Table Of Contentk-REFLEXIVITY DEFECT OF THE IMAGE OF A GENERALIZED
DERIVATION
TINARUDOLF
Abstract. LetX beafinite-dimensionalcomplexvectorspaceandletk beapositiveinteger.
Anexplicitformulaforthek-reflexivitydefectoftheimageofageneralizedderivationonL(X),
thespaceofalllineartransformationsonX,isgiven. Usinglatter,wealsostudythek-reflexivity
defect of the image of an elementary operator of theform ∆(T)=ATB−T (T ∈L(X)).
4
1
0
2 1. Introduction
n
a Let X be a finite-dimensional complex vector space and let L(X) be the space of all linear
J
transformations on X. Let k be a positive integer and denote by F the set of all elements in
8 k
2 M of rank k or less. The k-reflexive cover of a non-empty subset S ⊆ L(X) is defined by
n
]
A
RefkS = {T ∈ L(X) : ∀ε> 0, ∀x1,...,xk ∈ X, ∃S ∈ S : kTxi−Sxik < ε, i= 1,...,k}.
F
.
h It is easy to see that RefkS is a linear subspace of L(X). A linear subspace S is said to
t
a be k-reflexive if RefkS = S. The k-reflexivity defect of a non-empty subset S is defined by
m
[ rdk(S) = dim(RefkS/S). Since X is finite dimensional rdk(S) = dim(RefkS)−dim(S) holds.
1 The annihilator of a non-empty subset S ⊆ Mn is defined by S⊥ = {C ∈ Mn : tr(CS) =
v
0 for all S ∈ S}, where tr(·) denotes the trace functional. It was shown in [3, 4] that
4
4
2 (1) RefkS = (S⊥∩Fk)⊥
7
.
1
holds. The latter obviously implies that a k-reflexive space is also j-reflexive for all j ≥ k.
0
4 Let A, B ∈ L(X) be invertible linear transformations and let S be a linear subspace of L(X).
1
: Let us denote ASB = {ASB : S ∈ S} and S⊺ = {S⊺ : S ∈ S}. It is well known that
v
i transformations of the type
X
r
a (2) S 7→ ASB = {ASB : S ∈ S} and S 7→ S⊺ = {S⊺ : S ∈S}
preserve the k-reflexivity defect. Hence, since X is a finite-dimensional complex vector space,
one can assume that X = Cn for some n ∈ N and L(X) may be identified with M , the algebra
n
2010 Mathematics Subject Classification. Primary 47L05; Secondary 15A21, 15A22.
Key words and phrases. k-reflexivity;k-reflexivitydefect; generalized derivations; elementary operator.
1
2 T.RUDOLF
of all n-by-n complex matrices. Throughout this paper we will be dealing with subspaces of M
n
which have the decomposition of the form
S11 ... S1N
. .
S = .. .. ,
SM1 ... SMN
where, for each pair of indices (i, j), S is a subspace of M , the space of all m -by-n
ij mi,nj i j
complex matrices, and M m = N n = n. It is not hard to see that for spaces of this type
i=1 i j=1 j
P P
one has
Refk(S11) ... Refk(S1N) M N
. .
(3) Refk(S) = .. .. and rdk(S)= XXrdk(Sij).
i=1j=1
Refk(SM1) ... Refk(SMN)
In particular, S is k-reflexive if and only if S is k-reflexive for every pair of indices i ∈
ij
{1,...,M}, j ∈ {1,...,N}.
2. Elementary operators
Let (A1,...,Ak) and (B1,...,Bk) be arbitrary pairs of n-by-n complex matrices. The ele-
mentary operator on Mn with coefficients (A1,...,Ak) and (B1,...,Bk) is defined by
∆(T) = A1TB1+...+AkTBk, T ∈ Mn.
If all A are pairwise linearly independent and if the same holds for all B (1 ≤ i ≤ k), then ∆
i i
is called elementary operator of length k. The simplest example of such operator is of course
two-sided multiplication. Namely, let A, B ∈ M and let ∆ be an elementary operator defined
n
by ∆(T)= ATB for T ∈ M . It is easy to see that the kernel and the image of ∆ are reflexive
n
spaces. In fact, if ∆ is an elementary operator of length k on M , then by [1, Proposition 1.1]
n
the space ker∆ is j-reflexive for every j ≥ k. It is reasonable to ask whether the same holds for
im∆ and we show that this is not generally the case.
Lemma 2.1. Let ∆ be an elementary operator on Mn with coefficients (A1, ..., Ak) and
(B1, ..., Bk), defined by ∆(T) = A1TB1 +A2TB2 +...+AkTBk. Then there exists an ele-
mentary operator ∆˜ such that (im∆) = ker∆˜.
