Spec. Matrices 2016; 4:80–100
Research Article
Open Access
Ajda Fošner* and Bojan Kuzma
Preserving zeros of Lie product on alternate
matrices
DOI 10.1515/spma-2016-0009
Received July 8, 2015; accepted December 13, 2015
Abstract: We study continuous maps on alternate matrices over complex field which preserve zeros of Lie
product.
Keywords: Alternate matrix, Lie product, general preserver
MSC: 15A86, 15A99, 15B99
1 Introduction and the main theorem
Linear preserver problem is the problem which concerns the characterization of linear operators on matrix
algebras as well as on more general rings and operator algebras that leave certain functions, subsets, relations, etc., invariant. It is one of the most active subjects in the last few decades (for surveys of the topic we
refer the reader to the papers [13, 14, 24]). This kind of problems arise in most parts of mathematics because
in many cases the corresponding results provide important information on the automorphisms of the underlying structures. In the last few decades a lot of results on linear preservers and also general (non-linear)
preservers have been obtained (see [16]).
One among the important preserver problems is classifying maps that preserve commutativity (see, for
example, [2, 3, 5, 18–22, 25–27] and references therein). Part of the rationale behind studying this kind of problems is the fact that in associative algebras quite a few elements do commute (for example, every polynomial
in x commutes with x). Even more importantly, the assumption of preserving commutativity can be considered as the assumption of preserving zero Lie products. Because of applications in quantum mechanics it is
also interesting to study the problem of characterizing general commutativity preserving maps. These are the
maps that preserve zeros of Lie product and are not assumed to satisfy any additional algebraic assumption
like additivity (see e.g. [17]). Here, let us also mention two papers [4] and [6] dealing with maps preserving
zeros of a usual product.
In the present paper we will be interested in describing commutativity preserving maps on the algebra
of all n × n alternate complex matrices. We will only assume that the map is injective and continuous but will
not presume it is linear. Let us remark that our results are in the spirit of Šemrl [26] who studied injective
continuous maps which preserve zeros of Lie product on complex matrices. Moreover, we emphasize that
without imposing some additional regularity conditions on the map, like continuity, we cannot hope for a
nice structural result (see also [25, 26]).
Let us list some mostly standard notation. Throughout, n ≥ 3 will be an integer and M n will be the algebra
of all n × n matrices over the field of complex numbers. Let E ij be the standard basis of M n . By Altn ⊂ M n we
will denote the subspace of all alternate matrices (i.e., matrices with the property A t = −A). Here A t denotes
*Corresponding Author: Ajda Fošner: Ajda Fošner, University of Primorska, Faculty of Management, Cankarjeva 5, SI-6000
Koper, Slovenia, E-mail: [email protected]
Bojan Kuzma: Bojan Kuzma, University of Primorska, FAMNIT, Glagoljaška 8, SI-6000 Koper
and Institute of Mathematics, Physics and Mechanics, Department of Mathematics, Jadranska 19, SI-1000 Ljubljana, Slovenia,
E-mail: [email protected]
© 2016 Ajda Fošner and Bojan Kuzma, published by De Gruyter Open.
This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 3.0 License.
Preserving zeros of Lie product on alternate matrices |
81
the transpose of a matrix A ∈ M n . For A = (a ij ) ∈ M n we will denote by A = (a ij ) = (a ij ) the conjugate of
A. As usual, we use the notation diag(a1 , . . . , a k ) to denote the k × k diagonal matrix with diagonal entries
a1 , . . . , a k .
We will study injective continuous maps ϕ with the property
AB = BA
ϕ(A)ϕ(B) = ϕ(B)ϕ(A).
=⇒
This condition can be written as
[A, B] = 0
[ϕ(A), ϕ(B)] = 0,
=⇒
where [A, B] = AB − BA is the Lie product. Note that alternate matrices are closed under the Lie product.
We will represent vectors x ∈ Cn as n × 1 complex matrices. Note that the standard basis e1 , e2 , . . . , e n of
n
C is the set of all n × 1 matrices having all entries equal to zero but one that is equal to one. If x, y ∈ Cn are
two linearly independent nonzero vectors, then xy t − yx t = x ∧ y is a rank two alternate matrix. Every rank
two alternate matrix can be written in this form.
A matrix A is said to be nonderogatory if every eigenvalue of A has geometric multiplicity one. Let us point
out that a matrix B ∈ M n commutes with a nonderogatory matrix A ∈ M n if and only if there is a complex
polynomial p such that B = p(A) (see [12, p. 135]).
The main idea behind our proof is to utilize the Brouwer’s invariance of domain theorem [10, p. 344]
stating that if U is an open subset of Rm and F : U → Rm is a continuous injective map, then F(U) is open.
In particular, there is no injective continuous map from Rk into Rm whenever m < k. We acknowledge that
the same idea was already used before, say in Petek and Šemrl’s characterization of continuous maps which
preserve adjacency of matrices in one direction (see [23]). Later, the same idea was used in [26] and [9].
The following is a basis for our arguments in Section 2 and Section 3. If x t Ax = 0 for every column vector x ∈ Cn , a complex matrix A is alternate. Each alternate matrix is congruent to a block-diagonal matrix.
More precisely, there exists an invertible matrix S such that
SAS t =
m
⨁︁
J ⊕ 0n−2m ,
i=1
where
(︃
J :=
0
−1
)︃
1
.
0
Consequently, the rank of each alternate matrix is an even integer and a minimal possible nonzero rank is 2.
∑︀
∑︀m
Moreover, since SAS t = m
i=1 E (2i−1) (2i) − E (2i) (2i−1) =
i=1 e 2i−1 ∧ e 2i , we have
A=
m
∑︁
xi ∧ yi
i=1
for some linearly independent vectors x i = S−1 e2i−1 and y i = S−1 e2i , i = 1, . . . , m.
We will need some further properties of alternate matrices over the complex field. We borrow the following notations and facts from the book by Gantmacher, [11, p. 14–21]. In matrices below, all non-specified
entries are zero. Firstly, given an integer k ≥ 2, let
⎛
⎞
1
k
∑︁
⎜
⎟
.
⎟ ∈ Mk
V k :=
E(k−i+1) i = ⎜
(1)
..
⎝
⎠
i=1
1
be an anti-diagonal k × k matrix and let
J k :=
k−1
∑︁
i=1
E i (i+1)
⎛
0
⎜
⎜
⎜
=⎜
⎜
⎝
⎞
1
..
.
..
.
..
.
⎟
⎟
⎟
⎟
⎟
1⎠
0
82 | Ajda Fošner and Bojan Kuzma
be an elementary k × k Jordan upper-triangular nilpotent. For odd integers q we also define an alternated
version of J q , that is, a matrix
q−1
J (q) :=
2
∑︁
E i (i+1) −
q−1
∑︁
E i (i+1)
i= q+1
2
i=1
⎛
0
⎜
⎜
⎜
⎜
⎜
⎜
⎜
=⎜
⎜
⎜
⎜
⎜
⎜
⎝
1
..
⎞
.
..
.
..
.
1
0
−1
..
.
..
.
..
.
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟ ∈ Mq .
⎟
⎟
⎟
⎟
⎟
−1⎠
0
Note that J (q) is an upper-triangular q × q nilpotent having precisely q−1
2 of +1 and the same number of −1
on the first upper-diagonal. Note also that J (q) and J q are similar. Moreover, let Idk denotes the k × k identity
matrix and, given an integer p and a scalar λ, we define the (2p) × (2p) matrix K (pp)
by
λ
K (pp)
λ
:=
1
2
(︀
(︃
√
)︀
Id2p − −1V2p ·
λ Idp +J p
0
0
−λ Idp −J p
)︃
√
(︀
)︀
· Id2p + −1V2p .
(2)
(pp)
Notice that K (pp)
is similar to K−λ
, hence by [11, Theorem 4, p. 9] they are also orthogonally similar, that is,
λ
there exists a matrix Q such that
(pp)
(pp) −1
K−λ
= QK−λ
Q
and
Q−1 = Q t .
(3)
Secondly, given an odd integer q we define
K (q) :=
1
2
(︀
√
√
)︀
(︀
)︀
Idq − −1V q · J (q) · Idq + −1V q .
(4)
When q = 1 we let K (1) be a zero 1 × 1 matrix.
Example 1.1. The exact appearance of matrices defined by formulas (3)–(4) will play no role in the present
paper. For convenience we only show K (44)
and K (7) , here i2 = −1.
λ
⎛
K (44)
λ
0
⎜ i
⎜
⎜
⎜ 0
⎜
0
1⎜
= ⎜
2⎜
⎜ 0
⎜
⎜ 0
⎜
⎝ −1
2iλ
⎛
K (7)
−i
0
i
0
0
−1
2iλ
−1
0
⎜
−1
⎜
⎜
⎜0
1⎜
= ⎜
0
2⎜
⎜0
⎜
⎜
⎝ −i
0
0
−i
0
i
−1
2iλ
−1
0
1
0
−1
0
−i
0
−i
0
0
−i
0
2iλ
−1
0
0
0
1
0
−1 − i
0
−i
0
0
0
1
−2iλ
0
−i
0
0
0
0
1+i
0
1−i
0
0
0
1
−2iλ
1
i
0
−i
0
0
i
0
−1 + i
0
1
0
1
−2iλ
1
0
0
i
0
−i
i
0
i
0
−1
0
1
⎞
−2iλ
1 ⎟
⎟
⎟
0 ⎟
⎟
0 ⎟
⎟,
0 ⎟
⎟
⎟
0 ⎟
⎟
i ⎠
0
⎞
0
⎟
i ⎟
⎟
0⎟
⎟
0⎟
⎟
0⎟
⎟
⎟
−1⎠
0
Let d ≥ 1 be an integer. By induction on d, it is easy to prove the following lemma.
Preserving zeros of Lie product on alternate matrices |
83
Lemma 1.2. Let q = 2k + 1 be an odd integer. Then
(︃
)︃
0(q−d)×d
D
(q) d
(J ) =
,
0d×d
0d×(q−d)
where
D=
⎧
2
d−1
d
⎪
⎪
⎨Idk−d+1 ⊕ diag(−1, (−1) , . . . , (−1) ) ⊕ (−1) Idk−d+1 ,
(−1)
⎪
⎪
⎩0 ,
d−k+1
2
2k−d+1
diag(−1, (−1) , . . . , (−1)
),
d≤k
k < d ≤ 2k .
d ≥ 2k + 1
√
√
√
√
)︀t
)︀
)︀2
One easily verifies that (Idk − −1V k = (Idk − −1V k is symmetric with (Idk − −1V k = −2 −1V k and
√
√
(︀
)︀ (︀
)︀
(q)
1
is similar to J q because they are both nilpotents with
2 Idk − −1V k · Idk + −1V k = Idk . Moreover, J
maximal nilindex. Hence, it is easy to see that K (pp)
and K (q) are alternate matrices which are similar to (J p ⊕
λ
J p ) and to J q , respectively. In a sense, these are the only possible cases. Namely, by [11, Corollary on p. 21],
⨁︀ 1 (p i p i )
any alternate matrix is orthogonally similar to a block-diagonal alternate matrix of the form ni=1
K λi
⊕
⨁︀n2 (q j )
t
K
.
