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Generalized Induced Norms
Article in Czechoslovak Mathematical Journal · August 2004
DOI: 10.1007/s10587-007-0049-5 · Source: arXiv
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arXiv:math/0407357v1 [math.FA] 21 Jul 2004
Generalized Induced Norms∗
S. Hejazian, M. Mirzavaziri and M. S. Moslehian
Abstract
Let k.k be a norm on the algebra Mn of all n × n matrices over C. An interesting
problem in matrix theory is that ”are there two norms k.k1 and k.k2 on Cn such that
kAk = max{kAxk2 : kxk1 = 1} for all A ∈ Mn . We will investigate this problem and
its various aspects and will discuss under which conditions k.k1 = k.k2 .
1
Preliminaries
Throughout the paper Mn denotes the complex algebra of all n × n matrices A = [aij ] with
entries in C together with the usual matrix operations. Denote by {e1 , e2 , · · · en } the standard
basis for Cn , where ei has 1 as its ith entry and 0 elsewhere. We denote by Eij the n × n
matrix with 1 in the (i, j) entry and 0 elsewhere.
For 1 ≤ p ≤ ∞ the norm ℓp on Cn is defined as follows:
ℓp (x) = ℓp (
n
X
xi ei ) =
i=1




(
n
X
i=1
|xi |p )1/p
1≤p<∞


 max{|x |, · · · , |x |}
1
n
p=∞
A norm k.k on Cn is said to be unitarily invariant if kxk = kUxk for all unitaries U and all
x ∈Cn .
∗
2000 Mathematics Subject Classification 15A60 (Primary) 47A30, 46B99 (Secondary).
Keywords and phrases. induced norm, generalized induced norm, algebra norm, the full matrix algebra,
unitarily invariant, generalized induced congruent.
1
By an algebra norm (or a matrix norm) we mean a norm k.k on Mn such that kABk ≤
kAkkBk for all A, B ∈ Mn . An algebra norm k.k on Mn is called unitarily invariant if
kUAV k = kAk for all unitaries U and V and all A ∈ Mn . See [2, Chapter IV] for more
information.
Example 1.1 The norm kAkσ =
n
X
|aij | is an algebra norm, but the norm kAkm =
i,j=1

2
 1 1 


 1 1  2
max{|ai,j | : 1 ≤ i, j ≤ n} is not an algebra norm, since k 
 km > k 
 km .
1 1
1 1
kABk
:
Remark 1.2 It is easy to show that for each norm k.k on Mn , the scaled norm max{ kAkkBk
A, B 6= 0}k.k is an algebra norm; cf. [1, p.114]
Let k.k1 and k.k2 be two norms on Cn . Then for each A : (Cn , k.k1 ) → (Cn , k.k2 ) we can
define kAk = max{kAxk2 : kxk1 = 1}. If k.k1 = k.k2 , then kIk = 1 and there are many
examples of k.k1 and k.k2 such that kIk =
6 1. This shows that given k.k on Mn , we cannot
deduce in general that there is a norm k.k1 on Cn with kAk = max{kAxk1 : kxk1 = 1}. Let
us recall the concept of g-ind norm as follows:
Definition 1.3 Let k.k1 and k.k2 be two norms on Cn . Then the norm k.k1,2 on Mn defined
by kAk1,2 = max{kAxk2 : kxk1 = 1} is called the generalized induced (or g-ind) norm via
k.k1 and k.k2 . If k.k1 = k.k2 , then k.k1,1 is called induced norm.
n
X
n
X
|ai,j | : 1 ≤ j ≤ n}, kAkR = max{ |ai,j | : 1 ≤ i ≤ n} and
j=1
i=1
√
∗
the spectral norm kAkS = max{ λ : λ is an eigenvalue ofA A} are induced by ℓ1 , ℓ∞ and
Example 1.4 kAkC = max{
ℓ2 (or the Eucleadian norm), respectively.
It is known that the algebra norm kAk = max{kAkC , kAkR } is not induced [ ] and it is
not hard to show that it is not g-ind too; cf. [1, Corollary 3.2.6]
We need the following proposition which is a special case of a finite dimensional version
of the Hahn-Banach theorem [5, p. 104]:
2
Proposition 1.5 Let k.k be a norm on Cn and y ∈Cn be a given vector. There exists a
vector y◦ ∈Cn such that y◦∗ y = kyk and for all x ∈Cn , |y◦∗x| ≤ kxk. (Throughout, ∗ denotes
the transpose) [3, Corollary 5.5.15])
In this paper we examine the following nice problems:
(i) Given a norm k.k on Mn is there any class A of Mn such that the restriction of the norm
k.k on A is g-ind?
