h n J h=l,2,...k n n 3 1 1 ~ Pj (x) ¥j = 1,2, ... k.. Hence p(x) = (v(Exx) , ... v ( E k x ) ) . (2) => (3) Let u=(u 1,u 2, ... u k) , a= (ax, ... a k) . Then sin c e p(x) = u+a we have p'^(x) - u i + a i =>(by (2)) v f E ^ ) = u. + a. . I f v(E.x) = 0 then u=0=a so we are done. -47 u. I f v(E.x) i 0, then l e t v(E.x) = w. > 0. Set y. = _L E.x X X X X • W *^ X a i _1_ and z i = w7 E i x * T n e n + z ± = w i ( u i + a i ) E - [ X . N O W s i n c e u- + a. = p.(x) = vfE.x) = w. th e r e f o r e y.+z. - E . x k k k ¥i = 1,2 , . . . k.=> Z y- + £- z- = Z E.x => y+z = x. d = l 1 1=1 1 i = l 1 u i P i ( u . ) p . ( E . x ) Now y = z ^ E i X => p-.Cy) = i i i i i - " W i p . ( U i ) v ( E i X ) v ^ x ) = p i(u j L) ¥i = 1,2, ... k => p(y) = u. S i m i l a r l y p(z) = a. (3) => (4) Let X E C n and u = (px (x) , ... Pj M M ,0 ,p^ + i (x) .'. .-^ pg'("x*)0-^ an€-?'a^ = -^ (-x*) •~~vr~'"4fi -,'0 , . .. -p/fx) , 0 ... 0 ) . Now p(x) = u+a =>' by hypothesis that there e x i s t s y,z E C n such that x = y+z,p(y) = u, p(z) = a. Now u = p(y) => P j ( y ) = u. = 0 => y e K.(p). Also a = p(z) => p^(z) = a- = 0 ¥i V j ~> z E W^(p). Thus x = y+z E K.. (p) + W.. (p) => C n C Kj(p) + W.. (p) . But we always have K.. (p) + W.. (p) C n. Therefore C" = the sum i s d i r e c t . n = K.(p) + W.(p). Since W.(p) = O K.(p) t h e r e f o r e 3 3 3 it j 1 k (4) => ( 1 ) . Z © W.(p) C K.Cp) => Z © W . ( p ) O K (p) © W (p) if j 1 3 i = l . . 3 3 = C n => Z © W. Cp) C C n . On the other hand x E C n => x E K.(p) i = l 1 ~ 3 © VL(p) => x E W(p) => C n C W ( p ) . Thus W(p) = C n and the theorem i s proved. Remark: K a u t o r o v i t c h , V u l i k h and Piu s k e r consider mappings which s a t i s f y axioms (V«l) to (V«3) and property (3) of Theorem 1. F i e d l e r and Ptak consider mappings which are obtained through the procedure 'indicated i n property (2) -of Theorem 1. Consequently, theorem 1 shows that a l l these mappings are e x a c t l y r e g u l a r v e c t o r i a l norms. P r o p o s i t i o n 7: Let p and q be r e g u l a r v e c t o r i a l norms of order k such that p < q. Then (1) K.(p) = K.(q) ¥j = 1,2, ... k (2) p and q are e q u i v a l e n t , i . e . , IV. (p) = W.. (q) ¥j = 1,2, ... Proof: Since p and q are r e g u l a r v e c t o r i a l norms t h e r e f o r e by Theorem 1 we have C n = K.(p) © W.(p) = K.(q) © W.(q) . I t was proved i n p r o p o s i t i o n 5 that K..(q) C (p) and W. (q) c W.(p). Since K. (q) © W.(q-) and K.(p) © w!(p) have the 3 3 3 3 3 3 same dimension, K.. (p) = K.. (q) and W.