Real eigenvalues of linear operator

**charikaar** · March 30th 2010, 04:45 AM

Let L : D--->H be a linear operator defined on a domain D in a complex Hilbert space H. Given that L is self-adjoint, show that its eigenvalues are real.

I know Lf=af where a is the eigen value of L and f is the eigenfunction. L is self-adjoint <Lf,g>=<f,Lg>
How do i show eigenvalues are real? many thanks

**tonio** · March 30th 2010, 06:12 AM

Originally Posted by charikaar

Let L : D--->H be a linear operator defined on a domain D in a complex Hilbert space H. Given that L is self-adjoint, show that its eigenvalues are real.

I know Lf=af where a is the eigen value of L and f is the eigenfunction. L is self-adjoint <Lf,g>=<f,Lg>
How do i show eigenvalues are real? many thanks

When working with orthonormal basis you know that $L^{*}=\overline{L^t}$ , and if $\lambda$ is an eignevalue of $L$ then $\overline{\lambda}$ is an eigenvalue of $L^{*}$ , so...

Tonio

**HallsofIvy** · March 30th 2010, 01:48 PM

Also, suppose $\lambda$ is an eigenvalue for self adjoint operator L. Then $Lv= \lambda v$ for some v with norm 1.
Now $\lambda= \lambda<v, v>= <\lambda v, v>= <Lv, v>$ . Since L is self adjoint, $<Lv, v>= <v, Lv>= <v, \lambda v>= \overline{<\lambda v, v>}= \overline{\lambda}<v, v>= \overline{\lambda}$ .

Thread: Real eigenvalues of linear operator

LinkBack

Thread Tools

Display

Real eigenvalues of linear operator

Similar Math Help Forum Discussions

[SOLVED] QM - Part 1 - Eigenvalues of the Hermitian Operator A are real

Eigenvalues of a linear operator

Eigenvalues of Linear operator

Eigenvalues of Linear operator

linear operator eigenvalues

Search Tags