Proof: Linear Algebra - U Invariant Under Every Operator on V

[evilpostingmong]

Homework Statement

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.

Homework Equations

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.

 

[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.

 

[evilpostingmong]

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.

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

 

[Dick]

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

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?

 

[evilpostingmong]

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?

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.

 

[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?

 

[evilpostingmong]

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?

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.

 

[Dick]

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.

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?

 

[evilpostingmong]

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?

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}.

 

[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.

 

[Dick]

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

 

[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.