More
precisely,
if
A
∈
M
is
alternate,
then
there
exists
a
matrix
Q
∈
M
with
Q
Q
=
Id
and
there
n
n
n
j=1
exist integers n1 ≥ 0, n2 ≥ 0, integers p1 , . . . , p n1 and q1 , . . . , q n2 , and scalars λ1 , . . . , λ n1 such that
QAQ−1 = QAQ t =
n1
⨁︁
i pi )
K (p
⊕
λi
i=1
n2
⨁︁
K (q j ) .
(5)
j=1
If n1 = 0 or n2 = 0, we omit the corresponding summand.
We are now ready to state our main result. Recall that a polynomial p is called odd if p(−λ) = −p(λ) for
each λ.
Main Theorem. Let n ≥ 15 and let ϕ : Altn → Altn be an injective continuous map that preserves zeros of Lie
product. Then there exists an orthogonal matrix Q ∈ M n and for each A ∈ Altn there is an odd polynomial p A (λ)
such that either
ϕ(A) = Qp A (A)Q t
for all A ∈ Altn , or
ϕ(A) = Qp A (A)Q t
for all A ∈ Altn .
Remark 1.3. We believe that the above theorem is true also for matrices of lower dimensions n = 5, . . . , 14,
but we have not proved this yet. Note also that in the case n = 4 we have at least one additional map. Namely,
it can be shown that a map Φ : Alt4 → Alt4 , which swaps positions (1, 4), (2, 3) and (4, 1), (3, 2) and leaves
the other entries intact, preserves zeros of Lie product.
2 Preliminaries
Let us denote by A′ = {X ∈ Altn : AX = XA} the alternate commutant (a.k.a. a centralizer) of a matrix A ∈
Altn and by Sp(A) the spectrum of a matrix A. We will start with a well-known Lumer–Rosenblum theorem [15]
(see also [1, p. 2] for a short proof valid in general Banach algebras).
Lemma 2.1. Suppose that A ∈ M n and B ∈ M m . If 0 ∉ Sp(A) + Sp(B) = {α + β : α ∈ Sp(A), β ∈ Sp(B)}, then
the equation AX + XB = 0 has only a zero solution.
We proceed with a series of lemmas.
84 | Ajda Fošner and Bojan Kuzma
Lemma 2.2. If A = K (q) with q odd, then dim A′ =
(q−1)
2 .
Proof. Note that A is nonderogatory. Hence, its ordinary commutant (i.e., {X ∈ M q : AX − XA = 0}) equals to
Poly(A), i.e., the set of all polynomials in A [12, Theorem 3.2.4.2.]. For the alternate commutant, we must pick
out those X ∈ Poly(A) which are alternate. Now, if X = p(A) = α0 Id +α1 A + · · · + α q−1 A q−1 is alternate, then
∑︀ q−1
2
clearly Y = α1 A + α3 A3 + α5 A5 + · · · + α q−2 A q−2 is also alternate and so is X − Y = i=0
α2i A2i . But note that
q−1
q−1
∑︀
∑︀
2
2
(X − Y)t = i=0
α2i (A t )2i = i=0
α2i (−A)2i = X − Y. So, α0 = α2 = · · · = α q−1 = 0. But then dim A′ = (q−1)
2 .
Lemma 2.3. Let p ≥ 1 be any integer. Then the following holds.
(︀
)︀′
= p.
(i) If λ ≠ 0, then dim K (pp)
λ
(︀
)︀′
= 2p.
(ii) If λ = 0 and p is even, then dim K (pp)
λ
(︀ (pp) )︀′
= 2p − 1.
(iii) If λ = 0 and p is odd, then dim K λ
Proof. Suppose that X ∈ Alt2p satisfies K (pp)
X − XK (pp)
= 0. Using the definition of K (pp)
, we arrive at
λ
λ
λ
(︀
(︀
)︀
(︀
)︀
(λ Idp +J p ) ⊕ (−λ Idp −J p ) · S−1 XS − S−1 XS · (λ Idp +J p ) ⊕ (−λ Idp −J p ) = 0
√
(︃
−1
)︀
for S = Id2p − −1V2p . We now decompose S XS =
X11
X21
X12
X22
(6)
)︃
into a 2 × 2 block matrix. Comparing the
individual blocks of (6) we derive four equations
(λ Idp +J p )X11 − X11 (λ Idp +J p ) = 0,
(λ Idp +J p )X12 + X12 (λ Idp +J p ) = 0,
(7)
(λ Idp +J p )X21 + X21 (λ Idp +J p ) = 0,
(λ Idp +J p )X22 − X22 (λ Idp +J p ) = 0.
The first and the last equations simplify into J p X11 = X11 J p and J p X22 = X22 J p , respectively. Since J p is
nonderogatory we see that X11 and X22 have to be polynomials in J p . This yields that
(︃
)︃
f11 (J p )
X12
−1
S XS =
,
X21
f22 (J p )
where f1 , f2 are complex polynomials and X12 , X21 satisfy (7). Now, recall that X is alternate. Due to S t = S,
√
(︀
)︀2
this gives (S−1 XS)t = S t · (−X) · (S−1 )t = −SXS−1 = −S2 (S−1 XS)S−2 . On the other hand, S2 = Id2p − −1V2p =
√
−2 −1V2p , and S−2 =
√
−1
2 V 2p .
(︃
With this in hand we establish
)︃ (︃
)︃t
t
X21
f11 (J p )
X12
=
f22 (J p )t
X21
f22 (J p )
(︃
)︃
f (J p )
X12
= −V2p 11
V2p
X21
f22 (J p )
(︃
)︃ (︃
)︃ (︃
0 Vp
f11 (J p )
X12
0
=−
·
·
Vp 0
X21
f22 (J p )
Vp
(︃
)︃
V p f22 (J p )V p
V p X21 V p
=−
.
V p X12 V p
V p f11 (J p )V p
f11 (J p )t
t
X12
Vp
0
)︃
(8)
Comparing the (1, 1) block, we get that f11 (J p ) = −V p f22 (J p )t V p . Since Z ↦→ V p f22 (Z)t V p is a flip map (i.e.,
reflection over anti-diagonal), we easily conclude V p f22 (J p )t V p = f22 (J p ). Hence, f11 (J p ) = −f22 (J p ).
In the case of off-diagonal blocks we are facing two options.
Preserving zeros of Lie product on alternate matrices |
85
Case 1. Assume λ ≠ 0. Then, by Lemma 2.1, the second equation and the third equation of (7) have only zero
solution. So, X12 = 0 = X21 and, therefore,
{︃
(︃
)︃
}︃
0
(pp) ′
−1 f 11 (J p )
(K λ ) = S
S : f11 ∈ C[λ]
(9)
0
−f11 (J p )
is of dimension p.
Case 2. Assume λ = 0. Then, the second equation of (7) simplifies into J p X12 + X12 J p = 0 with a solution
X12 =
n−1
∑︁
αj
n−j
∑︁
j=0
(−1)i E i (i+j) .
i=1
However, X being alternate further forces −X12 = (V p X12 V p )t , by comparing the (2, 1) entries of (8). Note that
the right-hand side is a flip map. It is now easy to verify that, with p even, one has α1 = 0 = α3 = · · · = α p−3 =
α p−1 , and, with p odd, one has α0 = 0 = α2 = · · · = α p−3 = α p−1 . Therefore, with p even, at block position
(1, 2) of X we have 2p linearly independent matrices, while with p odd we have p−1
2 linearly independent
matrices. We can duplicate the calculations also for X21 . We already know that the diagonal blocks are of the
form described in (9). Combined, we have
⎧
⎨p + 2 · (︀ p )︀ = 2p ,
p is even
(pp) ′
(︁2 )︁
.
dim(K λ ) =
⎩p + 2 · p−1
= 2p − 1 , p is odd
2
Let n ≥ 3 and A ∈ Altn . We would like to estimate the dimension of A′ . Using orthogonal similarity, we may
assume that A is alternate and block-diagonal, say
A=
n1
⨁︁
K (q i ) ⊕
i=1
n2
⨁︁
(p j p j )
K0
j=1
⊕
n3
⨁︁
k pk )
K (p
,
λ
k
k=1
(︀ )︀
with q i ≥ 1 odd, p j , p k ≥ 1, and λ1 , . . . , λ n3 nonzero. Let us write X = X ij ∈ A′ into block form. Then it is
(︀
)︀′
(︀ (p p ) )︀′
(︀ k p k ) )︀′
easy to see that the diagonal blocks satisfy X ii ∈ K (q i ) , X jj ∈ K0 j j , and X kk ∈ K (p
. By the previous
λk
lemmas we have
qi − 1
,
2
{︃
2p j ,
(p p )
dim(K0 j j )′ = d(p j ) =
2p j − 1 ,
dim(K (q i ) )′ =
p j even
p j odd
,
k pk ) ′
dim(K (p
) = pk .
λ
k
Thus, the dimension of diagonal blocks of matrices from A′ equals
n3
n2
n1
∑︁
∑︁
q i − 1 ∑︁
+
d(p j ) +
pk .
2
i=1
j=1
k=1
It remains to calculate the dimension of off-diagonal blocks of matrices from A′ .
First, consider X ii′ , 1 ≤ i < i′ ≤ n1 . This block must satisfy the equation K (q i ) X ii′ = X ii′ K (q i′ ) . Note that
(q i )
K is similar to an elementary Jordan upper-triangular matrix J q i . It is easy to see that the dimension of the
solutions of the above equation equals the dimension of the solutions of a similar equation J q i Y = YJ q i′ . Also,
one easily computes that this dimension is min{q i , q i′ }.
(p p )
Next, consider X ij , 1 ≤ i ≤ n1 , 1 ≤ j ≤ n2 . This block must satisfy the equation K (q i ) X ij = X ij K0 j j .
(p p )
Note that K0 j j is similar to J p j ⊕ J p j , so the dimension of the solutions of the above equation equals the
dimension of the solutions of the equation J q i Y = Y(J p j ⊕ J p j ). Writing Y = (Y1 |Y2 ) we easily reduce this case
to the previous case and find that the dimension of the solutions is 2 min{q i , p j }.
86 | Ajda Fošner and Bojan Kuzma
(p ′ p ′ )
(p p )
Further, consider X jj′ , 1 ≤ j < j′ ≤ n2 . This block must satisfy the equation K0 j j X ij = X ij K0 j j . Arguing
as above, we reduce this equation to the equation (J p j ⊕ J p j )Y = Y(J p j′ ⊕ J p j′ ), which easily establishes that the
dimension of the solutions equals 4 min{p j , p j′ }.
Now, consider X ik or X jk , 1 ≤ i ≤ n1 , 1 ≤ j ≤ n2 , 1 ≤ k ≤ n3 . These blocks must be zero because in the
equation they are multiplied with a nilpotent matrix on the left and with an invertible matrix on the right
side. By the theory of elementary operators, this yields that X ik = 0 = X jk .
At the end, consider X kk′ , 1 ≤ k < k′ ≤ n3 . If λ k = λ k′ , then we easily reduce this case to the previous cases
to find that the dimension of these blocks equals 4 min{k, k′ }. On the other hand, if λ ≠ λ′ , then X kk′ = 0.
Consequently,
dim A′ ≤
n3
n2
n1
∑︁
∑︁
∑︁
∑︁
q i − 1 ∑︁
+
d(p j ) +
pk +
min{q i , q i′ } +
2 min{q i , p j }
2
′
1≤i≤n
j=1
i=1
+
∑︁
k=1
4 min{p j , p j′ } +
1≤j<j′ ≤n2
∑︁
1≤i<i ≤n1
1
1≤j≤n2
(10)
4 min{p k , p k′ }.
