Int. Journal of Math. Analysis, Vol. 3, 2009, no. 5, 231 - 238
Optimal Distance Vectors Kallol Paul Department of Mathematics Jadavpur University Kolkata 700032,India
[email protected] S. Ghosh Dastidar Department of Mathematics VidyaSagar Evening College Kolkata, India Abstract We study those vectors which are at a farthest distance from being an eigenvector of a matrix defined over the real field. Using the concept of stationary distance vectors studied by Das[2] and Lagrange multipliers we find the necessary and sufficient condition for a vector to be at an optimal distance from being an eigenvector. We also give examples to compute the optimal distance vectors for a real matrix.
Mathematics Subject Classification: 15A99; 47A75; 47C99 Keywords: Eigenvector, matrix
1
Introduction
Suppose T is a bounded linear operator on a complex Hilbert space H with inner product h, i. Then for a unit vector f , kT f − hT f, f if k gives the deviation of the vector f from being an eigen vector. Geometrically T f − hT f, f if is the component of Tf perpendicular to f. The important point here is how much a unit vector can move away from being an eigenvector and what is the length of that movement. The supremum of length of all such movements was studied by Bjorck and Thomee[1], Garske[3], Prasanna[8], Fujii and Prasanna[5], Fujii and Nakamoto[4] Furuta et al[6], Das[2] amongst many others. One of the authors Paul[7] studied the same problem with reference to generalized eigenvalue problem. Das[2] used the notion of stationary vectors to characterize the vectors f for which kT f − hT f, f if k is maximum or minimum. We
232
K. Paul and S. Ghosh Dastidar
define these vectors f as the optimal distance vectors. The notion of optimal distance vectors plays an important role in Operator Theory . We here study those vectors which are at a farthest distance from being an eigenvector. In this paper, we use Lagrange multipliers to compute the set of optimal distance vectors for matrices on the real field. Seddighin[9] used the concept of Lagrange multipliers to compute antieigenvectors of a real matrix. This idea of using Lagrange multipliers looks natural but for a general matrix the resulting equations are non-linear and hard to solve, even in the case of 2x2 matrices. To study the optimal distance vectors we use the concept of stationary vectors studied by Das in [2], the definition of which is given below : Definition 1.1 Stationary vector. Let φ(f ) be a functional of a unit vector f ∈ H. Then φ(f ) is said to have a stationary value at f if the function wg (t) of a real variable t, defined as wg (t) = φ(
f + tg ) kf + tgk
has a stationary value at t=0 for any arbitrary but fixed vector g ∈ H. The vector f is then called a stationary vector.
2
Structure of optimal distance vectors
For the sake of completeness we here give the necessary and sufficient condition for f to be an optimal distance vector. We write φ(f ) = kT f − hT f, f if k2 and wg (t) = k T (
f + tg f + tg f + tg f + tg ) − hT ( ), ( )i k2 kf + tgk kf + tgk kf + tgk kf + tgk
where g is an arbitrary but fixed vector of H. If f is a stationary vector then wg′ (0) = 0 and so we get ¯ (T ∗ − λ)(T − λ)f = kT f − hT f, f if k2 f Das[2] proved that for a self adjoint positive operator T M −m supkf k=1 kT f − hT f, f if k = 2
,
where m and M are the least and greatest eigen values of T. Fujii and Prasanna[5] proved that for a bounded linear operator supkf k=1 kT f − hT f, f if k ≥ wT
233
Optimal distance vectors
where wT is the radius of the smallest circle containing the numerical range of T. We here find the supremum for an n × n real matrix. If T = [tij ], 1 ≤ i ≤ n, 1 ≤ j ≤ n is an n×n matrix on the real field and f = (x1 , x2 , . . . xn ) is a vector in Rn , then the functional φ(f ) = kT f − hT f, f if k2 = kT f k2 − | hT f, f i |2 becomes n n n n J(f ) =
X X
(
i=1 j=1
X X
tij xj )2 − (
(
tij xj )xi )2
i=1 j=1
So in order to find the optimally distance vectors for an n × n real matrix we need to calculate the optimum values of n n X X
(
i=1 j=1
n n X X
tij xj )2 − (
(
tij xj )xi )2
i=1 j=1
under the condition x21 + x22 + . . . + x2n = 1. We use Lagrange multipliers to compute unit optimal distance vectors. The following theorem shows that even in the case of a 2 × 2 matrix it is not easy to find the optimal distance vectors and optimal distance algebraically. Theorem 2.1 Let T =
a b c d
bers. Then the unit vector f =
!
be a 2 × 2 matrix on the field of real num-
x y
!
