Let A and B be positive operators on H. It is clear that A + B is a positive operator. In general, the composition operator AB may not be a positive operator. For example, let H = K² and A(x(1), x(2)) = (x(1) + x(2), x(1) + 2x(2)), B(x(1), x(2)) = (x(1) + x(2), x(1) + x(2)). Then AB(x(1), x(2)) = (2x(1) + 2x(2), 3x(1) + 3x(2)) Ox= for all (x(1), x(2)) = K². Note that A and B are positive operators. But AB is not a positive operator, since it is not self-adjoint and if x = (-4, 3), then (AB(x), x) = -1. One can show that every positive operator A has a unique positive square root C which commutes with 3 BE BL(H) whenever A commutes with B. One can deduce that if A and B are positive operators and AB = BA, then AB is a positive operator. Request explain 43 how? 00
Let A and B be positive operators on H. It is clear that A + B is a positive operator. In general, the composition operator AB may not be a positive operator. For example, let H = K² and A(x(1), x(2)) = (x(1) + x(2), x(1) + 2x(2)), B(x(1), x(2)) = (x(1) + x(2), x(1) + x(2)). Then AB(x(1), x(2)) = (2x(1) + 2x(2), 3x(1) + 3x(2)) Ox= for all (x(1), x(2)) = K². Note that A and B are positive operators. But AB is not a positive operator, since it is not self-adjoint and if x = (-4, 3), then (AB(x), x) = -1. One can show that every positive operator A has a unique positive square root C which commutes with 3 BE BL(H) whenever A commutes with B. One can deduce that if A and B are positive operators and AB = BA, then AB is a positive operator. Request explain 43 how? 00
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Let A and B be positive operators on H. It is clear that A + B
is a positive operator. In general, the composition operator AB may
not be a positive operator. For example, let H = K² and
A(x(1), x(2)) = (x(1) + x(2), x(1) + 2x(2)),
B(x(1), x(2)) = (x(1) + x(2), x(1) + x(2)).
Then
AB(x(1), x(2)) = (2x(1) + 2x(2), 3x(1) + 3x(2))
Ox=
for all (x(1), x(2)) = K². Note that A and B are positive operators.
But AB is not a positive operator, since it is not self-adjoint and if
x = (-4, 3), then (AB(x), x) = -1. One can show that every positive
operator A has a unique positive square root C which commutes with
3
BE BL(H) whenever A commutes with B. One can deduce that if
A and B are positive operators and AB = BA, then AB is a positive
operator.
Request explain 43 how?
00
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