1≤k<k′ ≤n3
Lemma 2.4. Let n ≥ 3 and A ∈ Altn . Then A′ = Altn if and only if A = 0.
Proof. If A = 0, then, of course, A′ = Altn . Now, suppose that A ≠ 0. According to (10), we can easily see that
and, therefore, A′ ≠ Altn .
in this case dim A′ < n(n−1)
2
Lemma 2.5. Let n ≥ 9 and A ∈ Altn be a nonzero matrix. Then A is of minimal rank if and only if dim A′ ≥
(n−2)(n−3)
+ 1.
2
Proof. Let A ∈ Altn be a nonzero alternate matrix. Using orthogonal similarity, we may write A in a blockdiagonal form as in the previous investigations.
Now, if rk A = 2 (here, rk A denotes the rank of the matrix A), then either A = K (3) or A = K (11)
, λ ≠ 0, or
λ
(22)
A = K0 . In the first case we have n1 = n − 2, q1 = 3, q2 = · · · = q n1 = 1, while n2 = 0 = n3 . By the remarks
+ 1. Likewise we argue in the second case, where
above, it is easy to see that in this case dim A′ = (n−2)(n−3)
2
(n−4)(n−5)
′
dim A′ = (n−2)(n−3)
+
1,
and
in
the
last
case,
where
dim
A
=
+ 2(n − 4) + 4 = (n−4)(n−1)
+ 4.
2
2
2
It remains to show that if rk A > 2, then the dimension of A′ is strictly smaller than (n−2)(n−3)
+ 1.
2
Let m be a dimension of all nilpotent blocks in A. Then we can write A = A1 ⊕ A2 ∈ Altm ⊕ Altn−m , where
A1 is nilpotent and A2 is invertible. It is easy to show that A′ = A′1 ⊕ A′2 ⊆ Altm ⊕ Altn−m which gives
dim A′ ≤ dim Altm + dim Altn−m =
m(m − 1) (n − m)(n − m − 1)
+
.
2
2
This is strictly below (n−2)(n−3)
+ 1 whenever 3 ≤ m ≤ n − 3 because n ≥ 9. Recall that if m = 2, then dim A′ ≤
2
(n−2)(n−3)
dim Alt2 +(dim Altn−2 −1) <
+ 1 since A′2 = Altn−2 is possible only when A2 = 0. Note also that we
2
cannot have m = n − 1 since A is alternate. Hence, the equality holds only if m ∈ {0, 1, n − 2, n}.
Case m = 0. Note that in this case A is invertible. Let us fix an eigenvalue λ ∈ Sp(A) and write A block⨁︀ p i p i
diagonally as A = A1 ⊕ A2 , where A1 =
K λ is the sum of alternate blocks with eigenvalues −λ, λ, while
Sp(A2 ) ∩ {−λ, λ} = ∅. Let m1 be the dimension of A1 . Observe that m1 ∉ {0, 1, n − 1} because A is invertible.
+ 1, as desired. The same is true if
If 3 ≤ m1 ≤ n − 3, then, arguing as above, we get that dim A′ < (n−2)(n−3)
2
either m1 = 2 (by Lemma 2.4, 0 ≠ A2 is of dimension n − 2 > 2) or m1 = n − 2 (in this case we just change
⨁︀s
∑︀
(p k p k )
the roles of A1 and A2 ). At the end, if m1 = n, then we can write A =
with sk=1 2p k = n and
k=1 K λ
1 ≤ p1 ≤ · · · ≤ p s . We already know (according to previous observations) that the k-th diagonal block of A′
has dimension p k and the (k, h), k < h, off-diagonal block of A′ has dimension min{2p k , 2p h }. Therefore,
dim A′ =
s
∑︁
k=1
pk +
∑︁
1≤k<h≤s
min{2p k , 2p h } = 2sp1 + 2(s − 1)p2 + · · · + 2p s −
n
.
2
Preserving zeros of Lie product on alternate matrices |
87
Note that sp1 ≤ p1 + · · · + p s = 2n and (s − 1)p2 ≤ p2 + · · · + p s = 2n − p1 ≤ 2n − 1 and, inductively, (s − k + 1)p k ≤
p k + · · · + p s = 2n − p1 − · · · − p k−1 ≤ 2n − (k − 1) for every 1 ≤ k ≤ s. Therefore,
(︀
)︀ n
dim A′ =2 sp1 + (s − 1)p2 + · · · + p s −
2
(︀ n
)︀ n
n
n
≤2 2 + ( 2 − 1) + · · · + ( 2 − s + 1) −
2
n
2
)︁ n n2
(︁∑︁
k − =
.
≤2
2
4
k=1
But this is strictly below
(n−2)(n−3)
2
+ 1 whenever n ≥ 9.
Case m = 1. In this case we can write A = 0 ⊕ A2 ∈ 0 ⊕ Altn−1 , where A2 is invertible. As above, we easily see
that dim A′ = dim A′2 < ((n−1)−2)((n−1)−3)
+ 1 ≤ (n−2)(n−3)
+ 1, as desired.
2
2
Case m = n − 2. In this case we can write A = A1 ⊕ K (11)
, where A1 is nonzero. Observe that A′ ⊂ Altn−2 ⊕ Alt2 .
λ
Moreover, the inclusion is strict because A1 ≠ 0. Hence, dim A′ < dim Altn−2 +1 = (n−2)(n−3)
+ 1, as claimed.
2
⨁︀n1 (q i ) ⨁︀n2 (p j p j )
Case m = n. In this case A is nilpotent and we can decompose it into A =
⊕
with
i=1 K
j=1 K 0
q1 ≥ q2 ≥ · · · ≥ q n1 odd and p1 ≥ p2 ≥ · · · ≥ p n2 .
(q1 )
⊕ A2 , where A2 contains all other nilpotent blocks of A. If X =
(︃ If q1 ≥ 5,
)︃ then we have A = K
X11 X12
∈ A′ , then X11 ∈ (K (q1 ) )′ , X22 ∈ A′2 ⊆ Altn−q1 , and K (q1 ) X12 = X12 A2 . To compute the diment
−X12
X22
⨁︀ 1 (q i )
sion of (1, 2) block in A′ , we decompose X12 into blocks according to the block structure of A2 = ni=2
K ⊕
⨁︀n2 (p j p j )
∑︀n1
K
.
By
previous
observations,
we
see
that
the
dimension
of
(1,
2)
block
equals
min
{
q
1 , qi } +
i=2
0
∑︀nj=1
∑︀n1
∑︀n2
′
2
j=1 2 min{ q 1 , p j } ≤
i=2 q i +
j=1 2p j = n − q 1 . Hence, we may estimate the dimension of A as follows
q1 − 1 (n − q1 )(n − q1 − 1)
+
+ (n − q1 ).
2
2
dim A′ ≤ dim(K (q1 ) )′ + dim Altn−q1 +(n − q1 ) =
This is a strictly decreasing function of q1 ≤ n and it is strictly smaller than (n−2)(n−3)
+ 1 whenever q1 > 3. So,
2
we are done in this case.
If p1 ≥ 3, then we can write A = K0(p1 p1 ) ⊕ A2 , where A2 contains all other nilpotent blocks of A. Proceeding
as above, we see that
n1
n2
∑︁
∑︁
dim A′ ≤ dim(K0(p1 p1 ) )′ + dim Altn−2p1 +(
2 min{p1 , q i } +
4 min{p1 , p j })
i=1
≤2p1 +
=(2p1 +
(n − 2p1 − 1)(n − 2p1 )
+(
2
n2
∑︁
2p j +
n1
∑︁
i=1
j=2
=n + (n − 2p1 ) +
n1
∑︁
j=2
2q i +
i=1
n2
∑︁
4p j )
j=2
n2
n1
∑︁
∑︁
(n − 2p1 − 1)(n − 2p1 )
2p j +
qi ) +
qi ) + (
2
j=2
i=1
(n − 2p1 − 1)(n − 2p1 )
.
2
Note that this is a strictly decreasing function of p1 ∈ [3, 2n ] and it takes a maximum at p1 = 3. Since n ≥ 9 it
follows that the maximum is strictly smaller than (n−2)(n−3)
+ 1. Hence, dim A′ < (n−2)(n−3)
+ 1.
2
2
Now, suppose that there are at least two nonzero blocks in the decomposition of A. By the above, we may
assume that all blocks in A are of dimension at most 4. First, let A be orthogonal similar to A = K0(22) ⊕ K0(22) ⊕
A2 . Then
dim A′ ≤4 + 4 +
n1
n2
n1
n2
∑︁
∑︁
∑︁
(n − 8)(n − 9) ∑︁
+(
2 min{2, q i } +
4 min{2, p j }) + (
2 min{2, q i } +
4 min{2, p j })
2
i=1
≤(4 + 4 +
n1
∑︁
i=1
qi +
n2
∑︁
j=2
j=2
i=1
n1
n2
n1
n2
∑︁
∑︁
∑︁
∑︁
(n − 8)(n − 9)
2p j ) + (
qi +
2p j ) + (
2q i +
4p j ) +
2
i=1
j=3
i=1
j=2
j=3
88 | Ajda Fošner and Bojan Kuzma
=n + (n − 8) + 2(n − 4) +
(n − 8)(n − 9)
.
2
It is easy to see that this is strictly smaller than (n−2)(n−3)
+ 1 for n ≥ 9.
2
Next, let A be orthogonally similar to K (3) ⊕ K0(22) ⊕ A2 . In this case
n
∑︁
(︀
)︀
dim A′ ≤ dim K (3) + dim K0(22) + dim Altn−7 + 2 min{3, 2} +
min{3, 1} + 2(n − 7)
i=7
(n − 7)(n − 8)
+ 4 + 3(n − 7)
=1 + 4 +
2
(n − 7)(n − 2)
+9
=
2
and it is easy to see that this is strictly smaller than (n−2)(n−3)
+ 1 for n ≥ 9.
2
Assume lastly that A is orthogonally similar to K (3) ⊕ K (3) ⊕ A2 . In this case
n
n
∑︁
(︀
)︀ ∑︁
dim A′ ≤ dim K (3) + dim K (3) + dim Altn−6 + min{3, 3} +
min{3, 1} +
min{3, 1}
i=6
i=6
(n − 6)(n − 7)
=1 + 1 +
+ 3 + 2(n − 6)
2
(n − 6)(n − 3)
=
+5
2
and this is strictly smaller than
(n−2)(n−3)
2
+ 1 for n ≥ 9. The proof is completed.
We will also require an upper bound for dimension of the alternate commutant of rank-four alternate matrices.
Lemma 2.6. Let n ≥ 15 and let A ∈ Altn be of rank-four. Then, dim A′ (n−3)(n−4)
2
Proof. If A ∈ Altn , then, from its Jordan decomposition and the fact that each block of a block-diagonal
alternate matrix is itself alternate and hence of even rank, we see that rk A = 4 if and only if, up to orthogonal
similarity, (i) A = K (22)
⊕ 0n−4 for λ ≠ 0, (ii) A = K0(33) ⊕ 0n−6 , (iii) A = K (5) ⊕ 0n−5 , (iv) A = K0(22) ⊕ K0(22) ⊕ 0n−8 ,
λ
(11)
(22)
(v) A = K λ ⊕ K0 ⊕ 0n−6 with λ ≠ 0, (vi) A = K (11)
⊕ K (11)
⊕ 0n−4 for λ, µ ≠ 0, (vii) A = K (3) ⊕ K (3) ⊕ 0n−6 ,
µ
λ
(2,2)
(11)
(3)
(3)
(viii) A = K ⊕ K0 ⊕ 0n−7 , or (ix) A = K ⊕ K λ ⊕ 0n−4 with λ ≠ 0.
