Quadratic Transformations on Matrices: Rank Preservers

Ž .JOURNAL OF ALGEBRA 179, 549]569 1996ARTICLE NO. 0024

Quadratic Transformations on Matrices:Rank Preservers

William Watkins

California State Uni ersity at Northridge, Northridge, California 91330-8313

Communicated by Gordon James

Received August 8, 1994

Ž .Let F be an algebraically closed field of characteristic not 2, and let X s Xi jbe the n = n matrix whose entries X are independent indeterminates over F.i j

Ž . Ž Ž .. Ž .Now let Q X s q X be another n = n matrix each of whose entries q X isi j i ja quadratic F-polynomial in the X . The main result in this paper is: for n G 5,i jŽ . Ž 2 . Ž Ž .. n= nQ X satisfies rank A s r implies rank Q A s r, for all A g F for r s 0,

1, and 2, if and only if there exist invertible matrices P , P in F n= n such that1 2Ž . 2 Ž . Ž 2 .teither Q X s P X P or Q X s P X P . Q 1996 Academic Press, Inc.1 2 1 2

INTRODUCTION AND RESULTS

Linear Rank Preser ers

Let F be an algebraically closed field and let F n=n be the algebra ofn = n matrices over F. A linear transformation L on F n=n is said to

n=n Ž .preser e a subset S of F , if L A g S whenever A g S. During thepast 35 years, hundreds of results about the structure of linear transforma-tions that preserve various sets of matrices have appeared in the literature.

Ž .The fundamental result involves the case where S s Rank 1 , the set ofmatrices of rank 1:

FUNDAMENTAL RESULT FOR LINEAR PRESERVERS. If L is a linear trans-formation on F n=n and L satisfies the condition

rank A s 1 implies rank L A s 1, for all A g F n=n , 1Ž . Ž . Ž .Ž .then there exist inërtible matrices P and P such that L has one of the forms1 2

L X s P XPŽ . 1 2

L X s P X tP .Ž . 1 2

w xIn 1959, Marcus and Moyls 5 proved the fundamental result forw xalgebraically closed fields of characteristic 0 and in 1967, Westwick 8

549

0021-8693r96 $12.00Copyright Q 1996 by Academic Press, Inc.

All rights of reproduction in any form reserved.

WILLIAM WATKINS550

extended the result to fields of positive characteristic. For an excellentsurvey of linear transformations that preserve rank and other algebraic

w xsets of matrices see 4, 7 .In this paper, we determine the structure of quadratic transformations

Ž .that satisfy a rank-preserving condition analogous to condition 1 .

Quadratic Transformations

Ž .Let X s X be the n = n matrix whose entries X are independenti j i jŽ . Ž Ž ..indeterminates over F. Let Q X s q X be another n = n matrix,i j

Ž .each of whose entries q X is a homogeneous quadratic F-polynomial ini jthe indeterminates X . By substituting the matrices A g F n=n for X, wei jcan regard Q as a transformation on F n=n. This is the definition of a

Ž Ž .quadratic transformation. Of course, if L X is an n = n matrix and eachentry is a homogeneous linear F-polynomial in the X , then in the samei jway, L can be regarded as a linear transformation on F n=n. Indeed, every

.linear transformation can be obtained in this way.Ž .Let P , P be matrices in the general linear group GL F . Consider1 2 n

the following two types of quadratic transformations:

Q X s P X 2PŽ . 1 22Ž .

t2Q X s P X P .Ž . Ž .1 2

These transformations satisfy the condition

rank A2 s rank Q A , for all A g F n=n . 3Ž . Ž . Ž .Ž .Ž .Our main result is that the quadratic transformations of the form 2 are

Ž .the only ones that satisfy condition 3 , provided certain conditions on FŽ .and n are met. Actually, the full strength of condition 3 is not necessary.

It suffices to assume only that the condition

rank A2 s r implies rank Q A s r , for all A g F n=n 4Ž . Ž . Ž .Ž .holds for r s 0, 1, and 2. Since the zero matrix is the only matrix whose

Ž . 2rank is 0, condition 4 for r s 0 is equivalent to the condition that A s 0Ž . n=nimplies Q A s 0 for all A g F .

THEOREM 1. Let F be an algebraically closed field of characteristic not 2Ž . n=nand let Q X be a quadratic transformation on F . Then

Ž . Ž .a For n G 5, condition 4 holds for r s 0, 1, and 2, if and only ifŽ . Ž .there exist matrices P , P g GL F such that Q X has one of the two1 2 n

forms

Q X s P X 2PŽ . 1 25Ž .

t2Q X s P X P .Ž . Ž .1 2

QUADRATIC TRANSFORMATIONS ON MATRICES 551

Ž . Ž .b For n G 5, condition 4 holds for r s 0 and 1, if and only if theren=n Ž . Ž .exist matrices C g F and P , P g GL F such that Q X has one of1 2 n

the two forms

22 2Q X s P X P q tr X y tr X CŽ . Ž .Ž .1 26Ž .

t 22 2Q X s P X P q tr X y tr X C.Ž . Ž . Ž .Ž .1 2

Ž . Ž .c For n G 4, condition 4 holds for r s 0, if and only if there existŽ . Ž .linear transformations L X and M X such that

Q X s tr X L X q M X 2 . 7Ž . Ž . Ž . Ž . Ž .

One way to view this result is in terms of a group action on thequadratic transformations. Consider the group generated by

tQ X ª Q X ,Ž . Ž .Ž .and

Q X ª P Q X P ,Ž . Ž .1 2

Ž . Ž . Ž .where P , P run through GL F . If Q X and Q X are in the same1 2 n 1 2Ž Ž .. Ž Ž .. n=n Ž .orbit, then rank Q A s rank Q A , for all A g F . Part a of the1 2

Ž . Ž .theorem states that if Q X satisfies condition 4 for r s 0, 1, 2, thenŽ . 2Q X is in the same orbit as X .The validity of the theorem for small values of n and for other fields has

Ž .not been investigated. However, part c of the theorem is valid for n s 2,Ž .and part b is not.

EXAMPLE 1. Let n s 2 and

2tr X 0Ž .Q X s . 8Ž . Ž .0 0

Ž . Ž . Ž .Then Q X satisfies 4 for r s 0, 1, but Q X does not haë either of theŽ .forms in 6 .

