A complete unitary similarity invariant for unicellular matrices

0.15 MB·English

by  Douglas Farenick

A Complete Unitary Similarity Invariant for Unicellular
Matrices
Douglas Farenick
1
1
Department of Mathematics and Statistics, University of Regina, Regina, Saskatchewan
0
2 S4S 0A2, Canada

n
a Tatiana G. Gerasimova
J
The Faculty of Mechanics and Mathematics, Kiev National Taras Shevchenko University,
5
2 Volodymyrska St, 64, Kiev-33, 01033, Ukraine

] Nadya Shvai
T
R The Faculty of Mechanics and Mathematics, Kiev National Taras Shevchenko University,
. Volodymyrska St, 64, Kiev-33, 01033, Ukraine
h
t
a
m
[

2
Abstract
v
3
4 We present necessary and sufficient conditions for an n n complex ma-
×
0 trix B to be unitarily similar to a fixed unicellular (i.e., indecomposable by
0
similarity) n n complex matrix A.
.
1 ×
1 Keywords: unitary similarity problem, unicellular matrix, Toeplitz matrix,
0
Volterra operator
1
: 2010 MSC: 15A21, 15A60, 47A05, 47A65
v
i
X
r
1. Introduction
a
A fundamental problem in matrix analysis is the unitary similarity prob-
lem [2, 9]: Under what necessary and sufficient conditions are two n n
×
complex matrices unitarily similar? A classical and purely algebraic solution
to this problem due to Specht [7, 10]: two n n complex matrices A and B
×
are unitarily similar if and only if
Traceω(A,A∗) = Traceω(B,B∗), (1)
for every word ω in two noncommuting variables x and y.
Preprint submitted to Linear Algebra and its Applications January 26, 2011
In many applications, the data one has about a particular matrix are not
based on the trace of the matrix, but rather on some other analytical infor-
mation: thespectrum or pseudospectrum, thenumerical rangeor polynomial
numerical hull, the singular values, a unitarily invariant norm, and so forth.
Our concern in the present paper is with a solution to the unitary similarity
problem that is based on a particular choice of unitarily invariant norm.
Let M be the space of all n n complex matrices; we denote the unitary
n
×
groupbyU . TwomatricesA,B M areunitarilysimilar, whichweexpress
n n
∈
by A B, if there is a U U such that B = U∗AU. The norm under study
n
∼ ∈
is defined by
A = spr(A∗A), (2)
k k
p
where sprX is the spectral radius of X M . The norm (2) has the property
n
∈
that U∗AU = A , for all A M and U U , and it coincides with the
n n
k k k k ∈ ∈
largest singular value of A. Moreover, if A M is considered as a linear
n
∈
transformation on the complex inner product space Cn with respect to the
standard inner product ξ,η = η∗ξ, for ξ,η Cn, then
h i ∈
A = max Aξ,η .
k k hξ,ξi=hη,ηi=1 |h i|
Let C[t] denote the ring of polynomials with complex coefficients. If
A B, then necessarily f(A) = f(B) for all f C[t]. Conversely, if
∼ k k k k ∈
A,B M are such that f(A) = f(B) for all f C[t], then A and B
n
∈ k k k k ∈
yield to the same matrix analysis:
(i) A and B have the same spectrum;
(ii)A zI andB zI havethesameconditionnumbers, forallnonspectral
− −
z in the complex plane;
(iii)AandB have thesamepolynomial numerical hulls and, inparticular,
the same numerical range;
(iv) A and B have the same spectral set;
(v) A and B have the same pseudospectrum.
Our first objective is to determine cases in which the condition f(A) =
k k
f(B) for all f C[t] is also sufficient for A B. In general it will not be
k k ∈ ∼
so, for if one takes any two nonzero projections (selfadjoint idempotents) P
and Q, then one has f(P) = f(Q) for all f C[t], independent of the
k k k k ∈
ranks of P and Q. Therefore, for questions concerning unitary similarity, the
hypothesis f(A) = f(B) for all f C[t] is relevant only for the analysis
k k k k ∈
of nonnormal matrices.
2
Definition 1.1. A matrix A M is said to be unicellular if A is not similar
n
∈
to a matrix B M of the form B = G H, for some square matrices G
n
∈ ⊕
and H of strictly smaller size than B.
Our use of the term unicellular matrix is motivated by the concept of
unicellular operator or transformation in operator theory. If A M is
n
∈
a unicellular matrix, then A is unicellular in the sense of [5, 9], [6, 2.5]
§ §
as a linear transformation on Cn. Unicellular matrices are also said to be
indecomposable by similarity.
In this paper we present two main results. The first, Theorem 2.1, states
that the unitary similarity class of any upper triangular unicellular Toeplitz
matrix R is determined by the values of f(R) for various f C[t]. If one
k k ∈
drops the requirement that R be Toeplitz, yet remain upper triangular and
unicellular, thenthevaluesof f(R) , forf C[t], areinsufficient toidentify
k k ∈
R up to unitary similarity (Proposition 3.1). But with our second main
result, Theorem3.2,weaugmentthecriterionslightlytoobtainnecessaryand
sufficient conditions that classify unicellular matrices up to unitary similarity
(see, also, Proposition 5.1).
2. Upper Triangular Toeplitz Matrices
Definition 2.1. A matrix R M is an upper triangular Toeplitz matrix if
n
∈
z z a z
0 1 2 n−1
 0 z z ·.·.·. ... 
0 1
R =  0 0 ... ... z  , (3)
 2 
 .. .. .. 
 . . . z 
 1 
 0 ... ... 0 z 
0
 