Now, it is easy to see that in each of the cases (i)–(ix), the dimension of an alternate commutant is a
quadratic polynomial
in n.
(︃
)︃ Namely, it is an elementary that, with an alternate block matrix X kept fixed,
′
X
*
(X ⊕ 0n−i )′ =
, where the dimension of a rectangular space of matrices * is a linear function of
−*t Altn−i
n. So, after some calculations we see that under (i), dim A′ = (n−9)n+24
, under (ii), dim A′ = (n−9)n+28
, under
2
2
(n−9)n+24
(n−9)n+40
(n−9)n+28
′
′
′
(iii), dim A =
,
under
(iv),
dim
A
=
,
under
(v),
dim
A
=
,
under
(vi),
dim
A′ ≤
2
2
2
(11) ′
(11)
dim(K λ ⊕ K λ ) = (n−9)n+28
, under (vii), dim A′ = (n−9)n+28
, under (viii), dim A′ = (n−9)n+32
, and under (ix),
2
2
2
(n−9)n+24
′
dim A =
.
From
here,
one
easily
finds
that
the
matrix
from
the
case
(iv)
has
the
maximal
dimension
2
of its alternate commutant among all matrices in (i)–(ix). This dimension equals (n−9)n+40
and,
since
n ≥ 15,
2
this is smaller than (n−3)(n−4)
,
as
claimed.
2
Lemma 2.7. The set of all diagonalizable alternate matrices is dense in Altn .
Proof. Given an odd integer q, the matrix
L(q) =
√
√
1
(Idq − −1V q )(−J (q) )t (Idq + −1V q ) ∈ M q
2
is alternate. Namely, if we multiply alternate matrix
K (q) =
√
√
1
(Idq − −1V q )J (q) (Idq + −1V q )
2
Preserving zeros of Lie product on alternate matrices |
89
with symmetric matrix V q on the left and on the right, and use V q J (q) V q = −(J (q) )t we find L(q) = V q K (q) V q .
Moreover, also X = (L(q) )q−2 is alternate because alternate matrices are closed under forming odd powers.
Hence, given ε > 0, note that a matrix K (q) + εX is alternate and it is similar to J (q) + ε(−(J (q) )t )q−2 = J (q) +
q+1
(−1) 2 ε(−E(q−1) 1 + E q2 ). By expanding the determinant on the first column, we see that the characteristic
polynomial of this matrix is equal to −2ελ − λ q . In particular, for a nonzero ε, the alternate matrix K (q) + εX
is diagonalizable because it has pairwise distinct eigenvalues and approaches K (q) as ε → 0.
Similar, given any integer p, the matrix
L(pp) :=
√
√
1
(Id − −1V2p ) (J p ⊕ (−J p ))t (Id2p + −1V2p )
2 2p
(11)
is alternate. If p is even, then the matrix X = (L(pp) )p−1 is alternate as well. For any scalar λ0 it is easy to see
that the characteristic polynomial of K (pp)
+ εX is the same as the characteristic polynomial of (λ0 Idp +J p +
λ0
t p−1
t p−1
ε(J p ) ) ⊕ (−λ0 Idp −J p − ε(J p ) ), which is equal to ((λ0 − λ)p − ε) · ((λ0 + λ)p − ε). Since the eigenvalues of the
first block are equal to the zeros of the polynomial (λ0 − λ)p − ε, which are pairwise distinct if ε ≠ 0, we see
that the first block is diagonalizable for every ε > 0. The same is true for the second block. Hence, K (pp)
+ εX
λ0
is diagonalizable alternate matrix whenever ε > 0.
If p is odd, then the matrix X = (L(pp) )p−2 is alternate because it is an odd power of an alternate matrix
(︀
)︀
(pp)
L . As above, we see that the characteristic polynomial of K (pp)
+ εX is equal to (λ0 − λ) (λ0 − λ)p−1 − 2ε ·
λ
0
(︀
)︀
(−1)(λ0 + λ) (λ0 + λ)p−1 − 2ε (decompose the determinant of each block in the matrix (λ0 Idp +J p + ε(J pt )p−2 ) ⊕
(−λ0 Idp −J p − ε(J pt )p−2 ), which is similar to K (pp)
+ εX, by the first column). Hence, we deduce that the matrix
λ0
K (pp)
+ εX is diagonalizable and it approaches K (pp)
as ε → 0. Consequently, the set of all diagonalizable
λ0
λ0
alternate matrices is dense in Altn .
For a nonzero vector x ∈ Cn we denote
L x := {x ∧ v : v ∈ Cn }.
It is easy to see that L x = L y precisely when x, y are linearly dependent. Namely, otherwise we could assume
x = e1 and y = e2 . But then e1 ∧ e3 ∈ L e1 is not in L e2 . It is also easy to show that L x ∩ L y = C(x ∧ y) whenever
L x ≠ L y .
In the proof of the main theorem we will also need the following lemmas.
Lemma 2.8. If A, B ∈ L x are linearly independent, then dim(A′ ∩ B′ ) ≥
(n−3)(n−4)
.
2
Proof. Write A = x ∧ y and B = x ∧ z. Since A, B are linearly independent, so are the vectors x, y, z. Thus,
there exists invertible P such that Px = e1 , Py = e2 , and Pz = e3 . Now, pick any alternate X ∈ A′ . Then,
(Xx)y t − (Xy)x t = XA = AX = x(X t y)t − y(X t x)t = −x(Xy)t + y(Xx)t . If we multiply this equation on the right by
P and on the left by P t , we get (PXx)e2t − (PXy)e1t = −e1 (PXy)t + e2 (PXx)t . Comparing both sides, we easily
obtain PXx = αe1 + βe2 and PXy = 𝛾 e1 − αe2 for some scalars α, β, 𝛾 . Therefore,
Xx = αx + βy,
Xy = 𝛾 x − αy.
Xx = α′ x + β′ z,
Xz = 𝛾 ′ x − α′ z
Likewise we derive
for any alternate X ∈ B′ . Since x, y, z are linearly independent, X ∈ A′ ∩ B′ implies that
Xx = αx,
Xy = 𝛾 x − αy,
Xz = 𝛾 ′ x − αz.
Inversely, given X with the above property, X clearly belongs to A′ ∩ B′ , provided it is alternate. To check the
alternateness, we let Q = P−1 . Then x = Qe1 . We can claim similar for y and z. Thus, the above equations
reduce into
Q t XQe1 = αQ t Qe1 , Q t XQe2 = 𝛾 Q t Qe1 − αQ t Qe2 ,
90 | Ajda Fošner and Bojan Kuzma
Q t XQe3 = 𝛾 ′ Q t Qe1 − αQ t Qe3 .
(︀ )︀
Note that X is alternate precisely when Q t XQ is. So, writing the symmetric matrix Q t Q as Q t Q = s ij , the
above three conditions give
⎛
⎞
αs11 𝛾 s11 − αs12 𝛾 ′ s11 − αs13 *1
⎜
⎟
⎜αs12 𝛾 s12 − αs22 𝛾 ′ s12 − αs23 *2 ⎟
⎟
Q t XQ = ⎜
′
⎜αs13 𝛾 s13 − αs23 𝛾 s13 − αs33 *3 ⎟ ,
⎝
⎠
..
..
..
.
.
.
F
where *i represents the remainder of the i-th row. Clearly, this is an alternate matrix precisely when αs11 =
0 = 𝛾 s12 − αs22 = 𝛾 ′ s13 − αs33 and 𝛾 s11 − αs12 = −αs12 , etc. So, α = 0 = 𝛾 = 𝛾 ′ , unless the upper-left 3 × 3 block
of Q t Q is zero. However, we are free to choose the alternate (n − 3) × (n − 3) lower-right block, represented by
linearly independent alternate matrices in this block, which gives our lower
F. There are precisely (n−3)(n−4)
2
estimate.
Remark 2.9. The upper bound for dim A′ ∩ B′ equals 3 + (n−3)(n−4)
. This is possible precisely when the upper2
left 3 × 3 block of Q t Q is zero. Then, X ∈ A′ ∩ B′ is equivalent to
⎛
⎞
0
0
0
−αs14
...
⎜ 0
0
0
−(𝛾 s14 − αs24 ) . . .⎟
⎜
⎟
⎜
⎟
′
t
0
0
0
−(
𝛾
s
−
αs
)
.
.
.
⎜
⎟.
14
34
Q XQ = ⎜
⎟
′
⎜αs14 𝛾 s14 − αs24 𝛾 s14 − αs34
⎟
⎝
⎠
F
..
..
..
.
.
.
This can happen, for example, if A = (e1 + ie2 ) ∧ (e3 + ie4 ) and B = (e1 + ie2 ) ∧ (e5 + ie6 ) (i.e., A2 = 0 = B2 =
AB = BA).
Lemma 2.10. Let Ω ⊂ Altn be a subset of alternate matrices such that rk A, rk B, rk(A + B) ≤ 2 for every A, B ∈
Ω. Then either Ω ⊆ L ^x for some nonzero vector ^x or there exists invertible P ∈ M n such that Ω ⊆ P(Alt3 ⊕0n−3 )P t
lies in at most three dimensional subspace.
Proof. Let us choose two linearly independent matrices A1 , A2 ∈ Ω and write them as A i = x i ∧ y i , i = 1, 2.
Since rk(A1 + A2 ) ≤ 2 we may assume that x1 = x2 := ^x. Namely, at least one among the triples x1 , y1 , x2 or
x1 , y1 , y2 must be linearly dependent. Without loss of generality, assume the first one is. Write x2 = αx1 + βy1
and note that x1 ∧ y1 = (αx1 + βy1 ) ∧ ( 1α )y1 = x2 ∧ ( 1α )y1 , if α ≠ 0. However, if α = 0, then we may as well
assume that β = 1 and we have x1 ∧ y1 = y1 ∧ (−x1 ) = x2 ∧ (−x1 ). In any case, A1 , A2 ∈ L x2 . Moreover,
with the help of some invertible matrix P we may further assume that ^x = e1 , y1 = e2 , and y2 = e3 , that is,
A1 = e1 ∧ e2 ∈ L e1 and A2 = e1 ∧ e3 ∈ L e1 . Suppose that there exists a matrix A3 = x3 ∧ y3 ∈ Ω \ L e1 . Due to
rk(A1 + A3 ) ≤ 2, either x3 or y3 is a linear combination of e1 , e2 . Without loss of generality we may assume
the former and write x3 = αe1 + βe2 . Clearly, β ≠ 0, otherwise A3 ∈ L e1 . Similarly, rk(A3 + A2 ) ≤ 2 implies
that at least one among x3 and y3 is a linear combination of e1 , e3 . But for x3 this is already impossible, so
y3 = 𝛾 e1 + δe3 . Consequently, A3 ∈ Alt3 ⊕0n−3 whenever A3 ∈ Ω \ L e1 .
