8.7 Eigenvalues and Eigenfunctions of Operator Equations
489
(T u, u) − λ(u, u) = (T u, u) − λu
2
= 0
then
λ =
(T u, u)
(u, u)
=
(T u, u)
u
2
> 0
Theorem 8.7.1 let T be a self-conjugate operator on the Hilbert space H, then all the
characteristic values of T are the real number, and the corresponding characteristic
functions of different characteristic values are orthogonal to each other.
Proof Let T u = λu, making the inner product of it and u, there is
(T u, u) = (λu, u) = λ(u, u) = λu
2
Because T is a self-conjugate operator, namely T = T
∗ , there is
(T u, u) = (u, T
∗ u) = (u, T u) = (u, λu) = λ(u, u) = λu
2
Therefore
λ(u, u) = λ(u, u)
Two sides are divided by (u, u), we get λ = λ, thus λ must be a real number.
Let λ 1 , λ 2 be two different characteristic values, u 1 , u 2 are the corresponding
characteristic functions, then there are
T u 1 = λ 1 u 1 , T u 2 = λ 2 u 2
Making the inner product of the former expression and u 2 , the inner product of
the later expression and u 1 , we obtain
(T u 1 , u 2 ) = λ 1 (u 1 , u 2 ), (T u 2 , u 1 ) = λ 2 (u 2 , u 1 )
Subtracting the two expressions, we get
(λ 1 − λ 2 )(u 1 , u 2 ) = 0
According to the assumption that λ 1 = λ 2 , thus there is (u 1 , u 2 ) = 0, namely u 1
and u 2 are orthogonal. Quod erat demonstrandum.
The minimal characteristic value and corresponding characteristic function of
finding T with variational method are introduced below.
Let λ 0 be the characteristic value of Eq. (8.7.1), u 0 is the corresponding characteristic function, then there is T u 0 − λ 0 u 0 = 0, making the inner product of it and
u 0 , we get
489
(T u, u) − λ(u, u) = (T u, u) − λu
2
= 0
then
λ =
(T u, u)
(u, u)
=
(T u, u)
u
2
> 0
Theorem 8.7.1 let T be a self-conjugate operator on the Hilbert space H, then all the
characteristic values of T are the real number, and the corresponding characteristic
functions of different characteristic values are orthogonal to each other.
Proof Let T u = λu, making the inner product of it and u, there is
(T u, u) = (λu, u) = λ(u, u) = λu
2
Because T is a self-conjugate operator, namely T = T
∗ , there is
(T u, u) = (u, T
∗ u) = (u, T u) = (u, λu) = λ(u, u) = λu
2
Therefore
λ(u, u) = λ(u, u)
Two sides are divided by (u, u), we get λ = λ, thus λ must be a real number.
Let λ 1 , λ 2 be two different characteristic values, u 1 , u 2 are the corresponding
characteristic functions, then there are
T u 1 = λ 1 u 1 , T u 2 = λ 2 u 2
Making the inner product of the former expression and u 2 , the inner product of
the later expression and u 1 , we obtain
(T u 1 , u 2 ) = λ 1 (u 1 , u 2 ), (T u 2 , u 1 ) = λ 2 (u 2 , u 1 )
Subtracting the two expressions, we get
(λ 1 − λ 2 )(u 1 , u 2 ) = 0
According to the assumption that λ 1 = λ 2 , thus there is (u 1 , u 2 ) = 0, namely u 1
and u 2 are orthogonal. Quod erat demonstrandum.
The minimal characteristic value and corresponding characteristic function of
finding T with variational method are introduced below.
Let λ 0 be the characteristic value of Eq. (8.7.1), u 0 is the corresponding characteristic function, then there is T u 0 − λ 0 u 0 = 0, making the inner product of it and
u 0 , we get