Notation

F an algebraically closed field with characteristic not 2F n the vector space of n-tuples over FF n=n the algebra of n = n matrices over F

Ž . n=nGL F the group of invertible matrices in FnŽ . n=nsl F the subspace of F of matrices with trace zeron

WILLIAM WATKINS552

Ž . n=n Ž .Rank r, s, . . . the set of matrices in F with ranks r, s, . . . Rank 0 s� 40

At the transpose of matrix AE the n = n matrix whose only nonzero entry is a 1 ini j

Ž .position i, j² :u, ¨ , . . . the F-linear span of vectors u, ¨ , . . .W H the orthogonal complement of W : F n with respect to the

dot productŽ . Ž .u m ¨ the matrix u ¨ , where u s u , . . . , u and ¨ si j 1 n

Ž . n¨ , . . . , ¨ are in F1 n

The tensor notation u m ¨ is appropriate for the matrix defined above,because F n=n is a model for the tensor space F n m F n. If u and ¨ are

Ž .Žnonzero, then u m u has rank one. We will use the fact that u m ¨ y m. Ž .w s ¨ ? y u m w in the sequel.

LEMMAS AND PROOFS

Lemmas

LEMMA 1. Let u, ¨ , y, w be ëctors in F n and suppose u m ¨ q y m w gŽ .Rank 0, 1 . Then, either u and y are linearly dependent or ¨ and w are linearly

dependent.

� 4 � 4Proof. If both sets u, y and ¨ , w are linearly independent, thenŽ .u m ¨ q y m w g Rank 2 .

Ž .LEMMA 2. Let MM be a subspace contained in Rank 0, 1 . Then thereexists ¨ g F n such that either MM : ¨ m F n or MM : F n m ¨ .

Proof. Let x m y , . . . , x m y be a basis for MM. Suppose x , x are1 1 k k 1 2² : ² :linearly independent. Then from Lemma 1, y s y . Now suppose1 2

y , y are linearly independent for some i G 3. Then again from Lemma 11 i² : ² : ² :x s x s x , which is impossible since x , x are linearly indepen-1 i 2 1 2

² :dent. Hence dim y , . . . , y s 1.1 k

n=n Ž .LEMMA 3. Let A, B g F and suppose A q tB g Rank 0, 1 , for allŽ .t g F. Then A, B g Rank 0, 1 .

Ž .Proof. Set t s 0 to get A g Rank 0, 1 . For B, notice that every 2 = 2subdeterminant of A q tB is zero. Thus the coefficients on t 2 are alsozero, i.e., all 2 = 2 subdeterminants of B are zero.

Ž . Ž . n=nLEMMA 4. Let n G 4 and M X be a linear map from sl F to Fnthat satisfies the condition

u ? ¨ s 0 implies M u m ¨ g Rank 1 , 9Ž . Ž . Ž .


n Ž .for all nonzero u, ¨ g F . Then there exist P , P g GL F such that one of1 2 nthe following holds:

M A s P AP , for all A g sl FŽ . Ž .1 2 n10Ž .

tM A s P A P , for all A g sl F .Ž . Ž .1 2 n

Ž . nProof. Assume condition 9 holds for all nonzero u, ¨ g F . Then forŽ H. n=neach nonzero u, M u m u is a subspace of F contained in Rank

Ž . n0, 1 . By Lemma 2, there exists w g F such that either

M u m uH : w m F nŽ .or

M u m uH : F n m w.Ž .

Similarly, for each ¨ g F n, there exists w g F n such that either

M ¨ H m¨ : w m F nŽ .or

M ¨ H m¨ : F n m w.Ž .

We call a subspace of F n=n of type A, if it is contained in w m F n, forsome w and of type B, if it is contained in F n m w, for some w.

Next we prove that one of the following statements is true:

M u m uH is of type A, for all nonzero u g F nŽ .11Ž .

M u m uH is of type B, for all nonzero u g F n .Ž .n Ž H.Let u , u be any two nonzero vectors in F and suppose that M u m u1 2 1 1

Ž H. nis of type A while M u m u is of type B. That is, there exist w , w g F2 2 1 2such that

M u m uH : w m F n 12Ž .Ž .1 1 1

M u m uH : F n m w . 13Ž .Ž .2 2 2

² :H Ž H .Pick linearly independent vectors u , u g u , u . Then M u mu is3 4 1 2 3 3either of type A or of type B and we may assume without loss of generalitythat

M uH mu : w m F n , 14Ž .Ž .3 3 3

Ž H . Ž H. Žfor some w . Since u m u g u mu l u m u , we have M u m3 1 3 3 3 1 1 1. Ž n. Ž n. Ž . Ž . Ž .u g w m F l w m F . But M u m u g Rank 1 , by 9 and so3 3 1 1 3

WILLIAM WATKINS554

Ž n. Ž n. � 4 ² : ² :w m F l w m F / 0 . It follows that w s w . In fact, we3 1 1 3may assume that w s w . Then1 3

n n ² :M u m u g F m w l w m F s w m w . 15Ž . Ž .Ž . Ž .2 3 2 1 1 2

Ž H . Ž H .Now consider M u mu . It cannot be of type A. If M u mu :4 4 4 4n Ž . Ž n. Ž n.w m F for some w , then M u m u g w m F l w m F . So4 4 1 4 1 4

² : ² : Ž . Ž n. Ž n . ²w s w . But then M u m u g w m F l F m w s w m1 4 2 4 1 2 1: Ž . Ž . Žw . This contradicts 15 , for if M u m u s a w m w and M u m2 2 3 1 2 2. Ž Ž ..u s b w m w , for some a , b g F. Then M u m b u y a u s 0,4 1 2 2 3 4

Ž Ž .. Ž . Ž .which is impossible since M u m b u y a u g Rank 1 , by 9 . Thus2 3 4Ž H .M u mu is a subspace of type B.4 4

Ž H . n Ž .Now we have M u mu : F m w , for some w . Thus M u m u g4 4 4 4 2 4Ž n . Ž n . ² : ² : Ž .F m w l F m w and we have w s w . Thus M u m u g2 4 2 4 1 4² : Ž . Ž . ² : Ž .w m w , from 12 and M u m u g w m w from 15 . So there1 2 2 3 1 2

Ž . Ž .are scalars l, m such that M u m u s lw m w and M u m u s1 4 1 2 2 3Ž . Ž . ŽŽ . Žmw m w . Since mu q lu ? u y u s 0, M mu q lu m u y1 2 1 2 3 4 1 2 3

.. Ž .u g Rank 1 . But4

M mu q lu m u y uŽ . Ž .Ž .1 2 3 4

s mM u m u y mM u m u q lM u m u y lM u m uŽ . Ž . Ž . Ž .1 3 1 4 2 3 2 4

s mM u m u y lM u m uŽ . Ž .1 3 2 4

s w m x q y m w ,1 4

Ž .for some nonzero vectors x, y. Since w m x q y m w g Rank 1 , either1 4² : ² : ² : ² : ² : Ž .w s y or x s w s w . In the first case M u m u g1 4 2 2 4² : ² : Ž . ² : ² : ² :w m w s w m w , which contradicts 15 . If x s w s w ,1 4 1 2 4 2

Ž . ² : Ž . Ž .then M u m u g w m w , which also contradicts 15 . Thus 11 is1 3 1 2Ž H .proved. Similarly, either M ¨ m¨ is a subspace of type A for all ¨ , or it

is of type B for all ¨ .Ž . Ž Ž ..tBy replacing M X with M X if necessary, we will assume that

M u m uH has type A for all u g F n . 16Ž . Ž .