for some z ,z ,...,z C.
0 1 n−1
∈
The set of all upper triangular Toeplitz matrices R M is denoted by
n
∈
UpperToepl .
n
The main theorem of this section is:
Theorem 2.1. Let R M be an upper triangular Toeplitz matrix (3) with
n
∈
z = 0. If A M is any matrix for which f(A) = f(R) , for all
1 n
6 ∈ k k k k
f C[t], then A R.
∈ ∼
3
Before moving to the proof of Theorem 2.1, let us consider one of its
consequences, namely Corollary 2.2 below, which is of interest in linear-
algebraic analysis. For any A M , the unital algebra AlgA generated by
n
∈
A is
AlgA = f(A) : f C[t] .
{ ∈ }
In particular, UpperToepl = AlgS, where
n
0 1 0 0
 0 0 1 ·.·.·. ... 
S =  0 0 ... ... 0  .
 
 .. .. .. 
 . . . 1 
 
 0 ... ... 0 0 
 
More generally, if R UpperToepl is of the form (3) and satisfies z = 0,
∈ n 1 6
then the range of R z I is clearly (n 1)-dimensional and so the kernel
0
− −
of R z I is 1-dimensional. Thus, there is an invertible X M for which
0 n
− ∈
S = X(R z I)X−1, the Jordan canonical form of R z I. Hence, the
0 0
− −
abelian algebras UpperToepl = AlgS and AlgR are isomorphic. Because
n
AlgR is a subalgebra of UpperToepl , they can be isomorphic only if they
n
are equal. Thus, if R UpperToepl satisfies z = 0, then R is called a
∈ n 1 6
generator of UpperToepl . (Consideration of the Jordan form shows that
n
this necessary condition on z is also sufficient for R UpperToepl to be a
1 ∈ n
generator of UpperToepl , but we do not require this fact.)
n
Corollary 2.2. If ̺ : UpperToepl M is a homomorphism such that
n → n
̺(X) = X , for every X UpperToepl , then there is a U U such
k k k k ∈ n ∈ n
that ̺ is given by ̺(X) = U∗XU.
Proof. Choose R UpperToepl of the form (3) with z = 0 and let A =
∈ n 1 6
̺(R). Thus, f(A) = ̺(f(R)), for all f C[t]. By hypothesis, f(A) =
∈ k k
̺(f(R)) = f(R) , for all f C[t]; therefore, Theorem 2.1 asserts that
k k k k ∈
A = U∗RU forsomeU U . BecauseRgeneratesUpperToepl , weconclude
∈ n n
that ̺(X) = U∗XU, for every X UpperToepl .
∈ n
We move now to the proof of Theorem 2.1.
4
2.1. Lemmas
Lemma 2.3. If
0 1 1 1
 0 1 ·.·.·. ... 
Q =  ... ... ...  , (4)
 