It remains to show that Ω ⊆ Alt3 ⊕0n−3 . In fact, pick any matrix A4 ∈ Ω and assume erroneously that
it is linearly independent of A1 , A2 , A3 . Then, as before, due to rk(A2 + A4 ), rk(A1 + A4 ) ≤ 2 we must have
A4 ∈ L e1 . So, A4 = e1 ∧ y4 , where y4 is linearly independent of e1 , e2 , e3 . But then, A3 + A4 = ( αβ e1 + e2 ) ∧
(β𝛾 e1 + βδe3 ) + e1 ∧ y4 is of rank four, a contradiction.
For nonzero vectors x, y ∈ Cn , we denote [x] = {αx : α ∈ C} and [x] + [y] = {αx + βy : α, β ∈ C}.
Lemma 2.11. If x, y are linearly independent vectors, then [z] ⊆ [x] + [y] precisely when the intersection (L x ∩
L z ) ∩ (L y ∩ L z ) is nonzero.
Preserving zeros of Lie product on alternate matrices |
91
Proof. Suppose [z] ⊆ [x] + [y]. Then z = αx + βy for some scalars α, β. If β = 0 then L z = L x and (L x ∩ L z ) ∩
(L y ∩ L z ) = L x ∩ L y = C(x ∧ y) ≠ 0. The same is true when α = 0. Now, if α and β are both nonzero, then
(L x ∩ L z ) = C(x ∧ z) = Cx ∧ (αx + βy) = C(x ∧ y) and (L y ∩ L z ) = Cy ∧ (αx + βy) = C(y ∧ x) = C(x ∧ y) ≠ 0.
Now, assume that (L x ∩ L z ) ∩ (L y ∩ L z ) ≠ 0. Then λ(x ∧ z) = µ(y ∧ z) for some nonzero scalars λ, µ. If
x, y, z were linearly independent, there would exist an invertible P such that Px = e1 , Py = e2 , and Pz = e3 .
Multiplying the above identity with P on the left and P t on the right side yields that λ(e1 ∧ e3 ) = µ(e2 ∧ e3 ),
a contradiction.
3 Proof of the main theorem
Let ϕ : Altn → Altn be an injective continuous map that preserves zeros of Lie product. Then for every
A ∈ Altn , we have
(︀
)︀′
ϕ(A′ ) ⊆ ϕ(A) .
In particular,
ϕ(Altn ) = ϕ(0′ ) ⊆ ϕ(0)′ ⊆ Altn .
Since ϕ is injective and continuous, it follows by the invariance of domain theorem that ϕ(0)′ cannot be
contained inside a proper linear subspace of Altn . This yields that ϕ(0)′ = Altn and, consequently, ϕ(0) = 0.
The remainder of the proof is divided into several steps.
Step 1. The set Ξ2 ⊆ Altn of rank two alternate matrices is invariant under ϕ. Moreover, ϕ(CA) ⊆ Cϕ(A)
for every A ∈ Ξ2 .
Let 0 ≠ A ∈ Altn . By Lemma 2.5 dim A′ ≥ (n−2)(n−3)
+ 1 if and only if rk A = 2. This yields that every
2
A ∈ Altn of rank two is mapped by ϕ into an alternate matrix of rank two. Namely otherwise, ϕ would map
A′ , which is a linear subspace of dimension at least (n−2)(n−3)
+ 1, continuously and injectively into ϕ(A)′ ,
2
(n−2)(n−3)
whose dimension is strictly smaller than
+ 1. But this is impossible by the invariance of domain
2
theorem. Therefore, for every A ∈ Ξ2 of rank two there exists B ∈ Ξ2 such that ϕ(A) = B.
It remains to prove ϕ(CA) ⊆ CB. If there was a nonzero complex number λ such that ϕ(λA) ∉ CB, then ϕ
would map A′ = (λA)′ injectively and continuously into B′ ∩ ϕ(λA)′ . Now, any X ∈ B′ ∩ ϕ(λA)′ also commutes
with B + ϕ(λA). Hence, if rk(B + ϕ(λA)) ≥ 4, then, by Lemma 2.5, dim B′ ∩ ϕ(λA)′ ≤ dim(B + ϕ(λA))′ (n−2)(n−3)
+ 1 ≤ dim A′ . However, if rk(B + ϕ(λA)) = 2, then Remark 2.9 implies dim(B′ ∩ ϕ(λA)′ ) dim A′ .
2
Each of these two cases contradicts Brouwer’s invariance of domain theorem.
Step 2. Let x be a nonzero vector. Then, rank(ϕ(x ∧ y1 ) + ϕ(x ∧ y2 )) ≤ 2 for each x ∧ y1 , x ∧ y2 ∈ L x .
Indeed, write ϕ(x ∧ y1 ) = x̃1 ∧ ỹ1 and ϕ(x ∧ y2 ) = x̃2 ∧ ỹ2 . Note that
(︀
)︀
ϕ (x ∧ y1 )′ ∩ (x ∧ y2 )′ ⊆ (x̃1 ∧ ỹ1 )′ ∩ (x̃2 ∧ ỹ2 )′ .
Now, suppose erroneously that rk(x̃1 ∧ ỹ1 + x̃2 ∧ ỹ2 ) > 2. Then, by Lemma 2.8, the space (x ∧ y1 )′ ∩ (x ∧ y2 )′
would be mapped injectively and continuously into a subspace (x̃1 ∧ ỹ1 )′ ∩
of dimension at least (n−3)(n−4)
2
(x̃2 ∧ ỹ2 )′ ⊆ (x̃1 ∧ ỹ1 + x̃2 ∧ ỹ2 )′ which by Lemma 2.6 has a strictly smaller dimension, a contradiction to the
invariance of domain theorem. Therefore, rk(x̃1 ∧ ỹ1 + x̃2 ∧ ỹ2 ) ≤ 2.
Step 3. For each L x there exists some L x̃ such that ϕ(L x ) ⊆ L x̃ .
In fact, if this would not be the case for some x, then the previous step together with Lemma 2.10 would
imply that ϕ maps the n −1 dimensional space L x continuously and injectively into at most three dimensional
subspace, a contradiction.
Step 4. Let x be a nonzero vector such that ϕ(L x ) ⊆ L x̃ ∩ L x̃′ for some vectors x̃ and x̃′ . Then x̃ and x̃′ must
be linearly dependent.
To see this, note that ϕ(L x ) ⊆ L x̃ implies dim(Lin ϕ(L x )) = dim L x̃ = n − 1 by the invariance of domain
theorem (here, Lin ϕ(L x ) denotes the linear span of the set ϕ(L x )). But then ϕ(L x ) ⊆ L x̃ ∩ L x̃′ implies that
L x̃ = Lin ϕ(L x ) = L x̃′ which is possible precisely when x̃, x̃′ are linearly dependent.
92 | Ajda Fošner and Bojan Kuzma
We can hence introduce a well-defined map φ on a projective space PCn by letting
φ : [x] ↦→ [x̃],
where x̃ satisfies ϕ(L x ) ⊆ L x̃ . To prove that this is a projectivity, assume that [z] ⊆ [x] + [y]. We can clearly
suppose that x, y are linearly independent, otherwise [x] = [y] = [z] and, hence, there is nothing to prove. By
Lemma 2.11, we must have (L x ∩ L z ) ∩ (L y ∩ L z ) ≠ {0}. Since ϕ is injective and ϕ(0) = 0, this further implies
(︀
)︀
{0} ≠ ϕ (L x ∩ L z ) ∩ (L y ∩ L z ) ⊆ (ϕ(L x ) ∩ ϕ(L z )) ∩ (ϕ(L y ) ∩ ϕ(L z )).
Thus, by Lemma 2.11, φ([z]) ⊆ φ([x]) + φ([y]).
Step 5. The map φ is injective.
Let x1 , x2 be two linearly independent vectors and suppose that φ([x1 ]) = φ([x2 ]) = [x̃] for some x̃.
Then, by the definition of φ and the invariance of domain theorem, ϕ(L x1 ) and ϕ(L x2 ) are open subsets of
L x̃ containing ϕ(0) = 0. Let x1 ∧ y ∈ L x1 ∖L x2 . As limλ→0 ϕ(λx1 ∧ y) = ϕ(0) = 0, we can find a small enough
nonzero complex number λ such that ϕ(λx1 ∧ y) ∈ ϕ(L x2 ), contradicting the injectivity of ϕ. Thus, φ is
injective.
Step 6. The span of lines in Im φ equals PCn . Hence, Im φ is not contained in a projective line.
To prove this, denote [ẽ1 ] := φ([e1 ]). Note that L e1 and L ẽ1 are both n−1 dimensional linear subspaces and
ϕ maps continuously and injectively the first one into the second one. So, by the invariance of domains, ϕ(L e1 )
is an open subset inside L ẽ1 that contains a zero matrix. Therefore, given any vector z, linearly independent
of ẽ1 , there exists some y such that ϕ(e1 ∧ y) = λẽ1 ∧ z for small enough nonzero scalar λ. Clearly, e1 ∧ y ≠ 0,
because ϕ(0) = 0. Therefore, e1 , y are linearly independent. Now, e1 ∧ y ∈ L e1 ∩ L y . Thus, λẽ1 ∧ z = ϕ(e1 ∧ y) ∈
ϕ(L e1 ∩ L y ) ⊆ ϕ(L e1 ) ∩ ϕ(L y ) ⊆ L ẽ1 ∩ L ỹ . But this is possible only when ỹ = αẽ1 + βz for some scalars α, β.
Moreover, β ≠ 0 because linear independence of e1 , y and injectivity of φ forces L ỹ ≠ L ẽ1 . We may now
assume that β = 1. So, given any z, linearly independent from ẽ1 , there is some y ≠ 0 and a scalar α such that
φ([y]) = [αẽ1 + z]. But then the image of φ spans PCn : it clearly contains [ẽ1 ] and it also contains [α i ẽ1 + e i ]
for every i (except for one possible exception, when ẽ1 , e i are linearly dependent). But then this n lines (or
n + 1 lines, if ẽ1 , e i are always independent) span the whole projective space.
^ and a field homomorphism σ : C → C such that for each
Step 7. There exists an invertible matrix Q
^ σ ∧ y σ )Q
^ t for appropriate scalar λ x∧y ∈ C.
nonzero alternate x ∧ y we have ϕ(x ∧ y) = λ x∧y Q(x
By a nonsurjective version of the fundamental theorem of projective geometry (see e.g. [7, Theorem 3.1])
there exists a field homomorphism σ : C → C and a σ-linear map T : Cn → Cn such that φ([x]) = [Tx]
for every line [x] ∈ PCn . Consequently, ϕ(L x ) ⊆ L Tx . Thus, from x ∧ y ∈ L x ∩ L y we derive that ϕ(x ∧ y) ∈
ϕ(L x ∩ L y ) = ϕ(L x ) ∩ ϕ(L y ) ⊆ C (Tx ∧ Ty). It follows that ϕ(x ∧ y) = 1β (Tx) ∧ Ty, where a scalar β depends
^ σ for some matrix Q
^ ∈ M n . Since Lin(Im φ) = Cn , the
on the operator x ∧ y. It is also easy to see that Tx = Qx
σ
σ
σ
σ
t
1
1
^ is invertible. Moreover, ϕ(x ∧ y) = Qx
^ ∧ Qy
^ = Q(x
^
^ .
matrix Q
∧ y )Q
β
β
Step 8. Homomorphism σ is either a complex conjugation or the identity.
