# Spectrum of Hermitian operator

## Homework Statement

Show that the spectrum $\sigma$ of a linear continuous Hermitian operator A on a Hilbert space H consist of real numbers, ie $\sigma(A)\subset \mathbb{R}$.

## Homework Equations

Well the spectrum of A are the elements $\lambda\in\mathbb{C}$ such $\lambda I - A$ 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 $\lambda = \alpha + i\beta$,

$$1.\quad \|(\lambda I - A)x\| \geq |\beta|\|x\|$$
$$2.\quad \mathrm{Im}\left( (\lambda I - A)x, x) = \beta\|x\|^2$$

I have shown 2. but I'm lost on showing the 1.
By writing 1. as an innerproduct and expanding I can arrive at

$$\|(\lambda I-A)x\|^2 = \|\lambda x\|^2 + \|Ax\|^2 - 2\alpha(Ax,x)$$

If I could show that $\|Ax\|^2 - 2\alpha(Ax,x) \geq 0$ 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!

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olgranpappy
Homework Helper
Just look at (x,Ax) where 'x' is an eigenvector of A.

For a linear operator in infinite dimensions, eigenvalues sometimes don't even exist. So you can't just consider x to be an 'eigenvector' and say $Ax = \lambda x$. That is elements in the spectrum are not just eigenvalues.

Beside I don't think it is even possible to show $\|Ax\|^2 - 2\alpha(Ax,x) \geq 0$, since there is not exactly a direct connection between $\alpha$ and everything else in that equation, so that number could be arbitrarily big and what could compensate for that to hold the inequality true?

So maybe there is an easier approach?

olgranpappy
Homework Helper
if the eigenvalue doesn't exist then how are you going to prove that it is real?

morphism
Homework Helper
It seems to me that you're not expanding it correctly: $(x,\lambda y) = \bar{\lambda} (x,y)$.

Personally I would rescale A to B=(A-$\alpha$)/$\beta$. Then A-$\lambda$I is invertible iff B-i is. Now since we're in a Hilbert space, an operator is invertible if it and its adjoint are bounded below. So proceed to show that B-i and (B-i)*=B+i are bounded below in the same spirit you were attempting to do (1); this time it's easier because there are less things to juggle around.

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I don't see how that should make it easier, and I assume you mean B-iI and A-alpha*I? since A is an operator. And I haven't heard about the statement that an operator is invertible if it and its adjoint are bounded below.

But the expansion goes like this

$$\|(\lambda I - A)x\|^2 = (\lambda x - Ax,\lambda x - Ax) = (\lambda x, \lambda x) + (Ax,Ax) - (Ax,\lambda x) - (\lambda x, Ax)$$

and

$$(Ax,\lambda x) + (\lambda x, Ax) = (Ax,x)(\lambda + \bar{\lambda}) = 2\alpha(Ax,x)$$

And so my calculation from before follows, what is going wrong here, I can't see it :(

morphism
Homework Helper
Ah - yes, your work is fine. At first I confused your "alpha" and "lambda". Sorry about that!

Try Cauchy-Schwarz to wrap things up: (Ax,x) is real, and thus less than or equal to ||Ax|| ||x||. If you still need help, post back.

(By the way, if you rescale, then you won't need to use Cauchy-Schwarz or worry about 'completing the square'. This is what I meant when I said "easier". But I guess this isn't very helpful if you don't know the theorem I'm referring to.)

dextercioby
Homework Helper
If A is linear, continuous and symmetric it can be extended by continuity to the whole Hilbert space, thus making it self adjoint. In this case you need to show that the spectrum of a bounded self-adjoint operator is purely real. The best full and clear proof for this result is, i believe, in Geroch's book "Mathematical physics", pages 298-299. Unlike other books on functional analysis, he assumes the boundedness of the self-adjoint operator so eveything applies.

bigubau: I heard about extension by continuity in class, but I'm afraid the text book we're using doesn't talk about it at all. Using "An Introduction to Hilbert Space" by N. Young.

morphism: I did consider Cauchy-Schwarz before but didn't succeed in getting anywhere. But now you mentioned completing the square so I tried that and got (nowhere?):

$$\|(\lambda I -A)x\|^2 \geq \left( |\lambda|\|x\| - \frac{\alpha}{|\lambda|}\|Ax\| \right)^2 + \|Ax\|^2\left(1-\frac{\alpha^2}{|\lambda|^2}\right)$$

(did complete square in maple so I don't think there are errors there :P). The last term is non-negative so

$$\|(\lambda I -A)x\| \geq |\lambda|\|x\| - \frac{\alpha}{|\lambda|}\|Ax\| \geq |\beta|\|x\| - \|Ax\|$$

So close...! But you can't proceed anywhere from here, looks like a dead end.. :(

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morphism
Homework Helper
$$\|(\lambda I-A)x\|^2 = \|\lambda x\|^2 + \|Ax\|^2 - 2\alpha(Ax,x)$$

$$\geq |\lambda|^2 \| x\|^2 - 2\alpha \|Ax\| \|x \| + \|Ax\|^2$$

$$= (\alpha^2 + \beta^2) \| x\|^2 - 2\alpha \|Ax\| \|x \| + \|Ax\|^2$$

$$= (\alpha \|x\| - \|Ax\|)^2 + \beta^2 \| x\|^2$$

doh ofcourse, using it that way, Thx!

Well we wanted to show that the spectrum is real, so we have some steps left.
In the problem it is stated that by (1) I get that $\lambda I-A$ (lets call it B from now on) is injective and have closed range. The injective part is easy, but I'm a little unsure about the closed range part.
If I consider any sequence (y_n) with y_n = Bx_n for every n, then since we're in a Hilbert space lim x_n exists, and there for lim y_n is also in the range by continuity of B, but is this correct? I didn't even use (1).