(p) = VL (q) . We now define the n o t i o n of tne dual Of a v e c t o r i a l norm. Let p be v e c t o r i a l norm on C n. Consider the mappings q j : C n + R such t h a t q^(y) = sup |y*x | , j = 1,2, . . . k) (y* i s a d j o i n t J x^O " P T T O xeWj (p) -1 of y) Fu r t h e r , d e f i n e the mapping p^: C n •+ R^ such that P dCy) - ( q i ( y ) , ... q k ( y ) ) * y e c n. -49-Now we.claim p so defined i s a v e c t o r i a l p seudo-norm on C n pf(ay) = q.(ay) = sup. \(ay)*x| . = sup l a y * x l  3 3 x^O p'cxj x?0 p. CxJ xeW..(p) 3 3 = |a|sup | y*.x 1 - |a|p-(y) => P d(«y) = |a| p d ( y ) . x / 0 p. f . J xeVL Op) 3 W A l s o p d ( y + z) = ( q i ( y + z ) , ... q k(y+z)) ¥y,z e C n. Now q.(y+z) = sup |(y+z)*x| = sup |y*x + z*x| 4 3 xfO p. txj x?<0 pTTT) xeW.(p) 3 sup |y*x| + sup 1z*x| = p.(y) + p . ( z ) . Thus x?0 P7C>0 X ^ O pTCx) 3 3 p^(y+z) <_ p^(y) + p d ( z ) = > p d i s a v e c t o r i a l pseudo-„n norm on C . D e f i n i t i o n : The mapping p^ defined above i s c a l l e d the dual of p. P r o p o s i t i o n 8: Let p be a v e c t o r i a l norm on C n. Then CD K.(p d) - O y p ) ) ¥ j = 1,2, . . . k C2) W j(p d) 3 ( K j f p ) ) (3) K(p a) = (W(p)) (4) p d i s ^ r e g u l a r v e c t o r i a l pseudo-norm. ( 5 ) p d i s v e c t o r i a l norm i f f p i s r e g u l a r v e c t o r i a l norm. -50-(2) w,(pd) = n K . c P d ) = n cw.cp)) = c z w.(p)) Proof: (1) K.(p d) = {y t C n | p d ( y ) = 0 } ={y e C n|q j Cy) = 0 } X - { y e C n|y*x = 0 } = (W..(p)) . a ( P d )  Now sin c e E W. (p) C K. (p) , ( E W. (p) )*** (K. (p) ) therefore i f j 1 ~ J i ^ - j 1 " 3 j • -4". (p d) D (K.(p) . . • X X X (3) K(p d) =nK.(p d) - . n.(W.(p)) = (EW.(p)) = (W(p)) . i i i (4) We have shown already that p d i s v e c t o r i a l pseudo-nc>rm. We show tha t W(p d) = C n. Now W(p d) = E W . ( p d ) O E(K.(p)) i 1 ~ i 1 X X . » (HK-Cp)) = { H J x e C n J-p.ixJ = Oj.} = C n => wc.pd.)S c n . i 1 i 1 But we always have W(p d) CZ C n. Therefore W(p d) = C n. (5) Using P r o p o s i t i o n (2) ( i i ) , we have p d i s v e c t o r i a l norm i f f K (p d) = 0 i f f (WCp))*1 = 0 i f f W(p) = C n i f f p i s r e g u l a r v e c t o r i a l norm. P r o p o s i t i o n 9: Let p and q be v e c t o r i a l norms on C n such that p 4 q . Then p d ^ q d . A T\ Proof: We have p.(y) = q-(y) - sup |y*x| ¥ y e C . 3 3 x?0 pTTO xeW.(p) 3 Since p(x) 4 q(x) t h e r e f o r e 1 >^  1 . Thus p d ( y ) ^ Pj (.xj q^ (.xj 3 sup |y*x| = q d(y) => p d > q d . XrO qTIxt 3 xeWj(p) J 51-P r o p o s i t i o n 10: Let p be a r e g u l a r v e c t o r i a l norm on C n such tha t W. (p) I I.(p) -Vi / j . Then d " X b) K. ( p d ) = -K..(p) c) Wj (pd-) = Wj'(p) - i r e . ,'pand~"p d -are-equivalent. Proof: (a) Using P r o p o s i t i o n 8 ( i i ) and P r o p o s i t i o n ^ , we have W.(p d) 3 ( K . ( p ) j ^ ( E © W.(p)) = H (W.