Now we prove that

M ¨ H m¨ has type B for all ¨ g F n . 17Ž . Ž .


Ž H. Ž H .If not, then both M u m u and M ¨ m¨ are of type A, for all vectorsn ² :Hu, ¨ . Let u , u be any nonzero vectors in F . Pick ¨ g u , u with1 2 1 2

¨ / 0. Then

M u muH : w m F n ,Ž .1 1 1

M u muH : w m F n ,Ž .2 2 2 18Ž .

M ¨ Hm¨ : w m F n ,Ž .n Ž . Žfor some w , w , w g F . Then M u m ¨ is a nonzero matrix in w m1 2 1 1

n. Ž n. ² : ² : ² : ² :F l w m F , which implies that w s w . Similarly, w s w .1 2n Ž . nThus there exists w g F , such that M u m ¨ g w m F , for all u, ¨ with

Ž Ž .. nu ? ¨ s 0. It follows that M sl F ; w m F . In particular,nŽ .dim Range M F n.

Ž . Ž . Ž .Let W s Rank 0, 1 l sl F . Since Rank 0, 1 is an irreducible homo-nŽ .geneous algebraic variety of affine dimension 2n y 1, it follows that

Ž .dim W s 2n y 2. From the previous paragraph, we have dim Range M FŽ . Ž 2 . Ž .n. Thus dim ker M G n y 1 y n. By hypothesis 9 , we have ker M l

� 4 w x Ž .W s 0 . It follows from the dimension theorem 3 that dim ker M q2 Ž 2 . Ž . 2dim W F n y 1. So n y 1 y n q 2n y 2 F n y 1 and thus n F 2.

Ž .This contradicts the hypothesis n G 4. Thus 17 is proved.Ž .The next step is to prove that there exist matrices P , P g GL F such1 2 n

that

M A s P AP , for all A g sl F . 19Ž . Ž . Ž .1 2 n

Ž . Ž Ž ..tIn view of the fact that M X may have been replaced by M X in theŽ . Ž .argument above, 19 is equivalent to 10 .

Define two transformations u and f on the lines of F n as

H ² : nM u m u : u u m FŽ .and

H n ² :M ¨ m ¨ : F m f ¨ .Ž .

We will prove that each transformation u , f is a bijection on the lines ofF n and that each preserves coplanarity. Then we can use the fundamental

w xtheorem of projective geometry 2 to conclude that each transformation isinduced by a semilinear transformation on F n.

² : ² X:To prove u is one-to-one, assume that u u s u u , for some nonzeroX n Ž H. n Ž X X H. nu, u g F . Then M u m u : w m F and M u m u : w m F , where

² : ² : ² X: ² X:H Ž H .w s u u s u u . Pick nonzero ¨ g u, u . Then M ¨ m¨ :n ² : ² : Ž . Ž X .F m y, where y s f ¨ . So both M u m ¨ and M u m ¨ are in

² : Ž . Ž X . Xw m y , say M u m ¨ s a w m y and M u m ¨ s a w m y, for some

WILLIAM WATKINS556

X ŽŽ X X. . Xnonzero scalars a , a . Then M a u y a u m ¨ s 0, which implies a u yX Ž . ² : ² X:a u s 0, by 9 . Thus u s u . So u is one-to-one.

n ² :Let e , . . . , e be the standard basis for F and suppose that u e s1 n i² : ² : ² : nu and f e s w , for some vectors u , . . . , u , w , . . . , w g F . Theni j j 1 n 1 n

Ž .for i / j, M e m e s b u m w , for some b g F. None of the scalarsi j i j i j i jŽ . Ž .b or the vectors u , ¨ is zero, since M e m e g Rank 1 . Also, u , . . . , ui j i j i j 1 n

are linearly independent. To prove this, suppose u s Ýn a u and let1 is2 i in Ž .x s Ý a rb e . Thenis2 i i1 i

n a iM x m e s M e m eŽ . Ž .Ý1 i 1bi1is2

n

s a u m wÝ i i 1is2

s u m w .1 1

² : ² : ² : ² :Thus u x s u . But u e s u , which contradicts the fact that u is1 1 1

one-to-one. Similarly, w , . . . , w are linearly independent.1 n² : nNext we show that u and f are onto. Let p be a line in F . We will

n ² : ² :find z g F , such that u z s p . Suppose p s Ýa u . If all but one ofi i² : ² :the a are zero, say a / 0 and a s ??? s a s 0, then u e s p .i 1 2 n 1

Thus we may assume that at least two of the a are nonzero, say a / 0,i 1Ž . Ž .a / 0. In that case, let x s a rb e q a rb e . Then2 1 13 1 2 23 2

a a1 2M x m e s M e m e q M e m eŽ . Ž . Ž .3 1 3 2 3b b13 23

s a u m w q a u m w1 1 3 2 2 3

s a u q a u m w .Ž .1 1 2 2 3

² : ² : Ž . Ž .Thus u x s a u q a u . Let y s b ra e y b ra e . Then1 1 2 2 13 1 1 23 2 2² :H Ž . Ž .y g x, e , . . . , e . Since x ? y s 0, we have M x m y s a u q a u3 n 1 1 2 2

n Ž .m r, for some nonzero r g F . Also, for i G 3, M e m y s u m r , fori i i

some nonzero vectors r g F n. Thusi

n n

M x q e m y s a u m r q a u m r q u m r g Rank 1 .Ž .Ý Ýi 1 1 2 2 i iž /ž /is3 is3


² :But u , . . . , u are linearly independent, so r g r , for i G 3. Thus1 n iŽ . Ž .r s g r, for some g g F. The g are nonzero since M e m y g Rank 1 ,i i i i i

for i G 3. Now we have

n na ai iM x q e m y s M x m y q M e m yŽ . Ž .Ý Ýi iž /ž /g gi iis3 is3

n a is a u q a u m r q u m g rŽ . Ž .Ž .Ý1 1 2 2 i ig iis3

n

s a u m rÝ i iž /is1

s p m r .