(ii) When a g-ind norm is unitarily invariant?
(iii) If a given norm k.k is g-ind via k.k1 and k.k2 , then is it possible to find k.k1 and k.k2
explicitly in terms of k.k?
(iv) When two g-ind norms are the same?
(v) Is there any characterization of the g-ind norms which are algebra norms?
2
Main Results
We begin with some observations on generalized induced norms.
Let k.k1,2 be a generalized induced norm on Mn obtained via k.k1 and k.k2 . Then
kEij k1,2 = max{kEij xk2 : kxk1 = 1} = max{kxj ei k2 : k(x1 , · · · , xj , · · · , xn )k1 = 1} =
αj kei k2 , where αj = max{|xj | : k(x1 , · · · , xj , · · · , xn )k1 = 1}. In general, for x ∈Cn and
1 ≤ j ≤ n, if Cx,j ∈ Mn is defined by the operator Cx,j (y) = yj x then kCx,j k1,2 = αj kxk2 .
Also if for x ∈Cn we define Cx ∈ Mn by Cx =
where α = max{|
Pn
Pn
j=1 Cx,j , then clearly kCx k1,2 = αkxk2 ,
j=1 yj | : k(y1 , · · · , yj , · · · , yn )k1 = 1}.
Now we give a partial solution to Problem (i) and useful direction toward solving Problem
(iii):
Proposition 2.1 Let k.k be an algebra norm on Mn . Then k.k is a g-ind norm on {A ∈
Mn : kAk = kA−1 k = 1}.
Proof. Put kxk1 = max{kCAx k : kAk = 1}, λ−1 = max{|
λkCx k.
3
n
X
i=1
xi | : kxk1 = 1} and kxk2 =
Then we have kCy k1,2 = max{kCy xk2 : kxk1 = 1} = max{|
kyk2λ−1 = kCy k.
n
X
i=1
xi |kyk2 : kxk1 = 1} =
It follows that for each y ∈Cn there is some x ∈Cn such that kCy xk2 = kCy kkxk1 =
kCy k max{kCDx k : kDk = 1}.
Now let A be invertible and kA−1 k = kAk = 1 and z = A−1 Cy x. Then λ−1 kBzk2 =
λ−1 kBA−1 Cy xk2 = λ−1 kDxk2 = kCDx k ≤ kC1y k kCy xk2 = kC1y k kAzk2 .
Now choose y so that kCy k = 1. Then kCBz k ≤ kCAz k for all B ∈ Mn . This implies that
kCAz k is an upper bound for the set {kCBz k : kBk = 1} and indeed kCAz k = max{kCBz k :
kBk = 1} = kzk1 . It follows that kAk = 1 = kCA( kzkz ) k = max{kCAu k : kuk1 = 1} =
1
max{kAuk2 : kuk1 = 1} = kAk1,2 .2
Let us now answer Question (ii).
Proposition 2.2 An induced norm k.k1,2 is unitarily invariant if and only if so are k.k1
and k.k2 .
Proof. Let U, V be unital operators and A be an arbitrary operator on Cn .
Suppose that k.k1 and k.k2 are unitarily invariant. Then
kUAV xk2
kAV xk2
kAyk2
kAyk2
= max
= max −1
= max
= kA1,2 .
x6=0
x6=0
y6=0 kV
y6=0 kyk1
kxk1
kxk1
xk1
kUAV k1,2 = max
Conversely, if k.k1,2 is unitarily invariant, then kUxk1 = max{kAUxk2 : kAk1,2 ≤ 1} =
max{kBxk2 : kU −1 Bk1,2 ≤ 1} = max{kBxk2 : kBk1,2 ≤ 1} = kxk1 and kUxk2 = α1 kCU x k =
1
kUCx k = α1 kUCx k = α1 kCx k = kxk2 .2
α
Modifying the proof of Theorem 5.6.18 of [3], we obtain a similar useful result for g-ind
norms:
Theorem 2.3 Let k.k1 , k.k2 , k.k3 and k.k4 be four given norms on Cn and
Ri,j = max{
kxki
: x 6= 0}, 1 ≤ i, j ≤ 4.
kxkj
Then
max{
kAk1,2
: A 6= 0} = R2,4 R3,1
kAk3,4
4
kAk1,1
2,2
: A 6= 0} = max{ kAk
: A 6= 0} = R1,2 R2,1 .
In particual, max{ kAk
kAk1,1
2,2
2 kAxk4 kxk3
2
= kAxk
.
.
. Hence kAk1,2 ≤
Proof. Let A be a matrix and x 6= 0. Then kAxk
kxk1
kAxk4 kxk3 kxk1
kAk1,2
R2,4 kAk3,4 R3,1 . Thus kAk
≤ R2,4 R3,1 .
