proof: V is an invariant subspace of Hermitian H


by buffordboy23
Tags: hermitian, invariant, proof, subspace
buffordboy23
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#1
Oct9-08, 03:49 PM
P: 540
1. The problem statement, all variables and given/known data

If [tex] \vec{x} [/tex] is an eigenvector of a Hermitian matrix H, let V be the set of vectors orthogonal to [tex] \vec{x} [/tex]. Show that V is a subspace, and that it is an invariant subspace of H.

3. The attempt at a solution

The Hermitian H must act on some linear space, call it K and of dimension N. This space has N linear independent vectors. As given, there exists an eigenspace with dimension 1, so V cannot have dimension greater than N-1, and as a consequence, any vector in the eigenspace cannot be a linear combination of the vectors from V. I understand that for V to be an invariant subspace, that when H acts on a vector within V, the resulting vector must always be an element of V.

I really need help on logically connecting and formalizing these ideas into a coherent proof. Any starting points would be of great benefit, since it has been over a year since I had any linear algebra coursework. Thanks.
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Kreizhn
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#2
Oct9-08, 04:16 PM
P: 743
Since you're talking about matrices, I would imagine you're dealing with a finite space with a well defined inner product.

To show that V is a subspace, show that it is closed under addition and scalar multiplication. That is, let y and z be two elements of V, then y and z are both orthogonal to x. Now show that y+z is orthogonal to x, and cz is orthogonal to x for any field element c (which will follow easily from the linearity of the inner product).

For the invariance part, show that in this case the Hermitian matrix preserves orthogonality. This can be done by using contradiction. That is, assume there is an y in V such that Hy is not orthogonal to x. Then the inner product is not zero: is there some special property of Hermitian matrices that you could apply here to get a contradiction?
buffordboy23
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#3
Oct9-08, 05:58 PM
P: 540
Okay, I understand the content of the first paragraph. The second paragraph confuses me though.

I know that H = H[tex]^{+}[/tex], which is just the complex conjugate transpose. This doesn't appear to help. Another property I know about Hermitian matrices is that their eigenvalues are real-valued--doesn't appear to help. The last property is that unique eigenvalues have corresponding eigenvectors that are orthogonal. This seems to be on the right track somewhat, but then the question is, if this is indeed the correct way to think about the problem, how do I show that the vectors of V are eigenvectors of H?

Dick
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#4
Oct9-08, 06:16 PM
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proof: V is an invariant subspace of Hermitian H


Ok, so x is the eigenvector and take v in V. So v.x=0. You want to show H(v) is also in V. Using that x is an eigenvector, what is v.H(x)? Using the fact H is hermitian, what's the relation between v.H(x) and H(v).x? Conclusion?
buffordboy23
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#5
Oct9-08, 06:54 PM
P: 540
Quote Quote by Dick View Post
Ok, so x is the eigenvector and take v in V. So v.x=0. You want to show H(v) is also in V. Using that x is an eigenvector, what is v.H(x)?
This is just c(v.x) = 0, where H(x) = cx and c is a scalar (eigenvalue), since v.x are orthogonal.


Quote Quote by Dick View Post
Using the fact H is hermitian, what's the relation between v.H(x) and H(v).x? Conclusion?
v.H(x) equals the complex conjugate of H(v).x. This first expression equals 0 as already shown, which implies that the second expression equals zero as well. The second expression is key since it shows that the action of H on v produces another vector that is always orthogonal to x. Therefore the subspace V must be invariant.

Thanks. This was very helpful. I was starting to lean towards the fact that the Jordan Canonical form of a Hermitian is a diagonal matrix, from which it follows that its action on a vector implies that the subspace is always invariant. This seemed to make sense from brief inspection of the theorems, but the concepts are foreign to me so its logical consistency is not exactly clear.
Dick
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#6
Oct9-08, 11:57 PM
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I don't think v.H(x) is the complex conjugate of H(v).x. I think they are actually equal. Not that it matters for questions of orthogonality. But I do think you are overcomplicating this.
buffordboy23
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#7
Oct10-08, 01:35 PM
P: 540
Quote Quote by Dick View Post
I don't think v.H(x) is the complex conjugate of H(v).x. I think they are actually equal. Not that it matters for questions of orthogonality. But I do think you are overcomplicating this.
I meant complex conjugate transpose. I probably was overcomplicating it.
Dick
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#8
Oct10-08, 02:12 PM
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Quote Quote by buffordboy23 View Post
I meant complex conjugate transpose. I probably was overcomplicating it.
H(v).x and v.H(x) are complex numbers. So they aren't complex conjugate transpose either. They are EQUAL.
buffordboy23
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#9
Oct10-08, 06:12 PM
P: 540
Your right. I wasn't clear enough I suppose.

v.H(x) is the scalar (inner) product of the vector v with the vector H(x). Therefore, v.H(x) = H(v).x. since H is hermitian. A number is assigned to a scalar product; in this case, the number is 0 for both scalar products, implying orthogonality.
Dick
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#10
Oct10-08, 06:15 PM
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Right. So it's proved, correct?
buffordboy23
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#11
Oct11-08, 09:12 AM
P: 540
yes.


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