STRUCTURE OF THE ANTIEIGENVECTORS OFA STRICTLY ACCRETIVE OPERATOR

A necessary and sufficient condition that a vector f is an antieigenvector of a strictly accretive operator A is obtained. The structure of antieigenvectors of selfadjoint and certain class of normal operators is also found in terms of eigenvectors. The Kantorovich inequality for selfadjoint operators and the Davis's inequality for normal operators are then easily deduced. A sort of uniqueness is also established for the values of Re(Af,f) and ‖Af‖ if the first antieigenvalue, which is equal to min Re(Af,f)/(‖Af‖‖f‖) is attained at the unit vector f.


I. INTRODUCTION
Let H be a complex Hilbert space with inner product A a strictly accretive operator on H and A" its adjoint Let A(O (Af,fff(llAfllllfll), f 0 and I.t(A) min Re ^(f) f H, f 0}.Krein [1] and Gustafson [2] have studied I.tl(A) and indicated how the knowledge of I.tl(A) can be useful in the study of the spectrum of certain integral operators, initial value problems and some other areas.Geometrically lxt(A) represents the (real) cosine of the largest (real) angle through which an arbitary non-zero vector f can be rotated by action of A. Gustafson calls it the first antieigenvalue of A and f the corresponding antieigenvector (See Remark, 4.3, however, at the end of Section 4).The exact value of I.tt(A) can be found in the case of a selfadjoint A. using the Kantorovich inequality [2,3,4].
For a class of normal operators, Davis [3] obtained an inequality from which I.t(A) can be found.He made use of the concept of a shell of a Hilbert space operator [5.6] to arrive at the results.Mirman [7] gave a method of estimation of k(A), the higher antieigenvalues of A, which is defined by Gustafson as follows gn(A) min Re(Af,f)/(llAfllllfll) f, 0, f I fo), :)   where f0o is the k t antieigenvector.
In the present note we put tx(A) in a greater and clearer perspective by studying the st'ucture of the vectors that make Re(Af,f)/(llAfllllfll), f 0. stationary. (For definition see Section 2).We call them stationary vectors.Clearly every antieigenvector is a stationary vector.We show that for selfadjoint and normal operators a suitable combination of two distinct eigenvectors yields.a stationary vector while the converse is true for all selfadjoint and a certain class of normal operators.Kantorovich inequality and Davis's inequality all follow as immediate consequences.If the first antieigenvalue, namely rain Re A(f) min Re(Af,f)/(llAfllllfll), f 0 is attained for a unit vector f then it is shown that the values of Re(Af,f) and IIAfll are unique.An equation characterizing vectors for which I^(f)l is stationary, is also given.A recursive variational scheme is used to define higher antieigenvalues.