(pj) -3 3 ~ i * j 1 i * j 1 , , • . J -f ) K ( p d) = W (p d) => W (p d) = (K.(p)) . i¥i 1 3 3 3 (b) By P r o p o s i t i o n 6, we have K.(p) = E © W.(p) and t h e r e f o r e 3 iti 1 by hypothesis we obta i n K.. (p) _J_W.(p) and si n c e by Theorem l ( i v ) L -L C n = K.(p) © WjCp) th e r e f o r e K (p) > W.(p) , W (p) = K.(p). d •' X • Now us i n g P r o p o s i t i o n 8 ( i ) we get K.. (p ) = W.. (p) = K.. (p) . (c) Using, (a) we have W. (p d) = (K.(p)) = W. (p) . P r o p o s i t i o n 11: Let p be a v e c t o r i a l norm and G an i n v e r t i b l e n x n complex ma t r i x . Then (1) the dual of the G-transform of p i s the ( G " 1 ) * - transform of the dual o f p. (2) I f G i s a u n i t a r y m a t r i x , then the dual of the G-transform of p i s equal to the G-transform of the dual of p. -52-Proof: (1) The dual of the G-transform of the j t h component of sup 1y*x | = sup xeWjCpg) b 3 x G iW^Cp) 3 p i s given by ( p d ) , ( y ) "  l y * x |y*x| b 3 / r i T TvT vain U irA = sup |y*G"}z| (where Gx = z ) t z?0 p , ( z j ' ZeWjtp) 3 = ( p d ) (G r l*y) (s i n c e (G" 1 *y) * = y*G'M , « ( P Q - 1 * ) (y) . Therefore p° = P G " i * i s the G - transform of P D . (2) Since G i s a u n i t a r y m a t r i x , t h e r e f o r e GG* = 1 => G* = G~* i=> (G~*J* = G. Thus by (1) the dual of the G-transform of p i s ^equal -,to-:'the«G-4transf0rm...of<-the«dual-*of. .p... Define the norm of a matrix A as the maximum of the norms [5,113] of the vectors Ax where the v e c t o r x runs over the set of a l l v e c t o r s whose norm i s 1 i . e . , ||A|| = max ||Ax|| . ||x||=1 Let A j 1 0, Then a v e c t o r x can be found w i t h | |x| | = 1 such t h a t Ax i 0 => ||Ax|| + 0. Thus ||A|| = max ||Ax|| 1 0 I|x||=1 A l s o l l a A M - max || aAx|| = |a| max | | A * | | = . | ct | ||A|| | | x | | _ - l l | x | | - l F urther ||A+B|| = max -||(A+B)x|| = max ||Ax+Bx|| 4 ||x||=1 I Ix|J =1 max I IAxl I + max I|Bx|I = I IA|I + I|B|I . 11*11=1 11*11=1 and I|AB|I » max ||ABx|| = max ||AxBx|| 4 IIxl1=1 IIxI 1=1 max ||Ax|| . max ||Bx|| - ||A|| ||B||. I |x ||=1 | | x ||=1 D e f i n i t i o n : The norm of a matrix constructed i n t h i s way i s c a l l e d the matrix.norm sub-ordinate to the vector norm and i s written as l u b v ( l e a s t upper bound of v e c t o r norm v ) . Next we introduce the ide a of a m a t r i c i a l norm sub-ordinate to a vectorial norm which i s a generalization of matrix norm subordinate to a vector norm. Let p be a regular vectorial norm on C n. Define the mapping l u b p : Mn"-*- M*. such t h a t "l»bpCA) = ( " i j C A ) ) ^ , . l f 2 > . . . k * A e H n where m..(A) = sup p.(Ax) x e cn. 1 3 x^o .