Thus

n a i ² :u x q e s pÝ i¦ ;g iis3

and u is onto.Next we prove that u and f coplanarity. Suppose that the three lines

² : ² : ² : ² : ² : ² :x , x , x are coplanar, say x : x q x . Then x s1 2 3 3 1 2 3² :Ha x q a x , for some a , a g F. Pick a nonzero vector y g x , x .1 1 2 2 1 2 1 2

Ž . Ž . Ž . XThen M x m y s a M x m y q a M x m y . Let z , z , z , y be vec-3 1 1 2 2 1 2 3² : ² : ² X: ² : Ž .tors satisfying z s u x and y s f y so that M x m y s d z mi i i i i

yX for some nonzero scalars d , i s 1, 2, 3. Then d z m yX s a d z m yX qi 3 3 1 1 1X ² : ² : ² :a d z m y . Thus d z s a d z q a d z so that z : z q z .2 2 2 3 3 1 1 1 2 2 2 3 1 2

² : ² : ² :In other words, u x : u x q u x . Thus, u preserves coplanarity.3 1 2By a similar argument, f also preserves coplanarity.

We have now shown that the maps u , f satisfy the hypotheses of thew xfundamental theorem of projective geometry 2 . Thus there are invertible

semilinear maps S, T on F n such that

² : ² : ² : ² : nu u s Su , f ¨ s T¨ , for all u , ¨ g F .

Thus, if u ? ¨ s 0 then

M u m ¨ s c u , ¨ Su m T¨ , 20Ž . Ž . Ž .

�Ž . n n 4where c is a map from u, ¨ g F = F : u ? ¨ s 0 to F. We now provethat the maps S and T are linear and finally that there exist P P g1 2

Ž . Ž .GL F such that 19 holds.n

WILLIAM WATKINS558

To show that S is linear, let l ª l be the automorphism of F associ-Ž .ated with S. From 20 we have

M e q le m e s c e q le , e S e q le m TeŽ . Ž . Ž .Ž .2 3 1 2 3 1 2 3 1

s c e q le , e Se m Te q lSe m Te .Ž .2 3 1 2 1 3 1

But M is linear, so

M e q le m e s M e m e q lM e m eŽ . Ž . Ž .Ž .2 3 1 2 1 3 1

s c e , e Se m Te q lc e , e Se m Te .Ž . Ž .2 1 2 1 3 1 3 1

Ž . ŽBut Se m Te , i s 2, 3, are linearly independent. So c e , e s c e qi 1 2 1 2. Ž . Ž . Ž .le , e and lc e , e s lc e q le , e , for all l g F. Thus lc e , e s3 1 3 1 2 3 1 3 1

Ž . Ž . Ž .lc e , e , for all l g F. It follows that c e , e s c e , e and that2 1 2 1 3 1l s l, for all l g F. Thus S is linear. Similarly, T is linear.

Ž . Ž .In the argument above, we proved that c e , e s c e , e . Likewise,2 1 3 1Ž . Ž . � 4 Ž . Ž . � 4c e , e s c e , e , for s f i, j , and c e , e s c e , e , for i f s, t .i s j s i s i t

Ž .Since n G 3, the scalars c i, s , i / s, have a common value, c. That is,

M e m e s cSe m Te ,Ž .i s i s

for i / s.Now we show that

M e m e y e m e s c Se m Te y Se m Te ,Ž . Ž .i i s s i i s s

Ž . Ž .for i / s. It suffices to show this result for i, s s 1, 2 .First let u s e q e , ¨ s e y e . Then u ? ¨ s 0 and so1 2 1 2

M u m ¨ s c u , ¨ Su m T¨Ž . Ž .s c u , ¨ Se m Te y Se m Te y Se m Te q Se m Te .Ž . Ž .1 1 2 2 1 2 2 1

On the other hand,

M u m ¨ s M e m e y e m e y M e m e q M e m eŽ . Ž . Ž . Ž .1 1 2 2 1 2 2 1

s M e m e y e m e y cSe m Te q cSe m Te .Ž .1 1 2 2 1 2 2 1

It follows that

M e m e y e m e s c u , ¨ Se m Te y Se m TeŽ . Ž . Ž .1 1 2 2 1 1 2 2

q yc u , ¨ q c Se m TeŽ .Ž . 1 2

q c u , ¨ y c Se m Te .Ž .Ž . 2 1


Next let uX s e y e , ¨ X s e q e . Then uX? ¨ X s 0 and so1 2 1 2

M uX , ¨ X s c uX , ¨ X SuX m T¨ XŽ . Ž .s c uX , ¨ X Se m Te y Se m Te q Se m Te y Se m Te .Ž . Ž .1 1 2 2 1 2 2 1

But also

M uX , ¨ X s M e m e y e m e q M e m e y M e m eŽ . Ž . Ž . Ž .1 1 2 2 1 2 2 1

s M e m e y e m e q cSe m Te y cSe m Te .Ž .1 1 2 2 1 2 2 1

Thus,

M e m e y e m e s c uX , ¨ X Se m Te y Se m TeŽ . Ž . Ž .1 1 2 2 1 1 2 2

q c uX , ¨ X y c Se m TeŽ .Ž . 1 2

q yc uX , ¨ X q c Se m Te .Ž .Ž . 2 1

By comparing the coefficients of the four linearly independent matricesŽ .Se m Te , i, j s 1, 2, in the two expressions for M e m e y e m e , wei j 1 1 2 2

Ž . Ž X X. Ž . Ž X X. Ž .get c u, ¨ s c u , ¨ and yc u, ¨ q c s c u , ¨ y c. Thus c u, ¨ sŽ X X.c u , ¨ s c and so

M e m e y e m e s c Se m Te y Se m Te .Ž . Ž .1 1 2 2 1 1 2 2

Ž . Ž .There exist matrices P , P g GL F such that cSu m T¨ s P u m ¨ P ,1 2 n 1 2n Ž . Ž .for all u, ¨ g F . We have proved that M e m e s P e m e P andi j 1 i j 2

Ž . Ž . Ž .M e m e y e m e s P e m e y e m e P , for i / j. Equation 19 fol-i i j j 1 i i j j 2lows.