⊥
Proof. Define ∆˜ (T) = B1TA1+B2TA2+...+BkTAk for T ∈ Mn. If T is an arbitrary matrix,
then tr(∆(T)C) = tr(T(B1CA1 + ... + BkCAk)) and therefore C ∈ (im∆)⊥ if and only if
∆˜(C)∈ (Mn)⊥ = {0}, that is, C ∈ ker∆˜. (cid:3)
k-REFLEXIVITY DEFECT OF THE IMAGE OF A GENERALIZED DERIVATION 3
Next, we introduce some notation. For k ∈ N and α ∈ C, let J (α) denote the Jordan block
k
of size k, i.e.,
α 1
J (α) = ... ... ∈ M .
k k
α 1
α
In the following example we show that for any n ≥ 3 there exists an inner derivation δ on M
n
such that imδ is not (n−1)-reflexive. Consequently, the image of such elementary operator of
length 2 is not 2-reflexive.
Example 2.2. Define δ(T) = J (0)T −TJ (0) for T ∈ M . By (1), every subspace of M is
n n n n
′
n-reflexive, hence rdn(imδ) = 0. It follows by Lemma 2.1 that (imδ)⊥ is simply {Jn(0)}, the
′
commutant of the Jordan block J (0). One can easily verify that {J (0)} is the algebra of all
n n
n×n upper triangular Toeplitz matrices which we will denote by T . Namely,
n
a1 a2 ... ... an
(imδ)⊥ = 0... .a.1. .a.2. ... ...... : a1, a2,...an ∈ C.
0... ... ...... .0.. aa21
By (1), imδ is not (n−1)-reflexive space, since (imδ)⊥ ∩Fn−1 ( (imδ)⊥. Note that (1) also
2 2
implies that rdk(imδ) = n−k for 1≤ k ≤ n−1. Indeed,dim(imδ) = n −dim((imδ)⊥)= n −n
2 2
and by (1) we have dim(Refk(imδ)) = n −dim((imδ)⊥ ∩Fk) = n −k for 1 ≤ k ≤ n−1.
3. Generalized derivations
Let ∆ be an elementary operator of length 2 on M , i.e., a linear transformation of the
n
form ∆(T) = A1TB1 +A2TB2 (T ∈ Mn), where A1, A2 and B1, B2 are two pairs of linearly
independent matrices. By [1, Proposition 1.1], one has rd (ker∆) = 0 for all k ≥ 2. In [5]
k
reflexivity of such elementary operator was studied and an explicit formula for the reflexivity
defect of its kernel was given. This motivates the main subject of this paper, that is the k-
reflexivity defect of the image of some special examples of elementary operators of length 2.
Let A, B ∈ M be arbitrary matrices. Define the generalized derivation on M with coef-
n n
ficients A and B by ∆ (T) = AT −TB, T ∈ M . Obviously, ∆ is an example of an elemen-
n
tary operator of length 2. Let Jp1(λ1) ⊕ ... ⊕ JpN(λN) be the Jordan canonical form of A,
where Ni=1pi = n and λ1,...,λN are not necessarily distinct eigenvalues of A. Similarly, let
P
Jr1(µ1)⊕...⊕JrM(µM) be the Jordan canonical form of B, where PMi=1ri = n and µ1,...,µM
are not necessarily distinct eigenvalues of B. Let R(i, j, k) be a non-negative integer defined by
min{p , r }−k : λ = µ and k < min{p , r },
i j i j i j
R(i, j, k) :=
0 : λ 6= µ or k ≥ min{p , r }.
i j i j
4 T.RUDOLF
Proposition 3.1. With the above notation, the k-reflexivity defect of im∆ can be expressed as
N M
rd (im∆) = R(i, j, k).
k XX
i=1j=1
In particular, im∆ is a k-reflexive space if and only if all roots of the greatest common divisor
of m and m of A and B, respectively, are of multiplicity at most k.