will be an optimal distance vector if it
satisfies the following equation: xy(a2 + c2 ) + y 2 (ab + cd) − (ax2 + bxy + cxy + dy 2 )(2axy + by 2 + cy 2 ) = xy(b2 + d2 ) + x2 (ab + cd) − (ax2 + bxy + cxy + dy 2 )(bx2 + cx2 + 2dxy) . . . (1.1). Proof: In order to find the unit optimal distance vector f =
x y
!
we have
to solve the following non-linear equations Jx + 2λx = 0 Jy + 2λy = 0 where J(x, y) = x2 (a2 +c2 )+y 2 (b2 + d2 )+2xy(ab+cd)−(ax2 +bxy+cxy+dy 2 )2 . . . (1.2) Eliminating λ from Jx + 2λx = 0, Jy + 2λy = 0, we get the required equation (1.1). This completes the proof of the theorem.
234
K. Paul and S. Ghosh Dastidar
We will see later the system of equations (1.1) with x2 + y 2 − 1 = 0 can be solved algebraically for some special matrices. However the system (1.1) with x2 + y 2 − 1 = 0 can be solved numerically for all matrices, as the following example shows.
Example 2.2 Find optimum value of kT f −hT f, f if k for the matrix T=
1 1 2 −1
Solution: Here J(x,y)= 5x2 + 2y 2 − 2xy − (x2 + 3xy − y 2 )2 . . . (1.3). Solving the system Jx + 2λx = 0 Jy + 2λy = 0 2 x + y2 − 1 = 0 numerically we get the following set of solutions : ( x = -0.957092 , y = 0.289784 ),( x = 0.957092 , y = -0.289784), ( x = 0.343724 , y = 0.939071)and ( x = 0.343724 , y = -0.939071 ). One can verify that for the first two solutions the maximum value of kT f − hT f, f if k is 2.302776 and for the last two solutions the minimum value of kT f − hT f, f if k is 0. Thus ( x = -0.957092 , y = 0.289784 ),( x = 0.957092 , y = -0.289784) are maximal distance unit vectors and ( x = -0.343724 , y = 0.939071), ( x = 0.343724 , y = -0.939071 ) are minimal distance unit vectors.
√ The problem can also be solved graphically. If we substitute y= 1 − x2 in (1.3) we get √ √ √ J(x, 1 − x2 ) = (3x2 + 2 − 2x 1 − x2 ) − (2x2 − 1 + 3x 1 − x2 )2
!
.
235
Optimal distance vectors
whose graph is given in figure (1.1), the y-axis represents the square of the length of the movement of a vector from√being an eigenvector. On the other hand, if we substitute y=- 1 − x2 in (1.3) we get √ √ √ J(x, − 1 − x2 ) = (3x2 + 2 + 2x 1 − x2 ) − (2x2 − 1 − 3x 1 − x2 )2 whose graph is given below in figure (1.2), the y-axis represents the square of the length of the movement of a vector from being an eigenvector. We next state a corollary the proof of which is based on some elementary calculations. !
a b Corollary 2.3 Let T = be a 2 × 2 matrix on the field of real c d ! x will be a optimal distance vector if it numbers. The unit vector f = y satisfies the following system of equations: ) px4 + qx3 y + rx2 y 2 + sxy 3 + ty 4 + ux2 + vxy + wy 2 = 0 (1.4) x2 + y 2 − 1 = 0 where the coefficients p, q, r, s, t, u, v, w are functions of a, b, c, d as follows: p = −(ac + ab), q = −(b2 + 2ad + 2bc + c2 − 2a2 ), r = −(3bd + 3cd − 3ab − 3ac), s = −(2d2 − 2bc − 2ad − b2 − c2 ), t = (bd + cd), u = (ab + cd), v = (b2 + d2 − a2 − c2 ) and w = −(ab + cd). System (1.4) can be solved easily for some special cases. If the operator T be such that!for which q = 0, s = 0, v = 0 then the unit optimal distance vector x satisfies f= y x2 =
2
y =
−(r − 2t + u − w) ±
q
−(r − u + w − 2p) ±
q
(r − 2t + u − w)2 − 4(p − r + t)(t + w) 2(p − r + t)
(r − u + w − 2p)2 − 4(p − r + t)(p + u) 2(p − r + t)
Corollary 2.4 For the matrix T= = b.
a b −b a
!
, supkf k=1 kT f − hT f, f if k
Proof: For this operator all the coefficients of the first equation of system (1.4) are zero. From the direct computation (by (1.2)) it follows that supkf k=1 kT f − hT f, f if k = b .
236
K. Paul and S. Ghosh Dastidar
Corollary 2.5 For the matrix T= whose components satisfy: x2 =
a b b −a
!
, the unit vectors f =
x y
!
