SPACES OF MATRICES WITH SEVERAL ZERO EIGENVALUES M. D. ATKINSON Let V be an w-dimensional vector space over some field F, \F\ ^ n, and let SC be a space of linear mappings from V into itself {SC ^ Horn (V, V)) with the property that every mapping has at least r zero eigenvalues. If r = 0 this condition is vacuous but if r = 1 it states that SC is a space of singular mappings; in this case Flanders [1] has shown that dim#"^n(n 1) and that if dim #" = n(n 1) then either SC = Hom(K, U) for some (n l)-dimensional subspace U of V or there exists vev,v ^0 such that SC - {X e Horn {V, V): vx = 0}. At the other extreme r = n, the case of nilpotent mappings, a theorem of Gerstenhaber [2] states that dim SC ^ n(n -1) and that if dim SC = $n(n -1) then SC is the full algebra of strictly lower triangular matrices (with respect to some suitable basis of V). In this paper we intend to derive similar results for a general value of r thereby providing some common ground for the above theorems. The condition on the cardinal of F plays an important role in our proofs although we do not know whether it is essential to the theorems. Flanders required also the constraint char F ^ 2 at one point in his proof; we do not require this and in fact our type of argument allows thefieldin Flanders' theorem to have arbitrary characteristic. THEOREM 1. Let V be an n-dimensional vector space over somefieldf, \F\ ^ n, and let SC be a subspace o/hom (V, V). Suppose that every mapping in SC has at least r zero eigenvalues. Then dim^t ^ %r(r Proof. Gerstenhaber's theorem proves our result in the case n r = 0 and pro vides the first step of an induction on M r. Assume now that n > r > 0 and that the theorem holds for spaces whose transformations all have more than r zero eigenvalues. Since \r(r+l) + n{n r 1) < \r{r l) + n(n r) we can suppose that SC contains a transformation with precisely r zero eigenvalues. By an appropriate choice of basis for V we may take this to be (represented by) A " A = 0 with Ay an rxr nilpotent matrix and A 2 non-singular. By taking A x in rational canonical form we may take it to be strictly lower triangular. From now on we shall ( J K\ I partitioned so that J is an r x r matrix. M NJ Consider any matrix BeSC of the form [ n 1. For all a e F [ n mt A J 3 \0 NJ \ 0 N + aa 2 J has at least r zero eigenvalues and, except for at most n r eigenvalues of A2~ l N,N + %A 2 is non-singular. Hence for all but at most n r values of a, J + <xa, is Received 24 May, 1979. [BULL. LONDON MATH. SOC, 12 (1980), 89-95]
90 M. D. ATKINSON nilpotent. However the coefficients of the characteristic polynomial of J + cnay are polynomials in a which, since A x is singular, are of degree at most r 1. The nonleading coefficients of the characteristic polynomial all vanish for at least r values of a (since F ^ n) and so are identically zero i.e. J + CLA^ is nilpotent for all a and in particular J is nilpotent. Let / be the space of nilpotent rxr matrices which arise in this way; then dim / < %r(r 1). Let J,,..., J, be a basis for / and consider the r 2 1-vectors l pq whosefc-th component is (J k ) pq. Since t ^ %r{r 1) we may choose a set L of %r{r 1) index pairs (p, q) such that {l P q ' (P> <l) e L] spans the space generated by all the l pq. Let U be the complementary set of 2 r (>* + l) index pairs. Then every rxr matrix J may be written uniquely as J = J L + J V where {Ji) pq = 0if{p,q)eU and {J v ) pq = 0 if (p, q) e L. By the choice of L we have the condition J e J and J L = 0 implies J = 0, or equivalently $C and J L = 0 implies J v = 0. (*) By expressing each matrix of 9C in the form I L^, u.. I we may define a linear \ M NJ map $ on 9C by + Jy K\ 0 fj L 0 Then, by (*), -{( " " ) e SC \v& the kernel of 0. Let Clearly dim SC = dim range (0) + dim kernel {(/>) ^ dim $ + dim ^ + dim Jf + dim X < ir(r -1)+ («-r) 2 +dim ^T +dim Jf and the proof will be completed when we have shown that dim M + dim X ^ r{n r). To do this we shall construct a non-singular bilinear form 6: j/ rn _ r x s/ n. rr -> F where ^>n _ r, ^n- r, r denote the full spaces of r x (n r), {n r)xr matrices over F and show that 6 vanishes on X x Ji. This will prove that ^# is contained in the annihilator of X which has dimension r(n r) dim X, giving dim Ji ^ r{n r) dim X as required. '0 Consider any matrices KeX,M ejf. Then there exist Y = I I = 0> i; ) e 5" ( J H\ = (x.j) e 9C. Since A" + a/1 + /5Y has at least r zero eigenvalues for all
