April
4
th 2013Beyond Gerstenhaber: spaes
of matries with spetral
onditions
Clément de Seguins Pazzis
Lyée Sainte-Geneviève, Versailles
Université de Versailles-Saint-Quentin-en-Yvelines
dsp.profgmail.om
http://dsp.prod.free.fr/reherhe.html
I. The adapted vetor method
II. The diagonal-ompatibility method
III. Beyond nilpotent subspaes
K
denotes an arbitrary (ommutative) eld(possibly nite),
M n ( K )
the vetor spae ofsquare matries with
n
rows,NT n ( K )
the one ofstritly upper-triangular matries.
E i,j is the
elementary matrix with all entries zero, exept an
entry
1
at the(i, j )
-spot.Two linear subspaes
V
andW
ofM n ( K )
aresimilar, and one writes
V ≃ W
, when∃P ∈ GL n ( K )
s.t.W = P V P − 1. A linear
subspae of
M n ( K )
is nilpotent when all itselements are. Example: linear subspaes of
NT n ( K )
...Problem : an one lassify the nilpotent
subspaes of
M n ( K )
?Bakground: Engel's theorem: every Lie
subalgebra of nilpotent matries of
M n ( K )
isIs every nilpotent subspae of
M n ( K )
triangularizable? YES for
n = 2
, NO forn > 2
. Alassial ounterexample: the matries of the form
0 0 a
0 0 b
−b a 0
are all nilpotent; no ommon eigenvetor!
Gerstenhaber's theorem (1958)
If
V
is a nilpotent subspae ofM n ( K )
, thendim V ≤ n(n − 1)
2
and equality holds i
V ≃ NT n ( K )
.M. Gerstenhaber, On nilalgebras and linear varieties of
nilpotent matries (I), Amer. J. Math. 80 (1958)
614-622.
Gerstenhaber requires
# K ≥ n
. For an arbitraryeld, proof ompleted by V.N. Serezhkin:
V.N. Serezhkin, Linear transformations preserving
nilpoteny (in Russian), Izv. Akad. Nauk BSSR, Ser.
Fiz.-Mat. Nauk 125 (1985) 46-50.
Simplied proof for:
•
the inequality;•
the ase of equality if# K > 2
;in
B. Mathes, M. Omladi£, H. Radjavi, Linear spaes of
nilpotent matries, Linear Algebra Appl. 149 (1991)
215-225.
An appliation of Gerstenhaber's theorem:
Nilpoteny preservers.
E.P. Botta, S. Piere, W. Watkins, Linear transformations
that preserve nilpotent matries, Pa. J. Math. 104
(1983) 39-46.
Motivation: solve the ase of equality. Let
V
nilpotent subspae of
M n ( K )
.V
seen as a spae of endomorphisms ofK n.
One needs a basis
(f 1 , . . . , f n )
ofK n in whih
every element of
V
is upper-triangular.Basi idea: rst, nd an appropriate
f n! There
should be no rank
1
matrix inV
with olumnspae
K f n!
Denition 1. A (non-zero) vetor
x ∈ K n is
V
-adapted when no rank 1
matrix of V
has
olumn spae
K x
.A. Existene of adapted vetors
Lemma 1. Every nilpotent subspae has an
adapted vetor.
Even better:
Lemma 2. Let
V
a nilpotent subspae ofM n ( K )
.Then, one of the vetors of the standard basis
(e 1 , . . . , e n )
isV
-adapted.Proof by indution on
n
. Casen = 1
obvious.Assume
n ≥ 2
and no vetor of the standard basisis
V
-adapted. DeneU
as the subspae ofV
onsisting of matries with last row zero, and
write every
M ∈ U
asM =
K (M ) C (M ) [0] 1 × (n − 1) 0
.
Thus,
K (U )
is a nilpotent subspae ofM n − 1 ( K )
.By indution, some
e i (1 ≤ i ≤ n − 1
) is
K (U )
-adapted. E.g.e 1 is K (U )
-adapted. But e 1
is not
V
-adapted! One ndsR 0 ∈ M 1,n ( K ) \ {0}
s.t.