^znrr xeW j(p) P j W .Then l u b p ( A ) i s a regular matricial norm on Mn . Let A f 0. Then s i n c e x f 0 we have Ax f 0 => nu.. (A) f 0 ¥i,j = 1,2, ... k,=> l u b p ( A ) f 0 which s a t i s f i e s (M-1) . Let a e C # Therefore m^^ (aA) = | a | m^ (A) f or a l l i , j = 1, 2, ... Hence Tub p (oA) = |a| lub p (A) which v e r i f i e s - ' • (M.2). To see (M«3) and (M*4) we have A,B, e M n lub u b p ( A + B ) = m.-CA+B)^. , 1 > 2 > _ k -54-S " P Vi ( (A+B)x) sup \ lx\ ^ x*° xeW. (p ) p j L X J ~ xeW.(p) P 4 ( A x ) + p , (Bx) p y r x j sup p . ( A x ) sup p fx) + x*° p fx) x M V ^ p ) n j t x J x £ W . ( p ) P j W p , (Bx) • - l u b p ( A ) + l u b p ( B ) . ^ W - C - i j W ) 1,1, ... k - 3 sup p i ( ( A B ) x ) •xeW. (p) *j sup p . ( A x ) <x^0 • x e W . ( p ) P j ^ - " sup p . ( B x ) X e W ^ p ) P j U j l u b (A) . l u b (B) S i n c e p i s a regular vectorial norm therefore lub^CA) i s a regular m a t r i c i a l no rm. ,>D(gTihl' ftl"o'n": ••Th^m^p'pifn-g"'*l"Ofb ' X i e a s t ^upperbouird o f p ) " d e f i n e d above i s c a l l e d the m a t r i c i a l norm s u b o r d i n a t e t o the vectorial norm p . Example 2 : C o n s i d e r the v e c t o r i a l norm p : C 3 ^ R2 d e f i n e d by p ( x i , x 2 , x 3 ) = ( |x ! | , | x 2 | + |x 3 |) f o r a l l ( x 1 , x 2 , x 3 ) e C 3. I t i s e a s y t o see t h a t K 2 ( p ) = W 1 ( p ) , K 1 ( p ) = W 2 ( p ) and C 3 = W i ( p ) © W 2 ( p ) . T h e r e f o r e p i s a regular vectorial norm D e f i n e the mapping l u b ^ : M 3 M 2 such t h a t l u b p ( A ) = m i l (A) m 12(A) m 2 1 ( A ) m 2 2 ( A ) where m. .= sup 1 3 xfO XeWjCp) *j P i ( A x ) P,U) Now px(x) = |x 2| and p x(Ax) = | a x x | \xx\ . Therefore mil (A) = | a n | . S i m i l a r l y m 1 2(A) = | a 1 2 | + |a 1 3|,m 2 1 = | a 2 i | +| a 3 1 | , m 2 2 (A) = | a 2 2 | + | a 2 3 | + | a 3 2 | +|a 3 3| . an a 1 2 ai3 Hence lub (A) = P a 2 11 +1 a 3 i a 22 I + l a 2 3 | +1 a 3 2 | +1 a 3 3 Theorem 2: Let p: C n -> be a r e g u l a r v e c t o r i a l norm on C n. ijBlin • —Mil,I  ••n.,1- II ill - . " T * Let G be i n v e r t i b l e n x n complex m a t r i x . Then the m a t r i c i a l norm subordinate t o the G-transform of p i s equal to the G-transform of lubp. In other words l u b p = ( l u b p ) G . Proof: Set p G = q. Since p i s a r e g u l a r v e c t o r i a l norm t h e r e f o r e by P r o p o s i t i o n 4 (v) q i s a r e g u l a r v e c t o r i a l norm and so C n = Wi(p) "© W 2(p) © © W k(p) C n = W x(q) © W 2(q) © © W k(q). Let Ei ,E2 , ... and Fj ,F2 , .. . F^ be the associated projections of the above decompositions. Again by Proposition 4 ( i i i ) Wj(q) = G"1!^ (p) . Now ¥ x e C n, we have Fj(x) e Wj Cq) = G~V.(p). Also E.. G x e W.. (p) therefore G ' Ej Gx e G ^  W..