A bounded operator A on a Hilbert space H was positive. These operators were symmetric, and as such constitute a natural generalization of nonnegative real diagonal matrices. The following result is thus both well known and not surprising: A positive operator has a unique positive square root (under operator composition)

Brown, D. R.

Omalley, M. J.

English

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https://ntrs.nasa.gov/search.jsp?R=19780013898 2020-03-22T04:58:49+00:00Z
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	 DEPARTMENT OF MATHEMATICS
	 NASA CR-
?^? UNIVERSITY OF HOUSTON
	 HOUSTON, TEXAS l ^`
^^'nded ^g2
(NASA-CR-151697) ON Nth ROOTS OF POSITIVE 	 N78-21 841
OPRRATORS (Houston ?Tniv.) 7 p HC A02/MF A01
CSCL 12A
Unclas
G3/64 14086
ON N-TH ROOTS OF
POSITIVE OPERATORS
BY
O.R. BROWN & M.J. O'MALLEY
II
ON Nth ROOTS OF POSITIVE OPERATORS
D.R. Brown and M.J. O'Malley
Department of Mathematics
University of Houston
I
^	 Houston, Texas 77004
I
Report #68
February 1978
CA- w:.
PRODUCIBILI`!'Y OF T114
W	 ORIGINAL PAGE IS POOR
ON N-th ROOTS OF POSITIVE OPERATORS
by D.R. Brown and M.J. O'Malleyl
A bounded operator A on a Hilbert space H is positive
provided <Ax,x % _> 0 for `all x 6 H. These operators are
symmetric, and as such constitute a natural. generalization of
non-negative real diagonal matrices. The following result is
thus both well known and not surprising:
Theorem: A positive operator has a unique positive square root
(under operator composition).
This may be established by integration of the correct
function, invoking the spectral theorem for self-adjoint operators.
A mnre accessible argument for those not acquainted with the mysteries
'	 of spectral measures may be found in Ll,p-.3171.
While square roots and their iterates seem to provide a sufficient
analytic tool for most purposes, it is also a (folk) theorem that
positive operators possess unique positive irt-h- roots for every
positive integer n. As iii the n	 2 case, existence follows from an
application of the ,spectral theorem; however, we give an argument in the
spirit of L11. The purpose in so doing is not to exercise the reader's
knowledge of induction, but rather to illustrate another use of the Law of
the Mean as a motivational instrument.
1) Both authors received partial support under NASA contract;NAS-9-15000.
n
i
e2.
REPRODUCIBILITY OF THi
ORIGINAL PAGE IS
Let I be the identity operator on 11, and let B(H)
denote the set, of bounded operators on 11. We will need
the following properties of positive operators:
(1) the relation on positive operators defined b_ A 5 B
if and only i,f B - A is positive, is reflexive,
transitive, and consistent with the notation vQ < A
for any positive A; moreover, this relation is pre-
served by operator addition and positive real scalar
multiplication, and reversed by negative scalar
multiplication.
(2) If A and B are positive and if AB	 BA, then AB is
positive,
(3) If 0 5 A	 I, then 0 5 -A < I.
(4) If 0 5 A, then A	 11AIII,	 so that	 (IIAII) lA < I, if A x 0.
I	 (5) If 0 ^ A	 I, then An < A for all positive integers n.
i
We also require;
Lemma. If {S ) is a sequence in BOO such that 0 <_ S	 S
n	 - n - n+l
I, then there exists S E B(H); such that {Snu)	 Su for
all u E H.
All of the conclusions above are verified by straightforward
arguments in Fl,pp. 317-320;. -
Theorem: Let A c B(H), 0 5 A, and let k be a positive integer.
Then there exists a unique positive oPerator 13 such that B k = A.
Proof: By (4) above, we need only consider the case in which A 	 Z.
_	 a
i3.	 REPRODUCIBILITY OF THE
ORIGINAL PAGE IS POOR
We first prove the existence of B.. Since the theorem is a tautology
for all operators when k	 1, we assume the existence of positive
(k-1)-st roots for all positive operators.
Under the momentary supposition that B exists, let
R = I - A and S = I - B. Then (I - S) k = I	 R, so that
k
(*)	 S = (1/k) 1' R + r E 2 ( r ) (-1)rSra
Clearly the existence of a positive operator satisfying this
implicit relation is necessary and sufficient to establish the
existence of the desired operator B. To this end, we define a
sequence of operators by S o
 = 0, Sn+1
	
(1/k) IR + r, r)(- 1)rSn1
Tn order to,show Sn
	
n+1S	 it suffices to show, under the assumption
0< S	 <_ S -< 'I	 that 0	 S	 S -
n-1 - n	 n+l	 n
l k C E (r)	 r Sr - Sr( / ) r=2
	 (-1) ( n
	 Sn-L)
i
To accomplish this, we digress to a consideratitn of the
^	
k
polynomial f(x)	 J 2 (z)(-1) xr	 (1-x) k + kx	 1. Since
Since f(x) = k C1 - (1
	 x) k-1 7 	 0 on [0,11, clearly f is
increasing on this interval. To translate this to operators, it is
necessary to examine the situation more carefully. By the Mean Value
!.
	
