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An operator means a bounded linear operator on Hilbert span. This paper proves the assertion made in its title. Following theorem yields the famous result AB + BA * = I = A * B + BA (1) Where A and B are the bounded linear operator on a Hilbert span H. where B * is self adjoint satisfying the above equation. After modification of this equation some interesting results are obtained. Theorem: If AB + BA * = I has solution B, then 0    (A) and   0 B   . Further 1 2 . B A   and 0   (Re B). Proof. If   (A) denotes the approximate point spectrum of A then* A Abar is complex conjugate. Let 0  , A   then   * 0 , A    and so there exist a sequence   n X of unit vector in H such that * 0 n A x  as n . Now AB + BA * = I* 1 , , n n n n x x AB BA x xor,* 1 , , n n n n ABx x BA x x* * , , 0 n n n n Bx A x A x Bx    as n . This is contradiction, hence   0 . A    Similarly we can prove that   0 . B    since B is a self adjoint, we have   BB  Hence   0 B   Next we suppose that xH be arbitrary then2 * , x AB BA x x* , , ABx x BA x x* * , , Bx A x A x Bx   * * Bx A x A x Bx   2 Bx Ax  i.e. 1 2 B A  Existence Of Necessary Condition For Normal Solution Operator Eqution DOI: 10.9790/5728-1302054046 www.iosrjournals.org 41 | Page i.e. 1 2 B A   Now we suppose that 0    (Re A), this means that 0    (Re A), hence there exists a sequence   n x of unit vectors in H such that Re   0 n A x  as n we have * * 2I AB BA A B BA* * 2 , , n n n n x x A A B B A A x xor,2 2 Re , Re , n n n n ABx x B Ax xor,,Re Re , 0 n n n n I Bx Ax Ax Bx    as n This is contradiction, hence 0    (Re A). Since (Re A) is self adjoint, hence   (Re A)= (Re A).   0 Re A    Theorem : If (i) has a solution B > 0 (i.e. (Bx, x) > 0 for all   / 0 x H  ). Then there is an inner product on H, equivalent to the inner product (.,.), such that Re A > 0 with respect to it. Proof. We define a new equivalent inner product on H by   , , ,forall , x y x By x y H   Since * , 0 B A Awe have* * B A A A A B* * , , B A A x x A A Bx xor,* * , , A A x Bx Bx A A x* , A A x Bx   or,* * , , A A x x A A x x    or, * * , , , , Ax x A x x A x x Ax x    or, * * , , , , Ax x Ax x A x x A x x    or, * Re , Re , Ax x A x x  (2) Again from the equation (1) we have* * 2I A A B B A Athen for any . x H* * 2 , , , x x A A Bx x B A A x xOr,2 * * 2 || || , , x Bx A A x A A x Bx* *

Existence of Necessary Condition for Normal Solution Operator Equation