LEMMA 5. Let F be an algebraically closed field of characteristic not 2,� n=n 2 4n G 4, and let VV s A g F : A s 0 be the algebraic set of matrices of

nilindex 2. Let II be the space of all homogeneous quadratic F-polynomials2Ž . Ž 2 . Ž .p X in the n indeterminates X such that p A s 0 for all A g VV . Theni j

2 2 Ž 2 . Ž .dim II s 2n and the 2n polynomials X , tr X X are a basis.2 i j i j

Ž . nProof. Suppose q X g II . Let u, ¨ be in F with u ? ¨ s 0. Then22 Ž .A s u m ¨ satisfies A s 0, so q u m ¨ s 0. Now consider u ? ¨ and

Ž .q u m ¨ as polynomials in the 2n coordinates of u and ¨. It follows fromŽ . Ž .'the Nullstellensatz that q u m ¨ g id u ? ¨ , where id u ? ¨ is theŽ .

principal ideal generated by u ? ¨ . But u ? ¨ is an irreducible polynomialŽ .and hence id u ? ¨ is a prime ideal. Thus,

q u m ¨ s u ? ¨ b u , ¨ ,Ž . Ž . Ž .

WILLIAM WATKINS560

Ž . Ž .for some bilinear polynomial b u, ¨ . Thus, if A g Rank 0, 1 , so thatn Ž . Ž . Ž .A s u m ¨ for some u, ¨ g F , then q A s tr A b A , where b is the

Ž . Ž .linear map satisfying b u m ¨ s b u, ¨ .XŽ . Ž . Ž . Ž . XŽ .Let q X s q X y tr X b X . Then q X is a homogeneous

Ž .quadratic polynomial with the property that A g Rank 0, 1 impliesXŽ . w x XŽ .q A s 0. Thus 1 q X is in the ideal generated by all 2 = 2 subdeter-

XŽ . Ž .minants of X. But since q X is quadratic, there are scalars c i, j, s, t g Fsuch that

qX X s c i , j, s, t X X y X X . 21Ž . Ž . Ž .Ž .Ý i s jt i t jsi-j , s-t

2 Ž . XŽ .Also, if A s 0, then tr A s 0 and q A s 0; so q A s 0. We will proveXŽ . 2that q X is an F-linear combination of the 2n quadratic polynomials

Ž 2 . Ž .X , tr X X .i j i jŽ .Break the sum on the right side of Eq. 21 into three parts,

Ž . Ž . Ž .0 1 2c i , j, s, t X X y X X s q q ,Ž . Ž .Ý Ý Ý Ýi s jt i t js

i-j , s-t

Žk . <� 4 � 4 <where Ý is taken over all pairs i - j, s - t with i, j l s, t s k, fork s 0, 1, 2.

Ž . � 4 � 4First we prove that c i, j, s, t s 0, whenever i, j and s, t are disjointso that ÝŽ0. s 0. This follows from the fact that A s E q E satisfiesi s jt

2 XŽ .A s 0 so that q A s 0. But all 2 = 2 subdeterminants of A are zeroŽ .except the one in rows i, j and columns s, t. Thus from Eq. 21 , we have

XŽ . Ž .q A s c i, j, s, t .Next consider the second sum, ÝŽ1.. This sum is taken over all selections

of two rows i, p and two columns s, p, where i, s, p are distinct. To avoidseparating ÝŽ1. into additional cases depending on whether i - p or p - iand whether s - p or p - s, we introduce a notational convenience. At

Ž .present, the scalars c i, j, s, t are defined only for pairs i - j, s - t. ForŽ . Ž . Ž .pairs i - j, s - t, define c j, i, s, t s yc i, j, s, t , c i, j, t, s s

Ž . Ž . Ž . Ž .yc i, j, s, t , and c j, i, t, s s c i, j, s, t so that c i, j, s, t is now definedfor all pairs i / j, s / t. Then

c i , j, s, t X X y X XŽ . Ž .i s jt i t js

s c j, i , s, t X X y X XŽ . Ž .js i t jt i s

s c i , j, t , s X X y X XŽ . Ž .i t js i s jt

s c j, i , t , s X X y X X .Ž . Ž .jt i s js i t


Thus

Ž .1 s c i , p , s, p X X y X X , 22Ž . Ž .Ž .Ý Ý i s p p i p p si , s , p

where the sum is taken over all i, s, p distinct.Ž . Ž . � 4Now we prove that c i, p, s, p s c i, q, s, q , for i / s and p, q f i, s .

Let A s xE q x 2E y E y xE q E . The only nonzero entries in Ap p p q q p qq i soccur in the 4 = 4 principal submatrix lying in rows and columns p, q, i, s,viz.,

2x x 0 0y1 yx 0 0 .

0 0 0 10 0 0 0

Then A2 s 0 and A has only four nonzero 2 = 2 subdeterminants:

A A y A A s xi s p p i p p s

A A y A A s yxi s qq iq qs

A A y A A s x 2i s p q iq p s

A A y A A s y1.i s q p i p qs

Ž . Ž . Ž . Ž . 2Thus 0 s q A s c i, p, s, p x y c i, q, s, q x q c i, p, s, q x yŽ . Ž . Ž .c i, q, s, p , for all x. It follows that c i, p, s, p s c i, q, s, q .

Ž . Ž .For each pair i / s, let d i, s be the common value of c i, p, s, p for� 4all p f i, s . Then

Ž .1 s d i , s X X y X X , 23Ž . Ž .Ž .Ý Ý Ý i s p p i p p spi/s

where the inner sum can be taken over all p, since X X y X X s 0i s p p i p p s� 4 Ž .whenever p g i, s . Equation 23 can be rewritten as

Ž .1 2s d i , s X tr X y X . 24Ž . Ž . Ž . Ž .Ž .Ý Ý i si si/s

Ž1. 2 Ž 2 . Ž .Thus Ý is a linear combination of the 2n polynomials X , tr X X .i j i jNow we turn to the third sum:

Ž .2 s c i , s, i , s X X y X X .Ž . Ž .Ý Ý i i s s i s s ii-s

WILLIAM WATKINS562

Ž .The scalars c i, s, i, s satisfy

c i , s, i , s y c j, s, j, s y c i , t , i , t q c j, t , j, t s 0, 25Ž . Ž . Ž . Ž . Ž .

for all i, j, s, t distinct. To prove this, let

A s xE q x 2E y E y xE q yE q y2E y E y yE .i i i j ji j j s s st t s t t

All nonzero entries of A occur in the 4 = 4 principal submatrix lying inrows and columns i, j, s, t, viz.,

2x x 0 0y1 yx 0 0

.20 0 y y0 0 y1 yy

Thus A2 s 0 and A has 16 nonzero 2 = 2 subdeterminants. Each of theseis of the form "x a y b, where 0 F a, b F 2. Only four of the subdetermi-nants are of the form "xy, namely

A A y A A s xyii s s i s si

A A y A A s yxyj j s s js s j

A A y A A s yxyii t t i t t i

A A y A A s xy.j j t t jt t j

XŽ .Thus the coefficient of xy in q A is

c i , s, i , s y c j, s, j, s y c i , t , i , t q c j, t , j, t .Ž . Ž . Ž . Ž .XŽ . Ž . Ž .And since q A s 0 for all x, y. Eq. 25 must hold. Abbreviate c i, j, i, j

Ž . Ž .to c i, j so that Eq. 25 becomes

c i , s q c j, t s c j, s q c i , t . 26Ž . Ž . Ž . Ž . Ž .