	 Theorem, given 0 _<. y < z 	 1, there exists a (unique) number c e ` (y,z)
such that
**)	 f (z) - ,f (y)	 f,' (c) (z - y)
k-11/(k-1)
Upon solving, c = 1	 C(1/k) rE0 (1 - y) k-r-1 (1 - z)r]
Returning to our operator problem, we Ash to ripply this
information to the sequence {S }. Since all members of this
n
family are polynomials in R = I - A, any two of them commute.
This is a property sufficient to permit imitation of equation (**)
with operators; let z = S 	 y = S n_ l . In this format, we use C
to represent the operator I - J, where J is (any) positive
k-1
(k-l)st root of the operator ( 1/k) rFo (I _ Sn-1) k--r-l (I _ Sn)r.
The following chain of equalities is easily calculated:
Sn+1	 S  = (1/k) • (f (Sn ) - f (Sn-1 ) )
(1/k){kFI - (I - C) k 11} •(Sn	 Sn-l)
- 
[I -^ (I - C)k-ll,(Sn
	
Sn-1)
CI - Jk-1  .(Sn	 Sn-1)
= CI - {(1/k) k-1 (I-S	 ) k-r- l (1-S ) r }7 •(S
	
J O
	  S
i
	
0	 n-]	 n	 n	 n-1
By application of remarks(2), (3) and (S), the assumption of
existence of (k-l)st roots, and the inductive hypothesis Sn-1	 Sn'F
`	 the latter operator product exists and is positive. Hence Sn -_Sn+l''
and the sequence {Sn } is increasing. Of course, the Law of the Mean
is not applicable in this setting, nor is it used other than to motivate
the choice of C. Indeed, the discerning reader will note that the
extremes of the chain above. may be shown to be equal without the
introduction of C. However, the rather unusual factorization of
Sn+1	 S 	 would be more difficult- to discover without the example
REPRODUCIBILITY OF THE
ORIGINAL PAGE IS POOR.
.`urn 	 d by the d(:--v.-.°. eve _n the real function situation.
To invoke.the Lemma and complete the proof of existence of
kth roots, it remains to show S	 I for all. n. Assuming
•	 n -
0 <S < I, we have kS m.+1 = R + 
r 2 (r)(-1)rSm = R - I + kS m + (I - Sm)k.
By remark (5), (I - Sm) k < I	 Sm ; therefore
R + kS - I + (I - S )k	 •R + kS - I + I - S
m	 m	 m	 m
s I + (k-1)Sm 5 kI. Hence
kSm+l :5 kI and Sm+l <_ I, as desired. Thus, the Lemma gives an
operator as in (*) and I	 S = B is a 1^ root of A.
In order to prove the uniqueness of a positive kth root of A,
we first observe that if T is any positive Irth root of A, then T
must perforce commute with A, hence with I - A =- R, hence with each
Sn , and thus with S and I - S = B. Let u E ti, v = (B-T)u.
k k	 k-1 k-r-1 r	 k-1	 k-r-1 r
i
Then 0 = < (B -T )u,v > _ ( r EI B	 T ) (B-T)u,v> = r}0 <B	 T v,v>.
,	 k-r-1 r	 k-r-1 r
'.	 Since B and T commute, 0 < B
	
T , whence < B	 T v,v > = 0,
r	 0,1,...,k-1. Let F
r 
be any positive (hence symmetric) square root
I'	 k-r-1 r	 2	 2
of B	 T	 Then J jF v( I `	 <F v,F v> = <F v,v> =•0, so that
r	 r r	 r
k-r-1 r	 2	 k-r-1 rF r v
	 0 and B	 T v = Fry = 0. Therefore B	 T (B-T)u	 0.	 l
or 
B Tk-rru B	 Tk-r-l 	
r = 0,1,...k-1. In particular, for
r k-1, BT
	 = Tk . Multiplying by T, we have Bk+l = BA = BTk v Tk+l
If k	 2, the argument above shows B y = 0 = Tv, whence
J j (B-T)u j 2	 2< (B-T) u,u > = < (B-T)v,u> = 0. Hence Bu	 Tu for all
u c H, and B is thus unique. Now assume all positive roots, of order
less than k, for positive operators are unique. If k	 2j, then
. 2	 2	 k	 k	 2(Bi) = B j = B	 T	 (Tj) , whence Bi	 Tj and thus B = T. If
k+1	 kllk is odd, we have shown above that B , T 	 so, by the even
6.
exponent argument, again B = T. This completes the proof.
REFERENCE
;. Schechter, Martin,, Principles of Functional Analysis,
Academic Press, New York, 1971.
University of Houston
Houston, Texas, 77004
U-H1GINAL PA(IL^ ,I;^ '


On Nth roots of positive operators

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