3,4
There are vectors y, z in Cn such that kyk2 = kzk2 = 1, kyk2 = R2,4 kyk4 and kzk3 =
R3,1 kzk1 . By Proposition 1.15, there exists a vector z◦ ∈ Cn such that |z◦∗ x| ≤ kxk3 and
z◦∗ z = kzk3 .
∗
kyk2
◦ zk2
2 kzk3
◦ zk2
Put A◦ = yz◦. Then kAkzk
= kyz
= kykkzk
= kyk2R3,1 . Hence kA◦ k1,2 ≥ kyk
R3,1 kyk4 =
kzk1
1
1
4
∗
∗
◦ xk4
4 |z◦ x|
◦ xk4
R2,4 .R3,1 kyk4. On the other hand, kAkxk
= kyzkxk
= kykkxk
≤ kyk4. Thus kA◦ k3,4 ≤ kyk4.
3
3
3
R3,1 kyk4
◦ k1,2
Hence kA
≥ R2,4 kyk
= R2,4 R3,1 .2
kA◦ k3,4
4
Corollary 2.4 (i) k.k1,2 ≤ k.k3,2 if and only if k.k1 ≥ k.k3 ,
(ii) k.k1,2 ≤ k.k1,4 if and only if k.k2 ≤ k.k4 .
1,2
: A 6= 0} = R2,2 R3,1 ≤ 1 and this if and
Proof. (i) k.k1,2 ≤ k.k3,2 if and only if max{ kAk
kAk3,2
only if R3,1 ≤ 1 or equivalently k.k3 ≤ k.k1 . The proof of (ii) is similar.2
The following corollary completely answers to Question (iv):
Corollary 2.5 k.k1,2 = k.k3,4 if and only if there exists γ > 0 such that k.k1 = γk.k3 and
k.k2 = γk.k4 .
kAk1,2
3,4
: A 6= 0} = 1 = max{ kAk
: A 6=
Proof. If kAk1,2 = kAk3,4 , then R4,2 R1,3 = max{ kAk
kAk1,2
3,4
kxk2
kxk1
1
0} = R2,4 R3,1 . Hence max{ kxk
: x 6= 0} = R2,4 = R13,1 = min{ kxk
: x 6= 0} ≤ max{ kxk
:
kxk3
4
3
kxk2
x 6= 0} = R1,3 = R14,2 = min{ kxk
: x 6= 0}. Thus there exists a number γ such that
4
kxk1
kxk2
= γ = kxk
.2
kxk4
3
Remark 2.6 It is known that each induced norm k.k is minimal in the sense that for any
matrix norm k.k, the inequality k.k ≤ k.k1,1 implies that k.k = k.k1,1 . But this is not true
for g-ind norms in general. For instance, put k.kα = ℓ∞ (.), k.kβ = 2ℓ2 (.) and k.kγ = ℓ2 (.).
Then k.kγ,β ≤ k.kα,β but k.kγ,β 6= k.kα,β .
5
The following theorem is one of our main theorems and provide a complete solution for
Problem (v):
Theorem 2.7 Let k.k1 and k.k2 be two norms on Cn . Then k.k1,2 is an algebra norm on
Mn if and only if k.k1 ≤ k.k2 .
Proof. For each A and B in Mn we have
kABxk2 ≤ kAk1,2 kBxk1 ≤ kAk1,2 kBxk2 ≤ kAk1,2 kBk1,2 kxk1 .
Hence kABk1,2 ≤ kAk1,2 kBk1,2 .
Conversely, let k.k1,2 be an algebra norm. Then for each A, B ∈ Mn we have kABk2 ≤
kAk1,2 kBk1,2 kxk1 . Let B be an arbitrary member of Mn . For Bx 6= 0, take M to be the
1
linear span of {Bx} and define f : (M, k.k1 ) →C by f (cBx) = ckBxk
. By the Hahn-Banach
kBxk2
Theorem, there is an F : (Cn , k.k1 ) →C with F |M = f and kF k = kf k = max{|f (cBx)| :
1
1
: |c|kBxk1 = 1} = kBxk
. Define A : (Cn , k.k1 ) → (Cn , k.k2) by
kcBxk1 = 1} = max{ |c|kBxk
kBxk2
2
Ay = F (y)Bx. Then kAk1,2 = max{kAyk2 : kyk1 = 1} = max{|F (y)|kBxk2 : kyk1 = 1} =
kBxk1
1, and kABxk2 = |F (Bx)|kBxk2 = |f (Bx)|kBxk2 = kBxk
kBxk2 = kBxk1 . Thus for all B,
2
kBxk1 = kABxk2 ≤ kAk1,2 kBk1,2 kxk1 = kBk1,2 kxk1 ,
or
kBxk1 ≤ kBk1,2 kxk1 .