(p) and since the above sum is direct, therefore G _ 1 E.. G = F.\ ¥j = 1 , 2 , ... k. Now (lub A).. = ( i . j ) element of matrix lub A. q i j y q = sup q.(Ax) = sup q-(F.Ax) xeW^q) x e C V f p ) q j W (since W.. (q) = G~Vj(p) and q^y) = q i(F j ty))^ = sup q.(F.AF.x) (since ¥ x e W.(q), F. (x) = x) xcG V ( p ) ^ j * J = sup .q.^E.GAG^E.Gx) / s i n c e F > . G " l E . G ) xfO * ^ j j ) GxeW^p) qj(G _ 1E^Gx) - s u p p^E.GAG^E.y) ( w h e r e y'- Gx, p.(Gx) - q.(x) yeW.(p) P j(E.y) i . e. , q, (G^x) = p^x) ) ; •= sup p. (E.GAG^y) , x y r0 1 1 (since ¥-y e W.(p) , E. (y) = y J yeW.(p) ~TT ' > = sup P^GAGSO (SINCE PI(E.X) . P. ( X ))^ yeW.. (p) p; (y) -57-= ( l u b p GAG" 1)^ . Thus ( l u b ^ A) = ( l u b ^ GAG"1) = ( l u b p ) G and the proof i s PG complete. Example 3: Consider p the v e c t o r i a l norm, and l u b p the m a t r i c i a l norm subordinate t o p, given by the example 2. 3 0 0 1 0 0 2 1 0 . Then Q" 1 = 1 -2 3 3 0 0 1 3 0 0 3 Take G = Now the G-transform of p is given by P G ( x i , x 2 , x 3 ) = p(Gx) ¥ x = ( X ! , X 2 , X 3 ) e C = p(3x, ,2x r + x 2 , x 3 ) = ( 3 | X l | ,|2 X l + x 2 | + | x 3 | ) . NowW 1(p G) = G^WjCp) = ( x i ,0,0) and W 2(p G) = G' 1W 2(p) » ( 0 ! X 2 , x 3 ) . Since ( x i ,x 2 ,x 3) = (xj ,0,0) © (0xx 2 ,x 3) t h e r e f o r e p G i s a regular vectorial norm Now by the above theorem lub (A) = lub (GAG"1) = pG • P mji(GAG - 1) mi 2(GAG _ 1) m 2 i ( G A G 1 ) m 2 2(GAG _ 1) - 5 8 -GAG"1 = 1 J 3an-6ai2 9ai2 9ai3 2ai i+a2 i-4ai 2 -2a22 6aj2+3a22 6ai3+3a23 a3i-2a32 3a 3 2 3a3  Now m^CGAG"1) = sup pjCGAG^x) = 1 j 3 a i l - 6 a 1 2 | = Pi(x) | a n - 2 a 1 2 | s i n c e x e Wx (p) . S i m i l a r l y m 2 1(GAG' 1) = - { | 2a 2 x+ a 2 i - 4 a 2 2 - 2 a 2 2 I + | a 31 - 2 a 3 2 |> 3 i m 1 2(GAG _ 1) = 3 max {|a 1 2| , | a i 3 | } m 2 2(GAG" 1) = max{|2a 12 +a 2 2|+|a 3 2|,|2a 1 3ta 2 3 l +|a33 S u b s t i t u t i n g these values i n (1) ,we get a m a t r i c i a l norm ''s'tfb'"byd!i*h-ait'e',"''t'o''™pQ. We have seen that s t a r t i n g w i t h a vector norm v on C n and the given direct-sum decomposition C n = Xi© ... © X^, we can now generate two m a t r i c i a l norms of order k on M n as f o l l o w s : -(1) Given the v e c t o r i a l norm p on C n induced by v and the given direct-sum decomposition, we form the m a t r i c i a l norm l u b p subordinate to p. (2) Given the matrix norm l u b v subordinate t o v, we form the m a t r i c i a l norm y induced by l u b v and the given decomposition of C n. -The f o l l o w i n g theorem gives the r e l a t i o n s h i p between these two m a t r i c i a l norms. - 5 9 -Theorem 3: Let E\ ,E 2 , ... be the p r o j e c t i o n s a s s o c i a t e d with the decomposition C n = Xi© ... © X k of C