 ... 1 
 
 0 
 
∞
then ( 1)k+1Qk = S.
−
X
k=1
∞ ∞
Proof. Clearly Q = Sk. Thus, I +Q = Sj = (I S)−1, whence
−
Xk=1 Xj=0
∞
I = (I S)(I +Q). That is, S = I (I +Q)−1 = ( 1)k+1Qk.
− − −
X
k=1
Lemma 2.4. Let Q M be given by (4). If
n
∈
0 1 a a
13 1n
 0 1 ·.·.·. ... 
A =  ... ... a 
 n−2,n 
 ... 1 
 
 0 
 
∞
has the property that ( 1)k+1Ak 1, then A = Q.
(cid:13) (cid:13)
− ≤
(cid:13)X (cid:13)
(cid:13)k=1 (cid:13)
(cid:13) (cid:13)
Proof. We proceed by(cid:13)induction on n(cid:13). The base case is n = 3. In this case,
∞ 0 1 a13 1
−
( 1)k+1Ak =  0 0 1  .
−
Xk=1 0 0 0
 
The first row of the matrix above has Euclidean length at most 1, since A has
norm at most 1. Thus, a = 1, implying that A = Q. This row condition
13
extends unchanged to the induction step.
5
Assume now the statement holds in n-dimensional space and consider A,
Q, and S as acting on Cn+1. Let A˜, Q˜, and S˜ denote the versions of A, Q,
and S that act on Cn, and let e ,...,e denote the canonical orthonormal
1 n
basis vectors in Cn. Hence, as a partitioned matrix, A has the form
 A˜ η 
A = ,
 
 0 0 0 
 ··· 
where
n−1
η = e + a e = [a , ,a ]T Cn.
n i,n+1 i 1,n+1 n−1,n+1
··· ∈
Xi=1
Because
∞ ∞
1 ( 1)k+1Ak ( 1)k+1A˜k , (5)
(cid:13) (cid:13) (cid:13) (cid:13)
≥ (cid:13)X − (cid:13) ≥ (cid:13)X − (cid:13)
(cid:13)k=1 (cid:13) (cid:13)k=1 (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
the induction hypoth(cid:13)esis yields A˜ = (cid:13)Q˜. H(cid:13)ence, using Lem(cid:13)ma 2.3, we obtain
∞
 