1
Now take N to be the linear span of {x} and define g : (N, k.k1 ) →C by g(cx) = ckxk
. By
kxk2
the Hahn-Banach Theorem, there is a G : (Cn , k.k1 ) →C with G|N = g and kGk = kgk =
1
1
max{|g(cx)| : kcxk1 } = max{ |c|kxk
: |c|kxk1 = 1} = kxk
. Define B : (Cn , k.k1) → (Cn , k.k2 )
kxk2
2
by By = G(y)x. Then kBk1,2 = max{kByk2 : kyk1 = 1} = max{|G(y)|kxk2 : kyk1 = 1} =
kxk2
kxk1
kxk2 kGk = 1, and kBxk1 = |G(x)|kxk1 = |g(x)|kxk1 = kxk
kxk1 = kxk21 .
2
So
Thus k.k1 ≤ k.k2 .2
kxk21
= kBxk1 ≤ kBk1,2 kxk1 = kxk1 .
kxk2
6
Proposition 2.8 Suppose that k.k1,2 is a g-ind norm and λ > 0. Then the scaled norm
λk.k1,2 is a g-ind algebra norm if and only if λ ≥ R1,2 .
Proof. Evidently, λk.k1,2 = k.kk.k1 ,λk.k2 . If k.k3,4 = λk.k1,2 = k.kk.k1 ,λk.k2 then Corollary
2.5 implies that there exists α > 0 such that k.k3 = αk.k1 and k.k4 = αλk.k2 . Now Theorem 2.7 follows that λk.k1,2 = k.k3,4 is an algebra norm if and only if αk.k1 ≤ αλk.k2 or
equivalently R1,2 ≤ λ.2
Proposition 2.9 Let k.k1 and k.k2 be two norms on Cn and 0 6= α, β ∈C. Define k.kα and
k.kβ on Cn by kxkα = kαxk1 and kxkβ = kβxk2 , respectively. Then k.kα and k.kβ are two
norms on Cn and k.kα,β = | αβ |k.k1,2.
Proof. We have kAkα,β = max{kAxkβ : kxkα = 1} = max{kβAxk2 : kαxk1 = 1} =
| αβ | max{kAyk2 : kyk1 = 1} = | αβ |kAk1,2 .2
The preceding proposition leads us ti give the following definition:
Definition 2.10 Let (k.k1 , k.k2) and (k.k3 , k.k4 ) be two pairs of norms on Cn . We say
that (k.k1 , k.k2 ) is generalized induced congruent (gi-congeruent) to (k.k3 , k.k4) and we write
(k.k1 , k.k2 ) ≡gi (k.k3 , k.k4 ) if k.k1,2 = γk.k3,4 for some 0 < γ ∈R.
Clearly ≡gi is an equivalence relation. We denote by [(k.k1 , k.k2 )]gi the equivalence class
of (k.k1 , k.k2 ). Proposition 2.9 shows that for each 0 < α, β ∈R we have (αk.k1 , βk.k2) ≡gi
(k.k1 , k.k2 ). Indeed, we have the following result:
Theorem 2.11 For each pair (k.k1 , k.k2 ) of norms on Cn we have [(k.k1 , k.k2)]gi = {(αk.k1, βk.k2 ) :
0 < α, β ∈R}.
We can extend the above method to find some other norms on Mn which are not necessarily gi-congruent to a given pair (k.k1 , k.k2 ):
Proposition 2.12 Let (k.k1 , k.k2 ) be a pair of norms on Cn and K, L ∈Mn be two invertible
matrices. Define kkK and kkL and Cn by kxkK = kKxk1 and kxkL = kLxk2 . Then kkK and
kkL are norms on Cn and kAkK,L = kLAK −1 k1,2 .
7
Proof. Clear and see also Lemma 3.1 of [4].2
Remark 2.13 Note that the case K = αI and L = βI gives Proposition 2.9.
References
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[2] Bhatia, Rajendra. Matrix analysis. Graduate Texts in Mathematics, 169. SpringerVerlag, New York, 1997.
[3] Horn, Roger A.; Johnson, Charles R. Matrix analysis. Cambridge University Press,
Cambridge, 1985.
[4] Li, Chi-Kwong; Tsing, Nam-Kiu; Zhang, Fuzhen. Norm hull of vectors and matrices.
Linear Algebra Appl. 257 (1997), 1-27.
[5] Rudin, Walter. Real and Complex Analysis, McGraw-Hill, Singapore, 1987.
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