1 a 1 b b , y2 = + √ 2 , xy = √ 2 − √ 2 2 2 2 2 a +b 2 2 a +b 2 a + b2
or
1 b b a 1 2 √ √ , y = + √ 2 − , xy = − 2 2 a + b2 2 2 a2 + b 2 2 a2 + b 2 √ are maximal distance vectors giving supkf k=1 kT f − hT f, f if k = a2 + b2 . ! x whose components satisfy : Also the unit vectors f = y x2 =
x2 = or x2 =
1 1 b a a 2 √ √ , y = , xy = + √ 2 − 2 2 a + b2 2 2 a2 + b 2 2 a2 + b 2
1 a a b 1 , y2 = + √ 2 − √ 2 , xy = − √ 2 2 2 2 2 a +b 2 2 a +b 2 a + b2
are minimal distance vectors giving inf kf k=1 kT f − hT f, f if k = 0. Proof: Using the system of equations (1.4) we obtain 2abx4 − 4(a2 − b2 )x3 y − 12abx2 y 2 + 4(a2 − b2 )xy 3 + 2aby 4 = 0 . Solving this we get the result. Finally we mention another corollary without proof. !
k 0 is an 2 × 2 elementary Jordan matrix then Corollary 2.6 If T = 1 k the unit vectors (±1, 0) are the maximal distance vectors giving sup kT f − hT f, f if k = 1.
kf k=1
Also (0, ±1) are the minimal distance vector giving inf kf k=1 kT f − hT f, f if k= 0. Lastly we find out numerically optimal distance vectors of a 3 × 3 real matrix.
1 −2 1 Example 2.7 T = 0 3 2 1 −1 1
237
Optimal distance vectors
Substituting the entries of the matrix in J(f) we get J(x, y, z) = (x−2y+z)2 +(3y+2z)2 +(x−y+z)2 −(x2 +3y 2 +z 2 −2xy+2xz+yz)2
x where f= y is a unit vector. Using Lagrange multipliers we get z 2x − 3y + 2z − (x2 + 3y 2 + z 2 − 2xy + 2xz + yz)(2x − 2y + 2z) − λx −3x + 14y + 3z − (x2 + 3y 2 + z 2 − 2xy + 2xz + yz)(−2x + 6y + z) − λy 2x + 3y + 6z − (x2 + 3y 2 + z 2 − 2xy + 2xz + yz)(2x + y + 2z) − λz x2 + y 2 + z 2 − 1
= = = =
Solving the above system numerically finally we get four optimal distance vectors
f1 =
f4 =
−0.199744 0.560195 0.803918 0.842347 0.284434 −0.457765
−0.842347 0.199744 , f2 = −0.560195 , f3 = −0.284434 and 0.457765 −0.803918
.
The vectors f1 and f2 correspond to the maximal distance vectors and so sup kT f −hT f, f if k = kT f1 −hT f1 , f1 if1 k = kT f2 −hT f2 , f2 if2 k = 2.675679 .
kf k=1
The vector f3 and f4 correspond to the minimal distance vectors and so inf kT f − hT f, f if k = kT f3 − hT f3 , f3 if3 k = kT f4 − hT f4 , f4 if4 k = 0 .
kf k=1
0.842347 −0.842347 It then follows that −0.284434 and 0.284434 are eigenvectors cor−0.457765 0.457765 responding to the eigenvalue -.218777 .
ACKNOWLEDGEMENTS. The author thanks Professor K. C. Das and Professor T. K. Mukherjee for their invaluable suggestions while preparing this paper.
0 0 0 0
238
K. Paul and S. Ghosh Dastidar
References [1] Bjorck G. and Thomee V., A property of bounded linear operators in Hilbert space, Arkiv. for Math. 4 (1963) 551-555. [2] Das K.C., Stationary distance vectors and their relation with eigenvectors, Science Academy Medals for Young Scientists-Lectures (1978) pp.44-52. [3] Garske G., An inequality concerning the smallest disc that contains the spectrum of an operator , Proc. Amer. Math. Soc., 78 (1980) 529-532. [4] Fujii M. and Nakamoto R., An estimation of the transcendental radius of an operator Math. Japonica 27 (1982) 637-638. [5] Fujii M. and Prasanna S., Translatable radii for operators, Math. Japonica 26 (1981) 653-657. [6] Furuta T., Izumino S. and Prasanna S., A characterization of centroid operators, Math. Japonica 27 (1982) 105-106. [7] Paul K., Translatable radii of an operator in the direction of another operator Scientae Mathematicae 2 (1999) 119-122. [8] Prasanna S., The norm of a derivation and Bjorck-Thomee-Istratescu theorem Math. Japonica 26 (1981) 585-588. [9] Seddighin M., Optimally rotated vectors, International Journal of Mathematics and Mathematical Sciences 63 (2003) 4015-4023. Received: July, 2008