SPACES OF MATRICES WITH SEVERAL ZERO EIGENVALUES 91 a, j3 the coefficient of a"" r " l pa r " l in det {X + aa + p Y- XI) is zero (see [2] p. 615). The total coefficient of A*"" 1 in this determinant is (apart from a sign) the sum of all (w-r + l)-rowed principal minors of X + aa + fiyand we require the coefficient of a"" 1 "" 1 /? in this sum. Consider a typical one of these minors If N'+uA'j where J' + OLA\ is the principal submatrix of X + a.a + /? Ywhose rows are indexed by the set S {1, 2,..., r) and N' + aa' 2 is the principal submatrix whose rows are indexed by the set T {r + l,...,n} and where \S u T\ = n r + 1. By direct calculation the coefficient of a"" 1 " 1 /? in this minor is, to within a sign, Z ij e S s,re T c ij D styitx sj where (C y ) is the matrix of cofactors of A^ and (D st ) is the matrix of cofactors of A' 2. We shall express such a coefficient as where x = (x xx x x x ) and Z is an r{n r)x r(n r) matrix whose entries depend upon (C, v ) and (D st ). If \S\ > 1 then, since A\ is strictly lower triangular, C y = Ofor i,j e S,i ^ j and the matrix Z then has the form /0 with zeros above the diagonal and where each diagonal 0 represents an (n r) x (n r) matrix of zeros. If \S\ = \, say S = {i}, then \T\ = n-r, C n = 1 and {D st ) = E is the matrix of cofactors of A 2. In this case the matrix Z has the form 0 0
92 M. D. ATKINSON where each diagonal block is an (n r) x (n r) matrix, E is the i-th diagonal block and there are zeros elsewhere. Summing over all appropriate S, T we obtain the coefficient of a n ~ r ~ x flx r ~ l in the form yzx 1 where Z = and has zeros above the diagonal. Since A 2 is non-singular so also is E and thus Z is non-singular. This matrix Z defines our non-singular bilinear form vanishing on Jf x M and so completes the proof of the theorem. In the next theorem we consider the case where dim SC is maximal and determine the possibilities for SC. We first prove LEMMA. Let V be an n-dimensional vector space over somefieldf, \F\ ^ n, and let SCbea subspace of'horn (V', V). Suppose that W t and W 2 are ^-invariant subspaces ofv such that. (i) dim W i = s lt dim V/W 2 = s 2, dim W l r\w 2 = t, (ii) SC induces spaces ofnilpotent mappings in each of W t and V/W 2i (iii) every transformation in SC has at least r zero eigenvalues. Then dim 3C < \r{r -1) + n(n - r) - {s x - t){n -s 2-1). Proof W t + W 2 has dimension dim W x + dim W 2 dim Wj n W 2 = s l +n s 2 t and we may choose a basis of it having the form e lv> e t> fl>""> fsi-t> Q\i " 9n-n-t such that e lt..., e t is a basis for W x r\ W 2, e x,..., e t, f lt..., f S]. t is a basis for W ly and e lt... t e ti g lt... i g n _ n _ t is a basis for W 2. We complete this to a basis of V with hi,-.., h S2 _ sl+t. With respect to this basis each transformation of SC has matrix N, 0 0 0 ) A x N 2 0 0 A 2 0 B 0 N, and N 2 are nilpotent t x t and (s x t) x (s t -^ t) matrices since SC restricted to W x is a space of nilpotent mappings. JV 3 is a nilpotent (s 2 s t +t)x {s 2 s l +1) matrix since SC induces nilpotent transformations on V/{W i + W 2 ). Since the whole matrix has at least r zero eigenvalues the (n s 2 t)x(n-s 2 -t) matrix B has at least r s 2 t zero
SPACES OF MATRICES WITH SEVERAL ZERO EIGENVALUES 93 eigenvalues. Now, by Theorem 1, the space of all I * I has dimension at most \Ai N 2 J 1), the space of all matrices N 3 has dimension at most and the space of all matrices B has dimension at most Hr-s 2 -t){r-s 2 -t-l) + (n-s 2 -t)(n-r). A routine calculation now shows that dim SC < \r(r -1) + n(n - r) - (s x - t){n -s 2 -t). THEOREM 2. Let V be an n-dimensional vector space over some field F, \F\ ^ n, and let SC be a subspace of Horn (V, V). Suppose that every mapping in 9C has at least r zero eigenvalues and that SC has dimension precisely \r(r l) + n(n r). Then V contains SEinvariant subspaces V lt V 2 with V^ ^ V 2 and dim V x +dim V/V 2 = r and such that S operates on each of V^ and V/V 2 as the full algebra of strictly lower triangular matrices (with respect to suitable bases). Moreover the subspaces V l, V 2 are uniquely determined by 3C unless r = n. Proof. We continue to use the machinery and notation A, L, U, (j), f, M, Jf, tf, 6 introduced in the proof of Theorem 1. From n-r) = dim 3C < dim J^ + dim JT + dim Ji + dimjt we obtain dimj^ = ^r(r-l), dim^k = (n-r) 2, dim Jl + dim X = r{n-r) and M, Jf are the full annihilators of each other with respect to the bilinear form 9. Moreover, identifying S with < I I: L e if > etc., we have range(0) = 2 e Jt e jr. l\ / J In particular range(0) contains a matrix ( I and so SC contains a matrix. This latter matrix is similar to I J and so we could have taken this matrix as our matrix A to begin with. Let us suppose that this had been done. With this choice of A the bilinear form ( 0(1/, V) is easily seen to be tr(uv). H J\»,»r) be any matrix in SC and consider the characteristic M NJ polynomial det (X + aa-xl). In this polynomial the coefficients of
