# Invariant Subspace

1. Jun 29, 2010

### sol66

So i'm trying to get an idea of what an invariant subspace is and so please let me know if my understanding is correct. Given that you have some vector subspace being a collection of a particular number of vectors with the the space denoted as |$$\gamma$$>. If you have some other collection of vectors, not necessairly being a subspace in itself ... however we'll say that this collection of vectors is denoted as |$$\beta$$>. If |$$\gamma$$> is to be |$$\beta$$> invariant, does that mean beta is contained in the vector space gamma?

Or for gamma to be beta invariant, does that mean the vector collection |$$\beta$$> must make up a vector subspace itself spanning the vector subspace |$$\gamma$$>? If this is the definition, must beta be a collection of orthogonal vector states?

I'm not sure which definition it should be, or if I'm even right with any of the definitions.

Thanks for the responses.

2. Jun 29, 2010

### Fredrik

Staff Emeritus
$|\gamma\rangle$ is the notation for a vector, not a subset or a subspace. Let's use a more standard notation. Let's call the vector space V, and let S be a subset and U a subspace. You're asking about V being "S invariant". I don't think this concept makes sense. But suppose that T:V→V is linear. Now U is said to be invariant under T (or an invariant subspace of T) if Tx is in U for all x in U. If G is a group of linear operators, U is said to be invariant under G if U is invariant under T for all T in G.

3. Jun 29, 2010

### sol66

So what your saying is if you operate on some vector x contained in your subspace V by the linear operator T and the resulting transformed vector is Tx is contained or spanned by the subspace V you have a T invariant subspace? As long as the transformation applied to all vectors contained in subspace V has that property that means your subspace V is T invariant? So is putting this in my own words, is this what you're saying? I'm not so accustomed to using such notation and descriptor words, thats why it's easy for me to get confused. Is this definition right?

4. Jun 29, 2010

### Matthollyw00d

It needs to be for all x. As you stated next.
Right so $$T(V) \subseteq V$$, then V is T invariant.

5. Jun 29, 2010

### sol66

Hey thanks for the help

6. Jun 29, 2010

### Fredrik

Staff Emeritus
A good example is a rotation around the z axis (in 3-dimensional space). The z axis is the only 1-dimensional invariant subspace, and the xy plane is the only 2-dimensional invariant subspace.

7. Jun 30, 2010

### sol66

In other words given that your z axis only acts like a z operator/obsevable, the subspace on the z axis is z invariant. While the xy plane has the x and y operator spanning it become x and y invariant? That makes sens.