- #1
P3X-018
- 144
- 0
Homework Statement
Show that the spectrum [itex] \sigma [/itex] of a linear continuous Hermitian operator A on a Hilbert space H consist of real numbers, ie [itex] \sigma(A)\subset \mathbb{R} [/itex].
Homework Equations
Well the spectrum of A are the elements [itex] \lambda\in\mathbb{C} [/itex] such [itex] \lambda I - A [/itex] is NOT invertible (a generalization of eigenvalues).
Hermitian operator A satisfies: (Ax, y) = (x, Ay) for all x,y in H. (.,.) is the inner product in H.
The Attempt at a Solution
It is stated in the problem that I can do it, by first showing the following 2 statements for [itex] \lambda = \alpha + i\beta [/itex],
[tex]1.\quad \|(\lambda I - A)x\| \geq |\beta|\|x\| [/tex]
[tex]2.\quad \mathrm{Im}\left( (\lambda I - A)x, x) = \beta\|x\|^2 [/tex]
I have shown 2. but I'm lost on showing the 1.
By writing 1. as an innerproduct and expanding I can arrive at
[tex] \|(\lambda I-A)x\|^2 = \|\lambda x\|^2 + \|Ax\|^2 - 2\alpha(Ax,x) [/tex]
If I could show that [itex] \|Ax\|^2 - 2\alpha(Ax,x) \geq 0 [/itex] I would have shown 1. But I can't seem to show that, does inequality even hold?
Or maybe there is another way to show 1., any hint would be great!