n. Let v: C n -*• R + be a vector norm on C n and p: C n •*• R^ be the v e c t o r i a l norm induced by ( X l , X 2 , ... X^} and v. F u r t h e r , l e t u : M n -»• be the m a t r i c i a l norm induced by {Xj ,X2 , ... X k } and matrix norm l u b v . Then ( i ) v(AB) < (l u b p A ) y ( B ) ¥ A,B e M n ( i i ) ( l u b p A ) < y(A) "< ( l u b p A ) p ( I n ) ¥ A e M n ( i i i ) I f l u b y E ^ = 1 ¥j = 1,2, ... k then v = l u b p i . e . , l u b p ( A ) = u(A). T r o ^ : . A E v B ) ) - . 1 J K 1 J < (^{AExB))^ + ... ( v ( A E k B ) ) i j . .... (1) Now by d e f i n i t i o n (y ( A ) ) . . = lub,, (E.AE.) = sup v(E.AE.x) 1 3 v 1 1 Xj0 —1—2 xeC 1 1 v(x) . . . . (2) Since E.. i s i d e n t i t y map f o r X^ t h e r e f o r e E.. (x) = x ¥ x e X^ . Also s i n c e p i s the induced v e c t o r i a l norm t h e r e f o r e p^(x) = vfE^x) and so p^(Ax) = v(E^Ax) = v(E^AEjX) . Hence (lub ( A ) ) . . = sup p.(Ax) sup v(E.AE.x) P XJ X 5 f o - i ~ x*0 i - J (3) xeXj Pj (x) xeX^. v(x) -60-Let O^x e X m where 1 4 m <_ k. Therefore from (2) and (3) we have (lub A) i m . ( w ( B ) ) ' m j > v(E.AE mx) v(E mBE.z) ..... (4) v(x) v(z) Now choose z e C n such that E mBE^z ¥ 0 and set x = E^BEj z. Then from (4) ( l u b p A ) . m ( y ( B ) ) m . > v ( E i A E m x ) = v(E.AE*BE j Z) v(z) v(z) = vCEjAEjjjBE^z) .... (5) v ( z ) Now replace A by AE mB i n ( 2 ) . We have from (5) s i n c e z e C n ( l u b p A ) i m ( y ( B ) ) m . > ( y ( A E 1 B ) ) i j . Hence from (1) we have ( y ( A B ) ) i ; j < ( l u b p A ) . m ( y C B ) ) m j + ... ( l u b p A ) i k ( y (B)) k j => y(AB) < ( l u b p A ) y ( B ) ¥ A,B e M n. ( i i ) From (3) and (2) we have C l u b p A ) ± j < ( y ( A ) ) i ; j => ( l u b p A ) 4 ( y ( A ) ) . Now i n (i) l e t . B := I n . Therefore y (A) 4 ( l u b p A ) y ( I n ) . Hence ( l u b p A ) < y(A) < ( l u b p A ) y ( I n ) ¥ A e M n. ( i i i ) Since l u b v E i 0 0 0 0 lub _E2 0 0 0 0 lub,E and we are given that lubyEj = 1 ¥• j = 1 , 2 , ... k t h e r e f o r e y ( I n ) = IyC Hence from ( i i ) we have (l u b p A ) < y(A) < (lub pA) => l u b p ( A ) = y (A) V A E M R. The f o l l o w i n g example shows that the i n e q u a l i t i e s ( i ) and ( i i ) can be s t r i c t - . Example 4: Let C 3 = X x © X 2 where Xi = {(c^-2ct ,0) | a E C }, X 2 = ( ( 0 , 6 , Y ) | 3 , Y E C }. Consider the vec t o r norm v: C 3 ->• R + defined by v ( a i , a 2 , a 3 ) = | a i | + | a 2 | + | a 3 | . ¥ a = ( a 1 , a 2 , a 3 ) E C 3. The v e c t o r i a l norm p induced by {X 1,X 2} and v i s (Theorem 1) p(xO = Xv'CEix) , v ( E 2 x ) ) ~V x E C 3 . Now E r ( x ) = ( a , - 2 a , 0 ) t h e r e f o r e v ^ x ) = 31a x|. S i m i l a r l y v ( E 2 x ) = I2ai+ a2|+ | a 3 | . Thus