# Linear algebra proof

1. Jun 10, 2009

### evilpostingmong

1. The problem statement, all variables and given/known data
Prove or give a counterexample: If U is a subspace of V that is invariant
under every operator on V, then U={0} or U=V.

2. Relevant equations

3. The attempt at a solution
Counterexample: Suppose dimV=n and dimU=m with m<n.
let u=u1+...+um in U. Since u1+...+um=u1+...+um+...+0n
we can apply any T and get T(u1)+...+T(um)+..+T(0n)
=c1u1+...+cmum+cm+1*0+...+cn*0=c1u1+...+cmum in U.

2. Jun 10, 2009

### Dick

It looks like you are claiming to prove that ANY proper subspace of V is invariant under any ANY operator T. Does that really seem likely?? Do you really think for ANY operator T, that T(u1)=c1*u1?? Show me an operator T and a subspace U where that is NOT true. Be concrete. Work in say R^3 with basis {e1,e2,e3}. Pick say, U=span{e1}. Define an operator T such that U is not invariant.

Last edited: Jun 10, 2009
3. Jun 10, 2009

### evilpostingmong

T(e1)=T(e1)+T(0)+T(0)=0*e1+ c1*0+c2*0=0 So T maps from U to {0}.

4. Jun 10, 2009

### Dick

But what's T(e1)?. Look, you can define a linear transformation by giving the values of T(e1), T(e2) and T(e3). Here's an example. Define S by S(e1)=e2, S(e2)=0 and S(e3)=0. Is U=span{e1} invariant under S?

5. Jun 10, 2009

### evilpostingmong

Oh okay e2 is not in U so you didn't map from U to U.
Now I will prove that U={0} or V.
For T to map from U to U under any transformation T,
let u=0 so T(u)=0 so all members in the rangeT are
0 and for the domain to match the range, U={0}.
So we mapped from {0} to {0}. And U can also be V
to be invariant under T since we map from V to V if
we let U=V.

6. Jun 10, 2009

### Dick

I think you got the point of the example S, but your 'proof' doesn't convey that. It looks like you just defined T to be the zero map, T(v)=0 for any v. ANY subspace is invariant under that map. Why? The point to the example was that given a subspace U, if I can find a vector v that is not in U, I can define a linear transformation T that maps a nonzero vector in U to v. Try and use the example to model a proof for a general U. You may want to pick a specially chosen basis for V. Can you start?

7. Jun 10, 2009

### evilpostingmong

Basis for V <v1....vn> basis for U <v1....vm>.
Let m<n.
Now, a possible T maps vi to vm+1 1$$\leq$$i$$\leq$$m. Since vm+1
is not within U's basis, U is not invariant under T.

8. Jun 10, 2009

### Dick

That's it! Simple, isn't it? Now for what choices of U is this impossible? I.e. what happens if m=n or m=0?

Last edited: Jun 10, 2009
9. Jun 10, 2009

### evilpostingmong

let m=n. Basis for U=<v1...vm>. Since U is a subspace of V, <v1...vm> is in the basis
of V. But since dimU=dimV, U=V. Now we can map T(vi) to vn or "lower" without falling out
of the basis for U.
let m=0. Basis for U=<0>. T(0)=c*0=0. Since 0 is in {0}, we mapped from {0} to {0}.

10. Jun 10, 2009

### Dick

I guess. But it's not so much that you "CAN map T(vi) to vn or "lower" without falling out
of the basis", you can always do that, it's that you CAN'T map T(vi) outside of the basis.

11. Jun 10, 2009

### Dick

That's what I was thinking you were thinking. Expressing yourself clearly takes some practice. Keep practicing.

12. Jun 10, 2009

### evilpostingmong

I deleted that message, sorry lol.
That's the thing with abstract concepts, I can understand, but can't express.
Alright, were done here.