Ž . Ž . Ž .Equipped with Eq. 26 and the fact that c i, j s c j, i , we prove thatŽ . Ž . Ž . Ž . Ž .there are scalars a 1 , . . . , a n such that c i, s s a i q a s for all

1� 4 Ž . Ž Ž . Ž . Ž ..i / s. For i f 1, 2 , define a i s c i, 1 q c i, 2 y c 1, 2 . Thus for21� 4 Ž . Ž . Ž Ž . Ž . Ž . Ž .i, s f 1, 2 , a i q a s s c i, 1 q c i, 2 y c 1, 2 q c s, 1 q2

Ž . Ž .. Ž . Ž . Ž . Ž .c s, 2 y c 1, 2 . But from Eq. 26 we have c i, 1 q c s, 2 s c i, s qŽ . Ž . Ž . Ž . Ž . Ž . Ž . Ž .c 1, 2 and c i, 2 q c s, 1 s c i, s q c 1, 2 . Thus a i q a s s c i, s .

1 1Ž . Ž Ž . Ž . Ž .. Ž . Ž Ž .Define a 1 s c 1, 2 q c 1, 3 y c 2, 3 and a 2 s c 1, 2 q2 2Ž . Ž .. Ž . Ž . Ž . Ž . Ž .c 2, 3 y c 1, 3 . Then a 1 q a 2 s c 1, 2 and for s G 3, a 1 q a s s

1 Ž Ž . Ž . Ž . Ž . Ž . Ž .. Ž .c 1, 2 q c 1, 3 y c 2, 3 q c s, 1 q c s, 2 y c 1, 2 . From Eq. 26 ,2


Ž . Ž . Ž . Ž . Ž . Ž . Ž .c 1, 3 q c s, 2 s c 2, 3 q c 1, s , so a 1 q a s s c 1, s . Similarly,Ž . Ž . Ž .a 2 q a s s c 2, s for s G 3.Now we have

1Ž .2 s a i q a s X X y X XŽ . Ž . Ž .Ž .Ý Ý i i s s i s s i2 i/s

1s a i q a s X X y X XŽ . Ž . Ž .Ž .Ý Ý i i s s i s si2 si

1s a i X X y X X q a s X X y X XŽ . Ž . Ž . Ž .Ý Ý Ý Ýi i s s i s s i i i s s i s s iž /2 s si i

12s a i X tr X y XŽ . Ž . Ž .Ž .Ý i ii iž2 i

q a s X tr X y X 2Ž . Ž . Ž .Ž .Ý s ss s /s

s a i X tr X y a i X 2 .Ž . Ž . Ž . Ž .Ý Ý i ii ii i

Thus ÝŽ2. is an F-linear combination of the 2 n2 polynomialsŽ 2 . Ž .X , tr X X .i j i j

2 Ž 2 . Ž .Now we prove that the 2n polynomials X , tr X X are F-linearlyi j i jindependent. Suppose

a i , j X 2 q b i , j tr X X s 0,Ž . Ž . Ž . Ž .i jÝ i jij

Ž . Ž . � 4for some scalars a i, j and b i, j in F. Suppose s / t. Pick k f s, t and2 Ž .substitute A s E q E for X. Then A s E and tr A s 0. So 0 ssk k t st

Ž .Ž . Ž . 2Ý a i, j E s a s, t . Now let A s E y E . Then A s E q Ei j st i j s s t t s s t tŽ . Ž .Ž . Ž . Ž .and tr A s 0. So 0 s Ý a i, j E q E s a s, s q a t, t . Thus, fori j s s t t i j

Ž . Ž . Ž . Ž .example, a 1, 1 s ya 2, 2 s a 3, 3 s ya 1, 1 . Since the characteristicŽ . Ž . Ž .of F is not 2, a 1, 1 s 0. Similarly, a i, i s 0 for all i. So a i, j s 0, for

all i, j and we have

b i , j tr X X s 0.Ž . Ž .Ý i jij

Ž .Ž . Ž .Substitute A s E q E for X. Then 0 s Ý b i, j E q E s b s, sss st i j s s st i jŽ . Ž .Ž .q b s, t . Finally, substitute A s E for X to get 0 s Ý b i, j E ss s i j s s i j

Ž . Ž .b s, s . It follows that b i, j s 0 for all i, j.

Ž . Ž .2 Ž .Proof of Example. Let Q X s tr X E be defined as in 8 . Suppose112=2 2 Ž . Ž .A g F and A g Rank 1 . Then A g Rank 1 and tr A / 0. Thus

WILLIAM WATKINS564

Ž . Ž . Ž .2Q X satisfies condition 4 , for r s 0, 1. Now suppose tr X E s112 Ž 2 Ž .2 . Ž . 2=2P X P q tr X y tr X C, for some P , P g GL F and C g F .1 2 1 2 2

Substitute E " E for X to get 4E s P P y 2C and 0 s P P q 2C.11 22 11 1 2 1 2Ž .Then 4E s 2 P P , which is impossible since P , P g GL F . The same11 1 2 1 2 2

Ž .2 Ž 2 . tcontradiction arises from the assumption that tr X E s P X P q11 1 2Ž 2 Ž .2 .tr X y tr X C.

Proof of Theorem

Ž . Ž . Ž . Ž .Proof of c . Let q X be any one of the entries q X of Q X . Fromi jŽ .the hypothesis, q X is in the space II of quadratic polynomials that2

� n=n 2 4annihilate the algebraic set VV s A g F : A s 0 . Thus by Lemma 5,Ž . Ž . Ž .there exist F-linear polynomials l and m, such that q X s tr X l X q

Ž 2 . Ž .m X . Equation 7 follows.2 Ž . Ž .The converse is easy. If A s 0, then tr A s 0. So if Q X has form 7 ,

Ž .then Q A s 0.

Ž . Ž . Ž .Proof of b . Suppose Q X has one of the forms in 6 ; then clearlyŽ . 2condition 4 holds for r s 0. For r s 1, let A be a matrix with A g

Ž . 2 Ž .2 Ž . 2 Ž . Ž 2 . tRank 1 . Then tr A s tr A . So Q A s P A P or Q A s P A P1 2 1 2Ž . Ž .and thus Q A g Rank 1 .