^ 1 ∧ (σ(α)e2 + e3 ))Q
^ t for appropriate nonzero scalars µ α .
To see this, consider ϕ(e1 ∧ (αe2 + e3 )) = µ α Q(e
The continuity of ϕ forces the continuity of the functions µ α σ(α) and µ α . This gives that σ is a continuous
homomorphism of C. Thus, it is either a complex conjugation or the identity. Replacing ϕ by the map X ↦→
ϕ(X), X ∈ Altn , if necessary, we may assume in the sequel that σ is the identity.
^ tQ
^ is a scalar multiple of the identity matrix.
Step 9. The matrix Q
To see this, choose any orthogonal matrix W. Given any distinct i, j, k, l, observe that W(e i ∧ e j )W t commutes with W(e k ∧ e l )W t . Hence,
0 = ϕ(W(e i ∧ e j )W t )ϕ(W(e k ∧ e l )W t ) − ϕ(W(e k ∧ e l )W t )ϕ(W(e i ∧ e j )W t )
(︀
t ^t ^
t ^t
t ^t ^
t ^ t )︀
^
^
= µ QW(e
i ∧ e j )W Q QW(e k ∧ e l )W Q − QW(e k ∧ e l )W Q QW(e i ∧ e j )W Q ,
where µ = λ W(e i ∧e j )W t λ W(e k ∧e l )W t ≠ 0. In particular,
t ^t ^
^ t QW(e
^
(e i ∧ e j )W t Q
k ∧ e l ) = (e k ∧ e l )W Q QW(e i ∧ e j ).
Preserving zeros of Lie product on alternate matrices |
93
^ t QW
^ corresponding to
Comparing the (i, k), (i, l), (j, k), and (j, l) positions, we derive that the entries of W t Q
t ^t ^
those positions vanish. Due to arbitrariness of i, j, k, l, we see that W Q QW is diagonal for
(︃ every
(︃ orthogonal
)︃)︃
1
1
t
1
^ Q
^ is diagonal. Furthermore, choosing W = √
W. Choosing W = Idn , we conclude that Q
⊕
2 −1 1
^t ^
Idn−2 , we (︃
see that
(︃ the entries
)︃)︃ at the positions (1, 1) and (2, 2) of Q Q are equal. Choosing in this succession
1 1
^ tQ
^ are equal.
⊕ Idn−2−i for i = 1, . . . , n − 2, we derive that all diagonal entries of Q
W = 0i ⊕ √1
2 −1 1
^ tQ
^ = λ Idn for some nonzero scalar λ, as desired. Therefore, Q := √1 Q
^ is an orthogonal
This implies that Q
λ
^ Q
^ t = µ X QXQ t for every rank two alternate matrix X.
matrix with ϕ(X) = µ X QX
λ
Step 10. According to above observations, we may replace ϕ by the map X ↦→ Q t ϕ(X)Q. The new map
is still continuous, preserves zeros of Lie product, and fixes every alternate matrix of minimal rank modulo
scalar multiples.
Step 11. If D is a diagonalizable alternate matrix then there exists an odd polynomial p D (λ) (i.e., p D (λ) =
−p D (−λ)) such that ϕ(D) = p D (D) (i.e., ϕ acts locally polynomially on diagonalizable alternate matrices).
To see this, let D be a diagonalizable alternate matrix. It follows from (5) that there exists an orthogonal
matrix W such that
{︃
W(α1 J) ⊕ (α2 J) ⊕ · · · ⊕ (α k J)W t ;
n even
D=
(︀
)︀ t
W (α1 J) ⊕ (α2 J) ⊕ · · · ⊕ (α k J) ⊕ 0 W ; n odd
(︃
)︃
0 1
for some scalars α i , where J is a 2 × 2 matrix of the form J =
. Clearly, D commutes with a minimal
−1 0
rank matrix A i = W(02i ⊕ J ⊕ 0n−2i−2 )W t for every i = 1, . . . , ⌊n/2⌋. Consequently, also ϕ(D) commutes with
ϕ(A i ) = 𝛾i A i ≠ 0, which forces ϕ(D) = W((β1 J) ⊕ (β2 J) ⊕ · · · ⊕ (β k J))W t , respectively,
(︃
)︃ ϕ(D) = W((β1 J) ⊕
J
J
(β2 J) ⊕ · · · ⊕ (β k J) ⊕ 0)W t for some scalars β i . Take a block alternate matrix X =
⊕ 0 and note that this
J J
matrix is of minimal rank. If, say α1 = α2 , then X commutes with D. Hence, also its image, ϕ(X) = 𝛾X X ≠ 0,
commutes with ϕ(D) which is possible only when β1 = β2 . Likewise we show that α i = α j forces β i = β j ,
i, j = 1, . . . , k.
It is known that there exists a polynomial p D (λ) with p D (α i ) = β i for i = 1, . . . , n. Hence, ϕ(D) = p D (D).
In other words, ϕ acts locally as polynomial on diagonalizable alternate matrices. Moreover, we will show
that these polynomials can be chosen to be odd (recall that a polynomial f (λ) is odd if f (−λ) = −f (λ), or
∑︀
equivalently, if f (λ) = i a2i+1 λ2i+1 for some scalars a2i+1 ∈ C). In fact, suppose that A and p(A) are both
alternate for some polynomial p(λ). Decompose p(λ) = p+ (λ)+ p− (λ) into a sum of even and odd parts. Clearly,
A2i are symmetric matrices for each nonnegative integer i. So, p+ (A) is a linear combination of symmetric
matrices and, hence, it is symmetric. Likewise one can show that p− (A) is alternate. So, the symmetric matrix
p+ (A) = p(A)−p− (A) is a difference of two alternate matrices p(A) and p− (A), whence p+ (A) = 0 = p(A)−p− (A).
Thus, we can assume that p(λ) = p− (λ) is an odd polynomial, i.e., p(−λ) = −p(λ).
Step 12. ϕ(X) ∈ X ′ for every alternate X.
In fact, by Step 11 we know that ϕ is a locally polynomial map on the set of all diagonalizable alternate
matrices. In particular, every diagonalizable alternate matrix A commutes with ϕ(A). According to Lemma
2.7, the set of all diagonalizable alternate matrices is dense in Altn . Thus, by the continuity of ϕ, every X ∈ Altn
commutes with ϕ(X). In particular ϕ(X) ∈ X ′ , and, if alternate X is nonderogatory, then ϕ(X) ∈ X ′ = Poly(X) ∩
Altn [12, Theorem 3.2.4.2.].
Step 13. Action of ϕ on a general alternate A.
As explained in (5), there exists an orthogonal Q such that
A = Qt
n1
(︁⨁︁
i=1
i pi )
K (p
⊕
λi
n2
⨁︁
)︁
(︀
)︀
K (q j ) Q = Q t A1 ⊕ A2 Q.
j=1
To ease the proof, we will continue in substeps. Also, we will temporarily modify ϕ into a map
ϕ Q (X) := Qϕ(Q t XQ)Q t ,
(12)
94 | Ajda Fošner and Bojan Kuzma
which clearly shares the same properties as ϕ. In particular, all steps until the present one hold for ϕ Q as well.
The benefit of introducing ϕ Q is that ϕ(A) is a polynomial in A if and only if ϕ Q (A1 ⊕ A2 ) is a polynomial in
block-diagonal A1 ⊕ A2 .
Substep 13.a. There exist odd polynomials f1 (λ), . . . , f n1 (λ) and odd polynomials h1 (λ), . . . , h n2 (λ) such that
n1
(︁⨁︁
ϕ Q (A1 ⊕ A2 ) =
i pi )
f i (K (p
)⊕
λi
i=1
n2
⨁︁
)︁
h j (K (q j ) )
(13)
j=1
To see this, choose continuous functions λ1 (ε), . . . , λ n1 (ε), nonzero and in absolute value pairwise distinct at each fixed ε > 0, such that limε↘0 λ i (ε) = λ i for i = 1, . . . , n1 . Define
A1 (ε) :=
n1
⨁︁
i pi )
,
K (p
λ (ε)
(︀
)︀
A(ε) := A1 (ε) ⊕ A2 .
i
i=1
By its definition (2) we see that, at each fixed ε > 0, every diagonal block of A1 (ε) is a nonderogatory matrix
with eigenvalues {−λ i (ε), λ i (ε) : i = 1, . . . , n1 }. Hence, due to 0 ≠ |λ i (ε)| ≠ |λ j (ε)| for i ≠ j, we derive that A(ε)
is nonderogatory at each fixed ε > 0 and so by Step 12 and the fact that (X1 ⊕ X2 )′ = X1′ ⊕ X2′ whenever block
matrices X1 , X2 have no eigenvalue in common (see Lemma 2.1 or [8, Theorem 2 and Remark on p. 127]), we
have:
n1
)︁
(︀
)︀ (︁⨁︁
′
i pi ) ′
ϕ Q (A(ε)) ∈ A(ε)′ = A1 (ε)′ ⊕ A′2 =
(K (p
)
⊕
A
(14)
2 .
λ (ε)
i
i=1
i pi ) ′
i pi ) ′
i pi )
Since λ i (ε) ≠ 0 for ε > 0, the equation (9) gives (K (p
) = (K (p
) . Also, the matrices K (p
are nonderogaλ=1
λ=1
λ i (ε)
tory, and therefore
i pi ) ′
i pi )
i pi )
(K (p
) = Poly(K (p
) ∩ Altn = Poly(K (p
) ∩ Altn .
λ=1
λ=1
λi
Combined with (14) we derive
ϕ Q (A(ε)) ∈
n1
(︁⨁︁
)︁
′
i pi )
Poly(K (p
)
⊕
A
;
2
λi
ε > 0.
i=1
Hence, by continuity and the fact that the set on the right side of the above equation is closed,
ϕ Q (A1 ⊕ A2 ) = lim ϕ Q (A(ε)) ∈
n1
(︁⨁︁
ε↘0
)︁
i pi )
Poly(K (p
) ⊕ A′2 .
λi
(15)
i=1
Since ϕ Q (A1 ⊕A2 ) is alternate, there are hence odd polynomials (see last paragraph of Step 11) f1 (λ), . . . , f n1 (λ)
such that
n1
(︁⨁︁
)︁
i pi )
ϕ Q (A1 ⊕ A2 ) ∈
f i (K (p
) ⊕ A′2 .
(16)
λi
i=1
⨁︀ 2 (q j )
Likewise, it follows from (4) that the block constituents of A2 = nj=1
K , i.e., matrices K (q j ) , are non(︀
)︀
⨁︀
′
derogatory. Hence, if X ∈ A′2 =
K (q j ) is partition into blocks conformally to A2 then its j-th diagonal
block satisfies
X jj ∈ (K (q j ) )′ = Poly(K (q j ) ) ∩ Altq j .