∞ S˜ ( 1)k+1A˜k−1η
( 1)k+1Ak = − .
X −  Xk=1 
k=1  
 
 0 0 0 
 ··· 
That is,
0 1 0 0
 ... 0 1 ·.·.·. ... ∗... 
∞ ( 1)k+1Ak =  ... ... 0 ...  . (6)
Xk=1 −  ... ... 1 
 ∗ 
 0 0 1 
 ··· ··· 
 0 0 0 
 ··· ··· ··· 
Similar to the case n = 3, we have from (5) that
∞
1 ( 1)k+1Ak (7)
(cid:13) (cid:13)
≥ (cid:13)X − (cid:13)
(cid:13)k=1 (cid:13)
(cid:13) (cid:13)
(cid:13) (cid:13)
6
for the matrix (6). But (7) holds for the matrix (6) only if the i-th entry in
the final column of the matrix (6) is 0, for 1 i n 1. Therefore, using
≤ ≤ −
A˜ = Q˜, we have (I˜ S˜)η = λe for some complex number λ. Hence,
n
−
η = λ(I˜ S˜)−1e = λ(I˜+S˜+S˜2+ +S˜n−2)e = λ(e +e + +e ).
n n n n−1 1
− ··· ···
But on the other hand,
n−1
η = e + a e ,
n i,n+1 i
Xi=1
which implies that λ = 1 and a = 1 for all 1 i n 1. Therefore,
i,n
≤ ≤ −
A = Q.
2.2. Proof of Theorem 2.1
Assume first that the matrix R in (3) has z = 0 and z = 1 for 1 j
0 j
≤ ≤
(n 1); that is, assume that R = Q, where Q has the form (4). Thus, the
−
hypothesis is that A M satisfies f(A) = f(Q) , for all f C[t].
n
∈ k k k k ∈
By the Spectral Radius Formula,
0 = sprQ = lim Qk 1/k = lim Ak 1/k = sprA,
k→∞k k k→∞k k
which implies that A is nilpotent. Without loss of generality, A may be as-
sumed to be in upper triangular form. Furthermore, using a diagonal unitary
similarity transformation, the entries a may assumed to be nonnegative,
i,i+1
for 1 i n 1. Indeed, since 1 = Qn−1 = An−1 = a a a ,
12 23 n−1,n
≤ ≤ − k k k k | ··· |
each a is nonzero; thus, we may assume that a > 0 for all i.
i,i+1 i,i+1
The numerical range, or field of values, W(X) of any X M is given
n
∈
analytically by
W(X) = z C : αz +β αX +β1 .
{ ∈ | | ≤ k k}
α,\β∈C
Hence, W(A) = W(Q). Let (X) = 1(X+X∗), foranyX M ,andobserve
ℜ 2 ∈ n
that 1 + (Q) = 1 ξ ξ, where ξ = n e Cn and ξ ξ denotes the
2 ℜ 2 ⊗ i=1 i ∈ ⊗
outer product ξξ∗ M of ξ (a columnPvector) with its conjugate transpose
n
∈
ξ∗. Thus, for every unit vector γ Cn, the real part of Qγ,γ satisfies the
∈ h i
inequality
1
( Qγ,γ ) .
ℜ h i ≥ − 2
7
BecauseAandQhavethesamenumericalrange, (A)hasthesameproperty
ℜ
above. Now, if P is the projection of Cn onto Span e ,e , for each 1
i i i+1
{ } ≤
i n 1, then P AP as a linear transformation on the range of P is given
i i i
≤ −
by
0 a
i,i+1 .
(cid:20) 0 0 (cid:21)
Therefore, the numerical range of P AP is a disc of radius 1a centered
i i 2 i,i+1
at the origin. Because W(P AP ) W(A) z C (z) 1/2 , we
i i
⊆ ⊂ { ∈ |ℜ ≥ − }
conclude that each a 1. However, under these conditions the equation
i,i+1
≤
1 = An−1 = a a a holds only if a = 1 for all 1 i n 1.
12 23 n−1,n i,i+1
k k ··· ≤ ≤ −
Hence, A has the structure given in the hypothesis of Lemma 2.4. Moreover,
by Lemma 2.3,
∞ ∞
1 = S = ( 1)k+1Qk = ( 1)k+1Ak .
(cid:13) (cid:13) (cid:13) (cid:13)
k k − −
(cid:13)X (cid:13) (cid:13)X (cid:13)
(cid:13)k=1 (cid:13) (cid:13)k=1 (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
Thus, A satisfies all of the(cid:13)hypotheses of(cid:13)Lemm(cid:13)a 2.4, yielding(cid:13)Q = A.
For the general case, we now suppose that R UpperToepl satisfies
∈ n
z = 0 and A M is such that f(A) = f(R) for every f C[t].
1 n
6 ∈ k k k k ∈
Therefore, the ideals J and J coincide, where for a given X M
A R n
∈
J = p C[t] : p(X) = 0 .
X
{ ∈ }
Because R is a generator of UpperToepl , there is a g C[t] such that
n ∈
Q = g(R). Let B = g(A). Thus, h(B) = h(Q) , for every h C[t].
k k k k ∈
By what we proved above, this yields B = U∗QU for some U U . As Q
n
∈
generates UpperToepl , there is an q C[t] such that R = q(Q). Hence,
n ∈
p(t) = t q(g(t)) J = J .
R A
− ∈
This implies that
0 = p(A) = A q(g(A)) = A q(B) = A U∗q(Q)U = A U∗RU ,
− − − −
which completes the proof.
3. Necessary and Sufficient Conditions for Unitary Similarity
If A M is unicellular — say with spectrum λ — and if B M is
n n
∈ { } ∈
any matrix for which f(A) = f(B) for all f C[t], then A and B are
k k k k ∈
8
similar, as the condition implies that σ(B) = σ(A) and that (B λI)n−1 = 0.
− 6
But, unlike the case for generators of the upper triangular Toeplitz matrices,
Aand B need not be unitarily equivalent (Proposition3.1 below). Therefore,
one can have an invertible matrix Z M with
n
∈
f(A) = Zf(A)Z−1 , for all f C[t],
k k k k ∈
and yet Z can fail to be unitary.
Proposition 3.1. If 0 < α < β, then the unicellular matrices
0 α 0 0 β 0
A =  0 0 β  and A′ =  0 0 α  (8)
0 0 0 0 0 0
   