Furthermore it says I can conclude from (2) that B has a dense range. But if the range is closed this would just require showing that B is surjective, right? And I'm completly clueless on this one, any hints?

morphism
Homework Helper
If I consider any sequence (y_n) with y_n = Bx_n for every n, then since we're in a Hilbert space lim x_n exists
This doesn't make a lot of sense to me. Did you mean to say that (y_n) is a Cauchy sequence? If so, then yes this is true, and precisely because we know (1), since we can use it to show that (x_n) is also Cauchy. Then the continuity of B will let us conclude that the image of B is complete and hence closed.

Furthermore it says I can conclude from (2) that B has a dense range. But if the range is closed this would just require showing that B is surjective, right? And I'm completly clueless on this one, any hints?
The point, I think, is to avoid manually proving that B is surjective. Since we know that B has closed range, proving that its range is dense in H will automatically prove that it's surjective.

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This doesn't make a lot of sense to me. Did you mean to say that (y_n) is a Cauchy sequence? If so, then yes this is true, and precisely because we know (1), since we can use it to show that (x_n) is also Cauchy. Then the continuity of B will let us conclude that the image of B is complete and hence closed.
True (y_n) would have to be assumed to be a Cauchy sequence, and then I wanted to show that its limit is in the range, by using that if y_n = Bx_n then lim y_n = B(lim x_n) and we know that lim x_n exist since our space is complete, and hence lim y_n would be in the range.
But what you say is more clear, showing that it is complete rather than closed.

The point, I think, is to avoid manually proving that B is surjective. Since we know that B has closed range, proving that its range is dense in H will automatically prove that it's surjective.
To show a set is dense you can show that the closure equals the space surrounding it, so in this case you would have to show that rg B = H. But showing this would be no different than showing surjectivity.
Are there easier ways of showing that its dense?

morphism
Homework Helper
True (y_n) would have to be assumed to be a Cauchy sequence, and then I wanted to show that its limit is in the range, by using that if y_n = Bx_n then lim y_n = B(lim x_n) and we know that lim x_n exist since our space is complete, and hence lim y_n would be in the range.
But what you say is more clear, showing that it is complete rather than closed.
lim B(x_n) = B(lim x_n) is true only if we know a priori that lim x_n exists. This is where we use (1):
$$\|x_n - x_m\| \leq \frac{1}{|\beta|^2} \|Bx_n - Bx_m\| \to 0$$

To show a set is dense you can show that the closure equals the space surrounding it, so in this case you would have to show that rg B = H. But showing this would be no different than showing surjectivity.
Are there easier ways of showing that its dense?
Yup: in a Hilbert space, we know that $\bar{S} = S^{\perp \perp}$. But the orthogonal complement of rgB is {y : <y,Bx>=0 for all x}, so in particular, for y in the orthocomplement, <y,By>=0 which implies that y=0 (by (2)). So the closure of the range is {0}^\perp = H.

Thanks for your great help morphism.
Now there is only one last thing left, which is to show $\lambda$ is real. Since B is bijective it has an inverse (in the set theoretic sense), so by (1) we get

$$\|BB^{-1}x\| = \|x\| \geq |\beta|\|B^{-1}x\|$$

And so the operator norm of B^{-1} is bounded by

$$\|B^{-1}\| \leq |\beta|^{-1}$$

So for $\beta \neq 0$ B is invertible? It's because I just considered the set-theoretic inverse of B and therefore unsure of weather this implies that B is invertible.
And this (if true) then implies that if $\lambda$ has an imaginary part then it is not in the spectrum, right?

morphism
Homework Helper
It's a consequence (and in fact an equivalent) of the open mapping theorem that the set theoretic inverse of a mapping between Banach spaces is bounded (note that we used the fact that $\beta \neq 0$ to prove that B is invertible). I believe this is sometimes called the "inverse mapping theorem" or the "bounded inverse theorem". The rest of what you said is spot on: if $\lambda$ has an imaginary part, then we're screwed.

So

$$\|B^{-1}\| \leq |\beta|^{-1}$$

Does make sense and does imply that B^{-1} is bounded and hence B is invertible?´

EDIT:
Wait we just showed B is bijective, so B^{-1} IS bounded as you say by the theorem, but we wanted to tell when it's NOT bounded. And you can't just conclude it's not bounded when beta = 0. Or can u?

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morphism
Homework Helper
I couldn't exactly see how you got

$$\|BB^{-1}x\| = \|x\| \geq |\beta|\|B^{-1}x\|$$

But the point is, you don't even need to worry about B^{-1} being bounded. The Hilbert spaceness gives you this for free.

Well I apply (1) for $x = B^{-1}z$ so everything follows. And the boundedness of the operator follows from by dividing through with ||x|| so I get

$$|\beta|\|B^{-1}\left(\frac{x}{\|x\|}\right)\| \leq 1$$

And the norm of an operator is defined as

$$\|B^{-1}\| = \sup_{\|x\|\leq 1}(B^{-1}x)$$

So

$$\|B^{-1}\|\leq\|B^{-1}\left(\frac{x}{\|x\|}\right)\| \leq |\beta|^{-1}$$

But didn't we WANT to show that it IS bounded with that choice of beta, so that we could conclude lambda isn't in the spectrum when it has imgainary part?

EDIT: I guess since I don't know the theorem I have to use another approach.

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morphism
If you don't know the theorem then your method looks like the best alternative, and it does seem correct! But really, it's one of the fundamental theorems of functional analysis. It gives us the boundedness of the inverse for free once we establish its existence (and we DID establish it for said choice of beta). So I guess the problem is fully solved now. 