A B
Proof. Let 0 denote the p × r zero matrix (p, r ∈ N) and let A and B be as before the
p,r
Proposition 3.1. For 1 ≤ i ≤ N and 1 ≤ j ≤ M define the following elementary operators on
M and M , respectively,
pi,rj rj,pi
∆ (T)= J (λ )T −TJ (µ ) (T ∈ M ),
pi,rj pi i rj j pi,rj
∆ (T)= J (µ )T −TJ (λ ) (T ∈ M ).
rj,pi rj j pi i rj,pi
Lemma 2.1 yields (im∆pi,rj)⊥ = ker∆rj,pi. If λi 6= µj, then ∆pi,rj is bijective and im∆pi,rj is a
k-reflexive space for every k ∈ N. Now assume that λ = µ . It is not hard to see that
i j
ker∆rj,pi = n(cid:16) 0rj,pi−rj T (cid:17) :T ∈ Trjo if rj ≤ pi,
T
ker∆ = : T ∈ T if r > p .
rj,pi 0rj−pi,pi pi j i
Let us denote d = min{p ,r } and D = max{p ,r }. Since transformations of the type (2)
i j i j
preserve k-reflexivity defect we can without any loss of generality assume that ker∆ =
rj,pi
n(cid:0) 0d,D−d T (cid:1):T ∈ Tdo and therefore dim(im∆) = d(D − 1). The structure of the space
ker∆ yields that ker∆ ∩F is a linear space with the following property. If k ≥ d, then
rj,pi rj,pi k
ker∆ ∩F = ker∆ . Otherwise, if 1 ≤ k < d, then
rj,pi k rj,pi
ker∆ ∩F = 0k,d−k T : T ∈T .
rj,pi k 0D−k,d−k 0D−k,k k
Therefore, im∆ is a k-reflexive space iff k ≥ d or λ 6= µ . Otherwise, if k < d and λ = µ ,
pi,rj i j i j
one gets dim(Refk(im∆pi,rj)) = dD−k. The result in general setting now follows by (3). (cid:3)
LetA, B ∈ M beasbeforetheProposition3.1. Let∆ :M → M beanelementaryoperator
n n n
defined by ∆(T) = ATB−T. Let R(i, j, k) be a non-negative integer defined by
1
min{p , r }−k : λ , µ 6= 0, λ = and k < min{p , r },
R(i, j, k) := i j i j i µj i j
0 : otherwise.
Corollary 3.2. With the above notation, the k-reflexivity defect of im∆ can be expressed as
N M
rd (im∆) = R(i, j, k).
k XX
i=1j=1
k-REFLEXIVITY DEFECT OF THE IMAGE OF A GENERALIZED DERIVATION 5
Proof. Define ∆ (T) = J (λ )TJ (µ ) − T for T ∈ M . By (3) we get rd (im∆) =
pi,rj pi i rj j pi,rj k
N M rd (im∆ ), hence it suffices to determine rd (im∆ ). If λ = µ =0, then it is
Pi=1Pj=1 k pi,rj k pi,rj i j
not hard to see that for T = (t ) ∈ M we have
uv pi,rj
0 t21 ... t2,rj−1
. . .
. . .
. . .
∆pi,rj(T) = −T + ,
0 tpi,1 ... tpi,rj−1
0 0 ... 0
therefore im∆ = M and rd (im∆ ) = 0 for every positive integer k. If λ = 0
pi,rj pi,rj k pi,rj i
and µ 6= 0, then im∆ = {XJ (µ ) : X ∈ im∆˜ } where ∆˜ : M → M
j pi,rj rj j pi,rj pi,rj pi,rj pi,rj
˜ −1
is a generalized derivation of the form ∆ (T) = J (0)T − TJ (µ ) . Thus we have
pi,rj pi rj j
˜
rd (im∆ ) = rd (im∆ ). By [2, Example 6.2.13] one can easily see that inverting ma-
k pi,rj k pi,rj
trices preserves the sizes of Jordan blocks, hence Proposition 3.1 yields rd (im∆ ) = 0.
k pi,rj
1
Similarly, if λ 6= 0 and µ = 0 or if λ 6= 0, µ 6= 0 and λ 6= , then again Proposition
i j i j i µj
1
3.1 yields rd (im∆ ) = 0. Now assume that λ , µ 6= 0 and that λ = . As before,
k pi,rj i j i µj
rd (im∆ ) = rd (im∆˜ ) where ∆˜ : M → M is a generalized derivation of the
k pi,rj k pi,rj pi,rj pi,rj pi,rj
˜ −1 ˜
form ∆ (T) = J (λ )T −TJ (µ ) . Now the Proposition 3.1 yields that rd (im∆ )= 0
pi,rj pi i rj j k pi,rj
˜
if k ≥ min{p ,r } and rd (im∆ ) = min{p , r } − k if k < min{p ,r }. By (3) one gets
i j k pi,rj i j i j
rd (im∆)= N M R(i, j, k). (cid:3)
k i=1 j=1
P P
Acknowledgements
The author is thankful to the Slovenian Research Agency for their financial support.
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University of Ljubljana, IMFM, Jadranska ul. 19, 1000 Ljubljana, Slovenia
E-mail address: [email protected]