p(x) = ( 3 | 0 l | , |2o'i + o 2| + I a 3 1) where p: C 3 -*• R 2 • Now si n c e t h i s v e c t o r i a l norm p i s the same as p G of example 3 ther e f o r e the m a t r i c i a l norm l u b p subordinate to p i s given by lub (A) = P m i 1 m 2 1 m 1 2 m 2 2 where m l l m i 2 m2 1 = \alx - 2 a 1 2 I , = 3 max. { I a 1 2 I , | a 1 3 | } t {I 2ai i+a 21 -4ai 2 -2a l22 a 3 i - 2a3 2| } f m 2 2 max. { I 2 a 1 2 + a 2 2 j+| a l32 I > 2ai o + a 1 3 T £ l 2 3 &3 3 I > •62 Now i t i s proved [5,P.109] that lutr (A) = max Z | a. . j i = l 3 where A=(a..) e M3 i s the matrix norm lub,, subordinate t o v. I J 3 v + Thus the m a t r i c i a l norm y:M3 ->• M2 induced by {Xi,X 2} and l u b y i s given by y(A) ( l u b v ( E i A E . ) ) i j = 1 > 2 ¥ A e M 3 . Let us evaluate the matrix y (A) . Now Exx = (01,-20!,0) t h e r e f o r e A E ^ a 3 i a i - 2 a 3 2 a ! a l l 0 l - 2 a 1 2 o i ] and s o E jA E i = a 2 i a i - 2 a 2 2 a i ( a n - 2 a 1 2 , - 2 ( a u - 2 a 1 2 ) ,0) . Thus l u b v E i A E i = 3 | a 1 1 - 2 a i 2 | > mn.. S i m i l a r l y s i n c e E 2AEj = (0 ,2 (ai 1-2ax 2 ) + a 2 1 - 2 a 2 2 , a 3 1 - 2 a 3 2 ) ^ t h e r e f o r e l u b v E 2AE! = | 2ai 1 + a 2 1 - 2 a 2 2 - 4 a i 2 | + | a 3 i - 2 a 3 2 | >=m2 x S i m i l a r l y l u b ^ A E , , = 3 max {2 | a i 2 | , | a x 3 | } >, m i 2 l u b y E 2 A E 2 = >ax{2|2a 1 2+a 2 2|+2|a 3 2|,|2ai 3+a 2 3|+|a 3 3|} ^ m 2 2 . Obviously y (A) > l u b p ( A ) . Remark: From ( i i i ) of Theorem 3 we see that when l u b v E j = l ¥ j = l , 2 , ... k then y= l u b p ,that i s the m a t r i c i a l norm y generated by the matrix norm l u b v subordinate to v e c t o r norm ( P r o p o s i t i o n 3, Chapter 1) i s equal to the m a t r i c i a l norm l u b p generated by the v e c t o r i a l norm p, where p i t s e l f i s generated from v e c t o r norm v (Theorem 1 ) . In f a c t i t i s easy to see tha t the f o l l o w i n g diagram i s commutative lub. y= lub. Example 5: Let C n = X x @ X 2 @ ... @ X^ be an orthogonal direct-sum decomposition of C n and l e t Ex ,E2 , ... E^ be the as s o c i a t e d p r o j e c t i o n s . Further let |J || be the e u c l i d e a n n norm on C n i . e . , | | x l l 2 = z I x • I 2 ¥ x £ C n. Then the i = l 1 . v e c t o r i a l norm on C n induced by {X 1,X 2, ... X^} and the n k eu c l i d e a n norm || | | i s given by (Theorem 1) p:C R + such that p(x) = (||Eix||, .... ||E kx||) ¥ x e C n. Now i t i s proved [5,P.109] that the matrix norm ||| subordinate t o the euclidean norm || || i s given by |||A J|| = /r(A*A) ¥ A e M n where A* i s the a d j o i n t o f matrix A and r(B) i s the s p e c t r a l r a d i u s of matrix B. Since { X l t X 2 , X^ } i s an orthogonal decomposition of C n t h e r e f o r e E* = E. ¥j = 1,2, ... k, and so E|E. = 3 3 3 3 E.E* = E * = E. ¥j = 1,2, ... k. Hence the m a t r i c i a l J 3 3 3 J norm subordinate to the v e c t o r i a l norm p i s (by d e f i n i t i o n ) l u b p A = ( H I E . A E . | | | ) i J . 