(17)
Moreover, recall from the proof of Lemma 2.7 that the matrices
A ε :=
n1
(︁⨁︁
i=1
where
i pi )
X (p
⊕
ε
n2
⨁︁
(q j )
Xε
)︁
,
⎧
⎪
X (1)
⎪
ε := 0,
⎪
⎪
√
⎪
⎨X (q j ) := K (q j ) + ε (Id −√−1V)(︀E
qj
ε
(q j −1) 1 − E q j 2 )(Idq j + −1V),
2
i pi )
i pi )
⎪
X (p
:= K (p
+ ε(L(pp) )p−1 ,
for p even
⎪
ε
⎪
λ i +ε
⎪
⎪
⎩X (p i p i ) := K (p i p i ) + ε(L(pp) )p−2 ,
for p odd
ε
λ i +ε
(18)
j=1
qj ≥ 3
,
Preserving zeros of Lie product on alternate matrices |
95
(L(pp) is defined by equation (11)) are alternate, diagonalizable, and converge to A1 ⊕ A2 as ε → 0. By Step 11,
∑︀
ϕ Q (A ε ) are odd polynomials (i.e., p(λ) =
a2i+1 λ2i+1 ) in A ε and, by the continuity of ϕ Q , they converge to
ϕ Q (A1 ⊕ A2 ) as ε → 0. Note that a polynomial in a block-diagonal matrix A ε is itself a block-diagonal matrix.
Therefore, ϕ Q (A1 ⊕ A2 ) is also a block-diagonal alternate matrix. It then follows from (16)–(17) that there exist
odd polynomials h1 (λ), . . . , h n2 (λ) which satisfied the desired equation (13).
Substep 13.b. To finish the proof that ϕ Q (A1 ⊕ A2 ) is a polynomial in A1 ⊕ A2 we have to show that whenever
in equation (13) we have (a) λ i = ±λ i′ or (b) λ i = 0 then there exists a common polynomial f (λ) such that,
under (a),
(p p )
(p p )
i pi )
i pi )
f i (K (p
) ⊕ f i′ (K λ ′ i′ i′ ) = f (K (p
) ⊕ f (K λ ′ i′ i′ )
λi
λi
i
i
respectively, under (b)
i pi )
i pi )
f i (K (p
) ⊕ h j (K (q j ) ) = f (K (p
) ⊕ f (K (q j ) ).
λi
λi
Also, we have to show that there exists a common polynomial h(λ) such that
this is shown the proof ends because it is well-known that
⨁︀
h j (K (q j ) ) =
⨁︀
h(K (q j ) ). Once
Poly(X1 ⊕ X2 ) = Poly(X1 ) ⊕ Poly(X2 )
whenever matrices X1 , X2 have no eigenvalue in common.
We thus have to observe three cases.
(p p i′ )
i pi )
Substep 13.c. Assume first that K (p
and K λ ′ i′
λi
i
(p i p i )
K−λ
i
is orthogonally similar to
i pi )
K (p
.
λi
whenever
share the same eigenvalue, that is, λ i = ±λ i′ . By (3) the block
Hence we may assume that in (12),
λ i = ±λ i′
then actually
λ i = λ i′ .
For the sake of simplicity, we will also assume that p i ≤ p i′ and that i = 1 and i′ = 2, that is, the first two
1 p1 )
diagonal blocks in (13) share the same eigenvalue. Then, by the definition of K (p
(see (2)), we see that
λ1
A1 ⊕ A2 commutes with alternate block matrices
(︃
)︃
0
Xk
Tk =
⊕ 0n−2p1 −2p2 ,
k = 1, 2,
−X kt
0
where
√
(︃
[0|I]
X1 = Id2p1 − −1V2p1 ·
0p1 ×p2
(︃
√
(︀
)︀
[0|0]
X2 = 12 Id2p1 − −1V2p1 ·
0p1 ×p2
1
2
(︀
)︀
0p1 ×p2
[0|0]
)︃
√
)︀
(︀
· Id2p2 + −1V2p2 ,
0p1 ×p2
[0|I]
)︃
√
(︀
)︀
· Id2p2 + −1V2p2 ,
and I denotes the identity matrix of the size p1 × p1 . Assume that p1 = p(︃
2 . In this case, let
)︃ X = X1 + X2 =
√
√
(︀
)︀
(︀
)︀
0
Id2p1
1
⊕ 0n−4p1 . Then
2 Id2p1 − −1V 2p1 Id2p1 Id2p1 + −1V 2p1 = Id2p1 and T = T 1 + T 2 =
− Id2p1
0
T 3 = −T. In particular, T is diagonalizable. Thus, ϕ Q maps T into an odd polynomial in T. But due to T 3 = −T
it is fixed by ϕ Q , modulo multiplication by a nonzero scalar. We infer that ϕ Q (A1 ⊕ A2 ) must also commute
with T. Hence, the first two diagonal blocks of ϕ Q (A1 ⊕ A2 ) are intertwined with a matrix X. Since X = Id2p1
1 p1 )
2 p2 )
we get that, in identity (13), f1 (K (p
) = f2 (K (p
). Since p2 = p1 and λ1 = λ2 we can hence take f (λ) = f1 (λ)
λ1
λ2
in this case.
On the other hand, if p1 p2 , then we have T13 = −T1 and T23 = −T2 so, by the above, T1 and T2 are
both fixed by ϕ Q , modulo scalars. Since ϕ Q preserves commutativity, ϕ Q (A1 ⊕ A2 ) commutes with every
linear combination of ϕ Q (T1 ) ∈ CT1 and ϕ Q (T2 ) ∈ CT2 . In particular, ϕ Q (A1 ⊕ A2 ) also commutes with
96 | Ajda Fošner and Bojan Kuzma
(︀
T = (T1 + T2 ). By taking the first two blocks from (13) and using definition (2) we see that f1 (λ1 Idp1 +J p1 ) ⊕
)︀
(︀
)︀
(−λ1 Idp1 −J p1 ) ⊕ f2 (λ2 Idp2 +J p2 ) ⊕ (−λ2 Idp2 −J p2 ) commutes with
(︃
)︃⎞
⎛
[0|I]
0
0
⎟
⎜
0
[0|I] ⎟
⎜
⎟
(︃
)︃
T1 + T2 = ⎜
t
⎜
⎟
[0|I]
0
⎝
⎠
−
0
0
[0|I]
(︃
)︃
(︀
)︀
[0|I]
0
Again,
intertwines
blocks
f1 (λ1 Idp1 +J p1 )
⊕
(−λ1 Idp1 −J p1 )
and
0
[0|I]
(︀
)︀
f2 (λ2 Idp2 +J p2 ) ⊕ (−λ2 Idp2 −J p2 ) . It is straightforward that
f1 (λ1 Idp1 +J p1 ) [0|I] = [0|I] f2 (λ2 Idp2 +J p2 );
λ2 = λ1 .
Since polynomials f1 (λ), f2 (λ) are odd and since we can assume that the degree of f i (λ) is smaller than the
degree of minimal polynomial of corresponding block, i.e., smaller than p i , we can write
⎛
⎞
f1 (λ1 )
0
α1 0 α2 . . . . . . .
⎜
⎟
f1 (λ1 ) 0 α1 0 α2 . . .⎟
f1 (λ1 Idp1 +J p1 ) = ⎜
⎝
⎠
..
..
..
.
.
.
for suitable scalar α i and likewise for f2 (λ1 Idp2 +J p2 ). It then easily follows that f1 (λ) and f2 (λ) coincide in the
first min{p1 , p2 } = p1 monomials. Hence, the present substep is done by taking f (λ) = f2 (λ).
Substep 13.d. From Substep 13.c it follows that ϕ fixes every square-zero alternate matrix, modulo scalars,
(︀⨁︀ (22)
)︀
because such matrices take the form O
K λ=0 ⊕ 0 O t for some orthogonal O.
Substep 13.e. Next, assume that A1 ⊕ A2 contains two blocks K (q j ) and K (q j′ ) . As above, we may assume that
q j ≤ q j′ and that j = 1 and j′ = 2. Moreover, for simplicity we will temporarily permute the diagonal blocks so
that, in the present substep only,
A1 ⊕ A2 = K (q1 ) ⊕ K (q2 ) ⊕ . . . .
If q1 = q2 , then A1 ⊕ A2 commutes with an alternate block matrix
)︃
(︃
0
Idq1
⊕ 0n−q1 −q2
T=
0
− Idtq1
which satisfies T 3 = −T. Hence, T is fixed by ϕ Q , modulo multiplication by a nonzero scalar. As in the first
case, it is easy to see that the first two blocks B1 = h1 (K (q1 ) ) and B2 = h2 (K (q2 ) ) of ϕ Q (A1 ⊕ A2 ) coincide.
Hence, B1 ⊕ B2 = h(K (q1 ) ) ⊕ h(K (q2 ) ) where h(λ) = h1 (λ).
(︃
)︃
0 D1
(q1 ) d
On the other hand, assume q2 = q1 + 2d for some integer d ≥ 1. Then, by Lemma 1.2, (J ) =
,
0 0
where D1 is a diagonal (q1 − d) × (q1 − d) matrix with ±1 on the diagonal. Clearly, if d ≥ q1 , then (J (q1 ) )d = 0.
Similarly,
(︃
)︃
0 D2
(q2 ) d
(J ) =
,
(19)
0 0
where D2 is a diagonal (q2 − d) × (q2 − d) matrix with ±1 on the diagonal. Obviously, D2 is a nonzero matrix
since q2 > d. Moreover, if we write q1 = 2k + 1, then q2 = q1 + 2d = 2(k + d) + 1, and, by formula in Lemma 1.2,
(︀
)︀
D2 = Idk+1 ⊕ diag −1, (−1)2 , . . . , (−1)d−1 ⊕ (−1)d Idk+1 .
(20)
Now it easily follows that, if we remove from D2 the first d and the last d rows and columns, we obtain D1 .
More precisely, D1 is a compression of the matrix D2 onto the subspace generated by {e d+1 , ..., e q1 } and
D1 = 0 if d ≥ q1 .
Preserving zeros of Lie product on alternate matrices |
If d is odd, then we define the matrix
(︃(︃ √ (︀
)︀d
− −1 J (q1 )
−1
T=S
^ t V q1
−V q2 X
^
X
√ (︀ (q ) )︀d
−1 J 2
)︃
97
)︃
⊕ 0n−q1 −q2
S,
√
(︀
)︀
^ = 0q ×2d |D
^ ,D
^ is the compression of −D2 onto the last q1 basis vectors, and S = (Idq1 + −1V q1 ) ⊕
where X
1
√
^ = −P(J (q2 ) )d , where P : Cq2 → Cq1 is a projection
(Idq2 + −1V q2 ) ⊕ Idn−q1 −q2 . Observe that (19) implies X
^ (q2 ) = P(J (q2 ) )d+1 maps basis vector e i into 0 for
onto coordinates (d + 1) through (q2 − d). By Lemma 1.2, XJ
q1
i = 1, . . . , (2d + 1) and into −(±)i e i−2d−1 ∈ C for i = (2d + 2), . . . , (2d + q1 ) = q2 , where
⎧
d+1
⎪
i > q2 − k
⎪
⎨(−1) ;
(±)i =
(−1)d+k+1−i ;
⎪
⎪
⎩1;
q2 − (d + k) < i ≤ q2 − k .
otherwise
̂︁ e
^ i = 0 for 1 ≤ i ≤ 2d and Xe
^ i = −(±)
On the other hand, Xe
i i−2d for i ≥ 2d + 1, where
⎧
d
⎪
i > q2 − k − 1
⎪
⎨(−1) ;
̂︁
d+k−i−1
(±)i = (−1)
; q2 − (d + k) − 1 < i ≤ q2 − k − 1 .