Ž . Ž .Conversely, suppose condition 4 holds for r s 0, 1. From part a , weŽ .know that there are linear transformations L and M such that Eq. 7

holds:

Q X s tr X L X q M X 2 .Ž . Ž . Ž . Ž .

Ž . Ž .Form of M X . We begin by proving that there exist P , P g GL F1 2 nsuch that one of the following holds:

M A s P AP , for all A g sl FŽ . Ž .1 2 n27Ž .

tM A s P A P , for all A g sl F .Ž . Ž .1 2 n

n ² :HLet u, ¨ g F be nonzero vectors with u ? ¨ s 0. Then dim u, ¨ G² :Hn y 2. We want to pick w g u, ¨ such that w ? w / 0. Suppose to

² :H ² :Hthe contrary, that w ? w s 0 for all w g u, ¨ . That means u, ¨ is atotally isotropic subspace of F n. But the maximum dimension for a totally

w xisotropic subspace is nr2 6 . Since n G 5, this is impossible. Let A s? @2 Ž . Ž . Ž .u m w q w m ¨ . Then A s w ? w u m ¨ g Rank 1 . Thus Q A s

Ž . Ž . Ž 2 . Ž . Ž . Ž .tr A L A q M A s w ? w M u m ¨ g Rank 1 . So we have provedthat for nonzero u, ¨ g F n,

u ? ¨ s 0 implies M u m ¨ g Rank 1 . 28Ž . Ž . Ž .Ž .Thus by Lemma 4, 27 holds.


Ž .Depending on which one of the two conditions in 27 holds, replaceŽ . Ž . Ž . y1 Ž . y1 y1 Ž . y1 y1 Ž . y1Q X , M X , L X with P Q X P , P M X P , P L X P or1 2 1 2 1 2

Ž y1Ž Ž .. y1 .t Ž y1Ž Ž .. y1 .t Ž y1Ž Ž .. y1 .twith P Q X P , P M X P , P L X P . Then Eq.1 2 1 2 1 2Ž .7 still holds and we may assume that

M A s A , for all A g sl F . 29Ž . Ž . Ž .n

Thus there exists a matrix C g F n=n such that

M X s X q tr X CŽ . Ž .Ž .and thus from 7

Q X s tr X L X q X 2 q tr X 2 C. 30Ž . Ž . Ž . Ž . Ž .

Ž .Form of L X . The next major step is to prove that there is a scalarl g F such that

L A s l A , for all A g sl F . 31Ž . Ž . Ž .n

We begin by proving that

² :L u m u g u m u , 32Ž . Ž .1 2 1 2

n Žwhenever u , u g F , u ? u / 0, u ? u / 0, and u ? u s 0. If L u m1 2 1 1 2 2 1 2 1. Ž .u s 0, there is nothing to prove, so assume L u m u / 0. Suppose2 1 2

u ? u / 0, u ? u / 0, and u ? u s 0. Since F is algebraically closed, we1 1 2 2 1 2² :Hmay assume that u ? u s u ? u s 1. Pick u g u , u with u ? u s1 1 2 2 3 1 2 3 3

Ž1. This is possible since the maximum dimension of a totally isotropicn . Ž .subspace of F is nr2 . Let L s L u m u , for i, j s 1, 2, 3. Now let s? @ i j i j

be in F and define a matrix A s su m u q u m u . Then A2 s s2 u m3 3 1 2 3Ž . Ž . Ž .u g Rank 0, 1 . By the hypothesis of the theorem, Q A g Rank 0, 1 .3Ž .But from 30 ,

Q A s sL A q A2 q s2 CŽ . Ž .s s2L q sL q s2 u m u q s2 C33 12 3 3

w xs s sL q L q su m u q sC .33 12 3 3

Ž .It follows that sL q L q su m u q sC g Rank 0, 1 , for all s / 0.33 12 3 3Ž .But since Rank 0, 1 is an algebraic set, sL q L q su m u q sC g33 12 3 3

Ž . Ž .Rank 0, 1 , for all s g F. In particular for s s 0 we have L g Rank 0, 1 .12Ž .But since L s L u m u / 0,12 1 2

L g Rank 1 . 33Ž . Ž .12

WILLIAM WATKINS566

Ž . 2 Ž .Now pick A s u m su q u . Then A s sA g Rank 0, 1 , for all1 1 2Ž . Ž .s g F. Thus Q A g Rank 0, 1 . But

Q A s sL A q sA q s2 CŽ . Ž .w xs s sL q L q su m u q u m u q sC .11 12 1 1 1 2

Reasoning as before, we have

L q u m u g Rank 0, 1 . 34Ž . Ž .12 1 2

Ž . Ž .From 33 , 34 , and Lemma 1, it follows that

L g u m F n j F n m u . 35Ž .Ž . Ž .12 1 2

Ž . 2Our next choice for A is A s ru m u q u m u q u . Then A s1 2 2 2 3Ž . Ž . Ž .ru q u m u q u g Rank 1 , for all r g F. Thus1 2 2 3

Q A s L A q A2 q CŽ . Ž .s r L q u m u q u q L q L q u m u q u m u q CŽ .Ž .12 1 2 3 22 23 2 2 2 3

g Rank 1 ,Ž .

Ž . Ž .for all r g F. By Lemma 3, L q u m u q u g Rank 0, 1 . Thus by12 1 2 3Lemma 1,

L g u m F n j F n m u q u . 36Ž . Ž .Ž . Ž .12 1 2 3

Ž . Ž .Conditions 35 and 36 imply that

L g u m F n . 37Ž .12 1

Ž . 2 Ž .Finally, let A s u q u m u q ru m u . Then A s u q u m1 3 1 1 2 1 3Ž . Ž .ru q u g Rank 1 . So2 1

Q A s r L q u q u m u q L q L q u q u m u q CŽ . Ž . Ž .Ž .12 1 3 2 11 31 1 3 1

g Rank 1 .Ž .