satisfy f(A′) = f(A) for all f C[t], but A′ A.
k k k k ∈ 6∼
Proof. Note that A′ = W∗ATW, where X XT denotes the transpose map
7→
and
0 0 1
W =  0 1 0  .
1 0 0
 
Because the norm is transpose invariant, f(A′) = f(AT) = f(A)T =
k k k k k k
f(A) , for all f C[t]. On the other hand, A A′ by Littlewood’s
k k ∈ 6∼
algorithm [8] because 0 < α < β. (One also can verify directly that the
equation UA′ = AU is impossible to satisfy with U U . Alternatively, the
3
∈
referee observed that the matrices A and A′ fail to satisfy Specht’s tracial
condition with the word ω(x,y) = xy2x2y; hence, A A′.)
6∼
Notation 3.1. If 1 k n and X = [x ]n M , then X = [x ]k
≤ ≤ ij i,j=1 ∈ n k ij i,j=1 ∈
M . That is, X is the leading k k principal submatrix of X.
k k
×
The failure of A and A′ in (8) to be unitarily similar is explained by the
fact that the norms of f(A ) and f(A′) do not always coincide, even though
2 2
f(A) = f(A′) for all f C[t]. This observation motivates our second
k k k k ∈
main result of the present paper.
Theorem 3.2. Assume that A M is an upper triangular matrix such that
n
∈
(a) a = a for all 1 i,k n, and
ii kk
≤ ≤
9
(b) a = 0, for all 1 i (n 1) (that is, the first superdiagonal of A
i,i+1
6 ≤ ≤ −
has only nonzero entries).
Then the following statements are equivalent for an upper triangular matrix
A′ M :
n
∈
1. f(A ) = f(A′) , for all f C[t] and 1 i n;
k i k k i k ∈ ≤ ≤
2. A′ = W∗AW for some diagonal unitary matrix W U .
n
∈
Proof. We need only prove that first statement implies the second.
There is a diagonal unitary W U such that the entries in the first
n
∈
superdiagonal of the upper triangular matrix W∗AW are positive; therefore,
without loss of generality we assume that a > 0 for 1 i (n 1). As
i,i+1
≤ ≤ −
we argued in the proof of Theorem 2.1, the condition f(A) = f(A′) , for
k k k k
all f C[t], implies that A′ has one point of spectrum, in this case λ = a .
11
∈
Therefore, by scalar translation X X λI we may assume without loss
7→ −
of generality that λ = 0. That is,
0 a a ... a
12 13 1n
 0 0 a ... a 
23 2n
A =  ... ... ...  , (9)
 
 0 an−1n 
 
 0 0 
 
where a > 0, for 1 ℓ n 1.
ℓ,ℓ+1
≤ ≤ −
To complete the proof of theorem, it is sufficient to prove that the entries
ofAin(9)arecompletelydeterminedfromthevaluesof f(A ) for1 i n
i
k k ≤ ≤
and all f C[t].
∈
We shall proceed by induction on n 3.
≥
Let n = 3. Thus,
0 a a
12 13
A =  0 0 a  . (10)
23
0 0 0
 
The value of a is determined via the fact that A = a , and so the value
12 2 12
k k
of a is determined from the equation a a = A2 . Using f(t) = t, we
23 12 23
k k
have
1
A 2 = a2 +a2 + a 2 + (a2 +a2 + a 2)2 4a2 a2 ,
k k 2 (cid:18) 12 23 | 13| q 12 23 | 13| − 12 23(cid:19)
10