7.4 Operators and Functionals
411
Thus, for any x ∈ H , there is
(x, T 1 T 2 y) = (x, T 2 T 1 y)
Getting T 1 T 2 = T 2 T 1 .
The sufficiency. If T 1 T 2 = T 2 T 1 , then T 1 T 2 must be the self-conjugate operator.
As a matter of fact, from the self-conjugacy of T 1 and T 2 and T 1 T 2 = T 2 T 1 , there is
(T 1 T 2 x, y) = (T 2 x, T 1 y) = (x, T 2 T 1 y) = (x, T 1 T 2 y)
Hence T 1 T 2 must be the self-conjugate operator. Quod erat demonstrandum.
Let H be the Hilbert space, X is a subset in H , T is a bounded linear operator of
X to H , if for any x ∈ X , there is always (T x, x) 0, then T is called the positive
operator. If for an arbitrary nonzero x ∈ X , there is always (T x, x) > 0, then T is
called the positive definite operator.
The positive definite operator can also be defined like this: Let T be the bounded
linear operator in the Hilbert space H , if for an arbitrary nonzero x ∈ H , there
exists the constant r > 0, such that (T x, x) r
2
(x, x) = r
2
x
2 holds, then the
operator T is called the positive definite operator. When T is the unbounded linear
operator, let T be the densely defined operator in the Hilbert space, D(T ) is the
domain of definition of T , if for an arbitrary non zero x ∈ D(T ), there exists the
constant r > 0, there is always (T x, x) > 0, then the operator T is called the positive
definite operator.
If T is a positive definite operator, then the equation T y = f (x) at most has only
one solution.
In fact, if there are two different solutions y 1 and y 2 , then there must be
T y 1 − f = 0, T y 2 − f = 0
Subtracting the former equation and the later equation, we get
T (y 1 − y 2 ) = 0
Thus
[T (y 1 − y 2 ), y 1 − y 2 ] = 0
This is inconsistent with T being positive definite operator, therefore the equation
T y = f has only one solution.
Example 7.4.1 If T the self-conjugate operator in H , λ is a real number, prove that
λT and T − λI are both self-conjugate operators.
Proof Because T is the self-conjugate operator, by the definition there is (T x, y) =
(x, T y), and λ is a real number, to give λ = λ. Thus
411
Thus, for any x ∈ H , there is
(x, T 1 T 2 y) = (x, T 2 T 1 y)
Getting T 1 T 2 = T 2 T 1 .
The sufficiency. If T 1 T 2 = T 2 T 1 , then T 1 T 2 must be the self-conjugate operator.
As a matter of fact, from the self-conjugacy of T 1 and T 2 and T 1 T 2 = T 2 T 1 , there is
(T 1 T 2 x, y) = (T 2 x, T 1 y) = (x, T 2 T 1 y) = (x, T 1 T 2 y)
Hence T 1 T 2 must be the self-conjugate operator. Quod erat demonstrandum.
Let H be the Hilbert space, X is a subset in H , T is a bounded linear operator of
X to H , if for any x ∈ X , there is always (T x, x) 0, then T is called the positive
operator. If for an arbitrary nonzero x ∈ X , there is always (T x, x) > 0, then T is
called the positive definite operator.
The positive definite operator can also be defined like this: Let T be the bounded
linear operator in the Hilbert space H , if for an arbitrary nonzero x ∈ H , there
exists the constant r > 0, such that (T x, x) r
2
(x, x) = r
2
x
2 holds, then the
operator T is called the positive definite operator. When T is the unbounded linear
operator, let T be the densely defined operator in the Hilbert space, D(T ) is the
domain of definition of T , if for an arbitrary non zero x ∈ D(T ), there exists the
constant r > 0, there is always (T x, x) > 0, then the operator T is called the positive
definite operator.
If T is a positive definite operator, then the equation T y = f (x) at most has only
one solution.
In fact, if there are two different solutions y 1 and y 2 , then there must be
T y 1 − f = 0, T y 2 − f = 0
Subtracting the former equation and the later equation, we get
T (y 1 − y 2 ) = 0
Thus
[T (y 1 − y 2 ), y 1 − y 2 ] = 0
This is inconsistent with T being positive definite operator, therefore the equation
T y = f has only one solution.
Example 7.4.1 If T the self-conjugate operator in H , λ is a real number, prove that
λT and T − λI are both self-conjugate operators.
Proof Because T is the self-conjugate operator, by the definition there is (T x, y) =
(x, T y), and λ is a real number, to give λ = λ. Thus