Ž . Ž .Again by Lemma 3, L q u q u m u g Rank 0, 1 and by Lemma 1,12 1 3 2

L g u q u m F n j F n m u . 38Ž . Ž .Ž .Ž .12 1 3 2

Ž . Ž . n Ž .Conditions 35 and 38 imply that L g F m u . Thus from 37 we get12 2² : Ž .L g u m u and 32 is proved.12 1 2

n Ž .Let e , . . . , e be the standard basis for F and define L s L E .1 n i j i jŽ .Recall that E s e m e . For i / j, we have e ? e s e ? e s 1 andi j i j i i j je ? e s 0. Thus there are scalars l g F such that L s l E for i / j.i j i j i j i j i j


All l are equal. To prove this, take distinct integers i, j, s, t fromi j� 41, 2, . . . , n . This is possible since n G 4. Let u s e q e and ¨ s e q e .i s j t

Ž . Ž .Then u ? u s ¨ ? ¨ s 2 / 0 and u ? ¨ s 0. Thus L u m ¨ s a u m ¨ , forsome a g F, and we have

l E q l E q l E q l E s a E q E q E q E .Ž .i j i j i t i t s j s j s t s t i j i t s j s t

Since E , E , E , E are linearly independent, l s l s l s l s a .i j i t s j s t i j i t s j s tŽ . Ž .Thus all l i / j are equal to a common value l and L E s lE , fori j i j i j

i / j.Ž . Ž .Now we prove that L A s l A, for all A g sl F . Since the matricesn

Ž . Ž .E , E y E , i / j, span sl F , it suffices to prove that L E y E si j i i j j n i i j jŽ .l E y E . So let u s te q e and ¨ s e y te , where t g F. Theni i j j i j i j

2 ' Ž . Žu ? u s ¨ ? ¨ s 1 q t / 0, for all t / " y 1 . Thus from 32 , L u m.¨ s b u m ¨ , for some b g F. That is,

tL y t 2L q L y tL s b tE y t 2E q E y tE ,Ž .i i i j ji j j i i i j ji j j

'for all t / " y 1 . Equivalently,

t L y L y t 2lE q lE s tb E y E y t 2bE q bE .Ž . Ž .i i j j i j ji i i j j i j ji

Ž .Comparing coefficients on powers of t we get L y L s b E y Eii j j i i j jŽ . Ž .and lE s bE . Thus l s b and L E y E s L y L s l E y E .i j i j i i j j i i j j i i j j

Ž . Ž . Ž .It follows that L A s l A, for all A g sl F . So 31 is proved.n

Ž . Ž . n=nForm of Q X . It follows from 31 that there exists a matrix D g Fsuch that

L X s l X q tr X D. 39Ž . Ž . Ž .

Ž .Thus from 30 ,

Q X s tr X l X q tr X D q X 2 q tr X 2 C. 40Ž . Ž . Ž . Ž . Ž .Ž .

Ž .We now prove that l s 0 and D q C s 0, from which part b of thetheorem follows immediately.

Ž . Ž Ž ..Suppose l / 0 and let B s Q E s lE q D q E q C from 40 .11 11 112 Ž . Ž . Ž .2Since E s E g Rank 1 , B g Rank 1 . Also, E q E s E g11 11 11 23 11

Ž . Ž .Rank 1 and so again from 40 ,

Q E q E s l E q E q D q E q C s B q lE g Rank 1 .Ž . Ž . Ž .11 23 11 23 11 23

Ž n. Ž n . Ž n.By Lemma 1, B g e m F j F m e . Similarly, B g e m F j2 3 2Ž n . nF m e , from which we conclude that B g e m F . But an identical4 2

WILLIAM WATKINS568

argument proves that B g e m F n. It follows that B s 0, which contra-3Ž .dicts B g Rank 1 . Thus l s 0.

Now we have

2 2 2Q X s tr X D q X q tr X C 41Ž . Ž . Ž . Ž .

and it remains to show that D q C s 0.2 Ž .Let A s E q tE . Then A s E q tE g Rank 1 , for all t g F.11 12 11 12

Ž . Ž . Ž . Ž .Thus, using 41 , we have Q A s E q tE q D q C g Rank 1 . This11 12Ž .means that all 2 = 2 subdeterminants of E q tE q D q C are zero11 12

and hence the coefficient of t in each of these 2 = 2 subdeterminants iszero. In other words, all entries in D q C are zero except possibly those inrow 1 or in column 2. Similarly, by choosing A s E q tE , all entries in11 13D q C are zero except possibly those in row 1 or in column 3. Thus allentries in D q C are zero except possibly those in row 1. A similarargument proves that all entries in D q C are zero except possibly those

Ž . Ž .in row 2. We conclude that D q C s 0 and from 41 that Q X has theform

22 2Q X s X q tr X y tr X C ,Ž . Ž .Ž .Ž .which proves part b of the theorem.

Ž . Ž .Proof of a . From part b , we may assume that

22 2Q X s X q tr X y tr X C. 42Ž . Ž . Ž .Ž .Suppose

A2 g Rank 2 implies Q A g Rank 2 , for all A g F n=n .Ž . Ž . Ž .

Ž .We prove first that C g Rank 0, 1 by showing that every 2 = 2 subdeter-minant of C is zero. Consider the subdeterminant in rows i, j and columns

� 4 � 4r, s. Pick p, q such that p / q and such that p, i, j and q, r, s are� 4subsets of distinct triples from 1, 2, . . . , n . Let A s E q tE q E ,p p p q qq

2 Ž . Ž .where A g F. Then A s E q 2 tE q E g Rank 2 . Thus, by 42 ,p p p q qqŽ . Ž .Q A s E q 2 tE q E q D g Rank 2 , where D s y2C. Thus allp p p q qq

Ž .3 = 3 subdeterminants of Q A are zero. In particular, the 3 = 3 subde-Ž .terminant of Q A in rows p, i, j and columns q, r, s is zero, i.e.,

2 t q d ) )p q

) d ddet s 0. 43Ž .i r i s

) d djr js


Ž .Thus, the coefficient on t in 43 is also zero, i.e., the 2 = 2 subdetermi-Ž .nant of D in rows i, j and columns r, s is zero. Hence D, C g Rank 0, 1 .

Ž .Now set t s 0 and consider the subdeterminant of Q A in rows p, q, iand columns p, q, j,

1 q d d dp p p q p j

d 1 q d d0 s det s d ,q p qq q j i j

d d di p iq i j

Ž .since D g Rank 0, 1 . So D s 0 and hence C s 0.

ACKNOWLEDGMENT

The author is grateful to A. Sethuraman for several helpful conversations.

REFERENCES

1. E. Arbarello, M. Cornalba, P. A. Griffiths, and J. Harris, ‘‘Geometry of Algebraic Curves,’’Vol. I, Springer-Verlag, New York, 1985.

2. E. Artin, ‘‘Geometric Algebra,’’ Interscience, New York, 1957.3. R. Hartshorne, ‘‘Algebraic Geometry,’’ Springer-Verlag, New York, 1977.

Ž .4. M. H. Lim, Rank and tensor rank preserves, Linear and Multilinear Algebra 33 1992 ,7]21.

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Quadratic Transformations on Matrices: Rank Preservers