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## 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!