R 0 [0] (n − 1) × n
∈ V
R 0 = h
0 · · · 0 ? i
sine
e 1 is K (U )
-adapted.
Then,
E 1,n ∈ V
.More generally, we have some
i 6= n
suh thatE i,n ∈ V
.More generally, for every
k ∈ [[1, n]]
, one ndsf (k) ∈ [[1, n]] \ {k}
withE f (k),k ∈ V
.One hooses an
f
-yle(i 1 , . . . , i p )
, i.e.i 1 , . . . , i p
distint in
[[1, n]]
andf (i 1 ) = i 2 , . . . , f (i p − 1 ) = i p
and
f (i p ) = i 1.
Then,
p
P
k=1
E f (i k ),i k non-nilpotent in V
! QED
B. Using adapted vetors to prove the
inequality statement
Without loss of generality,
e n is V
-adapted. For
M ∈ V
, one splits
M =
K (M ) C (M ) L(M ) a(M )
with
K (M )
an(n − 1) × (n − 1)
matrix.Dene
W
as the set ofM ∈ V
of the formM =
K (M ) [0] (n − 1) × 1 L(M ) a(M )
.
Then,
K (W )
is a nilpotent subspae ofM n − 1 ( K )
.As
e n is V
-adapted, rank theorem yields
dim K (W ) = dim W .
Thus,
dim V = dim C (V ) + dim K (W ).
By indution
dim V ≤ (n − 1) + (n − 1)(n − 2)
2 = n(n − 1)
2 ·
Remark 1. By indution, one an nd a
permutation matrix
P
s.t.P V P − 1 ontains no
non-zero lower-triangular matrix!
Proposition 3. Let
V
be a linear subspae ofM n ( K )
in whih no matrix has a non-zeroeigenvalue. Then
V
has an adapted vetor.Corollary 4. Let
V
be a linear subspae ofM n ( K )
in whih no matrix has a non-zeroeigenvalue. Then
dim V ≤ n(n 2 − 1) ·
Similar proofs as for nilpotent subspaes.
Corollary 5. Let
V
be an ane subspae ofinvertible matries of
M n ( K )
. Thendim V ≤ n(n 2 − 1) ·
Proof. Denote by
V
the translation vetor spae ofV
. One may assume thatI n ∈ V
. Then,αI n − M = α(I n − α − 1 M )
non-singular for allM ∈ V
and allα ∈ K \ {0}
.No matrix of
V
has a non-zero eigenvalue.C. de Seguins Pazzis, On the matries of given rank in a
large subspae, Linear Algebra Appl. 435-1 (2011)
Proposition 6. Let
V
a linear subspae ofM n ( K )
in whih every matrix has≤ 1
eigenvaluein
K
.Then
V
has an adapted vetor unlessn = 2
andchar( K ) = 2
.Theorem 7. With the same assumptions,
dim V ≤ 1 + n(n 2 − 1) ·
Counter-example:
sl 2 ( K )
withchar( K ) = 2
.C. de Seguins Pazzis, Spaes of matries with a sole
eigenvalue, Lin. Multilin. Alg. 60 (2012) 1165-1190.
onsiders matries with rank
1
and trae zero:Proposition 8. Let
V
a linear subspae ofM n ( K )
in whih every matrix has≤ 2
eigenvaluesin
K
.If
n > 2
andchar( K ) 6= 2
, thenV
has an adaptedvetor.
Theorem 9. With the same assumptions,
dim V ≤ 2 + n(n 2 − 1) ·
C. de Seguins Pazzis, Spaes of matries with few
eigenvalues, arXiv preprint.
Remark 2. For the at most
3
eigenvalues hypothesis, the existene of an adapted vetormay fail (as for at most
2
eigenvalues whenchar( K ) = 2
).method
A. Setup
Aim: now that we have found an adapted vetor,
ontinue the analysis of the ase of equality.
Hypotheses:
V
a nilpotent subspae of dimensionn(n − 1)
2
withe n as adapted vetor. Same notation
as in the proof of inequality. Then,
dim C (V ) = n−1
anddim K (W ) = (n − 1)(n − 2)
2 ·
By indution
K (W ) ≃ NT n − 1 ( K )
. Changing therst
n − 1
basis vetors, one may assumeK (W ) = NT n − 1 ( K ).