STATIONARY VALUES OF Re ^(f)
let A be a strictly accretive operator on H and Let B (A + A')/2 and C A'A.Both the operators B and C are obviously positive definite.Re ^(f) represents the cosine of the real angle of rotation of the vector f under the action of A. Re %(0 is said to have a stationary value at a vector f (4 0) if the function ws(t) of a real variable t, defined by (Bf + tB,f + t) (2.1) w s (t)= (Of+ tCg,f + tg) f + tg, f+ tg) has a stationary value at 0 for any arbitrary but fixed vector g.In other words we must.have w' (q) 0 for all g.For lIfll set (Bf,f) b and (Cf,f) c2.With these notations.we e that Re (n(O is stationary at f iff 2c2Re (BLg) b Re (CLg) bc Re (f.g) 0.
(2.2) Since g is arbitrary, we have the following theorem.THEOREM 2.1.Let A(f), b and c be defined as above.A unit vector f is a stationary vector of Re %(0 iff c2(A + A')f-bA'Af-bc2f 0 (2.3) Equation (2.3) obviously characterizes the vectors for which Re q)A(0 is stationary, in particular.a minimum or a maximum.COROLLARY 2.1.ff for a stationary vector f, Af A'f then f is a linear combination 3. STATIONARY VALUF OF ^(I') FOR SELFADJOINT OPERATORS For a selfadjoint operator A we can obtain the structure of the stationary vectors of RA(f) (which is obviously equal to ^(f) in this case), in particular, of the antieigenvectors in terms of the eigenvectors.It will follow that tl(A) 2 (m + M) "l, where as usual tn, M are the least and the greatest eigenvalues of A respectively.This is the Kantorovich inequality (see [3]  and [4]).We now state THEOREM 3.1.A unit vector f is a stationary vector of a selfadjoint operator A iff there exism two eigenvectors whose appropriate linear combination (in the sense given below) yields f.PROOF.If f is a stationary vector, in particular an antieigenvector then it satisfies (2.3).
Conversely, let f Ixie + ojej, with o 2 = J and I ,1 where e and e are any two eigenvectors of A corresponding to eigenvalues ., and So b (Af,f) %,il(xi, = + we can (2.3)see that the equation (2.3) is satisfied by f.This completes the proof.
PROOF.One need only to observe that I.tt(A) is stationary value of ,(0 and 2/'%i / + %i 2 / (./ + /.) assumes the least value for greatest /2, which in this case is M/m.
Hence b2/c 2 will be a minimum for I(f,e3)l I(f,e4)l 0, and then I.tt(A)=l//" Note that in this case b Re %i and c = I,il.
EXAMPLE 4.3.This is the most important example in this section.It shows that unlike the selfadjoint case, a linear combination of more than two eigenvectors corresponding to dinstinct eigenvalues may exist for the attainment of the minimum of Re ^(f).Let us consider the normal operator A such that Af (1 + i) (f,e,)el +1/2 (1 + i) (f,e=)e + (2 + i) (f, e3)e with I(f, el)l 2 + I(f,e)l +l(f,e)l =1, 4-Si-.
cy b Re (Af,0 I(f,=01 + i I(f,e=)l + 2 I(f,=)l + I{f,)l we must have k -I or -1/3.The case k =-1 must be ruled out as in that case l(f,e a) = > 1.
For k -1/3, we have Re A(f) 2/3.Let l(f,e,)l 1/12, J(f.e2)l 10/12 andl(f.%) 10 12" So that unit vector f -el+e 2 + % will be the fhst antieigenvector of A. However it is possible to have a combination of only two eigenvectors corresponding to two eigenvalnes 2 for which the minimum in question is attained.Set I<f,)l o, I<f,)[ and I<f,,)I . Clearly f -et+e 2 will be a required vector.
THEOREM 4.1 Let A be normal with a complete orthonormal set of eigenvectors ei and corresponding eigenvalues i such that for any unit vector f E H, Af E (f,ei)ei.
If Re ^(f) is stationary at f, and f is not an eigenvector of A, then either f is a linear combination of two eigenvectors corresponding to two conjugate eigenvalues or there exists a linear combination of two igenveetors corresponding to two eigenvalnes with unequal real parts and moduli, say g, such that R ^(g) Re ^(f) and Re ^i s stationary at g. Further, we have the relation if ke and lae i are the distinct eigenvalues referred to above and are such that X * la and X cosO.p cos.
PROOF.If Rc A(f) is stationary at f, we must have c2(A+A)f--bA'Af--bc2f 0 (4.1)Substitute f Y (f,c,)c,, Af Z (f, ei)c and A*f Z X (f,c)c in (4.1) to get 2c 2 Re q b Il bc 2 0 (4.2) if (f,e i) 0 As in example 4.1, (4.2) may be satisfied if Af Zf, b Re ,, c IX[ 2. If f is not an eigenvcctor then (4.2) may be satisfied by only one pair (Re ki, I [2) in which case, since f is not an eigenvector, (f,ci) 0 for at least two ei's corresponding to two distinct ki's.So, the only way this can happen is that Af k(f,et)el + X(f,e2)c2 where the rcnumbering of e's is obvious.Here b Re Z, c [[2.This is the ease in example 4.2.
If (4.2) is satisfied by more than one pair (Re 7q, [hi 12) then as in example 4.3, the stationary vector f .maybe a linear combination of more than two eigenvectors corresponding to distinct cigenvalucs.We shall show, however, that there always exist a suitable linear combination g of two cigenvcctors corresponding to two cigenvalucs of distinct moduli and real parts, such that Re (A(g) Re oA(f) and Re 0^i s stationary at g. Let Z, kc ' and kj I.te i, , It and k cosO-[t cos satisfy (4.2).We win f'md and such that g oe, + oqe I1' +lml 1, b Re (Af,f) Rc(Ag,g), c 2 (Af,A0 (Ag,Ag) and (4.1) is satisfied for g.Now b i1 o,e + I1 o,, d I1 +11?.Sic I,1 / I,1= 1, w= gt But since 2c2, co b), bc 2 0 and 2c21x cost bl.tbc 0, the value of e 12 is the same.Hence a g is obtained.A simple calculation further shows that co --, cosO) b 2 4Zt.t (1 cosO , cos _)t (4.3) and that completes the proof.REMARK 4.1.It may be noted thin (4.3) is equivalent to the inequality of Davis [3] when b2/c is a minimum of Re ^(f) without the restrictive conditions (in part (ii) of his Theorem).Hence all cigenvalucs Z with RcZ lying between the above limits and ImP.determined from (4.2) may contribute corresponding eigenvectors to yield .a stationary vector of Re ^(O.
REMARK 4.3.Since rain Re ^(O may be attained even for an eigenvector as in example 4.1, Gustafson's choice to call it an anticigenvaluc does not sm to be a happy one. 5. FURTHER PROPERTIES AT min ^(0 AND VARIATIONAL CHARACTERIZATION Consideration of the second derivative of ws(t), as defined in equation (2.1), shows that ws"(0) >_ 0 is a necessary condition for a unit vector f to be the first antieigenvector.A rather lengthy calculation shows that then 4c2Re (Bf-bf,g) + 2bc(Bg,g) b(C,,g) b2c(g,g) >-0 (5.1) for arbitary g e H, and b,c defined as before.The quantities b and c are unique in the sense of the following theorem.THEROM 5.1.If the rain #^(O is attained for a unit vector f with b (Bf,f) and c (Cf,0, then there cannot exist unit vectors g such that (Bg,g) kb and (Cg,g) kc2, k 1.
PROOF.If such a g exists then without loss of generality we can have Re (Bf-bf,g) 0.
Higher anticigcnvalucs may be defined in a recursivc variational manner as follows Bt (A) min Re #A(O Re A(fl) fH t+t (A) min Re^( f) f,L fl fo f .L Bft Bf, Simplification in the selfadjoint case is obvious where eigenvectors get eliminated pairwise.Gustafson [2] defines imaginary and total anticigcnvalues by rain lm ^(0 and rain respectively.While the equation characterizing imaginary antieigenvectors will be similar to our equation (2.3), that for total antieigenvectors will be slightly different.If f is the unit vector for which rain I^(f)l is attained, then we must have 2J (b=A= bA)f IblCf-Ibl=f--0 where Ax (A + A*)/2, Ay (A A*)/2i, C A*A b (Af,f),b,, Re (Af,f), by =Im (At,f) and c (Cf,f).

REMARK 4. 2 .
The points of the line 2c2X bY bc 0 lying in the region Y > X are such that c 2-c 2 (I (b/c) 2) c2+ c2(I () )