94 M. D. ATKINSON cc"~ r, OL n ~ r k,...,<x n ~ r k r ~ l are all zero and are plainly equal to the coefficients of X, X,..., k r ~ l in det (H-XI). Thus H is nilpotent. Let Jf be the space of matrices H arising in this way. Then %r(r 1) = dim S ^ dim 2tf ^ \r{j 1) by Gerstenhaber's theorem and so dim Jf = /-(r-l). By Gerstenhaber's theorem again M may be reduced to the full algebra of strictly lower triangular matrices with a similarity matrix (T 0\ 7. Therefore by transforming SC by the similarity matrix I I we can take Jf to be the full algebra of strictly lower triangular matrices. Having done this we may compute the coefficient of a"~ r ~ 1 A r ~ 1 in det (X + aa-xi). This yields tr (JM) = 0. (H K\ Consider any matrix Y=[ I in SC. Then for all MtM there exists N 0 A 1 e 9C and we have tr (JM) = 0 and, since Y+Ze%, tr ((K + J)M) = 0. Hence tr (KM) = 0 and K is in the annihilator of Jt with respect to 6 i.e. K e X. Hence contains ( ) for all strictly lower triangular H and all N. Now we show that either every matrix in SC has first row zero or that every matrix in X has r-th column zero. Suppose that this is not so, in which case we may find some matrix X = (x u ) which has neither first row nor r-th column zero, say x ly x,- r ^ 0 for some i,7 > r. As shown above #" contains also the matrix I I where H has ones in the (t + l,t) position, t 1, 2,..., r, and zeros elsewhere and N is any matrix with a single one in every row and column except for the i-th row and j-th column and zeros elsewhere. But then x^x,,. is the coefficient of a"" 2 in det (X + aw) and so is zero since r ^ 1, a contradiction. In terms of matrices the existence part of the theorem asserts that the matrices of SC are simultaneously similar to matrices where L y, L 2 are strictly lower triangular matrices of degrees r {, r 2 with r t + r 2 = r. We can now prove this statement by induction taking as an inductive hypothesis that the statement is true for smaller values of n and all appropriate values ofr. In the case where the matrices of SC have zero first row we consider the space of (n 1) x (n 1) matrices SC obtained by deleting the first row and first column. It is clear that each matrix of SC has at least r 1 zero eigenvalues and hence ^ %r-\){r-2) + {n-\){n-r). On the other hand from the definition of SC, $* dim^-(n-l) = Hr-l){r-2) + {n-l)(n-r). Thus equality holds and the inductive hypothesis can be applied to St. Hence, for some non-singular (n -1) x (n -1) matrix T the matrices of T~ X SCT have the form
SPACES OF MATRICES WITH SEVERAL ZERO EIGENVALUES 95 with L\, L 2 strictly lower triangular of degree r x 1, r 2 and r x 1 + r 2 = r 1. But then taking S = I I the matrices of S~ 1 %'S have the required form. In the case where the matrices of 9C have zero r-th column we use a similar argument taking St to be the space obtained by deleting the r-th row and r-th column from the matrices of SC. Finally, returning to the language of vector spaces and linear mappings, suppose that U x, U 2 are two subspaces of V which have all the stated properties of V X,V 2. Let W x = U x + V x, W 2 = U 2 n F 2,dim W x = s Xi dim V/W 2 = s 2,dim W x n W 2 = t. Then SC induces nilpotent mappings in both W x and V/W 2 and therefore 2C, W x, W 2 satisfy all the conditions of the lemma. It follows that ir(r -1) + n{n - r) ^ \r(r -1) + n{n - r) - (s x - t)(n -s 2-1). But (s 1 t)(n s 2 t) is non-negative and so is zero. Hence s x = tor n = s 2 + t. From s x = t it follows that W x < W 2 i.e. U x + V x ^ U 2 n V 2 and hence U x = V x, U 2 = V 2 since otherwise all transformations in 9 would have more than r zero eigenvalues, contradicting Theorem 1. From n = s 2 + tit follows that W 2 ^ W x and SC is a space of nilpotent matrices i.e. r = n. References 1. H. Flanders, "On spaces of linear transformations with bounded rank", J. London Math. Soc, 37 (1962), 10-16. 2. M. Gerstenhaber, "On nilalgebras and linear varieties of nilp&tent matrices I", Amer. J. Math., 80 (1958), 614-622. Department of Computing Mathematics, University College, Cardiff.