Then, one proves:
Lemma 10.
e 1 is V T-adapted.
Proof. Let
C ∈ M n,1 ( K )
olumn matrix withh
C [0] n × (n − 1)
i ∈ V
. AsK (W ) = NT n − 1 ( K )
,C =
[0] (n − 1) × 1 α
.
As
e n is V
-adapted, α = 0
.
Then, one use a new splitting for
V
:M =
a ′ (M ) L ′ (M ) C ′ (M ) K ′ (M )
with
K ′ (M )
an(n − 1) × (n − 1)
-matrix.Dene
W ′ as the set of all M ∈ V
with
L ′ (M ) = 0
. Then:
• K ′ (W ′ )
nilpotent subspae ofM n − 1 ( K )
;•
Ase 1 is V T-adapted,
dim K ′ (W ) = (n − 1)(n 2 − 2) ·
By indution,
K ′ (W ′ ) ≃ NT n − 1 ( K ).
Lemma 11. The last vetor of the standard basis
is
K ′ (W ′ )
-adapted.Proof. Let
L ∈ M 1,n − 1 ( K )
s.t.K ′ (W ′ )
ontains
[0] (n − 2) × (n − 1) L
, i.e. for someC ∈ M n,1 ( K )
,
C [0] (n − 1) × (n − 1)
? L
∈ V
Then
C = 0
sineK (W ) = NT n − 1 ( K )
. Then,L = 0
sinee n is V
-adapted.
Lemma 12 (Diagonal-ompatibility). There is a
matrix
Q =
I n − 2 [0] (n − 2) × 1 [?] 1 × (n − 2) 1
s.t.K ′ (W ′ ) = Q NT n − 1 ( K )Q − 1.
Main ideas:
Point 1:
No generality is lost in assuming that
Qe n − 1 = e n − 1; this is based on the fat that e n − 1
is
K ′ (W ′ )
-adapted.Point 2:
For every
U ∈ NT n − 2 ( K )
,
U [0] (n − 2) × 1 [?] 1 × (n − 2) 0
∈ K ′ (W ′ ).
Follows from
K (W ) = NT n − 1 ( K )
.For every
L ∈ M 1,n − 2 ( K )
,V
ontainsA L =
0 L 0
[0] (n − 2) × 1 [0] (n − 2) × (n − 2) [0] (n − 2) × 1
f (L) ϕ(L) 0
(this uses
K (W ) = NT n − 1 ( K )
).For every
C ∈ M n − 2,1 ( K )
,V
ontainsB C =
0 [0] 1 × (n − 2) 0 ψ (C ) [0] (n − 2) × (n − 2) C
g(C ) [0] 1 × (n − 2) 0
(this uses
K ′ (W ′ ) = NT n − 1 ( K )
).For every
U ∈ NT n − 2 ( K )
,V
ontainsE U =
0 [0] 1 × (n − 2) 0
[0] (n − 2) × 1 U [0] (n − 2) × 1 h(U ) [0] 1 × (n − 2) 0
.
Finally,
V
ontains a matrix of the formJ =
? [0] 1 × (n − 2) 1
[?] (n − 2) × 1 [?] (n − 2) × (n − 2) [0] (n − 2) × 1
? [?] 1 × (n − 2) ?
.
Then, one suessively proves:
•
There is a salarλ
suh thatϕ(L) = λL
andψ(C ) = −λC
for allL
andC
.•
One replaesV
withP − 1 V P
whereP =
1 [0] 1 × (n − 2) 0
[0] (n − 2) × 1 I n − 2 [0] (n − 2) × 1 λ [0] 1 × (n − 2) 1
,
and thus one an assume
λ = 0
.•
Then, one proves thatf = 0
,g = 0
andh = 0
.•
One shows thatV
ontainsE 1,n (this uses J
).
One onludes that
V
ontainsNT n ( K )
, whihompletes the proof.
C. de Seguins Pazzis, On Gerstenhaber's theorem for
spaes of nilpotent matries over a skew eld, Linear
Algebra Appl. 438-11 (2013) 4426-4438.