⎪
⎪
⎩1;
otherwise
Apply J (q1 ) and use the fact that J (q1 ) e i−2d = e i−2d−1 if i − 2d ≤ q12−1 and J (q1 ) e i−2d = −e i−2d−1 if i − 2d >
is then a straightforward calculation that
^ = XJ
^ (q2 ) .
J (q1 ) X
q1 −1
2 .
It
(21)
Next, it is easy to see that T is alternate. Assume first d < q1 . Then the block structure of STS−1 equals
d
⎛
q1 − d
0
⎜ 0
d
⎜
⎜
⎜ (D′′ )f
d
⎜ 2
⎜
q1 − d ⎜ 0
⎜
⎝ 0
d
d
0
q1 − d
d
d
^2
− −1D
0
0
0
0
0
√
0
^ 2 )f
(D
0
0
0
0
0
0
q1 − d
^2
−D
0
√
−1D′2
0
0
0
0
^2
−1D
0
0
√
d
⎞
0
−D′′2 ⎟
⎟
⎟
0 ⎟
⎟ ⊕ 0n−q1 −q2 .
⎟
0 ⎟
√
⎟
−1D′′2 ⎠
0
∑︀
^2 ∈
Here, X f = VX t V, V = i E i (n−i+1) , is a flip map, i.e., reflection over anti-diagonal, and D′2 , D′′2 ∈ M d , D
′
′′
^
M q1 −d are appropriate diagonal matrices such that D2 = D2 ⊕ D2 ⊕ D2 . It follows from (20) that
^ 2 )f = − D
^ 2.
(D
We claim that T 2 = 0. In fact, STS−1 maps the basis vectors e q1 +1 , . . . , e q1 +d already into zero. It further maps
the vectors e k , k = 1, . . . , d, into ±e q1 +k which are annihilated by another application of STS−1 . Moreover,
√
it maps e q1 +d+k into ± −1e q1 +k for k = 1, . . . , d, which are annihilated by the second application of STS−1 .
√
Next, STS−1 maps the vectors e d+k , k = 1, . . . , q1 − d into ±( −1e k + e q1 +d+k ), which are further mapped by
STS−1 into
√
√
√
±( −1STS−1 e k + STS−1 e q1 +d+k ) = ±(− −1e k+q1 + −1e q1 +k ) = 0.
√
√
√
Finally, the vectors e q1 +2d+k , k = 1, . . . , q1 are mapped to STS−1 e q1 +2d+k = ±(e k − −1e q1 +d+k ) = ± −1( −1e k +
e q1 +d+k ), which, as we showed above, are annihilated by the second application of STS−1 . This shows T 2 = 0
if d < q1 . By the similar analysis we can show that T 2 = 0 also when d ≥ q1 (the block structure of STS−1 is
^ 2 = 0). Consequently, by Substep 13.d,
same as before except that now, D
ϕ Q (T) ∈ CT.
98 | Ajda Fošner and Bojan Kuzma
(q1 )
⊕ J (q2 ) commutes with
(︃ We next prove
)︃ that A1 ⊕ A2 commutes with T. It suffices to show that J
^
0
X
^ = XJ
^ (q2 ) and J (q2 ) V q2 X
^ t V q1 = V q2 X
^ t V q1 J (q1 ) . The
. This is equivalent to the fact that J (q1 ) X
t
^
−V q2 X V q1 0
first equality follows from (21) while the second one is equivalent to the first one which is seen by applying
the bijective operation Z ↦→ (V q1 ZV q2 )t to the first equality, and in the resulting
^ q2 )t = (V q1 (XJ
^ (q2 ) )V q2 )t
(V q1 (J (q1 ) X)V
^ q2 )t = (V q1 J (q1 ) V q1 · V q1 XV
^ q2 )t = (V q1 XV
^ q2 )t · (V q1 J (q1 ) V q1 )t =
rewrite the left-hand side into (V q1 J (q1 ) XV
t
t
(q1 )
^
(V q2 X V q1 ) · (−J ) and likewise rewrite the right-hand side.
It follows that ϕ Q (A1 ⊕ A2 ) commutes
Poly(T) = CT as well. This is equivalent to the fact
(︃ with ϕ Q (T) ∈ )︃
^
0
X
that h1 (J (q1 ) ) ⊕ h2 (J (q2 ) ) commutes with
for appropriate odd polynomials h1 (λ) and h2 (λ)
^ t V q1 0
−V q2 X
^ =
of degrees smaller than q1 or q2 respectively. Actually, this is further equivalent to the fact that h1 (J (q1 ) )X
(q2 )
(q1 ) ^
(q2 )
^
^
Xh2 (J ). Recall that J X = XJ , hence it easily follows that for each s = 0, 1, . . . , q1 , we have
(︀
)︀
^ = 0q ×2d |W s = X(J
^ (q2 ) )s ,
(J (q1 ) )s X
1
where W s ∈ M q1 is strictly upper-triangular matrix with ±1 on s-th super diagonal and zeros elsewhere. Thus,
writing
h1 (x) = α1 x + α3 x3 + · · · + α q1 −2 x q1 −2 , h2 (x) = β1 x + β3 x3 + · · · + β q2 −2 x q2 −2
∑︀
∑︀
^ = Xh
^ 2 (J (q2 ) ) implies α i W i =
then h1 (J (q1 ) )X
β j W j . Since matrices W1 , W3 , . . . , W q1 −2 are linearly independent while W q1 = W q1 +2 = · · · = 0 we deduce α i = β i for i = 1, . . . , q1 − 2. This shows that h1 (J (q1 ) ) =
h2 (J (q1 ) ) and we are done as before.
The case when d is even can be treated in a similar way. We just have to take the block matrix
)︃
)︃
(︃(︃ √ (︀
)︀d+1
^
X
− −1 J (q1 )
−1
T=S
√ (︀ (q ) )︀d+1 ⊕ 0n−q1 −q2 S,
^ t V q1
−V q2 X
−1 J 2
^ =
where S is similar as above except that now X
(︃
⃒ 0
(︁
⃒
(q1 −1)×1
0q1 ×2d ⃒
01×1
^′
D
01×(q1 −1)
)︃
)︁
with diagonal matrix
^ ′ being a compression of −D′2 onto the last (q1 − 1) basis vectors, where diagonal matrix D′2 comes from
D
(︃
)︃
0 D′2
(q2 ) d+1
(J )
=
. Similarly as in the previous case, we derive that T 2 = 0 and that T commutes with
0 0
A1 ⊕ A2 . So, we are done as above.
Substep 13.f. Before continuing with the last case we will show that every cube-zero alternate matrix is fixed
by ϕ Q , modulo scalars. So, let T be an alternate matrix with T 3 = 0. Since every cube-zero alternate matrix
consists of square-zero blocks and cube-zero blocks we may write without loss of generality that T = (K (2,2) ⊕
. . . ⊕ K (2,2) ) ⊕ (K (3) ⊕ . . . ⊕ K (3) ) ⊕ 0. Now, by subssteps 13.c and 13.e, we know that ϕ Q (T) has a similar blockdiagonal structure with each block being an odd polynomial of the corresponding block. That is, ϕ Q (T) =
(α1 K (2,2) ⊕ . . . ⊕ α r K (2,2) )⊕(β1 K (3) ⊕ . . . ⊕ β s K (3) )⊕0. Furthermore, from substep 13.c we know that α1 = . . . = α r
and from subsetep 13.e we learn that β1 = . . . = β s . Hence, ϕ Q (T) = α(K (2,2) ⊕ . . . ⊕ K (2,2) ) ⊕ β(K (3) ⊕ . . . ⊕
(2,2)
K (3) ) ⊕ 0. It only remains to show that α =⎛β. To this end, we may
⊕ K (3) ⊕ 0 and
⎞ assume that T = K
(︃
)︃
0 −1
0
i
0
X
⎜
⎟
(2,2)
(3)
1
i
ϕ Q (T) = αK
⊕ βK ⊕ 0. Using X = ⎝0 0
. Since
0⎠, define alternate Y =
2 + 2
−X t 0
0 0 − 21 + 2i 0
(rank Y , rank Y 2 , rank Y 3 ) = (4, 2, 0), it is easy to see that Y is similar to K (3) ⊕ K (3) ⊕ 0 and, hence, fixed
by ϕ Q , modulo scalars. Moreover, Y commutes with T. Therefore, Y has to commute with ϕ Q (T) as well. But
then scalars α and β have to be equal. Hence, every cube-zero alternate matrix is fixed by ϕ Q , modulo scalars.
Preserving zeros of Lie product on alternate matrices |
99
i pi )
Substep 13.g. Suppose lastly that K (p
and K (q j ) share the same eigenvalue, i.e, λ i = 0. Again for simplicity
λi
we temporarily permute the diagonal blocks so that, in the present substep only,
A1 ⊕ A2 = K (q j ) ⊕ K0(p i p i ) ⊕ . . . .
Define an alternate block matrix
(︃
T=
0
−X t
X
0
)︃
⊕ 0n−q j −2p i ,
where
√
√
)︀
(︀
)︀
^ · Id2p i + −1V2p i
Idq j − −1V q j · X
(︁
^ is a q j × 2p i matrix defined as follows. If q j ≤ p i , then we take X
^ = 0q ×(p −q )
and X
j
i
j
X=
1
2
(︀
(︀
)︀
D = Idk0 ⊕ diag (−1), (−1)2 , . . . , (−1)k0 −1 ,
k0 =
D
)︁
0q j ×p i , where
q j +1
2 .
)︃
[D|0p i ×p i ]
, where
0(q j −p i )×2p i
(︃
^=
If q j > p i , then we take X
D=
{︃
Idp i ;
p i ≤ k0 :=
Idk0 ⊕ diag (−1), (−1)2 , . . . , (−1)
(︀
)︀
p −k0
i
;
p i > k0 :=
q j +1
2
q j +1
2
.
(︀
)︀
^=X
^ J p i ⊕ (−J p i ) , so A1 ⊕ A2 commutes with T. Let us show that
An easy exercise gives that J (q j ) X
T 3 = 0.
Firstly, recall that T 2 = −(XX t ⊕ X t X) is block-diagonal. We claim that XX t = 0. To see this, note that
√
√
(Id2p i − −1V2p i )2 = −2 −1V2p i and that V q2j = Idq j . This yields that
XX t =
√
√
√
^ 2p i X
^ t V q j · V q j (Idq j − −1V q j ).
−1(Idq j − −1V q j ) · XV
^ · V2p i X
^ t V q j = 0 which gives the claimed XX t = 0. Then also X t XX t = 0, so T 3 = 0.
Now, it is easy to see that X
Consequently, by substep 13.f, T is fixed by ϕ Q , modulo a scalar. Since A1 ⊕ A2 commutes with T, then
ϕ Q (A1 ⊕ A2 ) commutes with ϕ Q (T) ∈ CT as well. Recall that
(︀
(︀
)︀
ϕ Q (A1 ⊕ A2 ) = h j K (q j ) ) ⊕ f i K0(p i p i ) ⊕ . . .
where h j and f i are odd polynomials of degrees less that q j and p i , respectively. It follows easily from ϕ Q (A1 ⊕
A2 )T = Tϕ Q (A1 ⊕ A2 ) that h j and f i coincide in the first min{q j − 1, p i − 1} monomials, as desired (here, we
^ = X(J
^ p i ⊕ (−J p i ))k ≠ 0 for k = 1, . . . , min{q j − 1, p i − 1}). The proof is completed.
use that (J (q j ) )k X
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