1 > 2 > m m f k ( / r ( ( E i A E j ) * E i A E j ) ) = ( /r(E|A*EjE iAE ;.) ) - ( /r(E . A * E i A E . j ) t J . 1 > 2 > ^ k ¥ A e M n. -65-B i b l i o g r a p h y 1. Deutsch, E., M a t r i c i a l norms. Numerische Math. 16 (1970) 73-84. 2. Deutsch, E., M a t r i c i a l norms and zeroes of po l y n o m i a l s . Linear algebra and i t s a p p l i c a t i o n s 3 (1970) 483-489. 3. Deutsch, E., On v e c t o r i a l norms and pseudo-norms. Proc. Amer. -Math, Soc. 28 (1971) 18-24. 4. Deutsch, E., On m a t r i c i a l norms subordinate to v e c t o r i a l norms. Math. L e i t s c h r i f t , 122 (1971) 142-150 . 5. Faddev, D.K. and Faddeeva, V.N., Computational methods o f l i n e a r algebra. Freeman § Co. San Francisco, WET. 6. F i e d l e r , M. and Patak, V. Generalized norms of matrices and the l o c a t i o n of spectrum. Czech. Math. J . 112 (87) (1962) 558-571. 7. Gantmacher, -F. *R. AppliG-ation^of-^theory-^o£-.matrices.. Interscience Publ ishers , New York, 1959. 8. Householder , A . S T h e theory of matrices i n numerical a n a l y s i s . B l a i s d e l l , New York, 1964 . 9. J o y a l , L a b e l l e and Rehman., On the l o c a t i o n o f the zeroes of the polynomials. Canad. Math. B u l l . 10 (1967) 53-63. 10. K a u t o r o v i t c h , L.V. The method of successive approximations f o r f u n c t i o n a l equations. Acta Math. 71 (1939) 63-97. 11. Marden, M. Geometry of polynomials. Math. Survey #3 Amer. Math. Soc. 1966. 12. O s t r o w s k i , A.M. On some m a t r i c i a l p r o p e r t i e s of operator matrices and matrices p a r t i t i o n e d i n t o b locks . J . Math. Anal. App. 2 (1961) 161-209. 13. Robert, F. Normes v e c t o r i e l l e s de vecteurs et de m a t r i c e s . Revue Franc, t r a i t m e n t de i n f o r m a t i o n ( C h i f f r e s ) 7 , (1964) 261-299 . . " ^ -66-14. Robert, F. Sur l e normes v e c t o r i e l l e s r e g u l i e r e s s u r un espace v e c t o r i e l de dimension f i n i e . C.R.  Acad. Sc. P a r i s 260 (1965) 5173 - 5176. 15. Simmons, G.F. I n t r o d u c t i o n t o topology and modern a n a l y s i s . McGraw-Hill Book Company, Inc. New York. (1963). 16. Varga, R.S. M a t r i x i t e r a t i v e a n a l y s i s . Prentice Hall. . Englewood C l i f f s (1962). 17. W i l f , H.S. Perron-Fsobenius theory and the zeroes of polynomials. Proc. Amer. Math. Soc. 12 (1961) 247-250. 

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