III. Beyond nilpotent subspaes
The adapted vetor method and the
diagonal-ompatibility methods yield theorems of
the same avor as Gerstenhaber's.
A. Large spaes of matries with no
non-zero eigenvalue
A matrix
P ∈ GL n ( K )
is non-isotropi whenX T P X 6= 0
for all non-zeroX ∈ K n. Two
matries
P
andQ
are sim-ongruent whenthere is a non-singular
R
and a non-zero salarλ
s.t.
P = λRQR T. Two quadrati forms ϕ
and ψ
are similar when there is a non-zero salar
λ
suh that
ψ
is equivalent toλ ϕ
.A n ( K )
is the spae of alln × n
alternatingmatries.
Theorem 13 (de Seguins Pazzis (2010)). Let
V
be a linear subspae of
M n ( K )
with dimensionn(n − 1)
2
, in whih no matrix has a non-zeroeigenvalue. Assume that
# K > 2
. Then, there arenon-isotropi matries
P 1 , . . . , P p respetively in
GL n 1 ( K ), . . . , GL n p ( K )
, s.t. V
is similar to the
spae of all matries of the form
P 1 A 1 [?]
.
.
.
[0] P p A p
with
A 1 ∈ A n 1 ( K ), . . . , A p ∈ A n p ( K )
.P 1 , . . . , P p are uniquely determined by V
up to
sim-ongruene.
Thus, one is redued to the lassiation of
non-isotropi bilinear forms up to similarity.
C. de Seguins Pazzis, Large ane spaes of non-singular
matries, Trans. Amer. Math. So. 365 (2013)
Theorem 14 (de Seguins Pazzis (2010)). Let
V
be an ane subspae of
M n ( K )
with dimensionn(n − 1)
2
, in whih all the matries arenon-singular. Assume
# K > 2
.Then, there are non-isotropi matries
P 1 , . . . , P p
respetively in
GL n 1 ( K ), . . . , GL n p ( K )
, suh thatV
is equivalent to the spae of all matries of theform
I n +
P 1 A 1 [?]
.
.
.
[0] P p A p
with
A 1 ∈ A n 1 ( K ), . . . , A p ∈ A n p ( K )
.The quadrati forms
X 7→ X T P 1 X
, ...,X 7→ X T P p X
are uniquely determined byV
up tosimilarity.
For a generalization to ane spaes of matries
with a lower bound on the rank:
C. de Seguins Pazzis, Large ane spaes of matries with
rank bounded below, Linear Algebra Appl. 437-2 (2012)
most one eigenvalue
Theorem 15. Let
V
be a linear subspae ofM n ( K )
in whih every matrix has1
eigenvalue inK
, anddim V = 1 + n(n 2 − 1) ·
Then,V ≃ K I n ⊕ NT n ( K )
exept in the followingsituations:
• char( K ) = 2
andn ∈ {2, 4}
;• char( K ) = 3
andn = 3
.In the exeptional ases stated above, all the
solutions are known.
C. de Seguins Pazzis, Spaes of matries with a sole
eigenvalue, op.it.
Remark 3. If
char( K ) 6 | n
, the result is a trivialonsequene of Gerstenhaber's theorem.
Conjeture 1. Let
V
be a linear subspae ofM n ( K )
in whih every matrix has≤ 1
eigenvaluein
K
anddim V = 1 + n(n 2 − 1) ·
Assume thatn ≥ 5
and
# K > 2
. Then,V = K I n ⊕ H
, where nomatrix of
H
has a non-zero eigenvalue inK
.C. Large spaes of matries with at
most two eigenvalues
For
char( K ) 6= 2
andn ≥ 3
, the full lassiationof spaes of matries of
M n ( K )
with≤ 2
eigenvalues in
K
and with the maximal dimension2 + n(n 2 − 1):
C. de Seguins Pazzis, Spaes of matries with few
eigenvalues, op.it.
→
Uses only low-level tools from linear algebra.→
Well-suited for proving very general results forproperties on matrix spaes with spetral
onditions.
→
Main drawbak: big mahinery, very longproofs!